Set_6_Sensitivity - Dr Maddah ENMG 500 Sensitivity Analysis...

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1 Dr. Maddah ENMG 500 11/19/07 Sensitivity Analysis 1 Changes in the RHS (b) Consider an optimal LP solution. Suppose that the original RHS ( b ) is changed from b 0 to b new . In the following, we study the affect of this change the optimal solution. First, note that changing b will not affect optimality. Recall that optimality is based on the condition 1 0, B j j j j c z c c B A for all variables x j (for max problems). Changing b will not affect z j c j . Hence, changing b will not affect optimality. However; changing b may affect feasibility. To check, whether the LP is still feasible after changing b from b 0 to b new compute x new = B -1 b new . If x new 0 (i.e., 0, new i x for all i ), then the LP is still feasible and the optimal solution is x new . If 0, new i x for some i , then the current basic solution is no longer feasible. Then, the dual simplex method is used to restore feasibility (leading to a new optimal basis). Note that when the LP remains feasible the rate of change of the optimal objective function, Z * ,as a function of a RHS value b i is y i * the corresponding dual variable. This follows by noting that * * i i y b Z and * * i i y b Z . Therefore, changing the b i by 1 unit will change Z * by y i * (assuming that this change will not affect feasibility ) Example 1. consider the LP max Z = 3 x 1 + 2 x 2 + 5 x 3 s.t. x 1 + 2 x 2 + x 3 430 3 x 1 + 2 x 3 460 x 1 + 4 x 2 420 x 1 , x 2 , x 3 0 Let S 1 , S 2 , S 3 be the slack variables corresponding to the three constraints. Upon solving the LP, the optimal simplex tableau is as follows.
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2 x 1 x 2 x 3 S 1 S 2 S 3 RHS Z 4 0 0 1 2 0 1350 x 2 -1/4 1 0 1/2 -1/4 0 100 x 3 3/2 0 1 0 1/2 0 230 S 3 2 0 0 -2 1 1 20 The optimal solution is x 1 * = 0, x 2 * = 100, x 3 * = 230, and Z * = 1350. Suppose that b is changed from 420 460 430 to 600 460 500 . What is the new optimal solution? 1 1 1 0 2 4 500 135 1 0 0 460 230 0 2 600 60 2 1 1 new new x B b . Since x new is feasible. Then, the new optimal solution is x *new = x new . That is, the new optimal solution is x 1 * new = 0, x 2 * new = 135, x 3 * new = 230, and the new optimal objective value is Z * new = c B x *new = 60 230 135 ) 0 5 2 ( = 1420. Note that Z *new can also be obtained using duality. From the simplex tableau, the optimal dual solution is y 1 * = 1, y 2 * = 2, y 3 * = 0. Then, Z *new = Z * + ( b 1 new b 1 ) y 1 * + ( b 2 new b 2 ) y 2 * + ( b 3 new b 3 ) y 3 * = 1350 + (500 430) 1 + (460 460) 2 + (600 420) 0 = 1420 . What if b is changed from 420 460 430 to 400 460 450 ? new 1 1 1 0 2 4 450 110 1 0 0 460 230 2 400 40 2 1 1 x B b .
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