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IEOR162_hw05_sol - IEOR 162 Fall 2011 Suggested Solution to...

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IEOR 162, Fall 2011 Suggested Solution to Homework 05 Problem 1 (Problem 4.5.2) We run two iterations to get - 2 - 3 0 0 0 1 2 1 0 6 2 1 0 1 8 0 - 2 0 1 8 0 3 2 1 - 1 2 2 1 1 2 0 1 2 4 0 0 4 3 1 3 32 3 0 1 2 3 - 1 3 4 3 1 0 - 1 3 2 3 10 3 An optimal solution to the original problem is ( x * 1 , x * 2 ) = ( 4 3 , 10 3 ) with objective value is z * = 32 3 . Note. You may enter x 2 instead of x 1 in the first iteration and see that it ends up with the same result. In general, you may always follow the smallest index rule by selecting the entering variable as the one with the smallest index. This will always allow you to find an optimal solution if one exists. Problem 2 (Modified from Problem 4.6.1) We first convert the original problem to z * = min 4 x 1 - x 2 s.t. 2 x 1 + x 2 8 x 2 5 x 1 - x 2 4 x 1 , x 2 0 so that all variables are nonnegative. Then we run one iteration to get - 4 1 0 0 0 0 2 1 1 0 0 8 1 1 0 1 0 5 1 - 1 0 0 1 4 - 4 0 0 - 1 0 - 5 2 0 1 - 1 0 3 0 1 0 1 0 5 1 0 0 1 1 9 Note that we enter a variable if it has a positive number in the objective row. An optimal solution to the new problem is ( x 0 1 , x 0 2 ) = (0 , 5). Therefore, an optimal solution to the original problem is (
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