1. Chapter 8 Integer Linear Programming n Types of Integer Linear Programming Models n Graphical and Computer Solutions for an All- Integer Linear Program n Applications Involving 0-1 Variables n Modeling Flexibility Provided by 0-1 Variables

2. Types of Integer Programming Models n An LP in which all the variables are restricted to be integers is called an all-integer linear program (ILP). n The LP that results from dropping the integer requirements is called the LP Relaxation of the ILP. n If only a subset of the variables are restricted to be integers, the problem is called a mixed-integer linear program (MILP). n Binary variables are variables whose values are restricted to be 0 or 1. If all variables are restricted to be 0 or 1, the problem is called a 0-1 or binary integer linear program.

3. Example: All-Integer LP n Consider the following all-integer linear program: Max 3 x 1 + 2 x 2 Max 3 x 1 + 2 x 2 s.t. 3 x 1 + x 2 < 9 s.t. 3 x 1 + x 2 < 9 x 1 + 3 x 2 < 7 x 1 + 3 x 2 < 7 - x 1 + x 2 < 1 - x 1 + x 2 < 1 x 1, x 2 > 0 and integer x 1, x 2 > 0 and integer

4. Example: All-Integer LP n LP Relaxation Solving the problem as a linear program ignoring the integer constraints, the optimal solution to the linear program gives fractional values for both x 1 and x 2. From the graph on the next slide, we see that the optimal solution to the linear program is: x 1 = 2.5, x 2 = 1.5, z = 10.5 x 1 = 2.5, x 2 = 1.5, z = 10.5

5. Example: All-Integer LP n LP Relaxation LP Optimal (2.5, 1.5) Max 3 x 1 + 2 x 2 Max 3 x 1 + 2 x 2 - x 1 + x 2 < 1 x2x2x2x2 x1x1x1x1 3 x 1 + x 2 < 9 1 3 2 5 4 1 2 3 4 5 6 7 x 1 + 3 x 2 < 7 x 1 + 3 x 2 < 7

6. Example: All-Integer LP n Rounding Up If we round up the fractional solution ( x 1 = 2.5, x 2 = 1.5) to the LP relaxation problem, we get x 1 = 3 and x 2 = 2. From the graph on the next slide, we see that this point lies outside the feasible region, making this solution infeasible.

7. Example: All-Integer LP n Rounded Up Solution LP Optimal (2.5, 1.5) Max 3 x 1 + 2 x 2 Max 3 x 1 + 2 x 2 - x 1 + x 2 < 1 x2x2x2x2 x1x1x1x1 3 x 1 + x 2 < 9 ILP Infeasible (3, 2) ILP Infeasible (3, 2) x 1 + 3 x 2 < 7 x 1 + 3 x 2 < 7 1 2 3 4 5 6 7 1 3 2 5 4

8. Example: All-Integer LP n Rounding Down By rounding the optimal solution down to x 1 = 2, x 2 = 1, we see that this solution indeed is an integer solution within the feasible region, and substituting in the objective function, it gives z = 8. We have found a feasible all-integer solution, but have we found the OPTIMAL all-integer solution? --------------------- The answer is NO! The optimal solution is x 1 = 3 and x 2 = 0 giving z = 9, as evidenced in the next two slides.

9. Example: All-Integer LP n Complete Enumeration of Feasible ILP Solutions There are eight feasible integer solutions to this problem: x 1 x 2 z x 1 x 2 z 1. 0 0 0 1. 0 0 0 2. 1 0 3 2. 1 0 3 3. 2 0 6 3. 2 0 6 4. 3 0 9 optimal solution 4. 3 0 9 optimal solution 5. 0 1 2 5. 0 1 2 6. 1 1 5 6. 1 1 5 7. 2 1 8 7. 2 1 8 8. 1 2 7 8. 1 2 7

10. Example: All-Integer LP ILP Optimal (3, 0) Max 3 x 1 + 2 x 2 Max 3 x 1 + 2 x 2 - x 1 + x 2 < 1 x2x2x2x2 x1x1x1x1 3 x 1 + x 2 < 9 x 1 + 3 x 2 < 7 x 1 + 3 x 2 < 7 1 3 2 5 4 1 2 3 4 5 6 7

11. Example: All-Integer LP n Partial Spreadsheet Showing Problem Data

12. Example: All-Integer LP n Partial Spreadsheet Showing Formulas

13. Example: All-Integer LP n Partial Spreadsheet Showing Optimal Solution

14. Special 0-1 Constraints n When x i and x j represent binary variables designating whether projects i and j have been completed, the following special constraints may be formulated: At most k out of n projects will be completed: At most k out of n projects will be completed:  x j < k  x j < k j j Project j is conditional on project i : Project j is conditional on project i : x j - x i < 0 x j - x i < 0 Project i is a corequisite for project j : Project i is a corequisite for project j : x j - x i = 0 x j - x i = 0 Projects i and j are mutually exclusive: Projects i and j are mutually exclusive: x i + x j < 1 x i + x j < 1

15. 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.

16. Example: Metropolitan Microwaves Initial Floor Space Exp. Rate Initial Floor Space Exp. Rate Product Line Invest. (Sq.Ft.) of Return Product Line Invest. (Sq.Ft.) of Return 1. TV/VCRs$ 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

17. Example: Metropolitan Microwaves Metropolitan has decided that they should not stock projection TVs unless they stock either TV/VCRs 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. If the company has $45,000 to invest and 420 sq. ft. of floor space available, formulate an integer linear program for Metropolitan to maximize its overall expected return.

18. Example: Metropolitan Microwaves n Define the Decision Variables x j = 1 if product line j is introduced; = 0 otherwise. = 0 otherwise.where: Product line 1 = TV/VCRs 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

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

20. Example: Metropolitan Microwaves n Define the Constraints 1) Money: 1) Money: 6 x 1 + 12 x 2 + 20 x 3 + 14 x 4 + 15 x 5 + 2 x 6 + 32 x 7 < 45 6 x 1 + 12 x 2 + 20 x 3 + 14 x 4 + 15 x 5 + 2 x 6 + 32 x 7 < 45 2) Space: 2) Space: 125 x 1 +150 x 2 +200 x 3 +40 x 4 +40 x 5 +20 x 6 +100 x 7 < 420 125 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 3) Stock projection TVs only if stock TV/VCRs or color TVs: stock TV/VCRs or color TVs: x 1 + x 2 > x 3 or x 1 + x 2 - x 3 > 0 x 1 + x 2 > x 3 or x 1 + x 2 - x 3 > 0

21. Example: Metropolitan Microwaves n Define the Constraints (continued) 4) Do not stock both VCRs and DVD players: x 4 + x 5 < 1 5) Stock video games if they stock color TV's: 5) Stock video games if they stock color TV's: x 2 - x 6 > 0 6) Introduce at least 3 new lines: 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: 7) Variables are 0 or 1: x j = 0 or 1 for j = 1,,, 7 x j = 0 or 1 for j = 1,,, 7

22. Example: Metropolitan Microwaves n Partial Spreadsheet Showing Problem Data

23. Example: Metropolitan Microwaves n Partial Spreadsheet Showing Example Formulas

24. Example: Metropolitan Microwaves n Solver Parameters Dialog Box

25. Example: Metropolitan Microwaves n Solver Options Dialog Box

26. Example: Metropolitan Microwaves n Integer Options Dialog Box

27. Example: Metropolitan Microwaves n Optimal Solution

28. Example: Metropolitan Microwaves n Optimal Solution Introduce: Introduce: TV/VCRs, Projection TVs, and DVD Players TV/VCRs, Projection TVs, and DVD Players Do Not Introduce: Do Not Introduce: Color TVs, VCRs, Video Games, and Home Computers Color TVs, VCRs, Video Games, and Home Computers Total Expected Return : Total Expected Return : $4,261 $4,261

29. 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 in the next slide. (An 'X' in the table indicates an unacceptable tailor-garment assignment.) 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

30. Example: Tina’s Tailoring 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. -------------------- This problem can be formulated as a 0-1 integer program. The LP solution to this problem will automatically be integer (0-1).

31. Example: 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 = Number of decision variables = [(number of garments)(number of tailors)] - (number of unacceptable assignments) - (number of unacceptable assignments) = [4(5)] - 3 = 17

32. Example: Tina’s Tailoring 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

33. Example: Tina’s Tailoring n Define the Constraints 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

34. Example: Tina’s Tailoring n Define the Constraints (continued) 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 Nonnegativity: x ij > 0 for i = 1,..,4 and j = 1,..,5

35. Cautionary Note About Sensitivity Analysis n Sensitivity analysis often is more crucial for ILP problems than for LP problems. n A small change in a constraint coefficient can cause a relatively large change in the optimal solution. n Recommendation: Resolve the ILP problem several times with slight variations in the coefficients before choosing the “best” solution for implementation.

36. End of Chapter 8