MS11 SM03 - Chapter 3 Linear Programming: Sensitivity...

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Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution Learning Objectives 1. Be able to conduct graphical sensitivity analysis for two variable linear programming problems. 2. Be able to compute and interpret the range of optimality for objective function coefficients. 3. Be able to compute and interpret the dual price for a constraint. 4. Learn how to formulate, solve, and interpret the solution for linear programs with more than two decision variables. 5. Understand the following terms: sensitivity analysis range of optimality dual price reduced cost range of feasibility 100 percent rule sunk cost relevant cost 3 - 1
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Chapter 3 Solutions: 1. Note: Feasible region is shown as part of the solution to problem 21 in Chapter 2. Optimal Solution: F = 25, S = 20 Binding Constraints: material 1 and material 3 Let Line A = material 1 = 2/5 F + 1/2 S = 20 Line B = material 3 = 3/5 F + 3/10 S = 21 The slope of Line A = -4/5 The slope of Line B = -2 Current solution is optimal for 4 2 30 5 F C - ≤ - ≤ - or 24 C F 60 Current solution is optimal for 40 4 2 5 S C - ≤ - ≤ - or 20 C S 50 2. Application of the graphical solution procedure to the problem with the enlarged feasible region shows that the extreme point with F = 100/3 and S = 40/3 now provides the optimal solution. The 3 - 2 S F
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Linear Programming: Sensitivity Analysis and Interpretation of Solution new value for the objective function is 40(100/3) + 30(40/3) = 1733.33, providing an increase in profit of $1733.33 - 1600 = $133.33. Thus the increased profit occurs at a rate of $133.33/3 = $44.44 per ton of material 3 added. Thus the dual price for the material 3 constraint is $44.44. 3. a. 8 4 4 8 x 2 x 1 2 6 10 2 6 0 10 Line B x 1 + x 2 = 10 Optimal Solution (3,7) Line A x 1 + 3 x 2 = 24 Feasible Region Optimal Value = 27 b. Slope of Line B = -1 Slope of Line A = -1/3 Let C 1 = objective function coefficient of x 1 C 2 = objective function coefficient of x 2 -1 - C 1 /3 -1/3 1 C 1 /3 C 1 /3 1/3 C 1 3 C 1 1 Range: 1 C 1 3 c. -1 -2/ C 2 -1/3 1 2/ C 2 2/ C 2 1/3 C 2 2 C 2 6 Range : 2 C 2 6 d. Since this change leaves C 1 in its range of optimality, the same solution ( x 1 = 3, x 2 = 7) is optimal. 3 - 3
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Chapter 3 e. This change moves C 2 outside its range of optimality. The new optimal solution is shown below. 8 4 4 8 x 2 x 1 2 6 10 2 6 0 10 Feasible Region 6 1 2 3 4 5 Alternative optimal solutions exist. Extreme points 2 and 3 and all points on the line segment between them are optimal. 4.
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MS11 SM03 - Chapter 3 Linear Programming: Sensitivity...

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