1. Lecture 6 – Integer Programming Models Topics General model Logic constraint Defining decision variables Continuous vs. integral solution Applications: staff scheduling, fixed charge, TSP Piecewise linear approximations to nonlinear functions

2. Maximize/Minimize z = c 1 x 1 + c 2 x 2 +    + c n x n {    } b i, i = 1,…, m s.t. a i 1 x 1 + a i 2 x 2 +    + a in x n 0  x j  u j, j = 1,…, n x j integer for some or all j =1,…, n Linear Integer Programming - IP

3.  An IP is a mixed integer program (MIP) if some but not all of the decision variables are integer.  If all decision variables are integer we have a pure IP.  A binary decision variable must be 0 or 1 (a yes-no decision variable).  If all decision variables are binary, then the IP is a binary IP (BIP)  Decision variables that are not required to be integer-valued are continuous variables Decision Variables in IP Models

4. (1)The LP divisibility assumption (fractional solutions are permissible) is not always valid. (2)Binary variables allow powerful new techniques like logical constraints. Why study integer programming?

5. Call Center Employee Scheduling Day is divided into 6 periods, 4 hours each Demand/period = {15, 10, 40, 70, 40, 35} Workforce consists of full-timers (FT) and part- timers (PT) –FT = 8-hour shift, $121.6/ shift –PT = 4-hr shift, $51.8/shift One PT = 5/6 FT In any period, at least 2/3 of the staff must be FT employees (this is a headcount constraint) Problem: Find minimum cost workforce

6. x t =# of full-time employees that begin the day at the start of interval t and work for 8 hours y t =# of part-time employees that are assigned interval t Min z =121.6( x 1 +  +x6+x6 ) + 51.8( y 1 +... +y6)+y6) s.t. x 1 + x 6 + 5 6 y1y1  15 x1x1 + x 2 + 5 6 y2y2  10... x5x5 + x 6 + 5 6 y6y6  35 x 1 + x 6  2 3 (x1(x1 + x 6 + y 1 )... x5x5 + x 6  2 3 (x5(x5 + y 6 ) x t  0, y t  0, t = 1,2,…,6 Decision variables: Call Center Employee IP Model

7. x = ( 7.06, 0, 40, 12.94, 27.06, 7.94 ) y = ( 0, 3.53, 0, 20.47, 0, 0 ) z = 12,795.2 Not feasible to IP model A correction method: round continuous solution x = ( 8, 0, 40, 13, 27, 8 ) y = ( 0, 3, 0, 21, 0, 0 ) z = 12,916.8 We do not know! Feasible – Yes, Optimal? We do not know! Optimal LP solution

8. z  LP = 12,795.2 is a lower bound on integer optimum Hence, the rounded solution is no more than ( 12,916.8 – 12,795.2 12,795.2 )  100% = 0.95% from the optimum. x = ( 10, 0, 40, 20, 20, 5 ) y = ( 10, 0, 0, 12, 0, 12 ) z IP = 12,795.2 Optimal solution is Here the optimal LP and IP objective functions have the same value. This is not commonly true.  How good is this solution?

9. Sometimes there is no “obvious” feasible solution that can be obtained by rounding The IP solution can be “far” from the LP rounded solution even when the rounded solution is feasible.                     optimal LP solution X1X1 X2X2 optimal IP solution iso-cost line   

10. The Days-Off Scheduling Problem Each employee works 5 days per week and is given 2 consecutive days off [ (5,7)-cycle problem] c j =weekly cost of pattern j per employee r i =number of employees required on day i x j =number of employees assigned to days-off pattern j Note:There are 7 days-off patterns; i, j = 1,…,7 Pattern i has days i and i + 1 off. For example, pattern 3 would be: (Mon, Tue, off, off, Thu, Fri, Sat, Sun)  (1, 1, 0, 0, 1, 1, 1)

11. Days-Off Mathematical Model subject to x j  0 and integer, j = 1,…,7; x 0 = x 7 Solve problem to get Minimum cost workforce

12. Compact Mathematical Model Minimize z = cx subject to x  0 and integer

13. Either-or constraints Either f 1 ( x 1,…, x n )  b 1 or f 2 ( x 1,…, x n )  b 2 or both IP formulation y = 0  first constraint must hold y = 1  second constraint must hold Optimization process will choose the y value. Let y  {0,1}; M = “large” number f 1 ( x 1,…, x n )  b 1 + M y f 2 ( x 1,…, x n )  b 2 + M (1 – y ) Logic Constraints

14. At least K out of N must be satisfied. N constraints f 1 ( x 1,…, x n )  b 1  f N ( x 1,…, x n )  b N N i =1  y i = K y i  {0, 1}, i = 1,…, N f 1 ( x 1,…, x n )  b 1 + M (1 – y 1 )  f N ( x 1,…, x n )  b N + M (1 – y N ) K out of N constraints must hold

15. A production system has N potential quality control inspection strategies. Management has decided that K of these strategies should be adopted. Example of K out of N Constraints

16. (Choice constraints for one region only) } Region 1 constraints } Region 2 constraints } Region 3 constraints y 1 + y 2 + y 3 = 1, y 1, y 2, y 3  {0,1} f 1 ( x 1,…, x n )  b 1 + M (1 – y 1 ) f 2 ( x 1,…, x n )  b 2 + M (1 – y 1 ) f 3 ( x 1,…, x n )  b 3 + M (1 – y 2 ) f 4 ( x 1,…, x n )  b 4 + M (1 – y 2 ) f 5 ( x 1,…, x n )  b 5 + M (1 – y 3 ) f 6 ( x 1,…, x n )  b 6 + M (1 – y 3 ) Compound Alternatives

17. n j =1 Min  h j ( x j ), where h j ( x j ) = { f j + c j x j if x j > 0 0 if x j = 0 and f j = setup cost, c j = per unit cost IP formulation: Min n j =1  ( c j x j + f j y j ) s.t. x j  My j, j = 1,…, n y j  {0,1}, j = 1,…, n x j  0, j = 1,…, n Fixed-Charge Problem

18. A company has m potential warehouse sites and n customers Data: d j : demand for customer j s i : capacity (supply) of warehouse i Decision variables: y i : build a warehouse at site i (1 = yes, 0 = no) x ij : shipments from warehouse i to customer j Example: Facility Location Problem

19. Min m  i =1 n  j =1 c ij x ij + m  i =1 fiyifiyi s.t. m i =1  x ij = d j j = 1,…, n n j =1  x ij  s i y i i = 1,…, m x ij  0, i = 1,…, m, j = 1,…, n y i  {0,1}, i = 1,…, m Satisfy each customer’s demand each warehouse can ship no more than its supply if it is built Facility Location IP Model

20. Problem:Open set of facilities and assign each customer to one facility such that cost is minimized. Cost could be a function of distance from facility to customer or could be based on a response time (e.g., locating fire stations). Uncapacitated Facility Location Problem

21. Indices/Sets potential facility locations, i  I with | I | = m customers, j  J with | J | = n Data f i : cost of opening a facility at location i c ij : cost of assigning customer j to facility i Decision variables y i : open facility at location i (1 = yes, 0 = no) x ij : assign customer j to location i (1 = yes, 0 = no) Notation

22. Min s.t. x ij  {0,1}, y i  {0,1}  i  I, j  J if y i = 0, x ij must be 0 for all j  J ; if y i = 1, up to n of the x ij can be 1.  Not computationally efficient   x ij = 1,  j  J iIiI   x ij  ny i,  i  I jJjJ iIiI i  I j  J  c ij x ij +  f i y i Weak Formulation

23. Min s.t.   x ij = 1,  j  J Each customer is assigned to exactly one facility Setup constraint Can assign customer j to facility i only if we open facility i. iIiI iIiI i  I j  J x ij  y i,  i  I, j  J x ij  {0,1} y i  {0,1},  i  I, j  J  c ij x ij +  f i y i Mathematically Equivalent Strong Formulation

24. 3 crews -- each must be assigned a sequence of flights that begins and ends in Dallas ( DFW ) Each flight leg must be covered Possible tours 123456789101112 DFW  LAX1111 DFW  DEN1111 DFW  SEA 11 1 1 LAX  CHI 22323 LAX  DFW 2 3 55 CHI  DEN334 CHI  SEA33334 DEN  DFW2445 DEN  CHI 2 2 2 SEA  DFW2 44 5 SEA  LAX 2 Cost ($10000)234675789989 Decision variables x j ( j =1,…,12) assign a crew to tour j (1 = yes, 0 = no) 2 44 2 Leg Set Covering Problem (Airline Crew Scheduling)

25. Min 2 x 1 + 3 x 2 + 4 x 3 + … + 8 x 11 + 9 x 12 s.t. x1x1 + x4x4 + x7x7 + x 10 (DFW  LAX) x2x2 + x5x5 + x 8 + x 11 (DFW  DEN) x3x3 + x 6 + x9x9 + x 12 (DFW  SEA) x4x4 + x7x7 + x9x9 + x 10 + x 12 (LAX  CHI) x1x1 + x6x6 + x 10 + x 11 (LAX  DFW)  x6x6 + x9x9 + x 10 + x 11 + x 12  1 (SEA  LAX) x1x1 + x2x2 + … + x 12 = 3 (assign 3 crews) x j  {0,1}, j = 1,…,12 Allows “dead heading”; i.e., multiple crews fly on 1 leg but only 1 crew works (all get paid).  1   side constraint Formulation (set covering problem)

26. 1 4 3 2 Problem:Find minimum distance tour that starts at city 1, visits every other city exactly once, and returns to city 1. Traveling Salesman Problem (TSP)

27. Decision variables x ij = 1 if tour includes arc ( i, j ) = 0 otherwise Initial formulation Min50 x 12 +44 x 13 + 25 x 14 +... +20 x 41 +45 x 42 +40 x 43 s.t. x 12 + x 13 + x 14 = 1 x 21 + x 31 + x 41 = 1 x 21 + x 23 + x 24 = 1 x 12 + x 32 + x 42 = 1... x 41 + x 42 + x 43 = 1 x 14 + x 24 + x 34 x ij  {0,1}, i  j This is incomplete because subtours are possible. = 1 1 3 2 4

28. Alternative formulation: x 13 + x 31  1 and x 24 + x 42  1 To generalize, let N = {1,…, n }, and let S  N SEC:  x ij  | S | – 1, 2  | S |  n /2 (ij)  S Subtour Elimination Constraints Let S 1 = {1,3} and S 2 = {2,4} and require at least 1 arc from S 1 to S 2 x 12 + x 14 + x 32 + x 34  1 Disallows the “2 loop” solution.

29. Example of SEC Let n = 10 and S = { 2, 5, 6, 9 }. Then | S | = 4. SEC: x 25 + x 26 + x 29 + x 52 + x 56 + x 59 + x 62 + x 65 + x 69 + x 92 + x 95 + x 96  4 – 1 = 3 In general, there are an exponential number of subtour elimination constraints. 2 5 6 9

30. Testing integrated circuits (ICs): A machine is used to test several batches of ICs of differing types. After each batch the machine must be reset. The changeover time depends on what type of IC was just tested and which type will be tested next. Changeover Times IC type 1234 1 -- 101715 220 -- 1918 35044--25 44540 20-- IC type Sequencing problems with sequence-dependent setup times can be modeled as a TSP

31. Graph for IC Testing Example Dummy node Sample path: 0  1  3  2  4  0 Cost of path: 0 + 17 + 44 + 18 + 0 = 79 44 17 18 0 0 4 3 2 0 1

32. If the cycle is repeated, IC1  IC3  IC2  IC4  IC1... then the total changeover time for one cycle is 17 + 44 + 18 + 45 = 124 [arc (4,1) added]. The goal is to sequence the testing order so that the throughput (i.e., minimize cycle time) is maximized for fixed batch sizes. Note that in this example, the “travel times” are not symmetric.

33. General Piecewise Linear Approximations Given: f j ( x j ), 0  x j  u j Let r = number of grid points Let ( d ij, f ij ) be i th grid point, i = 1,…, r

34. Linear Transformation for j th Variable Let x j =   i d ij and f j ( x j ) =   i f ij where   i = 1,  i  0, i = 1,…,r Not sufficient to guarantee solution is on one of the line segments. r i =1 r i =1 r i =1

35. Additional Constraints for Piecewise Linear Approximation Requirement:No more than two  i can be positive; also  i ’ s must be adjacent; i.e.,  i and  i +1  1 ≤ y 1  i ≤ y i -1 + y i, i = 2,…,r–1  r ≤ y r -1 y 1 + y 2 + · · · + y r -1 = 1 y i = 0 or 1, i = 1,...,r–1

36. What You Should Know About Integer Programming How to convert a problem statement in an IP model. How to define the decision variables. How to convert logic statements into constraints. How to formulation fixed charge problems, scheduling problems, covering problems,TSP, piece-wise linear approximation to nonlinear functions.