LinearProgramming x
1
.a)
Introducing the slack variables:
SIMPLE
X
TABLE
|
BASIC |
X | y |
Z |
Q |
SOLUTION |
||||||||||||||||||
| 1 |
0 |
2 |
24 |
||||||||||||||||||||
|
4 |
28 |
||||||||||||||||||||||
|
-2 |
6 |
44 |
|||||||||||||||||||||
|
-1 |
-6 |
b)
The highest negative number in the last row is -6. The pivot is determined by division
Pivot is therefore and now becomes z since it has the least ratio
BASIC
X
y
Z
Q
SOLUTION
1
0
2
1
0
0
0
24
z
1
0
0
0
7
1
-2
6
0
0
1
0
44
Q
-2
-1
-6
0
0
0
1
0
BASIC
X
y
Z
Q
SOLUTION
-1
0
0
0
0
10
z
1
0
0
0
7
x
-5
0
0
0
2
Q
–
2
0
0
0
1
42
The pivot now becomes =x since it has the least ratio
BASIC
X
y
Z
Q
SOLUTION
-1
0
0
0
0
10
z
-1
2
0
-2
1
0
0
8
x
1
10
0
0
3
-2
0
-4
Q
–
2
0
0
0
1
42
BASIC
X
y
Z
Q
SOLUTION
0
-6
0
1
-2
1
0
12
z
0
-2
1
0
0
0
8
x
1
10
0
0
3
-2
0
-4
Q
0
7
0
0
3
-1
1
40
c)
Determining y using Q
From the table we get the variable values as:
The objective function is
The pivot now is =y
2. a)
The problem is balanced, because the total supply is equivalent to the demand
Demand = 20+56+40=116
Supply=34+57+25=116
b)
2
S
3
S
1
S
2
S
3
S
12
2
24
1
=
=
S
7
4
28
2
=
=
S
3
.
7
6
44
3
=
=
S
2
S
1
S
4
1
2
1
4
1
3
S
z
Q
Q
z
S
S
z
S
S
6
6
2
3
3
1
1
+
=
–
=
–
=
1
S
2
1
2
1
–
4
1
2
1
4
1
2
1
–
2
3
–
1
2
1
2
3
3
S
0
0
0
0
6
2
0
,
,
,
,
,
44
6
2
28
4
2
24
2
3
2
1
3
2
1
3
2
1
=
+
+
+
–
–
–
³
=
+
+
–
=
+
+
+
=
+
+
S
S
S
z
y
x
Q
S
S
S
z
y
x
S
z
y
x
S
z
y
x
S
z
x
4
5
.
0
2
28
25
.
0
7
20
5
.
0
10
3
1
–
=
–
=
=
=
=
=
S
z
S
1
S
2
3
x
Q
Q
z
x
z
x
S
S
5
.
0
5
.
0
1
1
–
=
+
=
–
=
1
S
1
S
12
,
8
,
4
,
40
1
=
=
–
=
=
S
z
x
Q
0
44
)
8
(
6
)
0
(
2
4
44
6
2
0
28
)
8
(
4
)
0
(
2
4
28
4
2
0
40
)
8
(
6
)
4
(
2
6
2
3
3
3
2
2
2
=
=
+
+
–
–
=
+
+
–
=
=
+
+
+
–
=
+
+
+
=
=
+
+
–
=
+
+
S
S
S
z
y
x
S
S
S
Z
y
x
y
y
Q
z
y
x
40
6
2
=
+
+
z
y
x
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