Chapter
9: Line Balancing (longest task approach)
Formulas
Line balancing
Available Time = Net time available excluding meeting time, lunch break etc…
Cycle Time =
!”#$%#&
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6
A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
-%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
)%*’
+,%-. /’01%2’3
Theoretical minimum number of workstations =
∑)#.5 )%*’
)6-#$ 78’2#-%6, )%*’
• To measure the effectiveness, we have to find the efficiency and idle time by using
the following formulas:
Efficiency =
∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
x 100
Idle Time = (Actual number of workstations x largest assigned cycle time)- ∑Task Time
Remember
• The theoretical number of workstation is not always possible to reach when tasks
are actually assigned to stations.
• Cycle time is the maximum time that a product is allowed at each workstation.
Chapter 9: Line Balancing (longest task approach)
Question 1. If the goal is to produce 180 units per hour. The tasks, task times,
and immediate predecessors for the tasks are as follows:
Task Task Time (in sec) Precedence
A 12 –
B 15 A
C 8 A
D 5 B,C
E 20 D
a) What is the available time?
b) What is the cycle time?
c) What is the theoretical minimum number of workstations?
d) Use the longest task time approach to Assign tasks to work station
Solution
a) One hour = 60 min, since the task time is given in seconds we have to convert the
available time to seconds. So the available time = 60*60 = 3600 sec
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
= @ABB
CDB
= 20 seconds per units
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
AB
FB
= 3 workstation
d) Workstations assignment
Task WS1/CT R T
A+C 20 20 – 20 = 0
Task WS2/CT RT
B+D 20 20 – 20 = 0
Task WS3/CT RT
E 20 20 – 20 = 0
Chapter 9: Line Balancing (longest task approach)
Question 2. There are 500 minutes available during the day, lunch break is 20
min, and the average daily demand has been 60 chairs. There are 5 tasks:
Task Task Time (min) Precedence
A 4 —
B 7 —
C 8 B
D 5 C
E 6 A
a) Draw a precedence diagram of this operation
b) What is the available time?
c) What is the cycle time for this operation?
d) What is the theoretical minimum number of workstations?
e) Use the longest task time approach to Assign tasks to work station
f) What is the idle time?
g) What is the efficiency percentage of the assembly line?
Solution
a) Precedence Diagram
a) Available time = 500 min – 20 min = 480 min
b) Cycle Time = ?26319-%6, !”#$%#&$’ )%*’
+,%-. /’01%2’3
=
GDB *%,
AB 1,%-.
= 8 min (we have to produce 1 chair every 8 min)
c) Theoretical minimum number of workstations = ∑)#.5 )%*’
E>9$’ )%*’
=
@B
D
= 3.5 stations = 4 workstations
A
4
C
8
D
5
E
6
B
7
S
Fin
Chapter 9: Line Balancing (longest task approach)
d)
Task WS1/CT RT
B 7 1
Task WS2/CT** RT
C 8 0
Task WS3/CT RT
D 5 3
Task WS4/CT RT
A 4 4
Task WS5/CT RT
E 6 2
**largest assigned cycle time
e) Idle Time = (Actual number of workstations x largest assigned cycle time)-
∑Task Time
= (5 * 8) – 30 = 10 min per cycle
f) Efficiency = ∑)#.5 )%*’
!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’
=
@B
H∗D
x 100 = 75 %
Chapter 9: Line Balancing (longest task approach)
Question 3. Develop a solution for the following line balancing problem, allowing
a cycle time of 5 minutes.
a) Calculate the theoretical minimum number of workstations.
b) Use the longest task time approach to Assign tasks to work
c) Does the solution have the minimum number of stations? Explain.
d) How much idle time is there?
e) What is the efficiency of this line?
Work Task Task Time (seconds) Task Predecessor(s)
A 70 –
B 60 A
C 120 B
D 60 –
E 240 C, D
F 100 A
G 190 E, F
Solution
CT = 300 sec
a) No. w/s = 2.8 = 3 REMEMBER ALWAYS ROUND UP
b) Line balancing
Task WS1/CT** R T
A+F+D+B 290 10
Task WS2/CT RT
C 120 180
Task WS3/CT RT
E 240 60
Task WS4/CT RT
G 190 110
c) No, the solution uses 4 stations, not 3
d) Idle time = 320 sec
e) Efficiency = 72%
Chapter 9: Line Balancing (longest task approach)
Question 4. A company is designing a product layout for a new product. It plans
to use this production line eight hours a day in order to meet a schedule of 400
units per day. The tasks necessary to produce this product are detailed in the
table below.
Task Predecessor Time (seconds)
A – 50
B A 36
C – 26
D – 22
E B, D 70
F C, E 30
a. What is the required cycle time?
b. What is the theoretical minimum number of workstations needed to meet the
schedule?
c. Use the longest task time approach to Assign tasks to work
d. What is the idle time?
e. What is the efficiency?
f. Does the solution have the minimum number of stations?
Solution
a) Cycle time = 72 sec
b) w/s = 3.25 = 4 REMEMBER ALWAYS ROUND UP
c) balance
Task WS1/CT** RT
A+D 72 0
Task WS2/CT RT
B+C 62 10
Task WS3/CT RT
E 70 2
Task WS4/CT RT
F 30 42
d) idle time = 54 sec
e) efficiency = 81.25%
f) yes
Chapter9: Process Layout Design (From-To-Matrix)
Formula
Minimize Cost =
“#$%
“&$% Xij Cij
n = total number of work centers or departments
i, j = individual departments
Xij = number of loads moved from department i to department j
Cij = cost to move a load between department i and department j (distance * cost)
Chapter 9: Process Layout Design (From-To-Matrix)
Question 1. Suppose we have 5 hospital departments.
Assume that workers can’t move diagonally (through walls).
Assuming it costs $3/unit moved each unit distance
Departments
A: Receptionist
B: Waiting room
C: X-Ray
D: Exam room
E: Nurse station
The current layout looks like this
Patient and staff moved (load) per hour between work areas are shown below:
A B C D E
A – 5 2 4 1
B – 3 0 2
C – 0 0
D – 5
E –
Note: Loads could be the average number of patients that move between
the departments each hour.
a. What is the hourly cost for this layout?
b. If the new layout looks like the chart below, what would be the hourly cost for the
new layout?
D
E
A
B C
A
B
D
E C
Chapter 9: Process Layout Design (From-To-Matrix)
Solution
Part a
Pair Load (X) Distance’s Cost (C) Cost (XC)
A-B 5 1x$3=$3 $15
A-C 2 3x$3=$9 $18
A-D 4 1x$3=$3 $12
A-E 1 2x$3=$6 $6
B-C 3 2x$3=$6 $18
B-D 0 2x$3=$6 $0
B-E 2 1x$3=$3 $6
C-D 0 2x$3=$6 $0
C-E 0 1x$3=$3 $0
D-E 5 1x$3=$3 $15
Hourly cost = 15+18+12+6+18+0+6+0+0+15 = $90
Part b:
Pair Load (X) Distance’s Cost (C) Cost (XC)
A-B 5 1x$3=$3 $15
A-C 2 2x$3=$6 $12
A-D 4 1x$3=$3 $12
A-E 1 2x$3=$6 $6
B-C 3 1x$3=$3 $9
B-D 0 2x$3=$6 0
B-E 2 1x$3=$3 $6
C-D 0 3x$3=$9 0
C-E 0 2x$3=$6 0
D-E 5 1x$3=$3 $15
Hourly cost = 15+12+12+6+9+0+6+0+0+15 = $75
Chapter 9: Process Layout Design (From-To-Matrix)
Question 2. A job shop has four departments, A, B, C, and D. Assume it costs
$1 to move 1 work piece 1 foot. Distances in feet between centers of the work
areas are show in the chart below:
A B C D
A – 5 10 7
B – – 6 8
C – – – 9
D – – – –
Workpieces (loads) moved per week between departments are shown below:
A B C D
A – 900 900 500
B – – 500 400
C – – – 700
D – – – –
What is the weekly total material handling cost of the layout?
Solution:
Pair Load (X) Distance’s Cost (C) Cost (XC)
A-B 900 5x$1=$5 $4500
A-C 900 10x$1=$10 $9000
A-D 500 7x$1=$7 $3500
B-C 500 6x$1=$6 $3000
B-D 400 8x$1=$8 $3200
C-D 700 9x$1=$9 $6300
The weekly total material handling cost of the layout:
Cost = 4500+9000+3500+3000+3200+6300 = $29,500
Chapter 9: Process Layout Design (From-To-Matrix)
Question 3. An insurance claims processing center has 4 work centers, any of
which can be placed into any of 4 physical departmental locations. Call the centers
1, 2, 3, and 4 the departments A, B, C, and D. The current set of assignments is
A-3, B-1, C-4, D-2
The (symmetric) matrix of departmental distances, in meters is shown below.
1 2 3 4
1 — 5 30 20
2 — 40 15
3 — 50
4 —
The matrix of work flow (estimated trips per day) among centers is shown below.
A B C D
A — 35 20 30
B — 100 60
C — 50
D —
The firm estimates that each meter costs approximately $4. What is the cost of the current
assignment?
Solution
1 2 3 4
B D A C
Matrix of departmental distances
1 (B) 2 (D) 3 (A) 4 (C)
1 (B) — 5 30 20
2 (D) — 40 15
3 (A) — 50
4 (C) —
Pair X C XC
A-B 35 30×4= 120 4200
A-C 20 50×4= 200 4000
A-D 30 40×4= 160 4800
B-C 100 20×4=80 8000
B-D 60 5×4=20 1200
C-D 50 15×4= 60 3000
$25,200
Chapter12: EOQ
D = Annual demand (units) S = Cost per order ($)
P = Cost per unit ($) I = Holding cost (%)
H = Holding cost ($) = I x P
Formula
𝐓𝐨𝐭𝐚𝐥
𝐚𝐧𝐧𝐮𝐚𝐥
𝐨𝐫𝐝𝐞𝐫𝐢𝐧𝐠
𝐜𝐨𝐬𝐭
(𝐬𝐞𝐭𝐮𝐩
𝐜𝐨𝐬𝐭) = 𝐃
𝐒
𝑸
(Note: when we use Q*, Q in the equation become Q*,)
𝐓𝐨𝐭𝐚𝐥
𝐚𝐧𝐧𝐮𝐚𝐥 𝐡𝐨𝐥𝐝𝐢𝐧𝐠
𝐜𝐨𝐬𝐭
(𝐜𝐚𝐫𝐫𝐢𝐧𝐠
𝐜𝐨𝐬𝐭) = 𝐐
𝐇
𝟐
(Note: when we use Q*, Q in the equation become Q*,)
EOQ = Q* =
𝟐
𝑫
𝑺
𝑯
Note: the unit of measure for “D” and “H” have to be identical (days, weeks, months,
year, etc)
Expected Number of Orders 𝐍 = 𝐃𝐞𝐦𝐚𝐧𝐝
𝐎𝐫𝐝𝐞𝐫
𝐐𝐮𝐚𝐧𝐭𝐢𝐭𝐲
= 𝐃
𝐐∗
Expected Time between Orders 𝐓 = 𝐍𝐮𝐦𝐛𝐞𝐫
𝐨𝐟
𝐰𝐨𝐫𝐤𝐞𝐢𝐧𝐠
𝐝𝐚𝐲𝐬
𝐩𝐞𝐫
𝐲𝐞𝐚𝐫
𝐄𝐱𝐩𝐞𝐜𝐭𝐞𝐝
𝐧𝐮𝐦𝐛𝐞𝐫
𝐨𝐟
𝐨𝐫𝐝𝐞𝐫𝐬
(𝐍)
Total cost (TC) = Total Setup cost + Total Holding cost
TC =
𝐃
𝐒
𝑸
+
𝐐
𝐇
𝟐
(Note: when we use Q*, Q in the equation become Q*,)
Reorder Point ROP = d x L
𝐝 =
𝐃
𝑵𝒖𝒎𝒃𝒆𝒓
𝒐𝒇
𝒘𝒐𝒓𝒌𝒊𝒏𝒈
𝒅𝒂𝒚𝒔
𝒊𝒏
𝒂
𝒚𝒆𝒂𝒓
Note: Holding cost might be given in terms of cost/unit/year or it might be given in the
form of rate%, in that case we need to find the holding cost by:
H = I* P
Chapter 12: EOQ
Question 1. Joe Henry’s machine shop uses 2500 brackets during the course of a
year. The following information is known about the brackets:
Annual Demand 2500 Brackets
Holding Cost per bracket per year: $1.50
Order cost per order: $18.75
Lead Time 2 days
Working days per year 250 days
a. Given the above information, what would be the economic order quantity (EOQ)?
b. What would be the annual holding cost?
c. What would be the annual ordering cost?
d. What is the total annual cost of managing the inventor
e. How many orders would be made each year?
f. What is the time between orders?
g. What is the reorder point (ROP)
Solution:
a. Q* = [
\
]
^
= [∗[_“∗ab.d_
a._
= 250 units
b. Total
holding
cost
carring
cost = rs
[
=
[_“∗a._
[
= $187.5
c. Total
ordering
cost
(setup
cost) = w
x
y
=
[_“∗ab.d_
[_`
= $187.5
d. TC = w
x
y
+
r
s
[
= $187.5+ $187.5 = $375
e. N = w{|}~
€{
r‚}~ƒ„ƒ…
= w
r∗
=
[_“
[_`
= 10 orders/year
f. T = †‚|‡{
ˆ‰
Šˆ‹{„~Œ
}…
Ž{
…{}
Ž{‘ƒ{
~‚|‡{
ˆ‰
ˆ{
(†)
= T = [_`
a`
= 250 days
g. 𝑅𝑂𝑃 = 𝑑
𝑥
𝐿𝑇
d = [_“
[_`
= 10 units/day
ROP = 10 x 2 = 20 units
Question 2. If the demand = 8,000 per month, S = $45 per order, and H = $2 per
unit per year, what is the economic order quantity?
Solution:
Note that annual demand is 12 * 8,000 = 96,000.
Q* =
[∗™š“`∗›_
[
= 2078
units
Chapter 12: EOQ
Question 3. If EOQ = 50 units, order costs are $5 per order, and carrying costs
are $2 per unit/year, what is the Annual Demand?
Q* =
[
\
]
^
=
50 =
𝟐𝒙𝑫𝑿𝟓
𝟐
(50)2
=
𝟐𝒙𝑫𝒙𝟓
𝟐
2500 = 5 D
D= 500 units
Question 4. If the unit cost is $20, and the annual holding cost is 10%. Annual
demand is 400 units, and the order cost is $1 per order. What is the optimal
quantity?
H = IP = 0.1 * 20 = $2
Q* =
𝟐𝑫𝑺
𝑯
=
𝟐𝒙𝟒𝟎𝟎𝒙
𝟏
𝟐
= 20 units
Question 5. Xyz company purchases 8,000 cables each year as components in
computers. The cost of carrying one cable in inventory for a year is $3 and the
ordering cost is $30 per order. What are:
a. The optimal order quantity
b. Annual holding cost
c. Annual set up cost
d. Total annual cost
e. The expected number of orders placed each year, and
f. The expected time between order? Assume that xyz company operates on a 200-
day working year
Note: below are the final answers for each question, in the exam you need to show the
steps
Solution:
a. 400 units
b. $600
c. $600
d. $1200
Chapter 12: EOQ
e. 20 orders
f. 10 working days
Question 6. Lead time for one of your fastest- moving products is 21 days.
Demand during this period averages 100 units per day. What would be an
appropriate reorder point?
Solution:
a) ROP = d x LT = 100 X 21 = 2100 units
Question 7. If the economic order quantity = 300, annual demand = 8,000 units,
and order costs = $45 per order, what is the annual per unit holding cost (H)?
Solution:
Q* =
[
\
]
^
300 =
[∗b“∗›_
^
(300)2 =
[∗b“∗›_
^
90000 =
d[`,“`
^
90,000 H = 720,000
H = $8/unit/year
Chapter 12: EOQ
Question 8. Kara Chubrik uses 1,500 per year of a certain subassembly that has
an annual per unit holding cost of $45 per unit. Each order placed costs Kara
$150. She operates 300 days per year, and has found that an order must be
placed with her supplier 6 working days before she can expect to receive that
order. For this subassembly, find
(a) Economic Order quantity
Q*
H
DS2
=
45
150*1500*2
=
100= units
(b) Total annual holding cost
Annual holding cost H
Q
2
=
45*
2
100
=
250,2$=
(c) Annual ordering cost
Annual ordering cost S
Q
D
=
150*
100
500,1
=
250,2$=
(d) Reorder point
ROP = d*LT
d =
𝟏𝟓𝟎𝟎
𝟑𝟎𝟎
= 5
ROP = 5×6 = 30 units
Chapter 12: EOQ
Question 9. Xyz company stocks books with the following characteristics:
Demand D = 19,500 units/year, Ordering cost = $25/order, Carrying cost =
$4/unit/year
a. Calculate the EOQ
b. What are the annual holding costs?
c. What are the annual ordering costs?
d. The expected number of orders placed each year, and
e. The expected time between orders? Assume that xyz company operates on a 300
day/year
Note: below are the final answers for each question, in the exam you need to show the
steps
Solution:
a. 493.7 workbooks
b. $987
c. $987
d. 39 order/year
e. 7.7 days
Chapter6s: Control Chart
Formulas
For X Bar Charts when we know
UCLx̅ = x̿ + z σx̅
LCLx̅ = x̿ – z σx̅
σx̅ = σ/√n
Where
x̿ = mean of the sample means or a target value set for the process
z = number of normal standard deviations
𝜎�̅� = standard deviation of the sample means
= population (process) standard deviation
n = sample size (subgroup number)
When we don’t know
X Bar Chart
UCLX̅ = x̿ + A2 R̅
LCLX̅ = x̿ – A2 R̅
R Chart
UCLR = D4 R̅
LCLR = D3 R̅
Factor for Computing Control Chart
Sample Size (n) Mean Factor A2 Upper Range D4 Upper Range D3
2 1.88 3.268 0
3 1.023 2.574 0
4 0.729 2.282 0
5 0.577 2.115 0
6 0.483 2.004 0
7 0.419 1.924 0.076
8 0.373 1.864 0.136
9 0.337 1.816 0.184
10 0.308 1.777 0.223
Chapter 6s: Control Chart
Question 1. Boxes of chocolates are produced to contain average of 14 ounces
(target value), with a standard deviation of 0.1 ounce. Set up the 3-sigma �̅�
chart for a sample size of 36 boxes. Determine the upper and lower control
limits
Solution:
�̿� = 14 ounces z = 3
𝜎�̅� = 𝜎/√𝑛
𝜎�̅� = 0.1/√36 = 0.0167
𝑈𝐶𝐿�̅� = �̿� + z 𝜎�̅�
𝑈𝐶𝐿�̅� = 14 + (3 * 0.0167) = 14.05 oz
𝐿𝐶𝐿�̅� = �̿� – z 𝜎�̅�
𝐿𝐶𝐿�̅� = 14 – (3 * 0.0167) = 13.95 oz
Question 2. The overall average on a process you are attempting to monitor is
50 kg. The process standard deviation is 1.72. Determine the upper and lower
control limits for a mean chart, if you choose to use a sample size of 5.
a) Determine the upper and lower control limits for the x-bar chart Set z=3
b) Determine the upper and lower control limits for the x-bar chart Set z=2.
Solution
n= 5 �̿� = 50 = 1.72 z = 3
a) 𝜎�̅� = (1.75/√5) = 0.78
𝑈𝐶𝐿�̅� = 50 + (3 * 0.78) = 52.34
𝐿𝐶𝐿�̅� = 50 – (3 * 0.78) = 47.66
b) 𝜎�̅� = (1.75/√5) = 0.78
𝑈𝐶𝐿�̅� = 50 + (2 * 0.78) = 51.56
𝐿𝐶𝐿�̅� = 50 – (2 * 0.78) = 48.44
Chapter 6s: Control Chart
Question 3. For the past 7 hours, seven samples of size 3 were taken from a
process, and their weights measured. The sample averages and sample ranges
are in the following table.
a) Use the R chart to compute the UCL & LCL
b) Use X bar chart to calculate the UCL & LCL X bar chart
c) Is the process in control?
d) What additional steps should the quality assurance team take?
Note: Round up/down to the nearest whole number
Hour sample 1 sample 2 sample 3
1 32 28 30
2 40 29 31
3 32 30 42
4 31 30 29
5 29 27 28
6 32 30 27
7 42 40 39
Solution
n = 3
From the Factor for Computing Control Chart Table:
A2 = 1.023
D4 = 2.574
D3 = 0
Day sample 1 sample 2 sample 3 R x̅
1 32 28 30 32-28= 4 �̅� =
32+28+30
3
= 30
2 40 29 31 40-29 =11 �̅� =
40+29+31
3
= 33
3 32 30 42 42- 30 = 12 �̅� =
32+30+4
2
3
= 35
4 31 30 29 31- 29 = 2 �̅� =
31+30+29
3
= 30
5 29 27 28 29- 27 = 2 �̅� =
29+27+28
3
= 28
6 32 30 27 32- 27 = 5 �̅� =
32+30+27
3
= 30
7 42 40 39 42- 39 = 3 �̅� =
42+40+39
3
= 40
�̅� =
4+11+12+2+2+5+3
7
= 6
x̿ =
30+33.33+34.67+30+28+29.67+40.33
7
= 32
Chapter 6s: Control Chart
a) R chart
UCLR = D4 R̅
UCLR = (2.574 * 6) = 15
LCLR = D3 R̅
LCLR = (0 * 6) = 0
b) X bar chart
UCLX̅ = x̿ + A2 R̅
UCLX̅ = 32 + (1.023 * 6) = 38
LCLX̅ = x̿ – A2 R̅
LCLX̅ = 32 + (1.023 * 6) = 26
c) The process is out of control because x̅ for sample 7 is greater than the UCL in
x-bar chart. For R chart indicates all of the samples are within control limit. We
can conclude that the process is out of control.
d) The quality assurance team should investigative sample 7 to find the root cause of
the problem.
Question 4. Sampling 4 pieces of precision-cut wire (to be used in computer
assembly) every hour for the past 12 hours has produced the following results:
Hour x̅ R
1 3.5 1.25
2 3 1
3 1.25 1.5
4 3.5 1.25
5 3 1.5
6 2.75 0.25
7 3 0.5
8 2.5 1.5
9 3 0.75
10 2.75 1.5
11 2.25 1
12 3 0.5
Chapter 6s: Control Chart
a. Determine the upper and lower control limits for the x-bar chart
b. Determine the upper and lower control limits for the R-chart
c. Is the process in control? Why?
Note: use 2 decimal places
Solution
First we need to find the x̿, you have to add all of the x̅ and divide them by the sample
number 12:
x̿ =
3.5+3+1.25+3.5+3+2.75+3+2.5+3+2.75+2.25+3
12
x̿ = 2.79
Now we need to find the R̅, (R = Highest – Lowes value for each sample) in this example R
is already given so all we need to do is to add all of the Rs and divide them by 12 (number
of samples)
R̅= 1.04
n = 4
From the Factor for Computing Control Chart Table:
A2 = 0.729
D4 = 2.282
D3 = 0
a. UCL & LCL For x-BAR CHART
UCLX̅ = x̿ + A2 R̅
= 2.79 + (0.729 * 1.04) = 3.55
LCLX̅ = x̿ – A2 R̅
= 2.79 – (0.729 * 1.04) = 2.03
b. UCL & LCL For R CHART
UCLR = D4 R̅
= 2.282 * 1.04 = 2.38
LCLR = D3 R̅
= 0 * 1.04 = 0
Chapter 6s: Control Chart
c. For x bar chart: x̅ for sample 3 is out of control, for R chart: all of the samples are within the
control limits. We can conclude that the process is currently out of control.
Question 5. A process that is considered to be in control measures an
ingredient in ounces. Below are the last 10 samples (each of size n=5) taken.
The population process standard deviation is 1.36
Samples
1 2 3 4 5 6 7 8 9 10
10 9 13 10 12 10 10 13 8 10
9 9 9 10 10 10 11 10 8 12
10 11 10 11 9 8 10 8 12 9
9 11 10 10 11 12 8 10 12 8
12 10 9 10 10 9 9 8 9 12
a) What is the standard deviation of the sample means (𝜎�̅� ) ?
b) If z = 3, what are the control limits for the mean chart
c) If isn’t given, what are the control limits for x bar chart?
d) What are the control limits for the range chart?
e) Is the process in control?
f) What additional steps should the quality manager take?
Note: use 2 decimal places
Solution:
a) Process (population) standard deviation () = 1.36,
b) Using x̅
x
x
UCL 10 3 0.61 11.83
LCL 10 – 3 0.61 8.17
Standard deviation of the sampling means
1.36
5
0.61
x
Chapter 6s: Control Chart
c) X bar chart
First you have to find the x ̅ and R for each sample, then X double bar and R bar.
Sample
x̅ 1 x̅ 2 x̅ 3 x̅ 4 x̅ 5 x̅ 6 x̅ 7 x̅ 8 x̅ 9 x̅ 10
10 10 10.2 10.2 10.4 9.8 9.6 9.8 9.8 10.2
x̿ =
10+10+10.2+10.2+10.4 +9.8+9.6+9.8+9.8+10.2
10
= 10
Sample
R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R 10
3 2 4 1 3 4 3 5 4 4
R̅ = 3.3
Since n = 5, A2 = 0.577 (check the chart)
x
x
UCL 10 3.3 0.577 11.90
LCL 10 – 3.3 0.577 8.10
d) R Chart
UCLR = 2.115(3.3) = 6.98
LCLR = 0(3.3) = 0
e) Yes, both mean and range charts indicate process is in control.
f) As the process is in control no further action is needed.
Chapter 6s: Control Chart
Question 6. Auto pistons at Wemming Chung’s plant in Shanghai are
produced in a forging process, and the diameter is a critical factor that must be
controlled. From sample sizes of 10 pistons produced each day, the mean and
the range of this diameter have been as follows:
Day Mean (mm) Range (mm)
1 156.9 4.2
2 153.2 4.6
3 153.6 4.1
4 155.5 5
5 156.6 4.5
a. What is the value of the mean of �̿� ?
b. what is the �̅� ?
c. What are the 𝐔𝐂𝐋�̅� and 𝐋𝐂𝐋�̅�?
d. What are the UCLR and LCLR?
(c) -chart:X
2
2
155.16 mm from the sample data
UCL 155.16 (0.308 4.48) 156.54 mm
LCL 155.16 (0.308 4.48) 153.78 mm.
x
x
X
X A
R
X A R
(d) UCLR = 1.777 x 4.48 = 7.96
LCLR = 0.223 x 4.48 = 0.99
156.9 153.2 153.6 155.5 156.6
(a) 155.16 mm
5
4.2 4.6 4.1 5.0 4.5
(b) 4.48 mm
5
X
R
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