Recitation9-1 - Fall 2010 Optimization I (ORIE 3300/5300)...

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Optimization I (ORIE 3300/5300) Recitation 9 An example of the revised simplex method Use the revised simplex method to solve the following LP: max 60 x 1 + 30 x 2 + 20 x 3 s.t. 8 x 1 + 6 x 2 + x 3 48 4 x 1 + 2 x 2 + 1 . 5 x 3 20 2 x 1 + 1 . 5 x 2 + 0 . 5 x 3 8 x 1 , x 2 , x 3 0 . We can write it in standard form as: max c T x s.t. Ax = b x 0 where A = 8 6 1 1 0 0 4 2 1 . 5 0 1 0 2 1 . 5 0 . 5 0 0 1 , b = 48 20 8 , c = 60 30 20 0 0 0 ,x = x 1 x 2 x 3 x 4 x 5 x 6 . Initial tableau The set of basic variables is B = [5 , 6 , 7] and the set of nonbasic variables is N = [1 , 2 , 3]. So, A B = 1 0 0 0 1 0 0 0 1 , A N = 8 6 1 4 2 1 . 5 2 1 . 5 0 . 5 , c B = 0 0 0 ,c N = 60 30 20 . Iteration 1 With this current basis, the basic variables are: ˆ x B = A - 1 B b = b = 48 20 8 . Step 1: y = ( A T B ) - 1 c B = 1 0 0 0 1 0 0 0 1 0 0 0 = 0 0 0 . Step 2: Choose an entering index k N such that c k > y T A k . Equivalently, choose k N such that ¯ c k = c k - y T A k > 0. If there are more than one such index, choose the one with the most positive value of ¯ c k . ¯ c T N = c T N - y T A N = (60 , 30 , 20) - (0 , 0 , 0) 8 6 1 4 2 1 . 5 2 1 . 5 0 . 5 = (60 , 30 , 20) . 1
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This note was uploaded on 11/11/2010 for the course ORIE 3300 at Cornell University (Engineering School).

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Recitation9-1 - Fall 2010 Optimization I (ORIE 3300/5300)...

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