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Lecture

Lecture. 3. MGMT 650 Sensitivity Analysis in LP Chapter 3. Example. Consider the following linear program: Min 6 x 1 + 9 x 2 ($ cost) s.t. x 1 + 2 x 2 < 8 10 x 1 + 7.5 x 2 > 30 x 2 > 2 x 1 , x 2 > 0.

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Lecture

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  1. Lecture 3 MGMT 650 Sensitivity Analysis in LP Chapter 3

  2. Example • Consider the following linear program: Min 6x1 + 9x2 ($ cost) s.t. x1 + 2x2 < 8 10x1 + 7.5x2 > 30 x2 > 2 x1, x2> 0

  3. The Management Scientist Output OBJECTIVE FUNCTION VALUE = 27.000 VariableValueReduced Cost x1 1.500 0.000 x2 2.000 0.000 Constraint Slack/SurplusDual Price 1 2.500 0.000 2 0.000 -0.600 3 0.000 -4.500

  4. The Management Scientist Output (continued) OBJECTIVE COEFFICIENT RANGES VariableLower Limit Current ValueUpper Limit x1 0.000 6.000 12.000 x2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES Constraint Lower LimitCurrent ValueUpper Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

  5. Example Optimal Solution According to the output: x1 = 1.5 x2 = 2.0 Objective function value = 27.00

  6. Range of Optimality • Question Suppose the unit cost of x1 is decreased to $4. Is the current solution still optimal? What is the value of the objective function when this unit cost is decreased to $4?

  7. The Management Scientist Output OBJECTIVE COEFFICIENT RANGES VariableLower Limit Current ValueUpper Limit x1 0.000 6.000 12.000 x2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES Constraint Lower LimitCurrent ValueUpper Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

  8. Range of Optimality • Answer • The 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 optimal solution will not change. • However, the optimal total cost will be affected • 6x1 + 9x2 = 4(1.5) + 9(2.0) = $24.00.

  9. Range of Optimality • Question How much can the unit cost of x2 be decreased without concern for the optimal solution changing? • The Management Scientist Output OBJECTIVE COEFFICIENT RANGES VariableLower Limit Current ValueUpper Limit x1 0.000 6.000 12.000 x2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES Constraint Lower LimitCurrent ValueUpper Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

  10. Range of Optimality • Answer The output states that the solution remains optimal as long as the objective function coefficient of x2 does not fall below 4.5.

  11. Range of Optimality and 100% Rule • Question If simultaneously the cost of x1 was raised to $7.5 and the cost of x2 was 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 this 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 this is a 3/4.5 = 66.7% change. • The sum of the change percentages is 25% + 66.7% = 91.7%. • Since this does not exceed 100% the optimal solution would not change.

  12. Range of Feasibility • Question If the right-hand side of constraint 3 is increased by 1, what will be the effect on the optimal solution? OBJECTIVE COEFFICIENT RANGES VariableLower Limit Current ValueUpper Limit x1 0.000 6.000 12.000 x2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES Constraint Lower LimitCurrent ValueUpper Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

  13. OBJECTIVE FUNCTION VALUE = 27.000 VariableValueReduced Cost x1 1.500 0.000 x2 2.000 0.000 Constraint Slack/SurplusDual Price 1 2.500 0.000 2 0.000 -0.600 3 0.000 -4.500 Range of Feasibility – Contd.

  14. Range of Feasibility • 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, there is no change in the optimal solution. • However, the objective function value increases by $4.50.

  15. Reduced Cost • Definition • How much the objective function coefficient of each variable would have to improve before it would be possible for that variable to assume a positive value in the optimal solution

  16. (Alternatively) LINDO output Report (of Diet Problem)

  17. Integer Programming Applications Chapter 8

  18. Example: Metropolitan Microwaves • Metropolitan Microwaves, Inc. is planning to expand its operations into other electronic appliances. The company has identified seven new product lines it can carry. Relevant information about each line follows on the next slide. Initial Floor Space Exp. Rate Product Line Invest. (Sq.Ft.) of Return 1. TV/DVDs $ 6,000 125 8.1% 2. Color TVs 12,000 150 9.0 3. Projection TVs 20,000 200 11.0 4. VCRs 14,000 40 10.2 5. DVD Players 15,000 40 10.5 6. Video Games 2,000 20 14.1 7. Home Computers 32,000 100 13.2

  19. Example: Metropolitan Microwaves • Metropolitan has decided that they should not stock projection TVs unless they stock either TV/DVDs or color TVs. • Also, they will not stock both VCRs and DVD players, and they will stock video games if they stock color TVs. • Finally, the company wishes to introduce at least three new product lines. • The company has $45,000 to invest and • 420 sq. ft. of floor space available • Determine Metropolitan’s optimal expansion policy to maximize its overall expected return.

  20. Problem Formulation • Define Decision Variables xj = 1 if product line j is introduced; = 0 otherwise. where: Product line 1 = TV/DVDs Product line 2 = Color TVs Product line 3 = Projection TVs Product line 4 = VCRs Product line 5 = DVD Players Product line 6 = Video Games Product line 7 = Home Computers

  21. Problem Formulation • Define the Decision Variables xj = 1 if product line j is introduced; = 0 otherwise. • Define the Objective Function Maximize total expected return: Max .081(6000)x1 + .09(12000)x2 + .11(20000)x3 + .102(14000)x4 + .105(15000)x5 + .141(2000)x6 + .132(32000)x7

  22. Problem Formulation • Define the Constraints 1) Budget: 6x1 + 12x2 + 20x3 + 14x4 + 15x5 + 2x6 + 32x7< 45 2) Space: 125x1 +150x2 +200x3 +40x4 +40x5 +20x6 +100x7< 420 3) Stock projection TVs only if stock TV/VCRs or color TVs: x1 + x2 > x3 or x1 + x2 - x3> 0 4) Do not stock both VCRs and DVD players: x4 + x5< 1 5) Stock video games if they stock color TV's: x2 - x6> 0 6) Introduce at least 3 new lines: x1 + x2 + x3 + x4 + x5 + x6 + x7> 3 7) Variables are 0 or 1: xj = 0 or 1 for j = 1, , , 7

  23. Expected return Introduce TV/DVDs Projection TVs DVD players Optimal Solution in LINDO

  24. Optimal Solution in Management Scientist

  25. Integer Programming Application inDistribution System Design • Dell Computers operates a plant in St. Louis; annual capacity = 30000 units • Computers are shipped to regional distribution centers located in Boston, Atlanta and Houston; anticipated demands = 30,000, 20,000, 20,000 respectively • Because of anticipated increase in demand, plan to increase capacity by constructing a new plant in one or more locations in Detroit, Columbus, Denver or Kansas City. • Develop a model for choosing the best plant locations • Determine the optimal amounts to transport from each plant to each distribution center such that all demand is satisfied.

  26. Formulation • Minimize 5x11+2x12+3x13+4x21+3x22+4x23+9x31+7x32+5x33+10x41+4x42+2x43+8x51+4x52+3x53+175000y1+300000y2+375000y3+500000y4 Subject to • x11+x12+x13-10000y1<=0 Detroit capacity • x21+x22+x23-20000y2<=0 Columbus capacity • x31+x32+x33-30000y3<=0 Denver capacity • x41+x42+x43-40000y4<=0 Kansas City capacity • x51+x52+x53 <=30000 St. Louis capacity • x11+x21+x31+x41+x51=30000 Boston demand • x12+x22+x32+x42+x52=20000 Atlanta demand • x13+x23+x33+x43+x53=20000 Houston demand • xij>=0 Non-negativity constraints • y1, y2, y3, y4 = {0,1} Integrality constraints

  27. Optimal Solution • Locate 1 new plant in Kansas City (capacity = 40000) {y4=1} • Supply 30000 from existing St. Louis plant to Boston DC {x51=30000} • Supply 20000 from existing St. Louis plant to Atlanta DC {x42=20000} • Supply 20000 from existing St. Louis plant to Houston DC {x43=20000}

  28. Example: Tina’s Tailoring • Tina's Tailoring has five idle tailors and four custom garments to make. • The estimated time (in hours) it would take each tailor to make each garment is shown below. • An 'X' indicates an unacceptable tailor-garment assignment. • Formulate an integer program for determining the tailor-garment assignments that minimize the total estimated time spent making the four garments. • No tailor is to be assigned more than one garment and each garment is to be worked on by only one tailor. Tailor Garment12345 Wedding gown 19 23 20 21 18 Clown costume 11 14 X 12 10 Admiral's uniform 12 8 11 X 9 Bullfighter's outfit X 20 20 18 21

  29. Formulation: Tina’s Tailoring • Define the decision variables xij = 1 if garment i is assigned to tailor j = 0 otherwise. Number of decision variables = [(# of garments)(# of tailors)] - (# of unacceptable assignments) = [4(5)] - 3 = 17 • Define the objective function Minimize total time spent making garments: Min 19x11 + 23x12 + 20x13 + 21x14 + 18x15 + 11x21 + 14x22 + 12x24 + 10x25 + 12x31 + 8x32 + 11x33 + 9x35 + 20x42 + 20x43 + 18x44 + 21x45

  30. Constraints: Tina’s Tailoring • Exactly one tailor per garment: 1) x11 + x12 + x13 + x14 + x15 = 1 2) x21 + x22 + x24 + x25 = 1 3) x31 + x32 + x33 + x35 = 1 4) x42 + x43 + x44 + x45 = 1 • No more than one garment per tailor: 5) x11 + x21 + x31 <= 1 6) x12 + x22 + x32 + x42 <= 1 7) x13 + x33 + x43 <= 1 8) x14 + x24 + x44 <= 1 9) x15 + x25 + x35 + x45 <= 1 Integrality: xij > {0,1} for i = 1, . . ,4 and j = 1, . . ,5

  31. Optimal Solution Estimated total time Wedding Gown to Tailor #5 Clown costume to Tailor #1 Admiral’s uniform to Tailor #2 Bullfighter’s outfit to Tailor #4

  32. Airline Crew Scheduling • Southeast Airlines needs to assign crews to cover all of its upcoming flights. • We will focus on assigning 3 crews based in San Francisco to the flights listed below. • Exactly 3 of the feasible sequences need to be chosen such that every flight is covered. • It is permissible to have more than 1 crew per flight, where the extra crews would fly as passengers, but union contracts require that the extra crews would still need to be paid for their time as if they were working. • Schedule one crew for every feasible sequence to minimize the total cost of the 3 crew assignments that cover all the flights.

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