This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
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.
 Fall '07
 S.Furino,B.Guenin

Click to edit the document details