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Assignment3Solution _2010

# Assignment3Solution _2010 - Fall 2010 Optimization I(ORIE...

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Unformatted text preview: Fall 2010 Optimization I (ORIE 3300/5300) Assignment 3 Solution Problem 2 Grading Scheme: 6 points for simplex, 2 points for picture, 2 points for finding other solution. -1 if not finding alternate solution directly from the final tableau. After introducing the slack variables, the initial tableau is as follows: z- x 1- 2 x 2- x 3 = 0 x 1 + x 4 = 3 x 2 + x 5 = 3 x 1 + x 2 + x 6 = 4- x 1 + x 3 + x 7 = 1 The current BFS is x = [0 , , , 3 , 3 , 4 , 1] T , and the corresponding point in the polyhedron is [ x 1 ,x 2 ,x 3 ] T = [0 , , 0] T . Using the most negative reduced cost rule, x 2 will enter the basis (its reduced cost is 2, the negative of its coefficient in the top row). x 5 will leave the basis according to the minimum ratio test (3 / 1 < 4 / 1, ratios 3 / 0 and 1 / 0 ignored). Thus, the next tableau is: z- x 1- x 3 + 2 x 5 = 6 x 1 + x 4 = 3 x 2 + x 5 = 3 x 1- x 5 + x 6 = 1- x 1 + x 3 + x 7 = 1 The current BFS is x = [0 , 3 , , 3 , , 1 , 4] T , and the corresponding point in the polyhedron is [ x 1 ,x 2 ,x 3 ] T = [0 , 3 , 0] T . Now,using the least index rule, x 1 enters the basis, and x 6 leaves the basis because 1 / 1 < 3 / 1. The next tableau is: z- x 3 + x 5 + x 6 = 7 x 4 + x 5- x 6 = 2 x 2 + x 5 = 3 x 1- x 5 + x 6 = 1 x 3- x 5 + x 6 + x 7 = 2 The current BFS is x = [1 , 3 , , 2 , , , 2] T , and the corresponding point in the polyhedron is [ x 1 ,x 2 ,x 3 ] T = [1 , 3 , 0] T . Now, x 3 enters the basis, and x 7 leaves the basis. The next tableau is: z + 2 x 6 + x 7 = 9 x 4 + x 5- x 6 = 2 x 2 + x 5 = 3 x 1- x 5 + x 6...
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Assignment3Solution _2010 - Fall 2010 Optimization I(ORIE...

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