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

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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, )%*’

the following formulas:

∑)#.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

• 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.

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

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

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

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

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

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

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

e) Idle Time = (Actual number of workstations x largest assigned cycle time)-

∑Task Time

f) Efficiency = ∑)#.5 )%*’

!9-1#$ ,1*&’2 6: ;625.-#-%6,. < $#2='.- #..%=,'3 9>9$’ -%*’

=

@B

H∗D

x 100 = 75 %

Question 3. Develop a solution for the following line balancing problem, allowing

a cycle time of 5 minutes.

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?

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

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%

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.

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