HW5Solution_2010 - Fall 2010 Optimization I (ORIE...

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Fall 2010 Optimization I (ORIE 3300/5300) Assignment 5 Solution Problem 2 (a) We add four slack variables and set up the first tableau: z - x 3 = 0 - x 1 + x 3 + x 4 = 0 x 1 + x 3 + x 5 = 4 - x 2 + x 3 + x 6 = 0 x 2 + x 3 + x 7 = 4 We choose x 3 to enter the basis. The ratio rule shows a tie between x 4 and x 6 as a leaving variable, and we choose x 4 by the min-index rule. The resulting tableau is: z - x 1 + x 4 = 0 - x 1 + x 3 + x 4 = 0 2 x 1 - x 4 + x 5 = 4 x 1 - x 2 - x 4 + x 6 = 0 x 1 + x 2 - x 4 + x 7 = 4 Now, the only choice for the entering variable is x 1 , and the only choice for the leaving vari- able is x 6 . The resulting tableau is: z - x 2 + x 6 = 0 - x 2 + x 3 + x 6 = 0 2 x 2 + x 4 + x 5 - 2 x 6 = 4 x 1 - x 2 - x 4 + x 6 = 0 2 x 2 - x 6 + x 7 = 4 Now, the only choice for the entering variable is x 2 . The ratio rule shows a tie between x 5 and x 7 as a leaving variable, and we choose x 5 by the min-index rule. After this, the final tableau looks like: z 0 . 5 x 4 + 0 . 5 x 5 = 2 x 3 + 0
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This note was uploaded on 01/25/2011 for the course ORIE 5300 taught by Professor Todd during the Fall '08 term at Cornell University (Engineering School).

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HW5Solution_2010 - Fall 2010 Optimization I (ORIE...

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