# sol6 - CO 350 Linear Optimization(Spring 2009 Solutions to...

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CO 350 Linear Optimization (Spring 2009) Solutions to Assignment 5 Questions 7–10 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

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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. 8. (a) From the x 1 -row of the simplex tableau, we see that all feasible solutions must satisfy x 1 + 1 4 x 3 + 5 4 x 4 - 1 2 x 6 = 9 2 . Since x j 0 for all j , all feasible solutions must satisfy the inequality: x 1 + 1 4 x 3 + 5 4 x 4 + - 1 2 x 6 9 2 .
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