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Unformatted text preview: Fall 2010 Optimization I (ORIE 3300/5300) Assignment 8 Solution Problem 1 We are starting with A = 3 2 1 1 0 2 1 1 0 1 , b = 7 4 , c = 2 3 2 0 0 T . Iteration 1: For the Basis B = [4 , 5] , we have A B = 1 0 0 1 , A 1 B = 1 0 0 1 , c B = , x B = 7 4 . Step 1: We first compute the vector y : y = ( A T B ) 1 c B = ( A 1 B ) T c B = 1 0 0 1 = . Step 2: To choose an entering index k , we note j c j A T j y 1 2 > 2 3 > 3 2 > so we can choose k = 2 as the entering index according to the most positive reduced cost. Step 3: We next compute the vector d : d = A 1 B A k = 1 0 0 1 2 1 = 2 1 . Step 4: To choose a leaving index by the min ratio test, we note t = min 7 2 , 4 1 = 3.5 ( i = 4 , 5 ) so we can choose r = B (1) = 4 as the leaving index. Step 5: We update the basis and the current values of the basic variables: x 2 t = 3 . 5 x 5 x 5 td 5 = 4 3 . 5 1 = 0 . 5 B B { k }\{ r } = [2 , 5] Iteration 2: For the Basis B = [2 , 5] , we have A B = 2 0 1 1 , A 1 B = 1 / 2 1 / 2 1 , c B = 3 , x B = 3 . 5 . 5 ....
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 Fall '08
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