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Unformatted text preview: Math464  HW 8 Due on Tuesday, Mar 23 1 Linear Optimization (Spring 2010) Brief solutions to Homework 8 1. The simplex method with default rules cycles (Table 1). Table 1: Simplex with default rules (Problem 1). rhs x 1 x 2 x 3 x 4 x 5 x 6 x 7 10 57 9 24 x 5 = 1 / 2 11 / 2 5 / 2 9 1 x 6 = 1/2 3 / 2 1 / 2 1 1 x 7 = 1 1 1 53 41 204 20 x 1 = 1 11 5 18 2 x 6 = 4 2 8 1 1 x 7 = 1 11 5 18 2 1 29 / 2 98 27/4 53/4 x 1 = 1 1 / 2 4 3 / 4 11/4 x 2 = 1 1/2 2 1 / 4 1/4 x 7 = 1 1 / 2 4 3/4 11 / 4 1 29 1815 93 x 3 = 2 1 8 3 / 2 11/2 x 2 = 1 1 2 1/2 5 / 2 x 7 = 1 1 1 20 9 21 / 2 141/2 x 3 = 2 4 1 1 / 2 9 / 2 x 4 = 1 / 2 1/2 1 1/4 5 / 4 x 7 = 1 1 1 22 93 21 24 x 5 = 4 8 2 1 9 x 4 = 1/2 3 / 2 1 / 2 1 1 x 7 = 1 1 1 10 57 9 24 x 5 = 1 / 2 11 / 2 5 / 2 9 1 x 6 = 1/2 3 / 2 1 / 2 1 1 x 7 = 1 1 For the lexicographic pivoting rule (Table 2), we could use the same criterion for the entering variable the nonbasic variable with the most negative reduced cost enters. The first two pivots will be the same as in Table 1. The candidates for the lexicographically smallest row are indicated on the right side (in appropriate detail if there is only one candidate, there is no need to scale that row just to find the...
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This note was uploaded on 02/15/2011 for the course EECS 6.231 taught by Professor Bertsekas during the Spring '10 term at MIT.
 Spring '10
 Bertsekas
 Optimization

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