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IEOR162_hw08_sol

# IEOR162_hw08_sol - IEOR 162 Fall 2011 Suggested Solution to...

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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 0 1 = c 1 + Δ be the new objective coefficient. Since 1 I N , we have ¯ c 0 N = c B B - 1 N - c N + Δ £ 1 0 0 / · = ¯ c N - £ Δ 0 0 / = £ 3 4 1 / - Δ £ 1 0 0 / = £ 3 - Δ 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 - Δ 4 1 / 0 Δ 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. If the profit becomes 7 cents, the new reduced cost will become £ - 1 4 1 / and we know the current basis is not optimal. We should then enter x 1 (produce more Type 1 candy bar since it

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