1
Simplex Method: Maximization Problem
1. After adding the slack variables, the standard LPP is given as
maximize P = 70x1
+
50x2
subject to 4x1 + 3x2 + x3
=
240
2x1 + x2 + x4
=
100
x1 , x2 , x3 , x4
≥
0
Tableau 1:
Basic
x3
x4
P
x1
4
2
−70
x2
3
1
−50
x1
0
1
0
x2
1
x3
1
0
0
x4
0
1
0
R.H.S.
240
100
0
Tableau 2:
Basic
x3
x1
P
1
2
−15
x3
1
0
0
x4
-2
R.H.S.
40
50
3500
1
2
35
Tableau 3:
Basic
x2
x1
P
x1
0
1
0
x2
1
0
0
x3
1
1
-2
15
x4
-2
3
2
5
R.H.S.
40
30
4100
Thus, P = 4100, x1 = 30 , x2 = 40
2. After adding the slack variables, the standard LPP is given as
maximize P = 10x1
+
5x2
subject to 4x1 + x2 + x3
=
28
2x1 + 3x2 + x4
=
24
x1 , x2 , x3 , x4
≥
0
Tableau 1:
Basic
x3
x4
P
x1
4
2
−10
x2
1
3
−5
1
x3
1
0
0
x4
0
1
0
R.H.S.
28
24
0
Tableau 2:
Basic
x1
x1
1
x4
0
P
0
Basic
x1
x2
P
x1
1
0
0
x2
x3
1
4
5
2
− 52
1
4
- 12
5
2
x4
0
R.H.S.
7
1
10
0
70
Tableau 3:
x2
0
1
0
x3
x4
3
10
- 15
1
- 10
2
5
2
1
R.H.S.
6
4
80
Thus, P = 80, x1 = 6 , x2 = 4
3. After adding the slack variables, the standard LPP is given as
maximize P = 70x1
+
50x2 + 35x3
subject to 4x1 + 3x2 + x3 + x4
=
240
2x1 + x2 + x3 + x5
=
100
x1 , x2 , x3 , x4 , x5
≥ 0
Tableau 1:
x1
4
2
−70
x2
3
1
−50
Basic
x4
x1
P
x1
0
1
0
x2
1
x3
-1
1
2
1
2
−15
0
Basic
x2
x1
P
x1
0
1
0
x2
1
0
0
Basic
x4
x5
P
x3
1
1
-35
x4
1
0
0
x5
0
1
0
R.H.S.
240
100
0
Tableau 2:
x4
1
0
0
x5
-2
x4
1
1
−2
15
x5
-2
1
2
35
R.H.S.
40
50
3500
Tableau 3:
x3
-1
1
-15
Tableau 4:
2
3
2
5
R.H.S.
40
30
4100
Basic
x2
x3
P
x1
1
1
0
Thus, P = 4550, x1 = 0 , x2 = 70,
x2
1
0
0
x3
0
1
0
x4
x5
− 12
1
2
− 12
15
2
R.H.S.
70
30
4550
3
2
55
5
x3 = 30
4. After adding the slack variables, the standard LPP is given as
maximize P = 2x1
+ x2
subject to 5x1 + x2 + x3
=
9
x1 + x2 + x4
=
5
x1 , x2 , x3 , x4
≥ 0
Tableau 1:
Basic
x3
x4
P
x1
5
1
−2
x2
1
1
−1
x3
1
0
0
x4
0
1
0
R.H.S.
9
5
0
Basic
x1
x1
1
x2
x3
x4
0
P
0
1
5
- 51
2
5
x4
0
R.H.S.
1
5
4
5
- 53
x1
1
0
0
x2
0
1
0
x3
Tableau 2:
9
5
16
5
18
5
1
0
Tableau 3:
Basic
x1
x3
P
1
4
- 14
1
4
x4
- 14
R.H.S.
1
4
6
5
4
3
4
Thus, P = 6, x1 = 1 , x2 = 4
5. After adding the slack variables, the standard LPP is given as
maximize P = 30x1
+
40x2
subject to 2x1 + x2 + x3
=
10
x1 + x2 + x4
=
7
x1 + 2x2 + x5
=
12
x1 , x2 , x3 , x4 , x5
Tableau 1:
3
≥ 0
Basic
x3
x4
x5
P
x1
2
1
1
−30
x2
1
1
2
−40
x3
1
0
0
0
x4
0
1
0
0
x5
0
0
1
0
R.H.S.
10
7
12
0
Tableau 2:
Basic
x3
x4
x2
P
x1
3
2
1
2
1
2
x2
0
x3
1
x4
0
x5
- 12
R.H.S.
4
0
0
1
- 12
1
0
0
1
2
1
0
−10
0
0
20
6
240
Tableau 3:
Basic
x3
x1
x2
P
Thus, P = 260, x1 = 2 , x2 = 5,
x1
0
1
0
0
x2
0
0
1
0
x3
1
0
0
0
x4
-3
2
-1
20
x5
1
-1
1
10
R.H.S.
1
2
5
260
x3 = 1
6. After adding the slack variables, the standard LPP is given as
maximize P = −x1
+
2x2
subject to − x1 + x2 + x3
=
2
−x1 + 3x2 + x4
=
12
x1 − 4x2 + x5
=
4
x1 , x2 , x3 , x4 , x5
≥ 0
Tableau 1:
Basic
x3
x4
x5
P
x1
-1
-1
1
1
x2
1
3
-4
−2
x3
1
0
0
0
x4
0
1
0
0
x5
0
0
1
0
R.H.S.
2
12
4
0
Basic
x2
x4
x5
P
x1
-1
2
-3
−1
x2
1
0
0
0
x3
1
-3
4
2
x4
0
1
0
0
x5
0
0
1
0
R.H.S.
2
6
12
4
Tableau 2:
4
Tableau 3:
Basic
x2
x1
x5
P
P = 7,
2
x1 = 3,
x1
0
1
0
0
x2
1
0
0
0
x3
- 12
- 32
- 12
x4
1
2
1
2
3
2
1
2
1
2
x5
0
0
1
0
R.H.S.
5
3
21
7
x2 = 5
Simplex Method: Dual Problem
1.
Minimize C = 40x1 + 12x2 + 40x3
Subject to
2x1 + x2 + 5x3 ≥ 20
4x1 + x2 + x3 ≥ 30
x1 , x2 , x3 = 0
2
A= 4
40
1
1
12
5
1
40
20
30
,
2
1
AT =
5
20
4
1
1
30
40
12
40
The Dual Maximization Problem is
Maximize P = 20y1 + 30y2
Subject to 2y1 + 4y2 ≤ 40
y1 + y2 ≤ 12
5y1 + y2 ≤ 40
y1 , y2 ≥ 0
We then add the slack variables y3 , y4 , y5 to obtain the following standard LPP.
Maximize
Subject to
P = 20y1 + 30y2
2y1 + 4y2 + y3 = 40
y1 + y2 + y4 = 12
5y1 + y2 + y5 = 40
y1 , y2 , y3 , y4 , y5 ≥ 0
Tableau 1 is given below;
Basic
y3
y4
y5
P
y1
2
1
5
-20
y2
4
1
1
-30
5
y3
1
0
0
0
y4
0
1
0
0
y5
0
0
1
0
RHS
40
12
40
0
The pivot column is the one corresponding to -30 in the last row. Thus y2 enters the basic set.
Quotients are 40/1 = 40, 12/1 = 12, 40/4 = 10, thus the pivot row is the one corresponding
to the smallest quotients so y3 leaves the basic set. The number 4 is the pivot coefficient.
After performing the necessary row operations we get Tableau 2 shown below
Basic
y2
y1
1/2
y4
y2
1
y3
1/4
y4
0
y5
0
0
-1/4
1
0
2
0
0
-1/4
15/2
0
0
1
0
30
300
1/2
y5
P
9/2
-5
RHS
10
The entering variable is y1 . The quotients are 30/(9/2) = 6.667; 2/(1/2) = 4; 10/(1/2) = 20.
Thus y4 leaves the basic set and the pivot coefficient is 1/2.
After performing the necessary row operations we get Tableau 3 shown below
Basic
y2
y1
y5
P
y1
0
1
0
0
y2
1
0
0
0
y3
1/2
-1/2
2
5
y4
-1
2
-9
10
y5
0
0
1
0
RHS
8
4
12
320
Since all the coefficients in the last row are positive, the solution given in Tableau 3 is the
optimal solution.
C = P = 320,
x1 = 5,
x2 = 10,
x3 = 0.
2.
Minimize
C = 21x1 + 50x2
Subject to 2x1 + 5x2 ≥ 12
3x1 + 7x2 ≥ 17
x1 , x2 ≥ 0
2
A= 3
21
5
7
50
12
17
,
2
A = 5
12
T
3 21
7 50
17
The Dual Maximization Problem is
Maximize P = 12y1 + 17y2
Subject to 2y1 + 3y2 ≤ 21
5y1 + 7y2 ≤ 50
y1 , y2 ≥ 0
We then add the slack variables y3 , y4 to obtain the following standard LPP.
6
Maximize
Subject to
P = 12y1 + 17y2
2y1 + 3y2 + y3 = 21
5y1 + 7y2 + y4 = 50
y1 , y2 , y3 , y4 , y5 ≥ 0
Tableau 1 is given below;
Basic
y3
y4
P
y1
2
5
-12
y2
3
7
-17
y3
1
0
0
y4
0
1
0
RHS
21
50
0
The pivot column is the one corresponding to -17 in the bottom row, so y2 is the entering
variable. The quotients are 50/5 = 10 and 21/3 = 7, so y3 is the leaving variable and the
number 3 is the pivot coefficient.
After performing the necessary row operations we get Tableau 2 shown below
Basic
y2
y1
2/3
y2
1
y3
1/3
y4
0
RHS
7
y4
1/3
0
-7/3
1
1
P
-2/3
0
17/3
0
119
The entering variable is y1 . The quotients are 1/(1/3) = 3; 7/(2/3) = 10.5 ;. Thus y4 leaves
the basic set and the pivot coefficient is 1/3.
After performing the necessary row operations we get Tableau 3 shown below
Basic
y2
y1
P
y1
0
1
0
y2
1
0
0
y3
5
-7
1
y4
2
3
2
RHS
5
3
121
Since all the coefficients in the last row are positive, the solution given in Tableau 3 is the
optimal solution.
C = P = 121,
x1 = 1,
x2 = 2
3.
Minimize
C = 2x1 + 10x2 + 8x3
Subject to
x1 + x2 + x3 ≥ 6
x2 + 2x3 ≥ 8
−x1 + 2x2 + 2x3 ≥ 4
x1 , x2 , x3 ≥ 0
7
1
1 1 6
0
1 2 8
A=
−1
2 2 4
2 10 8
,
1
1
AT =
1
6
0
1
2
8
−1 2
2 10
2 8
4
The Dual Maximization Problem is
Maximize P = 6y1 + 8y2 + 4y3
Subject to
y1 − y3 ≤ 2
y1 + y2 + 2y3 ≤ 10
y1 + 2y2 + 2y3 ≤ 8
y1 , y2 , y3 ≥ 0
We then add the slack variables y4 , y5 , y6 to obtain the following standard LPP.
Maximize
Subject to
P = 6y1 + 8y2 + 4y3
y1 − y3 + y4 = 2
y1 + y2 + 2y3 + y5 = 10
y1 + 2y2 + 2y3 + y6 = 8
y1 , y2 , y3 ≥ 0
Tableau 1 is given below;
Basic
y4
y5
y6
P
y1
1
1
1
-6
y2
0
1
2
-8
y3
-1
2
2
-4
y4
1
0
0
0
y5
0
1
0
0
y6
0
0
1
0
RHS
2
10
8
0
Pivot coefficient is 2, and y2 enters, whilst y6 leaves the basic set.
After performing the necessary row operations we get Tableau 2 shown below
Basic
y4
y5
y2
P
y1
1
1/2
1/2
-2
y2
0
0
1
0
y3
-1
1
1
4
y4
1
0
0
0
y5
0
1
0
0
y6
0
-1/2
1/2
4
RHS
2
6
4
32
Comparing the quotients obtained using the column corresponding to -2, we see that y4 leaves
and y1 enters the basic set.
After performing the necessary row operations we get Tableau 3 shown below;
8
Basic
y1
y5
y2
P
y1
1
0
0
0
y2
0
0
1
0
y3
-1
3/2
3/2
2
y4
1
-1/2
-1/2
2
y5
0
1
0
0
y6
0
-1/2
1/2
4
RHS
2
5
3
36
Since all the coefficients in the last row are positive, the solution given in Tableau 3 is the
optimal solution.
C = P = 36,
x1 = 2, x2 = 0, x3 = 4.
4.
Minimize C = 16x1 + 9x2 + 21x3
Subject to
x1 + x2 + 3x3 ≥ 12
2x1 + x2 + x3 ≥ 16
x1 , x2 , x3 ≥ 0
1 1
A= 2 1
16 9
3
1
21
12
16
,
1
1
AT =
3
12
2 16
1 9
1 21
16
The Dual Maximization Problem is
Maximize P = 12y1 + 16y2
Subject to
y1 + 2y2 ≤ 16
y1 + y2 ≤ 9
3y1 + y2 ≤ 21
y1 , y2 ≥ 0
We then add the slack variables y3 , y4 , y5 to obtain the following standard LPP.
Maximize
P = 12y1 + 16y2
Subject to y1 + 2y2 + y3 = 16
y1 + y2 + y4 = 9
3y1 + y2 + y5 = 21
y1 , y2 ≥ 0
Tableau 1 is given in the table below
Basic
y3
y4
y5
P
y1
1
1
3
-12
y2
2
1
1
-16
9
y3
1
0
0
0
y4
0
1
0
0
y5
0
0
1
0
RHS
16
9
21
0
Comparing the quotients obtained using the column corresponding to -16, we see that y3 leaves
and y2 enters the basic set. The number 2 is the pivot coefficient.
After performing the necessary row operations we get Tableau 2 shown below
Basic
y2
y4
y5
P
y1
1/2
1/2
5/2
-4
y2
1
0
0
0
y3
1/2
-1/2
-1/2
8
y4
0
1
0
0
y5
0
0
1
0
RHS
8
1
13
128
Comparing the quotients obtained using the column corresponding to -4, we see that y4 leaves
and y1 enters the basic set. The number 1/2 is the pivot coefficient.
After performing the necessary row operations we get Tableau 3 shown below
Basic
y2
y1
y5
P
y1
0
1
0
0
y2
1
0
0
0
y3
1
-1
2
4
y4
-1
2
-5
8
y5
0
0
1
0
RHS
7
2
8
136
Since all the coefficients in the last row are positive, the solution given in Tableau 3 is the
optimal solution.
C = P = 136, x1 = 4, x2 = 8, x3 = 0.
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