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IEOR162_hw05_sol - IEOR 162 Spring 2012 Suggested Solution...

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IEOR 162, Spring 2012 Suggested Solution to Homework 05 Problem 1 (Modified from Problem 4.5.2) (a) 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 . (b) We run two iterations to get - 2 - 3 0 0 0 1 2 1 0 6 2 1 0 1 8 - 1 2 0 3 2 0 9 1 2 1 1 2 0 3 3 2 0 - 1 2 1 5 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 . The optimal solution is identical to the one we found in Part (a). (c) The route is depicted as the thick arrows in Figure 1. It starts at point A (0 , 0), passes through point B (4 , 0), and then stop at point C ( 10 3 , 4 3 ). Point C is the unique optimal solution. (d) The route is depicted as the dotted arrows in Figure 1. It starts at point A (0 , 0), passes through point D (0 , 3), and then stop at point C ( 10 3 , 4 3 ). Point C is the unique optimal solution.
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