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Unformatted text preview: IEOR 162, Fall 2011 Suggested Solution to Homework 08 Problem 1 (Modified from Problem 6.3.6) For this problem, we have A = 1 1 1 1 0 2 3 1 0 1 , c = 3 7 5 0 0 / , and b = 50 100 , where A is the constraint matrix, b is the RHS vector, and c is the objective coefficient vector. Note that we obtain A and c from the standard form, not the original program. In the optimal tableau, we choose I B = { 3 , 2 } and I N = { 1 , 4 , 5 } as the index sets of basic and nonbasic variables. 1 It then follows that B = 1 1 1 3 , N = 1 1 0 2 0 1 , c B = 5 7 / , c N = 3 0 0 / , and c N = 3 4 1 / . (a) Let c 1 = c 1 + be the new objective coefficient. Since 1 I N , we have c N = c B B 1 N c N + 1 0 0 / = c N 0 0 / = 3 4 1 / 1 0 0 / = 3 4 1 / as our new reduced cost. Note that we use 1 0 0 / since we define x 1 as the first nonbasic variable. We need this to be nonnegative, so 3 4 1 / 3 . Therefore, the current basis remains optimal if the profit of the Type 1 candy bar is no greater than 3 + 3 = 6 cents per unit....
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 Fall '07
 Zhang

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