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Unformatted text preview: 1 Linear Programming Sensitivity Analysis 2 Reminders Read Chapter 4 No class on Monday, September 5, because of Labor Day. Help session Tuesday, September 6, from 6 to 8 pm. Homework Due in Julies Office (519 Krannert) by 2:30 pm Wednesday September 7th. Be sure to include the name of your partner on the homework and the section you are both in. Quiz Monday September 12. Closed book, closed notes May use 8.5x11 double sided formula sheet Calculator and ruler allowed Covers material on HW 1. 3 Sensitivity Analysis: Postoptimality Analysis Many of the input parameters are only estimates and need to be refined if the model output is sensitive to small changes in these parameters. Possible future changes in a dynamic problem environment need to be easily analyzed without resolving the model. When certain parameters in the model represent managerial policy decisions, postoptimality analysis provides guidance to management about the impact of altering these policies. 4 Sensitivity Analysis Sensitivity analysis is to determine how the optimal solution and optimal objective value are affected by changes in the model input data (parameters). We investigate some possible change of objective function coefficient righthand side (RHS) of constraint to see how it influences the optimal solution and optimal objective value. 5 Changes of Objective Function Coefficients If an objective function coefficient changes, as long as the changed coefficient is not too far from the original coefficient, the optimal solution is still optimal for the changed model. In other words, as long as the changed coefficient is located in some range (interval) that contains the original coefficient, the optimal solution remains optimal for the new model. For each objective function coefficient, the range of numbers for the coefficient over which the optimal solution will remain optimal is called the range of optimality . 6 Example 1: CatchBig Problem LP model: Max 5 x1 + 7 x2 s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1, x2 > Optimal solution: x1 = 5 , x2 = 3 . Optimal objective value: 46 7 Example 1: Graphical Solution x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 2x1 + 3x2 < 19 x1 x1 + x2 < 8 objective function line x1 < 6 optimal x1 = 5, x2 = 3 8 Example 1: Effect of Changing Objective Coefficients Changing slope of objective function 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 Feasible Region 1 2 3 x2 Changing a coefficient in the objective function changes the slope of the objective function line....
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 Spring '08
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