Lecture 6 – Integer Programming Models

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

Linear Integer Programming - IP Maximize/Minimize z = c1x1 + c2x2 +    + cnxn } { £ s.t. ai1x1 + ai2x2 +    + ainxn 0 £xj£uj, j = 1,…,n xj integer for some or all j =1,…,n bi, i = 1,…,m ³ =

Decision Variables in IP Models · 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

Why study IP? • LP divisibility assumption (fractional solutions are permissible) is not always valid. • (2) Binary variables allow powerful new techniques like logical constraints.

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

Call Center Employee IP Model Decision variables: xt = # of full-time employees that begin the day at the start of interval t and work for 8 hours yt = # of part-time employees that are assigned interval t . . . . . . Min z = 121.6(x1 + +x6 ) + 51.8(y1 + +y6) 5 ³ s.t. x1 + x6 + y1 15 2 6 x1 + x6³ (x1 + x6 + y1) 3 5 . . ³ x1 + x2 + y2 10 . 6 . 2 . ³ . x5 + x6 (x5 + x6 + y6) 3 5 ³ x5 + x6 + y6 35 6 xt³ 0, yt³ 0, t = 1,2,…,6

Optimal LP solution 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 Feasible – Yes, Optimal? We do not know!

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

X2    optimal LP solution         X1 iso-cost line optimal IP solution             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.

The Days-Off Scheduling Problem Each employee works 5 days per week and is given 2 consecutive days off [ (5,7)-cycle problem] cj = weekly cost of pattern j per employee ri = number of required on day i xj = number of employees assigned to days-off pattern j Note: There are 7 days-off patterns; i, j = 1,…,7

Days-Off Mathematical Model subject to xj 0 and integer, j = 1,…,7; x0 = x7 Solve problem to get Minimum cost workforce

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

Logic Constraints Either-OrConstraints Either f1(x1,…,xn) £b1 or f2(x1,…,xn) £b2 or both IP formulation Let yÎ {0,1} f1(x1,…,xn) £b1 + My f2(x1,…,xn) £b2 + M(1 – y) y = 0  first constraint must hold y = 1  second constraint must hold Optimization process will choose the y value.

f1(x1,…,xn) £b1 + M(1 – y1)  fN(x1,…,xn) £bN + M(1 – yN) N Syi = K i=1 yiÎ {0, 1}, i = 1,…,N Kout ofN constraints must hold N constraints f1(x1,…,xn) £b1  fN(x1,…,xn) £bN At least K out of N must be satisfied

Example of K out of N Constraints • A production system has N potential qualitycontrol inspection strategies • Management has decided that K of these strategies should be adopted

Compound Alternatives (Choice constraints for one region only) } Region 1 constraints f1(x1,…,xn) £b1 + M(1 –y1) f2(x1,…,xn) £b2 + M(1 –y1) f3(x1,…,xn) £b3 + M(1 –y2) f4(x1,…,xn) £b4 + M(1 –y2) } Region 2 constraints f5(x1,…,xn) £b5 + M(1 –y3) f6(x1,…,xn) £b6 + M(1 –y3) } Region 3 constraints y1 + y2 + y3 = 1,y1, y2, y3Î {0,1}

IP formulation: n å (ckxk + fkyk) Min k=1 xk£Myk , k = 1,…,n ykÎ {0,1}, k = 1,…,n xk³ 0, k = 1,…,n s.t. Fixed-Charge Problem Min åhk(xk) where hk(xk) ={ n fk + ckxk if xk > 0 0 if xk = 0 k=1 fk = setup cost, ck = per unit cost

Example:Facility Location Problem A company hasm potential warehouse sites and n customers Data: dj : demand for customer j si : capacity (supply) of warehouse i Decision variables: yi: build a warehouse at site i (1 = yes, 0 = no) xij : shipment from warehouse i to customer j

Facility Location IP Model m n m å å å cijxij + fiyi Min j =1 i =1 i =1 m åxij =djj = 1,…,n s.t. Satisfy each customer’s demand i =1 n åxij£siyii = 1,…,m each warehouse can ship no more than its supply if it is built j =1 xij³ 0,i = 1,…,m, j = 1,…,n yiÎ {0,1}, i = 1,…,m

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

Indices/Sets potential facility locations, iÎI with |I| = m customers, jÎJ with |J| = n Data fi : cost of opening a facility at location i cij : cost of assigning customer j to facility i Decision Variables yi : open facility at location i (1 = yes, 0 = no) xij : assign customer j to location i (1 = yes, 0 = no)

Weak formulation å å cijxij + å fiyi Min iÎI iÎIjÎJ åxij = 1, "j Î J s.t. iÎI { xij must be 0 for all jÎJ if yi = 0; up to n of the xij can be 1 if yi = 1. åxij£nyi, "iÎ I jÎJ xijÎ {0,1}, yiÎ {0,1} "iÎ I, jÎ J  Not computationally efficiency

Mathematically Equivalent Strong Formulation å å cijxij + å fiyi Min iÎI iÎIjÎJ Each customer is assigned to exactly one facility Setup constraint åxij = 1, "jÎJ s.t. iÎI xij£yi, "i Î I, jÎ J xijÎ {0,1} yiÎ {0,1} , "iÎ I, jÎ J Can assign customer j to facility ionly if we open facility i.

Set Covering Problem (Airline Crew Scheduling) • 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 Leg 1 2 3 4 5 6 7 8 9 10 11 12 DFW  LAX 1 1 1 1 DFW DEN 1 1 1 1 1 1 1 1 DFW  SEA 2 2 3 2 3 LAX  CHI LAX DFW 2 3 5 5 CHI  DEN 3 3 4 3 3 3 3 4 CHI  SEA DEN  DFW 2 4 4 5 2 2 DEN  CHI 2 5 4 4 SEA  DFW 2 2 4 4 SEA  LAX 2 2 Cost ($10000) 2 3 4 6 7 5 7 8 9 9 8 9 Decision Variables xj (j =1,…,12) assign a crew to tour j (1 = yes, 0 = no)

Formulation (Set covering problem) … 2x1 + 3x2 + 4x3 + + 8x11 + 9x12 Min s.t. ³ 1 (DFW  LAX) x1 + x4 + x7 + x10 x2 + x5 + x + x11 (DFW  DEN) ³ 1 8 x3 + x + x9 + x ³ 1 (DFW  SEA) 6 12 x4 + x7 + x9 + x + x12 (LAX  CHI) ³ 1 10 x1 + x6 + x + x ³ 1 (LAX  DFW) 10 11    (SEA  LAX) ³ 1 x6 + x9 x10 + x + x12 + 11 … x1 + + + (assign 3 crews) x2 x12 = 3 xjÎ {0,1}, j = 1,…,12 side constraint Allows “dead heading”; i.e., multiple crews fly on 1 leg but only 1 crew works (all get paid).

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

Decision Variables xij = 1 if tour includes arc (i, j ) = 0 otherwise Initial Formulation + . . . + 20x41 + 45x42 + 40x43 Min 50x12 + 44x13 + 25x14 s.t. x + x + x = 1 12 13 14 xijÎ {0,1}, i¹j x + x + x = 1 21 31 41 x + x + x = 1 21 23 24 1 x + x + x = 1 . 42 12 32 . . 4 2 x + x + x = 1 41 42 43 x + x + x = 1 3 34 14 24 This is incomplete because subtours are allowed.

Add Subtour Elimination Constraints Let S1 = {1,3} and S2 = {2,4} and require at least 1 arc from S1 to S2 x12 + x14 + x32 + x34³ 1 Disallows the “2 loop” solution. Alternative formulation: x13 + x31£ 1andx24 + x42£ 1 To generalize, let N = {1,…,n}, and let SN SEC: Sxij£ |S| – 1, 2 £ |S| £n/2 (ij)S

Example of SEC Let n = 10 and S = { 2, 5, 6, 9 }. Then |S| = 4. SEC: x25 + x26 + x29 + x52 + x56 + x59 + x62 + x65 + x69 + x92 + x95 + x96£ 4 – 1 = 3 6 2 5 9 In general, there are an exponential number ofsubtour elimination constraints.

Sequencing problems with sequence-dependent setup times can be modeled as a TSP • 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 change-over time depends on what type of IC was just tested and which type will be tested next. Change-Over Times IC Type 1 2 3 4 -- 1 10 17 15 IC Type 2 20 19 18 -- 3 50 44 -- 25 20 -- 4 45 40

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

If the cycle is repeated, • IC1  IC3  IC2  IC4  IC1 . . . • then the total change-over time for one cycle is • 17 + 44 + 18 + 45 = 124. • 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.

General Piecewise Linear Approximations Given: fj(xj), 0 xjuj Let r = number of grid points Let (dij, fij) be ith grid point, i = 1,…,r

Linear Transformation for jth Variable r i=1 Let xj = Si dij and fj(xj) = Si fij where S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

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 a1 ≤ y1 ai ≤ yi-1+ yi, i = 2,…,r–1 ar ≤ yr-1 y1 + y2 + · · · + yr-1 = 1 yi = 0 or 1, i = 1,...,r–1

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.

Lecture 6 – Integer Programming Models