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Readings

Readings. Readings. Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution. Overview. Overview. Overview. Proportion Constraints. Proportion Constraints. Proportion Constraints. Overview

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Readings

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  1. Readings • Readings • Chapter 3 • Linear Programming: Sensitivity Analysis and Interpretation of Solution

  2. Overview • Overview

  3. Overview

  4. Proportion Constraints Proportion Constraints

  5. Proportion Constraints Overview Proportion Constraints restrict the size of one decision variable as a proportion of another. For example, B > 0.1J or B - 0.1J >0 says that variable B is at least 10% of variable J.

  6. Proportion Constraints Question: Slater & Gordon Lawyers is ready to award contracts for printing their annual reports. Past reports were printed by Johnson Printing and Lakeside Litho. A new firm, Benson Printing, is interested in doing some of the printing. The quality of Lakeside Litho has been high, with only 0.5% of their reports being unacceptable. Johnson Printing has also been high, with only 1% of reports unacceptable. Because Slater & Gordon has no experience with Benson Printing, they estimate their unacceptable rate at 10%. Slater & Gordon needs to determine how many reports should be printed by each firm to obtain 75,000 acceptable reports.

  7. Proportion Constraints To ensure Benson Printing will receive some of the contract, management specifies that Benson must print at least 10% of the number printed by Johnson. To spread the risk of unacceptable reports, the reports assigned to Benson, Johnson, and Lakeside must not exceed 30000, 50000, and 50000, respectively. To reward their long-standing relationship, Lakeside will be assigned at least 30,000 reports. The unit cost per report (not just acceptable reports) is $2.45 for Benson, $2.50 for Johnson, and $2.75 for Lakeside.

  8. Proportion Constraints Formulate and solve a linear program to determine how many copies should be assigned to each printer to minimize the cost of 75,000 acceptable reports. Suppose management reconsiders the requirement that Lakeside be awarded at last 30,000 reports. What affect does that have?

  9. Proportion Constraints Answer: First, formulate the problem: • Define decision variables: • B = number of copies done by Benson Printing • 0.9 B is the number of acceptable-quality reports (10% defective). • J = number of copies done by Johnson Printing • 0.99 L is the number of acceptable-quality reports (1% defective). • L = number of copies done by Lakeside Litho • 0.995 B is the number of acceptable-quality reports (0.5% defective).

  10. Proportion Constraints • Minimize (cost) 2.45B + 2.5J + 2.75L subject to these constraints: • B <30,000; J <50,000; L <50,000 (volume should not exceed) • 0.9B + 0.99J + 0.995L = 75,000 (useful reports, = means no disposal) • B - 0.1J >0 (Benson at least 10% of Johnson); L >30,000 (Lakeside min.)

  11. Proportion Constraints • Minimize (cost) 2.45B + 2.5J + 2.75L subject to these constraints: • B <30,000; J <50,000; L <50,000 (volume should not exceed) • 0.9B + 0.99J + 0.995L = 75,000 (useful reports, = means no disposal) • B - 0.1J >0 (Benson at least 10% of Johnson); L >30,000 (Lakeside min.)

  12. Proportion Constraints Hence, answer questions about Minimize (cost) 2.45B + 2.5J + 2.75Ls.t. • B <30,000; J <50,000; L <50,000 (volume should not exceed) • 0.9B + 0.99J + 0.995L = 75,000 (useful reports, = means no disposal) • B - 0.1J >0 (Benson at least 10% of Johnson); L >30,000 (Lakeside min.) • How many copies should be assigned to each printer? Rounding off to the nearest integer, B = 4181, J = 41806, L = 30000 (we later see that rounding is not always optimal) • Suppose management reconsiders the requirement that Lakeside be awarded at last 30,000 reports. What affect does that have? For every unit increase in the Lakeside minimum (up to 50,000), costs increase $0.221. For every unit decrease in the minimum (down to 21,106), costs decrease $0.221.

  13. Portfolio Selection with Risk Indices Portfolio Selection with Risk Indices

  14. Portfolio Selection with Risk Indices Overview Portfolio Selection Problems with Risk Indices use one specific measure of the risk of a portfolio. When the risk index is part of a linear risk function, that measure makes risk minimization a linear-programming problem. (But the leading measures of risk are variance or standard deviation, which are not linear functions.)

  15. Portfolio Selection with Risk Indices Question: Fidelity Investments has been authorized to invest up to $1,000,000 for a new client, in two investment funds: a stock fund and a money-market fund. • Each unit invested in the stock fund costs $100 and has an expected return of 5%; each unit in the money-market fund costs $50 and has an expected return of 2%. • The client wants to minimize risk subject to the constraint that the expected return on the investment be at least $4,000. Fidelity computes a risk index that a) is a linear function of the units invested in each fund, b) is 6 for a unit invested in the stock fund, and c) is 5 for a unit invested in the money market. • At least 100 units need to be in the money-market fund. How many units should be invested in each fund?

  16. Portfolio Selection with Risk Indices Answer: • Verbal Statement of the Objective Function Minimize risk index. • Verbal Statement of the Constraints Total investment < $1,000,000. Expected return > $4,000. Money-market units invested > 2. • Definition of the Decision Variables x1 = number of units invested in the stock fund. x2 = number of units invested in the money market fund. Min 6x1 + 5x2 (Risk index) s.t. 100x1 + 50x2 < 1,000,000 (Funds available) 5x1+ x2 > 4,000 (Expected return) x2 > 100 (Minimum money market) x1, x2 > 0

  17. Portfolio Selection with Risk Indices Min 6x1+ 5x2 s.t. 100x1+ 50x2<1,000,000 5x1+ x2> 4,000 x2> 100 x1> 0 and x2>0

  18. Portfolio Selection with Risk Indices Min 6x1+ 5x2 s.t. 100x1+ 50x2<1,000,000 5x1+ x2> 4,000 x2> 100 x1> 0 and x2>0 • Conclusions: • 780 units should be invested in the stock fund, and 100 units in the money-market fund. • Suppose the risk index on the money-market fund increased to 5.5. Is the previous solution still optimal?The output states that the solution remains optimal for any increase in the objective function coefficient of x2 (the number of money-market units). Range of optimality

  19. 100% Rules 100% Rules

  20. 100% Rules Overview 100% Rules expand the range of optimality and the range of feasibility to simultaneous changes in two or more parameters. That is, the rule indicates when simultaneous changes in two or more objective function coefficients will not cause a change in the optimal solution. And the rule indicates when simultaneous changes in two or more right-hand-side constants will not cause a change in any of the dual prices.

  21. 100% Rules Question: Wild Oats Bicycles is introducing two new lightweight bicycle frames, the Deluxe and the Professional, to be made from special aluminum and steel alloys. • The anticipated unit profits are $10 for the Deluxe and $15 for the Professional. • Each week, a supplier delivers 100 pounds of the aluminum alloy and 80 pounds of the steel alloy. • The number of pounds of each alloy needed per frame is as follows: How many Deluxe and how many Professional frames should Wild Oats produce each week? Aluminum AlloySteel Alloy Deluxe 2 3 Professional 4 2

  22. 100% Rules Answer: • Verbal Statement of the Objective Function Maximize total weekly profit. • Verbal Statement of the Constraints Total weekly use of aluminum alloy < 100 pounds. Total weekly use of steel alloy < 80 pounds. • Definition of the Decision Variables x1 = number of Deluxe frames produced weekly. x2 = number of Professional frames produced weekly. Max 10x1 + 15x2 (Total Weekly Profit) s.t. 2x1 + 4x2 < 100 (Aluminum Available) 3x1 + 2x2 < 80 (Steel Available) x1, x2 > 0

  23. 100% Rules Max 10x1+ 15x2 s.t. 2x1 + 4x2<100 3x1+ 2x2<80 x1> 0 and x2>0

  24. 100% Rules Max 10x1+ 15x2 s.t. 2x1 + 4x2<100 3x1+ 2x2<80 x1> 0 and x2>0 • Conclusions: • Wild Oats should produce 15.0 Deluxe and 17.5 Professional frames each week, for a profit of $412.50. • Suppose the unit profit on deluxe frames increases to $20.Is the previous solution still optimal?The output states that the solution remains optimal as long as the objective function coefficient of x1 (number of deluxe frames) is in the range of optimality, between 7.500 and 22.500. Because 20 is within that range, the optimal solution will not change if the unit profit on deluxe frames increases to $20. And what is the value of the objective function when that unit profit increases to $20? The optimal profit changes to $20x1 + $15x2 = $20(15) + $15(17.5) = $562.50.

  25. 100% Rules One 100% rule is that simultaneous changes in objective function coefficients will not change the optimal solution as long as the sum of the percentages of the change divided by the corresponding maximum allowable change in the range of optimality for each coefficient does not exceed 100%.

  26. 100% Rules Max 10x1+ 15x2 s.t. 2x1 + 4x2<100 3x1+ 2x2<80 x1> 0 and x2>0 If simultaneously the profit on Deluxe frames were raised to $16 and the profit on Professional frames were raised to $17, would the current solution (15.0 Deluxe and 17.5 Professional frames ) remain optimal? If c1 = 16, the amount c1 changed is 16 - 10 = 6 . The maximum allowable increase is 22.5 - 10 = 12.5, so this is a 6/12.5 = 48% change. If c2 = 17, the amount that c2 changed is 17 - 15 = 2. The maximum allowable increase is 20 - 15 = 5 so this is a 2/5 = 40% change. The sum of the change percentages is 88%. Since that does not exceed 100%, the current solution remains optimal.

  27. 100% Rules The other 100% rule is that simultaneous changes in right-hand sides will not change the dual prices as long as the sum of the percentages of the changes divided by the corresponding maximum allowable change in the range of feasibility for each right-hand side does not exceed 100%.

  28. 100% Rules • Suppose Wild Oats has to pay for all 100 pounds of the aluminum alloy regardless of whether they use each pound. What is the maximum amount the company should pay for 50 extra pounds of aluminum? • Answer: Because aluminum is a sunk cost, the dual price is the value of each unit of extra aluminum. The dual price for aluminum is $3.125 per pound and the maximum allowable increase is 60 pounds (to 160 pounds). Because 50 is in this range, the $3.125 is valid. Thus, the value of 50 additional pounds is = 50($3.125) = $156.25. Range of feasibility

  29. Negative Dual Prices Negative Dual Prices

  30. Negative Dual Prices Overview Negative Dual Prices measure the rate of deterioration (negative improvement) in the objective function value per unit increase in a right-hand side constant.

  31. 100% Rules Question: Use the Management Scientist to solve the following cost-minimization problem: Min 6x1 + 9x2 ($ cost) s.t. x1 + 2x2 < 8 10x1 + 7.5x2 > 30 x2 > 2 x1, x2> 0

  32. Negative Dual Prices Min 6x1 + 9x2 ($ cost) s.t. x1 + 2x2 < 8 10x1 + 7.5x2 > 30 x2 > 2 x1, x2> 0 Solution: The optimal x1 = 1.5 and optimal x2 = 2.0, with the minimized objective function value = $27.00.

  33. Negative Dual Prices • Suppose the unit cost of x1 is decreased in the objective function 6x1 + 9x2 from $6 to $4. Is the current solution still optimal? What is the value of the objective function when this unit cost is decreased to $4? • Answer: That Management Scientist output states that the solution remains optimal as long as the objective function coefficient of x1 is between 0 and 12. Because 4 is within this range, the current solution is still optimal. However, the optimal cost will be affected: 4x1 + 9x2 = 4(1.5) + 9(2.0) = $24.00. • How much can the unit cost of x2 be decreased while keeping the current solution optimal? • Answer: The current solution remains optimal as long as the objective function coefficient of x2 does not fall below 4.5.

  34. Negative Dual Prices • If simultaneously the cost of x1 were raised to $7.5 and the cost of x2 were reduced to $6, would the current solution remain optimal? • Answer: If c1 = 7.5, the amount c1 changed is 7.5 - 6 = 1.5. The maximum allowable increase is 12 - 6 = 6, so that is a 1.5/6 = 25% change. If c2 = 6, the amount that c2 changed is 9 - 6 = 3. The maximum allowable decrease is 9 - 4.5 = 4.5, so that is a 3/4.5 = 66.7% change. The sum of those percentages is 25% + 66.7% = 91.7%. Because that does not exceed 100%, the current solution remains optimal.

  35. Negative Dual Prices • If the right-hand side of constraint 3 is increased by 1, what will be the effect on the optimal objective-function value? • Answer: A dual price represents the improvement in the objective function value per unit increase in the right-hand side. A negative dual price indicates a deterioration (negative improvement) in the objective, which in this problem means an increase in total cost because we're minimizing. Since the right-hand side remains within the range of feasibility (0 to 4), there is no change in the dual price. However, the objective function value increases by $4.50.

  36. Negative Dual Prices • If the right-hand side of Constraint 2 increases by 5 and the right-hand side of Constraint 3 decreases by 1, what will be the effect on the optimal objective-function value? • Answer: The increase of 5 in Constraint 2 is 20% of the maximum allowable increase of 25, and the decrease of 1 in Constraint 3 is 50% of the maximum allowable decrease of 2. The sum of the change percentages is 20% + 50% = 70%. Because that does not exceed 100%, the dual prices do not change. So the optimal objective-function value changes by 0.600 (5) + 4.500 (-1) = -1.500, which is a decrease (improvement) of 1.500.

  37. Blending with Sensitivity Analysis Blending with Sensitivity Analysis

  38. Blending with Sensitivity Analysis Overview Blending Problems with Sensitivity Analysis helps production managers decide how cost is affected by changing product characteristics (shade tolerance, calories, vitamin content, …).

  39. Blending with Sensitivity Analysis • Question: Bluegrass Farms, located in Lexington, Kentucky, has been experimenting with a special diet for racehorses. The feed components available for the diet are: • A standard horse feed product • An enriched oat product • A new vitamin and mineral feed additive. • Each feed component combines ingredients A, B, C. • Nutritional value and cost data for feed components: • Feed Component Standard Enriched Oat Additive • Ingredient A 0.8 0.2 0.0 • Ingredient B 1.0 1.5 3.0 • Ingredient C 0.1 0.6 2.0 • Cost per pound $0.25 $0.50 $3.00

  40. Blending with Sensitivity Analysis • The minimum daily diet requirements for each horse are • 3 units of Ingredient A • 6 units of Ingredient B • 4 units of Ingredient C • To control the weight of the horses, the total daily feed for a horse should not exceed 6 pounds. • What is the cheapest way to satisfy the daily diet • requirements?

  41. Blending with Sensitivity Analysis • S = number of pounds of the standard horse feed product. • E = number of pounds of the enriched oat product. • A = number of pounds of the vitamin and mineral feed additive.

  42. Blending with Sensitivity Analysis Objective: minimize cost from blending S standard feed, E enriched oats, and A additive. Min 0.25 S + 0.50 E + 3 A s.t. 0.8 S + 0.2 E > 3 1.0 S + 1.5 E + 3.0 A > 6 0.1 S + 0.6 E + 2.0 A > 4 S + E + A < 6 S, E, A >0 Minimum requirement of A Minimum requirement of B Minimum requirement of C Maximum weight constraint Non-negativity constraints

  43. Blending with Sensitivity Analysis Min 0.25 S + 0.50 E + 3A s.t. 0.8 S + 0.2 E > 3 (A) 1.0 S + 1.5 E + 3.0 A > 6 (B) 0.1 S + 0.6 E + 2.0 A > 4 (C) S + E + A < 6 (weight) S, E, A >0 • Minimize cost ($5.973 per horse) from blending S = 3.514 pounds of standard feed, E = 0.946 pounds of enriched oats, and A = 1.541 pounds of additive.

  44. Blending with Sensitivity Analysis • How much could be saved by decreasing the Ingredient A minimum from 3 to 2? • The range of feasibility of the first constraint is 1.143 to 3.368, so the decrease from 3 to 2 is in that range. • The dual price of -1.216 means saving $1.216 when the right-hand side of constraint 1 decreases by 1. • The negative dual price means the objective function improves (lowers) with a right-hand side decrease. Min 0.25 S + 0.50 E + 3A s.t. 0.8 S + 0.2 E > 3 (A) 1.0 S + 1.5 E + 3.0 A > 6 (B) 0.1 S + 0.6 E + 2.0 A > 4 (C) S + E + A < 6 (weight) S, E, A >0

  45. BA 452 Quantitative Analysis End of Lesson A.7

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