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Linear Prog COP4355_Part_16

# Linear Prog COP4355_Part_16 - Chapter 6 Goal Programming 68...

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68 Chapter 6 Goal Programming Chapter 6 Goal Programming 6.1 MAX Z: 4X 1 + 3X 2 + 5X 3 – 50V 1 - 50V 2 Subject to: X 1 + X 2 + X 3 - V 1 + S 1 = 10 2X 1 + X 2 + 2X 3 + V 2 – S 2 = 50 (X i , S i , V i 0) Solution: X 1 = 0, X 2 = 0, X 3 = 35, V 1 = 15, S 1 = 0, V 2 = 0, S 2 = 0, Z MAX = 125 (Note that the maximum value of Z is computed without the violation variables’ effect, since their weights are arbitrary.) This solution violates the first constraint by 15 units. 6.2 MAX Z: 2X 1 + X 2 + 4X 3 + 5X 4 – 50V 1 – 50V 2 – 50V 3 – 50V’ 3 S.T.: 3X 1 + 2X 2 + X 3 + 4X 4 + V 1 – S 1 = 100 X 1 + 2X 2 + 3X 3 + X 4 – V 2 + S 2 = 50 2X 1 + 3X 2 + 4X 3 + 2X 4 + V 3 – V’ 3 = 40 (X 1 , X 2 , X 3 , X 4 , S i , V i 0) Solution: X 1 = 0, X 2 = 0, X 3 = 9.1, X 4 = 22.7, V 1 = 0, S 1 = 0, V 2 = 0, S 2 = 0, V 3 = 0, V’ 3 = 41.8, Z MAX = 149.1. (Note that we compute Z without the effect of violation variables.) This solution violates the equality constrain (#3) by 41.8 units on the side away from the origin. 6.3 MAX Z: -100V 1 + 100S 1 -100V 2 + 100S 2 Subject to: 4X 1 + 2X 2 + 8X 3 + V 1 – S 1 = 100 X 1 + 12X 2 + X 3 + V 2 – S 2 = 100 X 1 + 2X 2 + 3X 3 18 5X 1 + X 2 + X 3 48 (X 1 , X 2 , X 3 , S i , V i 0) Solution: X 1 = 0 X 2 = 48 X 3 = 0 V 1 = 4 S 1 = 0 Z 1 = 96 V 2 = 0 S 2 = 476 Z 2 = 576 6.4 MAX Z: -500V 1 + 500S 1 – 100V 2 + 100S 2 Subject to: 2X 1 + 5X 2 + 8X 3 + V 1 – S 1 = 100 X 1 + 2X 2 + 5X 3 + V 2 - S 2 = 100 3X 1 + 5X 2 + X 3 60 X 1 + 6X 2 + 4X 3 64 8X 1 + X 2 + 2X 3 20 (X 1 , X 2 , X 3 , S i , V i 0) Solution: X 1 = 0 X 2 = 0 X 3 = 16 V 1 = 0 S 1 = 28

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Solutions Manual and Workbook 69 V 2 = 20 S 2 = 0 MAX Z: -100V 1 + 100S 1 -500V 2 + 500S 2 Subject to: 2X 1 + 5X 2 + 8X 3 + V 1 – S 1 = 100 X 1 + 2X
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