Module5 Handouts

Module5 Handouts - 8/13/2010 LearningModule5

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8/13/2010 1 Learning Module 5 Interpretation and Sensitivity Analysis of Software Solution Output Learning Objectives: Module 5 Interpret the reduced cost in the solution output Identify and interpret the slack and surplus variables in the MS software output Identify and interpret the significance of binding constraints in the MS software output. Write a linear programming model in standard form . Identify and interpret the dual price in the MS software output. Interpret the range for a right hand side in the MS software output. Understand shadow prices and their relation to dual prices. Write an LP model using index notation. Reduced Cost Concept Maximize: ( , ) 3 9 . . PSD S Dst = + Suppose that in the PAR Golf Bag model the profit for the standard bags had been $3 instead of $10, but otherwise the problem is identical. .1 0.7 1 630 .2 0.5 0.8333 600 .3 1 0.6667 708 .4 0.1 0.25 135 ,0 CS D D D D SD +≤
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8/13/2010 2 Solution to Modified PAR Golf Bag Problem MAX 3S+9D s.t. 1) 0.7S+1D<630 2) 0.5S+0.83333D<600 3) 1S+0.66667D<708 4) 0.1S+0.25D<135 OPTIMAL SOLUTION Objective Function Value = 4860.00000 Variable Value Reduced Costs S 0.00000 0.60000 D 540.00000 0.00000 Constraint Slack/Surplus Dual Prices 1 90.00000 0.00000 2 150.00180 0.00000 3 347.99820 0.00000 4 0.00000 36.0000 OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit S No Lower Limit 3.00000 3.60000 D 7.50000 9.00000 No Upper Limit RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit 1 540.00000 630.00000 No Upper Limit 2 449.99820 600.00000 No Upper Limit 3 360.00180 708.00000 No Upper Limit 4 0.00000 135.00000 157.50000 Reduced Cost for Decision Variables The reduced cost for a decision variable tells you how much the coefficient of the decision variable in the objective function must be improved before the optimal value of that decision variable will be positive. Reduced cost for a DV in a MAX model =0.6 means the coefficient of the DV in the objective must increase by more than 0.6 before the optimal value of that DV will be non zero.
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Module5 Handouts - 8/13/2010 LearningModule5

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