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

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Unformatted text preview: 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 , , 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 } 12 10 4 { x 2 ,x 4 ,x 5 } 3- 2 1 { x 2 ,x 3 ,x 5 } 5 2 2 3 2 { x 2 ,x 3 ,x 4 } 4- 4- 6 { x 1 ,x 4 ,x 5 } 6 4- 2 { x 1 ,x 3 ,x 5 } 10- 8- 6 { x 1 ,x 3 ,x 4 } 4 4 6 { x 1 ,x 2 ,x 5 } 2...
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This note was uploaded on 11/03/2011 for the course IEOR 162 taught by Professor Zhang during the Fall '07 term at University of California, Berkeley.

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

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