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

2. 2 2 Example n Consider the following linear program: Min 6 x 1 + 9 x 2 ($ cost) Min 6 x 1 + 9 x 2 ($ cost) s.t. x 1 + 2 x 2 < 8 s.t. x 1 + 2 x 2 < 8 10 x 1 + 7.5 x 2 > 30 10 x 1 + 7.5 x 2 > 30 x 2 > 2 x 2 > 2 x 1, x 2 > 0 x 1, x 2 > 0

3. 3 3 OBJECTIVE FUNCTION VALUE = 27.000 Variable Value Reduced Cost Variable Value Reduced Cost x 1 1.500 0.000 x 1 1.500 0.000 x 2 2.000 0.000 x 2 2.000 0.000 Constraint Slack/Surplus Dual Price Constraint Slack/Surplus Dual Price 1 2.500 0.000 1 2.500 0.000 2 0.000 -0.600 2 0.000 -0.600 3 0.000 -4.500 3 0.000 -4.500 The Management Scientist Output

4. 4 4 The Management Scientist Output (continued) OBJECTIVE COEFFICIENT RANGES OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit Variable Lower Limit Current Value Upper Limit x 1 0.000 6.000 12.000 x 1 0.000 6.000 12.000 x 2 4.500 9.000 No Limit x 2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit Constraint Lower Limit Current Value Upper Limit 1 5.500 8.000 No Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 2 15.000 30.000 55.000 3 0.000 2.000 4.000 3 0.000 2.000 4.000

5. 5 5 Example Optimal Solution According to the output: x 1 = 1.5 x 1 = 1.5 x 2 = 2.0 Objective function value = 27.00

6. 6 6 Range of Optimality n Question Suppose the unit cost of x 1 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. 7 7 The Management Scientist Output OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit Variable Lower Limit Current Value Upper Limit x 1 0.000 6.000 12.000 x 1 0.000 6.000 12.000 x 2 4.500 9.000 No Limit x 2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit Constraint Lower Limit Current Value Upper Limit 1 5.500 8.000 No Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 2 15.000 30.000 55.000 3 0.000 2.000 4.000 3 0.000 2.000 4.000

8. 8 8 Range of Optimality n Answer The output states that the solution remains optimal as long as the objective function coefficient of x 1 is between 0 and 12. The output states that the solution remains optimal as long as the objective function coefficient of x 1 is between 0 and 12. Because 4 is within this range, the optimal solution will not change. Because 4 is within this range, the optimal solution will not change. However, the optimal total cost will be affected However, the optimal total cost will be affected 6 x 1 + 9 x 2 = 4(1.5) + 9(2.0) = $24.00.6 x 1 + 9 x 2 = 4(1.5) + 9(2.0) = $24.00.

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

10. 10 Range of Optimality n Answer The output states that the solution remains optimal as long as the objective function coefficient of x 2 does not fall below 4.5.

11. 11 Range of Optimality and 100% Rule n Question If simultaneously the cost of x 1 was raised to $7.5 and the cost of x 2 was reduced to $6, would the current solution remain optimal? n Answer If c 1 = 7.5, the amount c 1 changed is 7.5 - 6 = 1.5. If c 1 = 7.5, the amount c 1 changed is 7.5 - 6 = 1.5. The maximum allowable increase is 12 - 6 = 6, The maximum allowable increase is 12 - 6 = 6, so this is a 1.5/6 = 25% change.so this is a 1.5/6 = 25% change. If c 2 = 6, the amount that c 2 changed is 9 - 6 = 3. If c 2 = 6, the amount that c 2 changed is 9 - 6 = 3. The maximum allowable decrease is 9 - 4.5 = 4.5, The maximum allowable decrease is 9 - 4.5 = 4.5, so this is a 3/4.5 = 66.7% change.so this is a 3/4.5 = 66.7% change. The sum of the change percentages is 25% + 66.7% = 91.7%. The sum of the change percentages is 25% + 66.7% = 91.7%. Since this does not exceed 100% the optimal solution would not change. Since this does not exceed 100% the optimal solution would not change.

12. 12 Range of Feasibility n 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 Variable Lower Limit Current Value Upper Limit Variable Lower Limit Current Value Upper Limit x 1 0.000 6.000 12.000 x 1 0.000 6.000 12.000 x 2 4.500 9.000 No Limit x 2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit Constraint Lower Limit Current Value Upper Limit 1 5.500 8.000 No Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 2 15.000 30.000 55.000 3 0.000 2.000 4.000 3 0.000 2.000 4.000

13. 13 Range of Feasibility – Contd. OBJECTIVE FUNCTION VALUE = 27.000 Variable Value Reduced Cost Variable Value Reduced Cost x 1 1.500 0.000 x 1 1.500 0.000 x 2 2.000 0.000 x 2 2.000 0.000 Constraint Slack/Surplus Dual Price Constraint Slack/Surplus Dual Price 1 2.500 0.000 1 2.500 0.000 2 0.000 -0.600 2 0.000 -0.600 3 0.000 -4.500 3 0.000 -4.500

14. 14 Range of Feasibility n Answer A dual price represents the improvement in the objective function value per unit increase in the right-hand side. 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. 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. 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. However, the objective function value increases by $4.50.

15. 15 Reduced Cost n 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 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. 16 (Alternatively) LINDO output Report (of Diet Problem)

17. 17 Integer Programming Applications Chapter 8

18. 18 Example: Metropolitan Microwaves nMetropolitan 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. 19 Example: Metropolitan Microwaves nMetropolitan has decided that they should not stock projection TVs unless they stock either TV/DVDs or color TVs. nAlso, they will not stock both VCRs and DVD players, and they will stock video games if they stock color TVs. nFinally, the company wishes to introduce at least three new product lines. nThe company has $45,000 to invest and n420 sq. ft. of floor space available nDetermine Metropolitan’s optimal expansion policy to maximize its overall expected return.

20. 20 Problem Formulation nDefine Decision Variables x j = 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. 21 Problem Formulation n Define the Decision Variables x j = 1 if product line j is introduced; = 0 otherwise. n Define the Objective Function Maximize total expected return: Max.081(6000) x 1 +.09(12000) x 2 +.11(20000) x 3 +.102(14000) x 4 +.105(15000) x 5 +.141(2000) x 6 +.132(32000) x 7

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

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

24. 24 Optimal Solution in Management Scientist

25. 25 Integer Programming Application in Distribution 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. Proposed LocationAnnual Fixed Cost ($)Annual Capacity Detroit175,00010,000 Columbus300,00020,000 Denver375,00030,000 Kansas City500,00040,000 Plant SiteBostonAtlantaHouston Detroit523 Columbus434 Denver975 Kansas City1042 St. Louis843

26. 26 Formulation  Minimize 5x 11 +2x 12 +3x 13 +4x 21 +3x 22 +4x 23 +9x 31 +7x 32 +5x 33 +10x 41 +4x 42 +2x 43 +8x 51 +4x 52 +3x 53 +175000y 1 +300000y 2 +375000y 3 +500000y 4 Subject to  x 11 +x 12 +x 13 -10000y 1 <=0Detroit capacity  x 21 +x 22 +x 23 -20000y 2 <=0Columbus capacity  x 31 +x 32 +x 33 -30000y 3 <=0 Denver capacity  x 41 +x 42 +x 43 -40000y 4 <=0Kansas City capacity  x 51 +x 52 +x 53 <=30000St. Louis capacity  x 11 +x 21 +x 31 +x 41 +x 51 =30000Boston demand  x 12 +x 22 +x 32 +x 42 +x 52 =20000Atlanta demand  x 13 +x 23 +x 33 +x 43 +x 53 =20000Houston demand  x ij >=0Non-negativity constraints  y 1, y 2, y 3, y 4 = {0,1}Integrality constraints

27. 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. 28 n Tina's Tailoring has five idle tailors and four custom garments to make. n The estimated time (in hours) it would take each tailor to make each garment is shown below. n An 'X' indicates an unacceptable tailor-garment assignment. n Formulate an integer program for determining the tailor- garment assignments that minimize the total estimated time spent making the four garments. n No tailor is to be assigned more than one garment and each garment is to be worked on by only one tailor. Tailor Tailor Garment 1 2 3 4 5 Garment 1 2 3 4 5 Wedding gown 19 23 20 21 18 Wedding gown 19 23 20 21 18 Clown costume 11 14 X 12 10 Clown costume 11 14 X 12 10 Admiral's uniform 12 8 11 X 9 Admiral's uniform 12 8 11 X 9 Bullfighter's outfit X 20 20 18 21 Bullfighter's outfit X 20 20 18 21 Example: Tina’s Tailoring

29. 29 Formulation: Tina’s Tailoring n Define the decision variables x ij = 1 if garment i is assigned to tailor j x ij = 1 if garment i is assigned to tailor j = 0 otherwise. = 0 otherwise. Number of decision variables = [(# of garments)(# of tailors)] - (# of unacceptable assignments) = [4(5)] - 3 = 17 Number of decision variables = [(# of garments)(# of tailors)] - (# of unacceptable assignments) = [4(5)] - 3 = 17 n Define the objective function Minimize total time spent making garments: Minimize total time spent making garments: Min 19 x 11 + 23 x 12 + 20 x 13 + 21 x 14 + 18 x 15 + 11 x 21 Min 19 x 11 + 23 x 12 + 20 x 13 + 21 x 14 + 18 x 15 + 11 x 21 + 14 x 22 + 12 x 24 + 10 x 25 + 12 x 31 + 8 x 32 + 11 x 33 + 14 x 22 + 12 x 24 + 10 x 25 + 12 x 31 + 8 x 32 + 11 x 33 + 9x 35 + 20 x 42 + 20 x 43 + 18 x 44 + 21 x 45 + 9x 35 + 20 x 42 + 20 x 43 + 18 x 44 + 21 x 45

30. 30 Constraints: Tina’s Tailoring n Exactly one tailor per garment: 1) x 11 + x 12 + x 13 + x 14 + x 15 = 1 1) x 11 + x 12 + x 13 + x 14 + x 15 = 1 2) x 21 + x 22 + x 24 + x 25 = 1 2) x 21 + x 22 + x 24 + x 25 = 1 3) x 31 + x 32 + x 33 + x 35 = 1 3) x 31 + x 32 + x 33 + x 35 = 1 4) x 42 + x 43 + x 44 + x 45 = 1 4) x 42 + x 43 + x 44 + x 45 = 1 n No more than one garment per tailor: 5) x 11 + x 21 + x 31 <= 1 6) x 12 + x 22 + x 32 + x 42 <= 1 7) x 13 + x 33 + x 43 <= 1 8) x 14 + x 24 + x 44 <= 1 9) x 15 + x 25 + x 35 + x 45 <= 1 Integrality: x ij > {0,1} for i = 1,..,4 and j = 1,..,5

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

32. 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. FlightFlight Leg123456789101112 1SF to LA1111 2SF to Denver1111 3SF to Seattle1111 4LA to Chicago22323 5LA to SF2355 6Chicago to Denver334 7Chicago to Seattle33334 8Denver to SF2445 9Denver to Chicago222 10Seattle to SF2445 11Seattle to LA22442 Cost (000’s) 234675789989  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.