1. Kerimcan OzcanMNGT 379 Operations Research1 Integer Linear Programming Chapter 8

2. Kerimcan OzcanMNGT 379 Operations Research2 Overview and Types of Integer Programming Models Overview:  Types of Integer Linear Programming Models  Graphical and Computer Solutions for an All- Integer Linear Program  Applications Involving 0-1 Variables  Modeling Flexibility Provided by 0-1 Variables An 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.

3. Kerimcan OzcanMNGT 379 Operations Research3 Example: All-Integer LP Consider the following all-integer linear program: Max 3x 1 + 2x 2 s.t. 3x 1 + x 2 < 9 x 1 + 3x 2 < 7 -x 1 + x 2 < 1 x 1, x 2 > 0 and integer 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

4. Kerimcan OzcanMNGT 379 Operations Research4 Example: All-Integer LP LP Relaxation LP Optimal (2.5, 1.5) Max 3 x 1 + 2 x 2 - x 1 + x 2 < 1 x2x2 x1x1 3 x 1 + x 2 < 9 1 3 2 5 4 1 2 3 4 5 6 7 x 1 + 3 x 2 < 7

5. Kerimcan OzcanMNGT 379 Operations Research5 Example: All-Integer LP 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 below, we see that this point lies outside the feasible region, making this solution infeasible. LP Optimal (2.5, 1.5) Max 3 x 1 + 2 x 2 - x 1 + x 2 < 1 x2x2 x1x1 3 x 1 + x 2 < 9 1 3 2 5 4 1 2 3 4 5 6 7 x 1 + 3 x 2 < 7 ILP Infeasible (3, 2)

6. Kerimcan OzcanMNGT 379 Operations Research6 Example: All-Integer LP 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 below. Complete Enumeration of Feasible ILP Solutions There are eight feasible integer solutions to this problem: x 1 x 2 z 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

7. Kerimcan OzcanMNGT 379 Operations Research7 Example: All-Integer LP ILP Optimal (3, 0) Max 3 x 1 + 2 x 2 - x 1 + x 2 < 1 x2x2 x1x1 3 x 1 + x 2 < 9 1 3 2 5 4 1 2 3 4 5 6 7 x 1 + 3 x 2 < 7

8. Kerimcan OzcanMNGT 379 Operations Research8 Computer Output INTEGER LINEAR PROGRAMMING PROBLEM MAX 3X1+2X2 S.T. 1) 3X1+1X2<9 2) 1X1+3X2<7 3) -1X1+1X2<1 OPTIMAL SOLUTION Objective Function Value = 9.000 Variable Value -------------- --------------- X1 3.000 X2 0.000 Constraint Slack/Surplus -------------- --------------- 1 0.000 2 4.000 3 4.000

9. Kerimcan OzcanMNGT 379 Operations Research9 Special 0-1 Constraints 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:  x j < k j  Project j is conditional on project i: x j - x i < 0  Project i is a corequisite for project j: x j - x i = 0  Projects i and j are mutually exclusive: x i + x j < 1

10. Kerimcan OzcanMNGT 379 Operations Research10 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. Initial Floor Space Exp. Rate 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 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.

11. Kerimcan OzcanMNGT 379 Operations Research11 Example: Metropolitan Microwaves 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. Define the Decision Variables x j = 1 if product line j is introduced; = 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 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

12. Kerimcan OzcanMNGT 379 Operations Research12 Example: Metropolitan Microwaves Define the Constraints 1) Money: 6x 1 + 12x 2 + 20x 3 + 14x 4 + 15x 5 + 2x 6 + 32x 7 < 45 2) Space: 25x 1 +150x 2 +200x 3 +40x 4 +40x 5 +20x 6 +100x 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

13. Kerimcan OzcanMNGT 379 Operations Research13 Example: Metropolitan Microwaves INTEGER LINEAR PROGRAMMING PROBLEM MAX 486X1+1080X2+2200X3+1428X4+1575X5+282X6+4224X7 S.T. 1) 6X1+12X2+20X3+14X4+15X5+2X6+32X7<45 2) 125X1+150X2+200X3+40X4+40X5+20X6+100X7<420 3) 1X1+1X2-1X3>0 4) 1X4+1X5<1 5) 1X2-1X6>0 6) 1X1+1X2+1X3+1X4+1X5+1X6+1X7>3 OPTIMAL SOLUTION Objective Function Value = 4261.000 Variable Value -------------- --------------- X1 1.000 X2 0.000 X3 1.000 X4 0.000 X5 1.000 X6 0.000 X7 0.000 Constraint Slack/Surplus -------------- --------------- 1 4.000 2 55.000 3 0.000 4 0.000 5 0.000 6 0.000

14. Kerimcan OzcanMNGT 379 Operations Research14 Example: Metropolitan Microwaves Optimal Solution Introduce: TV/VCRs, Projection TVs, and DVD Players Do Not Introduce: Color TVs, VCRs, Video Games, and Home Computers Total Expected Return: $4,261

15. Kerimcan OzcanMNGT 379 Operations Research15 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' in the table indicates an unacceptable tailor-garment assignment.) Tailor Garment 1 2 3 4 5 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 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.

16. Kerimcan OzcanMNGT 379 Operations Research16 Example: Tina’s Tailoring This problem can be formulated as a 0-1 integer program. The LP solution to this problem will automatically be integer (0-1). Define the decision variables x ij = 1 if garment i is assigned to tailor j = 0 otherwise. Number of decision variables = [(number of garments)(number of tailors)] - (number of unacceptable assignments) = [4(5)] - 3 = 17 Define the objective function Minimize total time spent making garments: Min 19x 11 + 23x 12 + 20x 13 + 21x 14 + 18x 15 + 11x 21 + 14x 22 + 12x 24 + 10x 25 + 12x 31 + 8x 32 + 11x 33 + 9x 35 + 20x 42 + 20x 43 + 18x 44 + 21x 45

17. Kerimcan OzcanMNGT 379 Operations Research17 Example: Tina’s Tailoring Define the Constraints Exactly one tailor per garment: 1) x 11 + x 12 + x 13 + x 14 + x 15 = 1 2) x 21 + x 22 + x 24 + x 25 = 1 3) x 31 + x 32 + x 33 + x 35 = 1 4) x 42 + x 43 + x 44 + x 45 = 1 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 ; j = 1,..,5

18. Kerimcan OzcanMNGT 379 Operations Research18 Example: Tina’s Tailoring OPTIMAL SOLUTION Objective Function Value = 55.000 Variable Value -------------- --------------- x11 0.000 x12 0.000 x13 0.000 x14 0.000 x15 1.000 x21 1.000 x22 0.000 x24 0.000 x25 0.000 x31 0.000 x32 1.000 x33 0.000 x35 0.000 x42 0.000 x43 0.000 x44 1.000 x45 0.000 Constraint Slack/Surplus -------------- --------------- 1 0.000 2 0.000 3 0.000 4 0.000 5 0.000 6 0.000 7 1.000 8 0.000 9 0.000

19. Kerimcan OzcanMNGT 379 Operations Research19 Cautionary Note About Sensitivity Analysis 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. Recommendation: Resolve the ILP problem several times with slight variations in the coefficients before choosing the “best” solution for implementation.