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Slides by John Loucks St. Edward’s University

Slides by John Loucks St. Edward’s University. Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution. Introduction to Sensitivity Analysis Graphical Sensitivity Analysis Sensitivity Analysis: Computer Solution Limitations of Classical Sensitivity Analysis.

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Slides by John Loucks St. Edward’s University

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  1. Slides by John Loucks St. Edward’s University

  2. Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution • Introduction to Sensitivity Analysis • Graphical Sensitivity Analysis • Sensitivity Analysis: Computer Solution • Limitations of Classical Sensitivity Analysis

  3. Introduction to Sensitivity Analysis • In the previous chapter we discussed: • objective function value • values of the decision variables • reduced costs • slack/surplus • In this chapter we will discuss: • changes in the coefficients of the objective function • changes in the right-hand side value of a constraint

  4. Introduction to Sensitivity Analysis • Performed after the objective solution is found in an LP problem • “How much can we change the problem before the optimal solution changes?” • the objective function coefficients • the right-hand side (RHS) values • Sensitivity analysis allows a manager to ask certain what-if questions about the problem.

  5. Graphical Sensitivity Analysis • For LP problems with two decision variables, graphical solution methods can be used to perform sensitivity analysis on • the objective function coefficients • the right-hand-side values for the constraints.

  6. Example Pivoting the solution line around optimal solution

  7. Example If the objective function line pivots outside Of the available area, we have To find the new optimal solution

  8. Example Boundaries of area (or “range of optimality”) defined by Constraints, therefore we can use their slopes to Calculate the boundaries

  9. Example 1: A Simple Maximization Problem Par Inc Solution Let s = number of standard golf bags produced d = number of deluxe golf bags produced Max 10s + 9d Subject to 7/10s + 1d <= 630 (Cutting and Dyeing) 1/2s + 5/6d <= 600 (Sewing) 1s + 2/3d <= 708 (Finishing) 1/10s + 1/4d <= 135 (Inspecting and Packing) s, d >= 0

  10. Par Inc Solution • The objective function will remain optimal if • slope of line b <= slope of optimal line <= slope of line a • (the “higher” slope is the one that is more counterclockwise)

  11. Par Inc Constraint Slopes • We need to find the slopes of the constraint lines • Optimal Solution at intersection of two constraints • C+D: 7/10s + 1d = 630 • F: 1s + 2/3d = 708 • Express in terms of D (or whatever your vertical axis is) • C+D: d = -7/10s + 630 • F: d = -3/2s + 1062

  12. Par Inc Constraints • Slope of the optimal line, in general form • X (horizontal) term over Y (vertical) term, representing the profit (in this case) Profit of standard bag (Cs)Profit of deluxe bag (Cd) • Therefore • -3/2 <= -Cs/Cd <= -7/10

  13. If we want to find the range for standard bags.. • -3/2 <= -Cs/Cd <= -7/10 • Lower Range • Sub in 9 for deluxe profit: -3/2 <= -Cs/9 • Elim. Negatives: 3/2 => Cs/9 • Times 9: 27/2 => Cs • Decimal, switch: Cs <= 13.5 • Upper Range • Sub in 9: -Cs/9 <= -7/10 • Elim. Negatives Cs/9 => 7/10 • Times 9: Cs => 63/10 • Decimal: Cs => 6.3

  14. Range of Optimality for Standard Bag Price • 6.3 <= Cs <= 13.5

  15. If we want to find the range for deluxe bags.. • -3/2 <= -Cs/Cd <= -7/10 • Flip: -2/3 <= -Cd/Cs <= -10/7 • Lower Range • Sub in 10 for std profit: -2/3 <= -Cd/10 • Elim. Negatives: 2/3 => Cd/10 • Times 10: 20/3 => Cd • Decimal, switch: Cs <= 6.67 • Upper Range • Sub in 10: -Cd/10 <= -10/7 • Elim. Negatives Cd/10 => 10/7 • Times 10: Cd => 100/7 • Decimal: Cs => 14.29

  16. Range of Optimality for Deluxe Bag Price • 6.67 <= Cd <= 14.29 • So what’s the problem with this method? • You can only change one value at a time, and not both!

  17. What if the standard bag profit is 18? The slope changes, but if it passes a certain value, the optimal value changes Objective function line, At old optimal point

  18. Right hand Sides (Constraints) • What if the amounts of production hours we have available change? • What if we have 10 extra hours cutting and dyeing time? • From this: 7/10s + 1d <= 630 • To this: 7/10s + 1d <= 640

  19. An increase in hours available Means that the constraint line Slides away from the origin Feasible region may expand as a result

  20. New Optimal Solution Results… • Optimal Solution now • s = 527.5 d = 270.75 • P = 7771.75 Change in P = 43.75 • Change in P per hour = 43.75/10 = 4.375 • “for every extra hour of cutting and dyeing, profit can increase by 4.375” • This change in profit is called the dual value • Marginal increase of profit when x changes by ONE point • Before other constraints become binding

  21. Sensitivity Analysis: Computer Solution

  22. Sensitivity Analysis: Computer Solution

  23. Sensitivity Analysis: Computer Solution

  24. Limitations of Classical Sensitivity Analysis • Can’t do simultaneous changes • Can only change one variable at a time • Provides no info on changing constraints • Ie if we can reduce production time per product • Advanced constraints can become difficult to analyze • Ie d >= 0.3s

  25. Electronic Communications Problem • If we have time

  26. End of Chapter 3

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