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# sol6 - Steel 1 Steel 2 Steel 3 Dummy Plant 1 60 40 28 120...

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AMS 341 Spring, 2010 Homework Set # 6 — Solution Notes #3, 6.7: The dual of the LP is: min z = 0 . 5 y 1 + 0 . 5 y 2 s . t . y 1 - y 2 ≥ - 1 2 y 1 + 3 y 2 5 y 1 , y 2 0 Using the formulas on page 310, we get y 1 = 0 . 4, the objective coeFcient of s 1 , y 2 = 1 . 4 the the coeFcient of s 2 , and so z = 0 . 5 · 0 . 4 + 0 . 5 · 1 . 4 = 0 . 9. #5, 6.7: Using the formulas on page 310, we get y 1 = 0, the objective coeFcient of s 1 , y 2 = 1 the coeFcient of s 2 . So the optimal to the dual is 6 y 1 +10 y 2 = 10, but the optimal to the primal is 20 / 3. This would contradict the strong duality theorem, which says that at the optimal solutions, opt of primal is equal to opt of dual. Section 7.1, 4a. Let x ij = number of tons of steel j produced at plant i. Then the supply at plant 1 is 40(60) / 20 = 120 tons, the supply at plant 2 is 40(60) / 16 = 150 tons, and the supply at plant 3 is 40(60) / 15 = 160 tons. We obtain the following cost and requirement table for the balanced transportation problem:
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Unformatted text preview: Steel 1 Steel 2 Steel 3 Dummy Plant 1 60 40 28 120 Plant 2 50 30 30 150 Plant 3 43 20 20 160 100 100 100 130 4b. In this situation we cannot de±ne the production capacity of a plant in terms of tons of steel produced; this is because, even if produced at the same plant, each type of steel requires a di²erent amount of time to produce. ³or instance, an attempt to determine the plant 1 supply constraint would yield 15 x 11 + 12 x 12 + 15 x 13 ≤ 2400 and this constraint is not of the required form for a supply constraint. #6, 7.1: Let y stand for the cost per check in di²erent sites. We obtain the following cost and requirement table for the balanced transportation problem: Vendor Salary Personnel DUMMY Site 1 5 4 2 10000 Site 2 3 4 5 6000 5000 5000 5000 1000...
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