hw04solutions - IEOR 162 Linear Programming Spring 2007...

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IEOR 162 Linear Programming Spring 2007 Homework 4 Solutions 3.12.4 (4 points) Data: T = the number of months p t = purchase price per 1000 bushels of wheat during month t s t = selling price per 1000 bushels of wheat during month t Variables: x t = amount of wheat (in 1000 bushel units) to purchase during month t y t = amount of wheat (in 1000 bushel units) to sell during month t I t = inventory of wheat (in 1000 bushel units) at the end of month t Formulation: max T t =1 ( s t y t - p t x t ) s.t. I 0 = 6 (1) I t = I t - 1 + x t - y t t = 1 , . . . , T (2) y t I t - 1 t = 1 , . . . , T (3) I t 20 t = 1 , . . . , T (4) x t , y t , I t 0 t = 1 , . . . , T (5) The objective maximizes the revenue earned from selling minus the costs of purchasing wheat. Constraint 1 sets our initial inventory to 6000 bushels as specified. Constraint set 2 indicates that the inventory on hand at the end of month t is equal to last month’s inventory plus whatever we buy minus whatever we sell. Constraint set 3 asserts that we cannot sell more than the amount we had on hand at the end of the previous month (stipulated in part b of the problem). Constraint set 4 ensures that we do not exceed our
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