Chap 7 WE - Worked Examples for Chapter 7 Example for...

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Worked Examples for Chapter 7 Example for Section 7.1 Consider the following linear programming problem. Maximize Z = -5 x 1 + 5 x 2 + 13 x 3 , subject to -1 x 1 + 1 x 2 + 3 x 3 20 12 x 1 + 4 x 2 + 10 x 3 90 and x 1 0, x 2 0, x 3 0. After introducing x 4 and x 5 as the slack variables for the respective constraints and then applying the simplex method, the final simplex tableau is Basic Variable Eq Coefficient of: Right Side Z x 1 x 2 x 3 x 4 x 5 Z (0) 1 0 0 2 5 0 100 x 2 (1) 0 -1 1 3 1 0 20 x 5 (2) 0 16 0 -2 -4 1 10 Now suppose that the right-hand side of the second constraint is changed from b 2 = 90 to b 2 = 70. Using the sensitivity analysis procedure described in Sec. 6.7 for Case 1, the only resulting change in the above final simplex tableau is that the Right Side entry for Eq. (2) changes from 10 to - 10. This revised tableau is shown below, labeled as Iteration 0 (for the dual simplex method). Application of the Dual Simplex Method Because x 5 = -10 < 0, our basic solution that was optimal is no longer feasible. However, since all the coefficients in Eq. (0) still are nonnegative, we can quickly reoptimize by applying the dual simplex method, starting with the Iteration 0 tableau shown below. Since x 5 is the only negative variable in this tableau, it is chosen as the leaving basic variable for the first iteration of the dual simplex method.
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To select the entering basic variable, we consider x 3 and x 4 , since they are the only nonbasic variables that have negative coefficients in Eq. (2). Taking the absolute values of the ratios of these coefficients to the corresponding coefficients in E. (0), 2 2 < 5 4 , so x 3 is selected as the entering basic variable. To use Gaussian elimination to solve for the new basic solution, we divide Eq.
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This note was uploaded on 04/25/2008 for the course ADG 111 taught by Professor Tomas during the Spring '08 term at Uni. Iceland.

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Chap 7 WE - Worked Examples for Chapter 7 Example for...

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