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IEOR162_hw06_sol

# IEOR162_hw06_sol - IEOR 162 Fall 2011 Suggested Solution to...

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IEOR 162, Fall 2011 Suggested Solution to Homework 06 Problem 1 (a) The basic feasible solution is x = ( x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) = (0 , 2 , 0 , 9 , 12 , 0). (b) It is not degenerate because all basic variables are positive. (c) The candidates are x 1 and x 6 as they have negative values in the objective row. Because x 1 ’s index is smaller, x 1 should be the entering variable. (d) The ratio test suggests that x 5 (in the first row) and x 4 (in the last row) both have the smallest ratio (3) and are candidates. Because x 4 ’s index is smaller, x 4 should be the leaving variable. Problem 2 (Modified from Problem 4.11.1) (a) The ten possible ways to choose two nonbasic variables to be 0 are listed in the table below. The first column indicates the bases. The last three bases correspond to a single basic solution. The basic variable that is equal to 0 is indicated with a box. Basis x 1 x 2 x 3 x 4 x 5 { x 3 , x 4 , x 5 } 0 0 12 10 4 { x 2 , x 4 , x 5 } 0 3 0 - 2 1 { x 2 , x 3 , x 5 } 0 5 2 2 0 3 2 { x 2 , x 3 , x 4 } 0 4 - 4 - 6 0 { x 1 , x 4 , x 5 } 6 0 0 4 - 2 { x 1 , x 3 , x 5 } 10 0 - 8 0 - 6 { x 1 , x 3 , x 4 } 4 0 4 6 0 { x 1 , x 2 , x 5 } 2 2 0 0 0 { x 1 , x 2 , x 4 } 2 2 0 0 0 { x 1 , x 2 , x 3 } 2 2 0 0 0 (b) We run three iterations to get - 3 - 7 0 0 0 0 2 4 1 0 0 x 3 = 12 1 4 0 1 0 x 4 = 10 1 1 0 0 1

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IEOR162_hw06_sol - IEOR 162 Fall 2011 Suggested Solution to...

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