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IEOR162_hw04_sol - IEOR 162 Spring 2011 Suggested Solution...

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IEOR 162, Spring 2011 Suggested Solution to Homework 04 Problem 1 Solution 1. First, we label the two refinery at Los Angeles and Chicago as refinery 1 and 2 and the two distribution points at Houston and New York City as distribution points 1 and 2. Then let the decision variables be x i = million barrels of capacity created for refinery i , i = 1 , 2, and y ij = million barrels of oil shipped from refinery i to distribution point j , i = 1 , 2, j = 1 , 2. For parameters, let P ij be the profit (in thousands) per million barrels of oil shipped from refinery i to distribution point j , C i be the unit cost (in thousands) of expanding capacity for one million barrel in refinery i , K i be the current capacity (in million barrel) in refinery i , and D j be the demand size (in million barrels) at distribution point j for all i = 1 , 2 and j = 1 , 2: P = 20 15 18 17 , C = 120 150 , K = 2 3 , D = £ 5 5 / . With the definitions of variables and parameters, we formulate the problem as max 10 2 X i =1 2 X j =1 P ij y ij - 2 X i =1 C i x i s.t. 2 X i =1 y ij D j j = 1 , 2 2 X j =1 y ij K i + x i i = 1 , 2 x i , y ij 0 i = 1 , 2 , j = 1 , 2 . The objective function consists of two parts, the 10-year total profit and the one-time expansion cost. The first constraint ensures that the total sales at each distribution point is at most the demand size. The second constraint ensures that the total production quantity at each refinery does not excess the (post-expansion) capacity. The last constraint is the nonnegativity constraint. Solution 2. We may alternatively define, for i = 1 , 2 and j = 1 , 2, w ij = million barrels of “original” capacity allocated to refinery i and distribution point j and z ij = million barrels of “additional” capacity allocated to refinery i and distribution point j . The formulation is max 10 2 X i =1 2 X j =1 P ij ( w ij + z ij ) - 2 X i =1 C i 2 X i =1 z ij s.t. 2 X i =1 ( w ij + z ij ) D j j = 1 , 2 2 X j =1 w ij K i i = 1 , 2 w ij , z ij 0 i = 1 , 2 , j = 1 , 2 . The two formulations are equivalent. To see this, note that x i = 2 j =1 z ij and y ij = z ij + w ij .
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