sol6 - CO 350 Linear Optimization (Spring 2009) Solutions...

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Unformatted text preview: CO 350 Linear Optimization (Spring 2009) Solutions to Assignment 5 Questions 710 7. (a) The first row is z + 1 + 5 12 x 4 + 0 + 1 12 x 5 + 0 + 5 12 x 6 = 14 + 1 4 with cut 5 12 x 4 + 1 12 x 5 + 5 12 x 6 1 4 . The second row is x 3 + 0 + 11 60 x 4 +- 1 + 11 12 x 5 +- 1 + 59 60 x 6 = 1 + 3 4 with cut 11 60 x 4 + 11 12 x 5 + 59 60 x 6 3 4 The third row is x 2 + 0 + 1 10 x 4 + 0 + 1 2 x 5 +- 1 + 9 10 x 6 = 1 + 1 2 with cut 1 10 x 4 + 1 2 x 5 + 9 10 x 6 1 2 The forth row is x 1 + 0 + 1 12 x 4 + 0 + 5 12 x 5 + 0 + 1 12 x 6 = 1 + 1 4 with cut 1 12 x 4 + 5 12 x 5 + 1 12 x 6 1 4 (b) For the first cut we have- 5 12 x 4- 1 12 x 5- 5 12 x 6 + s =- 1 4 , the leaving variable is s . Compute min 17 12 12 5 , 1 12 12 1 , 5 12 12 5 = 1 1 , so x 5 (or x 6 ) enters. For the second cut we have- 11 60 x 4- 11 12 x 5- 59 60 x 6 + s =- 3 4 the leaving variable is s . Compute min 17 12 60 11 , 1 12 12 11 , 5 12 60 59 = 1 11 , 1 the entering variable is x 5 . For the third cut we have- 1 10 x 4- 1 2 x 5- 9 10 x 6 + s =- 1 2 the leaving variable is s . Compute min 17 12 10 1 , 1 12 2 1 , 5 12 10 9 = 1 6 , the entering variable is x 5 . For the last cut we have- 1 12 x 4- 5 12 x 5- 1 12 x 6 + s =- 1 4 min 17 12 12 1 , 1 12 12 5 , 5 12 12 1 = 1 5 , again the entering variable is x 5 . (c) From the initail ILP we have x 4 = 10- x 1- 5 x 3 , x 5 = 1- x 1- x 2 + x 3 and x 6 =- 6 x 1 + 5 x 2 , substituting into the cuts we get 1. 3 x 1- 2 x 2 + 2 x 3 4, 2. 7 x 1- 4 x 2 2, 3. 6 x 1- 4 x 2 1 and 4. x 1 1....
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This note was uploaded on 04/11/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Fall '07 term at Waterloo.

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sol6 - CO 350 Linear Optimization (Spring 2009) Solutions...

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