Readings

1. Readings • Readings • Chapter 7 • Integer Linear Programming

2. Overview • Overview

3. Overview

4. Overview Tool Summary • Do not make integer restrictions, and maybe the variables at an optimum will be integers. • First Example: Pi = (integer) number of producers in month i. • Use compound variables: • First Example: Pi = number of producers in month i • Use dynamic or recursive constraints: • First Example: Define the constraint that the number of apprentices in a month must not exceed the number of recruits inthe previous month: A2 - R1< 0; A3 - R2< 0 • Constrain one variable to be a proportional to another variable: • First Example: Define the constraint that each trainer can train two recruits: 2T1 - R1> 0; 2T2 - R2> 0 • Use inventory variables: • Second Example: P2 + I1–I2 = 150 (production-net inventory = demand)

5. Tool Summary Tool Summary • Use binary variables to model fixed cost constraints. • For example, consider a linear programming problem with a constraint • M1<15 • Introduce binary variable U1 and constraints: • M1>5 U1 • 15U1> M1 • Then, U1 = 1 if M1 > 0, and either M1 = 0 or M1>5. • Do not try to solve an integer linear program by rounding off (up or down) a solution to the problem without integer constraints. Rather, graph all feasible integer solutions and use iso-value linear, or use the Management Scientist module.

6. Rounding Off • Rounding Off

7. Rounding Off Overview Rounding Off solutions in continuous variables to the nearest integer (like 2.67 rounded off to 3) is an unreliable way to solve a linear programming problem when decision variables should be integers.

8. Rounding Off • A LP in which all the variables are restricted to be integers is called an all-integer linear program (ILP). • The LP that results from dropping the integer requirements is called the LP Relaxation of the ILP. • If only a subset of the variables are restricted to be integers, the problem is called a mixed-integer linear program (MILP). • 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.

9. Rounding Off • Solve the following all-integer linear program, and compare that solution to the problem where the decision variables do not have to be integers: Max 3x1 + 2x2 s.t. 3x1 + x2< 9 x1 + 3x2< 7 -x1 + x2< 1 x1, x2> 0 and integer

10. Rounding Off Max 3x1 + 2x2 s.t. 3x1 + x2< 9 x1 + 3x2< 7 -x1 + x2< 1 x1, x2> 0 and integer • LP Relaxation. If we drop the integer constraints, we can graph the optimal solution to the linear program. And the optimal solution has fractional values: x1= 2.5, x2= 1.5, Max 3x1 + 2x2 = 10.5 x2 5 -x1 + x2< 1 3x1 + x2< 9 4 Max 3x1 + 2x2 3 LP Optimal (2.5, 1.5) 2 x1 + 3x2< 7 1 x1 1 2 3 4 5 6 7

11. Rounding Off Max 3x1 + 2x2 s.t. 3x1 + x2< 9 x1 + 3x2< 7 -x1 + x2< 1 x1, x2> 0 and integer • Rounding Up. If we round up the fractional solution (x1 = 2.5, x2 = 1.5) to the previous relaxed LP problem, we get x1 = 3 and x2 = 2. But the graph shows those values are infeasible. x2 5 -x1 + x2< 1 3x1 + x2< 9 4 Max 3x1 + 2x2 3 ILP Infeasible (3, 2) LP Optimal (2.5, 1.5) 2 x1 + 3x2< 7 1 x1 1 2 3 4 5 6 7

12. Rounding Off Max 3x1 + 2x2 s.t. 3x1 + x2< 9 x1 + 3x2< 7 -x1 + x2< 1 x1, x2> 0 and integer • Rounding Down. By rounding the non-integer solution down to x1 = 2, x2 = 1, we have a feasible solution, with objective function 3x1 + 2x2 = 8. But that solution is not optimal for the integer program. x2 -x1 + x2< 1 5 3x1 + x2< 9 4 Max 3x1 + 2x2 3 ILP Optimal (3, 0) x1 + 3x2< 7 • Optimal ILP solution. The optimal ILP solution (3,0) is not the closest feasible point to the non-integer solution (2.5,1.5). 2 1 x1 1 2 3 4 5 6 7

13. Rounding Off • Exhaustive Search. One way to solve the integer linear program is to evaluate the objective function at each feasible solution. There are 8 alternative feasible solutions in Example 1. Max 3x1 + 2x2 s.t. 3x1 + x2< 9 x1 + 3x2< 7 -x1 + x2< 1 x1, x2> 0 and integer x1x23x1 + 2x2 1. 0 0 0 2. 1 0 3 3. 2 0 6 4. 3 0 9 optimal solution 5. 0 1 2 6. 1 1 5 7. 2 1 8 8. 1 2 7

14. Rounding Off Max 3x1 + 2x2 s.t. 3x1 + x2< 9 x1 + 3x2< 7 -x1 + x2< 1 x1, x2> 0 and integer

15. Rounding Off Max 3x1 + 2x2 s.t. 3x1 + x2< 9 x1 + 3x2< 7 -x1 + x2< 1 x1, x2> 0 and integer ILP Optimal (3, 0)

16. Sensitivity Analysis with Integer Variables • Sensitivity Analysis with Integer Variables

17. Sensitivity Analysis with Integer Variables Overview Sensitivity Analysis with Integer Variables is more important than Sensitivity Analysis with Continuous Variables because a small change in a constraint coefficient can cause a relatively large change in the optimal solution, and in the objective-function value of the optimal solution.

18. Sensitivity Analysis with Integer Variables • Sensitivity analysis often is more crucial for ILP problems than for LP problems. • A small change in a constraint coefficient can cause a relatively large change in the optimal solution, and the objective-function value of the optimal solution. • Recommendation: Resolve the ILP problem several times with slight variations in the coefficients before choosing the “best” solution for implementation.

19. Sensitivity Analysis A small change in the constraints (adding the red feasible area) can cause a jump in the objective function at the optimum, from 9 to 13. x2 5 Max 3x1 + 2x2 = 9 at old optimum (3,0) 4 3 Max 3x1 + 2x2 = 13 at new optimum (2,2) 2 1 x1 1 2 3 4 5 6 7

20. Assignment with Valuable Time • Assignment with Valuable Time

21. Assignment with Valuable Time Overview Assignment Problems with Valuable Time minimize the total time of assigning workers to jobs. Minimizing total time is appropriate when each worker has the same value of time.

22. Assignment with Valuable Time Question: The NPD Group is a market research firm with three clients that each request the firm conduct a sample survey. Four statisticians can be assigned to these three projects; however, all four are busy, and therefore can handle only one client. The following are the number of hours required for each statistician to complete each job. The differences in time are based on experience and ability of the statisticians.

23. Assignment with Valuable Time What is the most natural objective function for the firm to optimize? Formulate the firm’s optimization problem. Are there any implicit assumptions in your formulation?

24. Assignment with Valuable Time • Answer: Linear programming formulation (supply inequality, demand equality). • Variables: Xij = 1 if Statistician i is assigned to Client j, else 0 • Objective (minimize time, assuming all time values equal): • Min 150X11 + 210X12 + 270X13 + 170X21 + 230X22 + 220X23 • + 180X31 + 230X32 + 225X33 + 160X41 + 240X42 + 230X43 • Supply Constraints (Statisticians to at most 1 Client): • X11 + X12 + X13 < 1, X21 + X22 + X23 < 1, • X31 + X32 + X33 < 1, X41 + X42 + X43 < 1 • Demand Constraints (each Client served): • X11 + X21 + X31 + X41 = 1 • X12 + X22 + X32 + X42 = 1 • X13 + X23 + X33 + X43 = 1

25. Assignment with Supply and Demand • Assignment with Supply and Demand

26. Assignment with Supply and Demand Overview Assignment Problems with Supply and Demand are Transportation Problems of suppliers to demanders except that each demand is assigned to exactly one supplier.

27. Assignment with Supply and Demand Question: Dow Chemical uses the chemical Rbase in production operations at five divisions. Only six suppliers of Rbase meet Dow’s quality standards. The quantity of Rbase needed by each Dow division and the price per gallon charged by each supplier are as follows:

28. Assignment with Supply and Demand The cost per gallon ($) for shipping from each supplier to each division are as follows:

29. Assignment with Supply and Demand Dowwants to diversify by spreading its business so that each division’s demand is assigned to exactly one supplier. Formulate the optimal assignment of suppliers to divisions as a linear-programming problem.

30. Assignment with Supply and Demand • Answer: Linear programming formulation (supply inequality, demand equality). • Variables: Xij = 1 if Supplier i is assigned to Division j, else 0 • Assignment Costs • The total cost is the sum of the purchase cost and the transportation cost. • Supplier 1 assigned to Division 1 (cost in $1000s): • Purchase cost: (40 x $12.60) = $504 • Transportation Cost: (40 x $2.75) = $110 • Total Cost: $614

31. Assignment with Supply and Demand • Assignment Costs: Cij = Cost of assigning Supplier i to Division j

32. Assignment with Supply and Demand • Linear programming formulation (supply inequality, demand equality). • Objective (minimize cost): Min SCijXij • Demand Constraints (since each division’s demand is assigned to exactly one supplier): • X11 + X21 + X31 + X41 + X51 + X61 = 1 • X12 + X22 + X32 + X42 + X52 + X62 = 1 • X13 + X23 + X33 + X43 + X53 + X63 = 1 • X14 + X24 + X34 + X44 + X54 + X64 = 1 • X15 + X25 + X35 + X45 + X55 + X65 = 1

33. Assignment with Supply and Demand • Optional: There is no mention of supply constraints, which are common in assignment problems. Here is what those common constraints would be in this problem. • Supply Constraints (Each supplier can supply at most 1 Division): • X11 + X12 + X13 + X14 + X15 < 1 • X21 + X22 + X23 + X24 + X25 < 1 • X31 + X32 + X33 + X34 + X35 < 1 • X41 + X42 + X43 + X44 + X45 < 1 • X51 + X52 + X53 + X54 + X55 < 1 • X61 + X62 + X63 + X64 + X65 < 1

34. BA 452 Quantitative Analysis End of Lesson B.3