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IEOR162_hw04_sol

# IEOR162_hw04_sol - IEOR 162 Fall 2011 Suggested Solution to...

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IEOR 162, Fall 2011 Suggested Solution to Homework 04 Problem 1 (Modified from 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 the table below. x 1 x 2 x 3 x 4 x 5 Extreme Point 0 0 100 80 40 H 0 100 0 - 20 40 (Infeasible) 0 80 20 0 40 D 0 N/S N/S N/S 0 (No solution) 50 0 0 30 - 10 (Infeasible) 80 0 - 60 0 - 40 (Infeasible) 40 0 20 40 0 E 20 60 0 0 20 G 40 20 0 20 0 F 40 40 - 20 0 0 (Infeasible) 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 x 1 + x 5 = 40 (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 points of the feasible region according to Figure 1, if the basic solution is feasible.

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