Question

A. Do 。me Simplex ituutin, Rt the table Bas Eql Coefficient ofRight Var No 2x1 X2 x34 side X3 1 0 5 10 9 な.
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Answer #1

MAXIMIZE: Z = X1 + X2 + 1 X3 + 2 X4

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MAXIMIZE: Z = 0 X1 + 0 X2 + 1 X3 + 2 X4 + 0 X5 + 0 X6 + 0 X7

subject to

-6 X1 -8 X2 + 0 X3 + 0 X4 ≤ 0
5 X1 + 10 X2 + 1 X3 + 0 X4 ≤ 60
4 X1 + 4 X2 + 0 X3 + 1 X4 ≤ 40

subject to

-6 X1 -8 X2 + 1 X5 = 0
5 X1 + 10 X2 + 1 X3 + 1 X6 = 60
4 X1 + 4 X2 + 1 X4 + 1 X7 = 40

X1, X2, X3, X4 ≥ 0

X1, X2, X3, X4, X5, X6, X7 ≥ 0

Tableau 1

0

0

1

2

0

0

0

Base

Cb

P0

P1

P2

P3

P4

P5

P6

P7

P5

0

0

-6

-8

0

0

1

0

0

P6

0

60

5

10

1

0

0

1

0

P7

0

40

4

4

0

1

0

0

1

Z

0

0

0

-1

-2

0

0

0

Tableau 2

0

0

1

2

0

0

0

Base

Cb

P0

P1

P2

P3

P4

P5

P6

P7

P5

0

0

-6

-8

0

0

1

0

0

P6

0

60

5

10

1

0

0

1

0

P4

2

40

4

4

0

1

0

0

1

Z

80

8

8

-1

0

0

0

2

Tableau 3

0

0

1

2

0

0

0

Base

Cb

P0

P1

P2

P3

P4

P5

P6

P7

P5

0

0

-6

-8

0

0

1

0

0

P3

1

60

5

10

1

0

0

1

0

P4

2

40

4

4

0

1

0

0

1

Z

140

13

18

0

0

0

1

2

The optimal solution value is Z = 140
X1 = 0
X2 = 0
X3 = 60
X4 = 40

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