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IEOR162_hw06

# IEOR162_hw06 - IEOR 162 Spring 2010 Suggested Solution to...

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IEOR 162, Spring 2010 Suggested Solution to Homework 06 Problem 4.1.1 The standard form is max 3 x 1 + 2 x 2 s.t. 2 x 1 + x 2 + x 3 = 100 x 1 + x 2 + x 4 = 80 x 1 + x 5 = 40 x j 0 j = 1 , ..., 5 . Problem 4.1.2 The standard form is min 50 x 1 + 100 x 2 s.t. 7 x 1 + 2 x 2 - x 3 = 28 2 x 1 + 12 x 2 - x 4 = 24 x j 0 j = 1 , ..., 4 . Problem 4.1.3 The standard form is min 3 x 1 + x 2 s.t. x 1 - x 3 = 3 x 1 + x 2 + x 4 = 4 2 x 1 - x 2 = 3 x j 0 j = 1 , ..., 4 . Problem 4.4.1 Since in the standard form we have five variables and three constraints, there should be three basic variables and two nonbasic variables in a basic solution. The ten possible ways to choose two (nonbasic) variables to be 0 are listed in Table 1. For each possibility, we try to solve the remaining three basic variables. Note that when x 1 and x 5 are chosen to be nonbasic, we can not find any solution that satisfies constraint 3 (the entries “N/S” means “no solution”). This happens because the constraint x 1 40 is parallel to the constraint x 1 0. The last column indicates the corresponding extreme point of the feasible region according

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