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Slides by JOHN LOUCKS St. Edward’s University

Slides by JOHN LOUCKS St. Edward’s University. INTRODUCTION TO MANAGEMENT SCIENCE, 13e Anderson Sweeney Williams Martin. Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution. Introduction to Sensitivity Analysis Graphical 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 INTRODUCTION TO MANAGEMENT SCIENCE, 13e Anderson Sweeney Williams Martin

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

  3. Introduction to Sensitivity Analysis • Sensitivity analysis (or post-optimality analysis) is used to determine how the optimal solution is affected by changes, within specified ranges, in: • the objective function coefficients • the right-hand side (RHS) values • Sensitivity analysis is important to a manager who must operate in a dynamic environment with imprecise estimates of the coefficients. • Sensitivity analysis allows a manager to ask certain what-if questions about the problem.

  4. Example 1 • LP Formulation

  5. Example 1 • Graphical Solution (objective function coefficient)

  6. Example 1 • Graphical Solution (objective function coefficient) ―3/2 <= slope of objective function <= ―7/10

  7. Objective Function Coefficients • The range of optimality for each coefficient provides the range of values over which the current solution will remain optimal. • Objective function coefficient’s range (range of optimality) is just for one variable given that all others are not changed • What if the coefficients are 13 and 8 for S and D respectively. but which is out of range of ―3/2 <= slope of objective function <= ―7/10

  8. Right-Hand Sides • Let us consider how a change in the right-hand side for a constraint might affect the feasible region and perhaps cause a change in the optimal solution. • The improvement in the value of the optimal solution per unit increase in the right-hand side is called the dual price. • The range of feasibility is the range over which the dual price is applicable. • As the RHS increases, other constraints will become binding and limit the change in the value of the objective function.

  9. Example 1 • Graphical Solution (Right Hand Side)

  10. Example 1 • Graphical Solution (Right Hand Side of Constraint 1) Intersection of constraints (3) & (4) : (474.545, 350.182) (1) 7/10*474.545 + 1*350.182 = 682.364 Intersection of S-axis & (3) : (708, 0) (1) 7/10*708 + 1*0 = 495.6

  11. Computer Solutions • Management Scientist

  12. Example 1 • Graphical Solution (Right Hand Side of Constraint 2) No upper limit Intersection of constraints (1) & (3) : (540, 252) (1) 1/2*540 + 5/6*252 = 480

  13. Dual Price • The improvement in the value of the optimal solution per unit increase in the right-hand side is called the dual price. • The dual price for a nonbinding constraint is 0. For >= constraints, dual price of 0 surplus is ― For <= constraints, dual price of 0 slack is + • A negative dual price indicates that the objective function will not improve if the RHS is increased. • The range of feasibility (range of RHS)is the range over which the dual price is applicable (not changed).

  14. Sensitivity Analysis: Computer Solution • Simultaneous Changes • Until now, the sensitivity analysis information is based on the assumption that only one coefficient changes • 100% rule • More than 2 objective coefficientsor more than 2 RHS • Optimal solution basis (positive valued decision variables) are not changedif sum of all the (changes / allowable changes) ratios is less than 1.

  15. Sensitivity Analysis: Computer Solution • Objective function coefficients • Ex1 : • Ex2 : • Ex3 : (not simultaneously binding)

  16. Sensitivity Analysis: Computer Solution • RHS values • Ex1 : • Ex2 : • Ex3 : (not simultaneously binding)

  17. Sensitivity Analysis: Computer Solution • RHS values Global optimal solution found. Objective value: 7390.898 Infeasibilities: 0.000000 Total solver iterations: 3 Variable Value Reduced Cost S 395.4528 0.000000 D 381.8189 0.000000 Row Slack or Surplus Dual Price 1 7390.898 1.000000 2 11.36416 0.000000 3 84.09247 0.000000 4 0.000000 8.727289 5 0.000000 12.72711

  18. Sensitivity Analysis: Computer Solution • RHS values Global optimal solution found. Objective value: 7353.117 Infeasibilities: 0.000000 Total solver iterations: 2 Variable Value Reduced Cost S 406.2477 0.000000 D 365.6266 0.000000 Row Slack or Surplus Dual Price 1 7353.117 1.000000 2 0.000000 4.374957 3 92.18853 0.000000 4 0.000000 6.937530 5 2.968579 0.000000

  19. Sensitivity Analysis: Computer Solution • RHS values Global optimal solution found. Objective value: 8536.745 Infeasibilities: 0.000000 Total solver iterations: 2 Variable Value Reduced Cost S 677.4988 0.000000 D 195.7509 0.000000 Row Slack or Surplus Dual Price 1 8536.745 1.000000 2 0.000000 4.374957 3 98.12555 0.000000 4 0.000000 6.937530 5 18.31241 0.000000

  20. Sensitivity Analysis: Second Example (p.110)

  21. Sensitivity Analysis: Second Example (p.110)

  22. Sensitivity Analysis: Second Example (p.110) • Dual price • The improvement of the objective function value per 1 unit increase of the RHS. • Total production requirement and Processing time are binding • Dual price of processing time is 1 • Dual price of total minimum (350) is -4 • Notes • Dual price is an extra cost. If the profit contribution is calculated considering the purchasing cost of the resource, the price we are willing to pay for that resource is purchasing cost + dual price for 1 unit.

  23. Sensitivity Analysis: Note and Comments • Degeneracy • Consider the available Sewing time is 480 which is calculated with 1/2*540 + 5/6*252

  24. Sensitivity Analysis: Note and Comments • Degeneracy • Consider the available Sewing time is 480 which is calculated with 1/2*540 + 5/6*252 Global optimal solution found. Objective value: 7668.000 Infeasibilities: 0.000000 Total solver iterations: 2 Variable Value Reduced Cost S 540.0000 0.000000 D 252.0000 0.000000 Row Slack or Surplus Dual Price 1 7668.000 1.000000 2 0.000000 4.375000 3 0.000000 0.000000 4 0.000000 6.937500 5 18.00000 0.000000 Objective Coefficient Ranges: Current Allowable Allowable Variable Coefficient Increase Decrease S 10.00000 3.500000 3.700000 D 9.000000 5.285714 2.333333 Righthand Side Ranges: Current Allowable Allowable Row RHS Increase Decrease 2 630.0000 0.000000 134.4000 3 480.0000 INFINITY 0.000000 4 708.0000 192.0000 0.000000 5 135.0000 INFINITY 18.00000

  25. Sensitivity Analysis: Note and Comments • Degeneracy • Consider the available Sewing time is 480 which is calculated with 1/2*540 + 5/6*252 • In the standard form number of variables (2+3=5),number of constraints 3. Thus, basic solution has (set 2 variables to 0, and solve simultaneous equations (연립방정식).Now, at the optimal solution 3 variables are 0. • Dual price of binding constraints is 0. • Constraint 2 (Sewing) has 0 slack, but dual price is 0 • Range of Feasibility (range of RHS) for constraints 2, 3 and 4 are only one direction. • 100% rule works only when sum of ratios are less than 100.

  26. Example 3 (more than 2 variables) • Consider the following linear program: Max 10S + 9D + 12.85L s.t.0.7S + 1D + 0.8L <630 0.5S + 5/6D + 1L< 600 1S + 2/3D + 1L < 708 0.1S + 0.25D + 0.25L < 135 x1, x2> 0

  27. Example 3

  28. Example 3 (more than 2 variables) • Interpretation • Deluxe model is not produced • Finishing (Constraint 3) and Inspection and Packaging (Constraint 4) are binding • Range of objective function for Deluxe is ― infinity < current 9 < 10.15 • Reduced cost : the amount that an objective function coefficient would have to improve in order for the corresponding decision variable becomes positive. Reduced cost of Deluxe is 1.15 = 1 * 0 + 5/6*0 + 2/3*8.1 + 0.25*19 – 9(sum of dual prices consumed to produce 1 unit of Deluxe)

  29. Example 3 (more than 2 variables) • Primal problem vs. Dual problem Max 10S + 9D + 12.85L 0.7S + 1D + 0.8L <630 0.5S + 5/6D + 1L< 600 1S + 2/3D + 1L < 708 0.1S + 0.25D + 0.25L < 135 S, D, L> 0 Min 630C + 600W + 708F + 135I 0.7C + 0.5W + 1F + 0.1I –R1 = 10 1C + 5/6W + 2/3F + 0.25I –R2 = 9 0.8C + 1W + 1F + 0.25I –R3 = 12.85 C, W, F, I, R1, R2, R3> 0

  30. Example 3 (more than 2 variables) • Primal problem vs. Dual problem • Primal problem maximize the total profit contribution with the constraints of limited available resources • Dual problemminimize the total cost allocation to resources with the constraints of guaranteeing the minimum profitability.

  31. Example 3 (more than 2 variables) • Alternative optimal solution (p.116 Fig. 3.7) • Profit contribution of Deluxe is 10.15 • Slack of Constraint 1 is 0, but the dual price is also 0. • Range of optimality (range of objective function coefficient) has one direction • If the primal problem has an alternative optima, the dual is degenerate and vice versa. • Extra constraint (p.117 Fig. 3.8) • Deluxe should be produces at least 30% of standard bag. D > 0.3S  –0.3S + D > 0 • Dual price –1.38 means that the total profit will decrease if Deluxe is produce 1 more than 30% of standard bag.

  32. Example 4 (Bluegrass Farms Problem, p.118) • Decision variables S = pounds of standard horse feed product to feed E = pounds of vitamin-enriched oat product to feed A = pounds of new vitamin and mineral feed additive Min 0.25S + 0.5E + 3A 0.8S + 0.2E + 0.0A > 3 1S + 1.5E + 3.0A> 6 0.1S + 0.6E + 2.0A > 4 1S + 1E + 1A < 6 S, E, A> 0

  33. Example 4 (Bluegrass Farms Problem, p.118)

  34. Example 4 (Bluegrass Farms Problem, p.118) • Interpretation • What’s the optimal decision? • What’s the optimal cost? • Which constraint has slack/surplus? • What the dual prices for binding constraints?0.919 of maximum weight meansif maximum weight requirement is increased, some cheaper product will be feed to meet the requirements of ingredients by allowing more weights • Explain with the ranges of objective functionWhat will happen if the standard horse feed product is free • Explain with the ranges of RHS

  35. Example 5(Electronic Communication Problem, p.123) • Maximize or minimize • What are the constraints How many? • Decision variables M = number of unit to produce for the marine equipment distribution channel B = number of units to produce for the business equipment distribution channel R = number of units to produce for the national retail chain distribution channel D = number of units to produce for the direct mail distribution channel

  36. Example 5(Electronic Communication Problem, p.123) • Model Formulation

  37. Example 5(Electronic Communication Problem, p.123)

  38. Example 5(Electronic Communication Problem, p.123)

  39. Example 5(Electronic Communication Problem, p.123) • Interpretation • What’s the optimal decision? • What’s the optimal cost? • What should be the profit for the direct mail channel in order to produce some for the direct model? • Which constraint has slack/surplus? • What the dual prices for binding constraints? • Explain with the ranges of objective function • Explain with the ranges of RHS What if the production requirement of 600 is changed to 601? How much of the advertising budget is allocated to business distributors?

  40. Ch.3 Homework • Q29 on p.149 • Formulate the model • Solve with Excel In Excel, you choose all options of 보고서 after 해찾기to get the output of sensitivity analysis • Solve with LINGO • Answer all questions on p.149 Q29. • Put all output answers in one file except Excel file and upload through mis3nt.gnu.ac.kr

  41. End of Chapter 3

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