is called an integer programming (IP) problem. model is called a mixed integer programming (MIP)

Transcription

1 INTEGER PROGRAMMING

2 Integer Programming g In many problems the decision variables must have integer values. Example: assign people, machines, and vehicles to activities in integer quantities. If this is the only deviation from linear programming, it is called an integer programming (IP) problem. If only some variables ibl are required to be integer, the model is called a mixed integer programming (MIP) San Francisco Police Dep. problem is an IP problem. Wyndor Glass Co. problem could be an IP problem; how? João Miguel da Costa Sousa / Alexandra Moutinho 261

3 Integer Programming g In integer programming the divisibility assumption must be dropped. Another area of application relates to problems involving yes or no decisions, which have binary variables. These IP problems are ecalled binary integer programming (BIP) problems. A small example of a typical BIP problem is given in the following. João Miguel da Costa Sousa / Alexandra Moutinho 262

4 Prototype example California Manufacturing Company is considering expansion, building a factory in Los Angeles, San Francisco or in both cities. One new warehouse can also be considered in a city where a new factory is being built. Maximum $10 million to invest. Objective: find feasible combination of alternatives that maximizes the total net present value. Decision Yes or no question Decision Net present Capital number variable value required 1 Build factory in Los Angeles? x 1 $9 million $6 million 2 Build factory in San Francisco? x 2 $5 million $3 million 3 Build warehouse in Los Angeles? x 3 $6 million $5 million 4 Build warehouse in San Francisco? x 4 $4 million $2 million João Miguel da Costa Sousa / Alexandra Moutinho 263

5 BIP model All decision variables have the binary form: x j 1 if decision j is yes, = j= 1,2,3,4 0 if decision j is no, Z = total net present value of these decisions. Maximize Z = 9x 1 + 5x 2 + 6x 3 + 4x 4. Constraints: 6x 1 + 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 3 x 1 and x 4 x 2 x j is binary, for j = 1,2,3,4.,3,4 Mutually exclusive alternatives Contingent decisions João Miguel da Costa Sousa / Alexandra Moutinho 264

6 BIP models Groups p of yes or no decisions often constitute groups of mutually exclusive alternatives: only one decision in the group can be yes. Occasionally, decisions of the yes or no type are contingent decisions: decision that depends upon previous ones. Software options for solving BIP, IP or MIP models: Excel MatLab LINGO/LINDO MPL/CPLEX João Miguel da Costa Sousa / Alexandra Moutinho 265

7 BIP applications Investment analysis, such as the California Man. Co. Site selection, of factories, warehouses, etc. Designing a production and distribution network, or more generally the entire global supply chain. Dispatching shipments, scheduling routes, vehicles and time period for departure and arrivals. Airline applications, as e.g. fleet assignment and crew scheduling. Scheduling interrelated activities, asset divestures, etc. João Miguel da Costa Sousa / Alexandra Moutinho 266

8 Formulation examples Example 1: making choices when decision variables are continuous. R&D Division of Good Products Co. has developed three possible new products. Requirement 1: from the three, at most two can be chosen to be produced. Each product can be produced in either of two plants. However, management has imposed another restriction: Requirement 2: just one of the two plants can be chosen as the producer of the new products. João Miguel da Costa Sousa / Alexandra Moutinho 267

9 Example 1 Production time used for each unit produced Product 1 Product 2 Product 3 Production time available per week Plant 1 3 hours 4 hours 2 hours 30 hours Plant 2 4 hours 6 hours 2 hours 40 hours Unit profit (10 3 $) Sales (units per potential week) Objectives: choose the products, the plant and the production rates of the chosen products to maximize total profit. João Miguel da Costa Sousa / Alexandra Moutinho 268

10 Formulation of the problem Similar to a standard product mix problem,, such as the Wyndor Glass Co. if we drop the two restrictions and require each product to use production hours in both plants. Let x 1, x 2, x 3 be the production rates of the respective products: Maximize Z= 5x + 7x + 3x subject to 3x + 4x + 2x x + 6x + 2x x1 7 x2 5 x3 9 x, x, x João Miguel da Costa Sousa / Alexandra Moutinho 269

11 Formulation of the problem For real problem, restriction 1 adds the constraint: Number of strictly positive variables (x 1, x 2, x 3 ) must be 2 This must be converted to an IP problem. It needs the introduction of auxiliary binary variables. Restriction 2 requires replacing the first two functional constraints by: Either 3x + 4x + 2x or 4x + 6x + 2x 40 must hold. od This again requires an auxiliary binary variable. João Miguel da Costa Sousa / Alexandra Moutinho 270

12 Auxiliary binary variables For requirement 1, three auxiliary binary variables (y 1, y 2, y 3 )are 3 introduced: y j 1 if xj > 0 can hold (can produce product j) = 0 if xj = 0 must hold (cannot produce product j) This is introduced in the model with the help of an extremely large positive number M, adding the constraints: x My 1 1 x2 My2 x3 My3 y1+ y2+ y3 2 y is binary, for j= 1,2, j João Miguel da Costa Sousa / Alexandra Moutinho 271

13 Auxiliary binary variables For requirement 2, another auxiliary binary variable y 4 is introduced: y 4 1 if 4 x1 + 6 x2 + 2 x3 40 must hold (choose Plant 2) = 0 if 3x1+ 4x2+ 2x3 30 must hold (choose Plant 1) This adds the constraints: 3x + 4x + 2x 30+ My x + 6x + 2x 40 + M(1 y ) y is binary João Miguel da Costa Sousa / Alexandra Moutinho 272

14 Complete model (MIP) Maximize Z= 5x1 + 7x2+ 3x3 subject to x1 7 x2 5 x3 9 x1 My1 0 x2 My2 0 x3 My3 0 y1 + y2 + y3 2 3x1 + 4x2+ 2x3 My4 30 4x + 6x + 2x + My 40+ M and xi 0, for i= 1,2,3 y is binary, for j = 1,2,3,4 j João Miguel da Costa Sousa / Alexandra Moutinho 273

15 Solution MIP problem with 3 continuous and four binary variables. Optimal solution: y 1 = 1, y 2 = 0, y 3 = 1, y 4 = 1, x 1 = 5.5, x 2 = 0, x 3 = 9. That is, produce products 1 and 3 with production rates 5.55 units per week and 9 units per week respectively, and choose Plant 2 for production. Resulting total profit is $54, per week. João Miguel da Costa Sousa / Alexandra Moutinho 274

16 Example: Southwestern Airways Southwestern Airways needs to assign three crews to cover all the upcoming flights. Table shows the flights in the first column. Other 12 columns show the 12 feasible sequences of flights for a crew. Numbers in each column indicate the order of the flights. Exactly three sequences must be chosen (one per crew). More than one crew can be assigned to a flight, but it must be paid as if it was working. Last row shows the cost of assigning a crew to a particular sequence of flights. João Miguel da Costa Sousa / Alexandra Moutinho 275

17 Data for Southwestern Airways Feasible sequence of flights Flight San Francisco to Los Angeles San Francisco to Denver San Francisco to Seattle Los Angeles to Chicago Los Angeles to San Francisco Chicago to Denver Chicago to Seattle Denver to San Francisco Denver to Chicago Seattle to San Francisco Seattle to Los Angeles Cost (1000 ) João Miguel da Costa Sousa / Alexandra Moutinho 276

18 Formulation of the problem Objective: minimize the total cost for the three crew assignments that cover all flights. 12 feasible sequences of flights: 12 yes or no decisions: Should sequence j be assigned to a crew? The 12 binary variables to represent the decisions are: 1 if sequence j is assigned to a crew x j = 0 otherwise João Miguel da Costa Sousa / Alexandra Moutinho 277

19 Formulation of the problem Minimize Z= 2x + 3x + 4x + 6x + 7x + 5x + 7x + 8x + 9x + 9x10 + 8x11 + 9x12 subject to x + x + x + x (SF to LA) x2+ x5+ x8+ x11 1 x3+ x6+ x9+ x12 1 x4 + x7 + x 9 + x10 + x12 1 x and x j is binary, 1 x6 x10 x11 1 x4+ x5+ x9 1 for j = 1,2,,12 x7 + x8 + x10 + x11 + x12 1 x2+ x4+ x5+ x9 1 x 5+ x 8+ x 11 1 x3+ x7 + x8+ x12 1 x + x + x + x + x x j j= 1 = 3 (assign three crews) João Miguel da Costa Sousa / Alexandra Moutinho 278

20 Solution One optimal solution is: x 3 = 1 (assign sequence 3 to a crew) x 4 = 1 (assign sequence 4 to a crew) x 11 = 1 (assign sequence 11 to a crew) And all other x j = 0. Total cost is $18,000. Another optimal solution is: x 1 = x 5 = x 12 = 1. João Miguel da Costa Sousa / Alexandra Moutinho 279

21 Discussion This example belongs to a class called set covering problems,, with a number of potential activities (e.g. flight sequences) and characteristics (e.g. flights). Objective: determine the least costly combination of activities that collectively possess each characteristic at least once. S i is the set of all activities that possess characteristic i. A constraint is included for each characteristic i: j S 1 In set partitioning i problems the constraint t is i j S j x x j j = 1 João Miguel da Costa Sousa / Alexandra Moutinho 280

22 Solving IP problems Are integer problems easy to solve? Difference to LP is that IP have far fewer solutions. IP problems have a finite number of feasible solutions. However: Finite it numbers can be astronomically large! With n variables a BIP problem has 2 n solutions, having exponential growth. LP assures that a CPF solution can be optimal, guaranteeing gthe remarkable efficiency of the simplex method. LP problems are much easier to solve than IP problems! João Miguel da Costa Sousa / Alexandra Moutinho 281

23 Solving IP problems Consequently, most IP algorithms incorporate the simplex method. This is called the LP relaxation. Sometimes, the solution of the LP problem is the solution of the IP problem, such as : Minimum cost flow problem, including transportation problem, assignment problem, shortest path problem and maximum flow problem. Special structures (see examples 2 and 3): mutually exclusive alternatives, contingent decisions or setcovering constraints can also simplify the problem. João Miguel da Costa Sousa / Alexandra Moutinho 282

24 Solving IP problems Primary determinants of computational complexity: 1. number of integer variables, 2. these variables are binary or general integer variables, 3. any special structure in the problem. This is in contrast to LP, where number of constraints is much more important than the number of variables. As IP problems are much more difficult than LP, we could apply LP and round the obtained solution... Yes? João Miguel da Costa Sousa / Alexandra Moutinho 283

25 Example 1 Minimize Z= x 2 subject to x1+ x2 0.5 x1 + x2 3.5 and x1, x2 0, x, x integers. 1 2 João Miguel da Costa Sousa / Alexandra Moutinho 284

26 Example 2 Minimize Z= x 1+ 5x 2 subject to x 1+ 10x 2 20 x1 2 and x, x 0, integers. 1 2 João Miguel da Costa Sousa / Alexandra Moutinho 285

27 Solving IP problems Thus, a better approach to deal with IP problems that are too large to be solved exactly are heuristic algorithms. Heuristics and metaheuristics are extremely efficient for very large problems, but do not guarantee to find an optimal solution. These algorithms will be discussed later. Most popular traditional method for solving IP problems is the branch and bound technique. João Miguel da Costa Sousa / Alexandra Moutinho 286

28 Branch and bound applied to BIP Pure IP problems can consider some type of enumeration procedure. This should be done in a clever way such that only a tiny fraction of the feasible solutions is examined. Branch and bound with a divide to conquer technique can be used. dividing (branching) the problem into smaller and smaller subproblems until it can be conquered conquering (fathoming) by bounding how good the best solution can be. If no optimal solution in subset: discard it. João Miguel da Costa Sousa / Alexandra Moutinho 287

29 Example: California Manuf, Co. Recall prototype example: Maximize Z = 9x 1 + 5x 2 + 6x 3 + 4x 4 subject to and (1) 6x 1 + 3x 2 + 5x 3 + 2x 4 10 (2) x 3 + x 4 1 (3) x 1 + x 3 0 (4) x 2 + x 4 0 (5) x j is binary, for j = 1, 2, 3, 4. João Miguel da Costa Sousa / Alexandra Moutinho 288

30 Branching Most straightforward way to divide the problem: fix the value of a variable: e.g. x 1 = 0 for one subset and x 1 = 1 for another subset. Subproblemp 1 (fix x 1 =0): Subproblem 2 (fix x 1 = 1): Maximize Z =5x 2 +6x 3 +4x 4 Maximize Z =9+5x 2 +6x 3 +4x 4 subject to subject to (1) 3x 2 +5x 3 +2x 4 10 (1) 3x 2 +5x 3 +2x 4 4 (2) x 3 + x 4 1 (2) x 3 + x 4 1 (3) x 3 0 (3) x 3 1 (4) x 2 + x 4 0 (4) x 2 + x 4 0 (5) x j is binary, for j = 2, 3, 4. (5) x j is binary, for j = 2, 3,4. João Miguel da Costa Sousa / Alexandra Moutinho 289

31 Branching Dividing (branching) into suproblems creates a tree with branches (arcs) for the All node. This is the solution tree or enumeration tree. Branching variable is the one used for branching. The branching continues or not after evaluating the subproblem. Other IP problems usually creates as many branches as needed. João Miguel da Costa Sousa / Alexandra Moutinho 290

32 Bounding A bound is needed for the best feasible solution of each of the subproblems. Standard way is to perform a relaxation of the problem, e.g. by deleting one set of constraints that makes the problem difficult to solve. Most common o is to require e integer variables, ab es,soso LP relaxation is the most widely used. João Miguel da Costa Sousa / Alexandra Moutinho 291

33 Bounding in example Example: for the whole problem, (5) is replaced by x j 1 and x j 0 for j=1,2,3,4. Using simplex: (x 1, x 2, x 3, x 4 ) = (5/6, 1, 0, 1), with Z = 16.5 Thus, Z 16.5 for all feasible solutions for BIP problem. Can be rounded to Z 16 (why?) LP relaxation for subproblem 1 (x 1 =0): (x 1, x 2, x 3, x 4 ) = (0, 1, 0, 1), with Z = 9 LP relaxation for subproblem 2 (x 1 =1): (x 1, x 2, x 3, x 4 ) = (1, 4/5, 0, 4/5), with Z = 16.5 João Miguel da Costa Sousa / Alexandra Moutinho 292

34 Fathoming A subproblem can be conquered (fathomed, i.e. search tree is pruned) in three ways: 1. When the optimal solution for the LP relaxation of a subproblem is integer, it must be optimal. Example: for x 1 =0, (x 1, x 2, x 3, x 4 ) = (0, 1, 0, 1), is integer. It must be stored as first incumbent (best feasible solution found so far) for the whole problem, along with value of Z: In the example Z * = 9. Z * = value of Z for first incumbent Subproblem 1 is solved, so it is fathomed (dismissed). João Miguel da Costa Sousa / Alexandra Moutinho 293

35 Fathoming 2. As Z * = 9, we should not consider subproblems with bound 9. Thus, a problem is fathomed when Bound Z * In Subproblem 2 that does not occur, the bound of 16 is larger than 9. However, it can occur for descendants. As new incumbents with larger values of Z * are found, it becomes easier to fathom in this way. 3. If the simplex method finds that a subproblem s LP relaxation has no feasible solution, the subproblem has no feasible solution and can be dismissed. João Miguel da Costa Sousa / Alexandra Moutinho 294

36 Summary of fathoming tests A subproblem is fathomed (dismissed) if Test 1: Its bound Z * or Test 2: Its LP relaxation has no feasible solutions or Test 3: Optimal solution for its LP relaxation is integer. If better, this solution becomes new incumbent, and Test 1 is reapplied for all unfathomed subproblems. João Miguel da Costa Sousa / Alexandra Moutinho 295

37 Fathoming in example Result of applying the three tests is in figure below. Subproblem 1 is fathomed by test 3. João Miguel da Costa Sousa / Alexandra Moutinho 296

38 BIP branch and bound algorithm Initialization: Set Z * =. Apply bounding, fathoming and optimization steps described below to the whole problem. If not fathomed, perform iteration. Steps for each iteration: 1. Branching: Among the remaining subproblems, select the one created most recently. Branch from this node by fixing the next variable as either 0 or Bounding: For each new subproblem, obtain its bound by applying its LP relaxation. Round down Z for resulting optimal solution. João Miguel da Costa Sousa / Alexandra Moutinho 297

39 BIP branch and bound algorithm 3. Fathoming: For each new subproblem, apply the three fathoming tests, and discard subproblems that are fathomed by the tests. Optimality test: Stop when there are no remaining subproblems. The current incumbent is optimal. Otherwise, perform another iteration. João Miguel da Costa Sousa / Alexandra Moutinho 298

40 Completing example Iteration 2. Remaining subproblems are for x 1 = 1. Subproblem 3 (fix x 1 = 1, x 2 = 0): Subproblem 4 (fix x 1 = 1, x 2 = 1): Maximize Z = 9 + 6x 3 + 4x 4 Maximize Z = x 3 + 4x 4 subject to subject to (1) 5x 3 + 2x 4 4 (2) x 3 + x 4 1 (1) 5x 3 + 2x 4 1 (2) x 3 + x 4 1 (3) x 3 1 (3) x 3 1 (4) x 4 0 (4) x 4 1 (5) x j is binary, for j = 3, 4. (5) x j is binary, for j = 3, 4. João Miguel da Costa Sousa / Alexandra Moutinho 299

41 Example LP relaxation is obtained by replacing (5) by 0 x j 1 j = 3, 4. Optimal solutions are: LP relaxation for subproblem 3: (x 1, x 2, x 3, x 4 ) = (1, 0, 0.8, 0), with Z = 13.8 LP relaxation for subproblem 4: (x 1, x 2, x 3, x 4 ) = (1, 1, 0, 0.5), with Z = 16 Resulting bounds: Bound for subproblem 3: Z 13 Bound for subproblem 4: Z 16 João Miguel da Costa Sousa / Alexandra Moutinho 300

42 Example All three fathoming tests fail, so both are unfathomed. João Miguel da Costa Sousa / Alexandra Moutinho 301

43 Iteration 3 Subproblem 4 has the larger bound, so next branching is done from (x 1, x 2 ) = (1, 1). Subproblem 5 Subproblem 6 (fix x 1 = 1, x 2 = 1, x 3 = 0): Maximize Z = x 4 subject to (1) 5x 3 + 2x 4 1 (fix x 1 = 1, x 2 = 1, x 3 = 1): Maximize Z = x 4 subject to (1) 2x 4 4 (2), (4) x 4 1 (2) x 4 0 (5) x 4 is binary (4) x 4 1 (5) x 4 is binary João Miguel da Costa Sousa / Alexandra Moutinho 302

44 Iteration 3 (cont.) LP relaxation: replace (5) by 0 x 4 1. Optimal solutions are: LP relaxation for subproblem bl 5: (x 1, x 2, x 3, x 4 ) = (1, 1, 0, 0.5), Z = 16 LP relaxation for subproblem 6: No feasible solutions. Bound for subproblem 5: Z 16 Subproblem 6 is fathomed by test 2, but not subproblem 5. João Miguel da Costa Sousa / Alexandra Moutinho 303

45 Iteration 3 (concl.) João Miguel da Costa Sousa / Alexandra Moutinho 304

46 Iteration 4 Node created most recently is selected for branching: x 4 = 0: (x 1, x 2, x 3, x 4 ) = (1, 1, 0, 0) is feasible, with Z = 14, x 4 = 1: (x 1, x 2, x 3, x 4 ) = (1, 1, 0, 1) is infeasible. ibl First solution passes test 3 (integer solution) and second passes test 2 (infeasible) for fathoming. First solution is better than incumbent, so it becomes new incumbent, with Z * = 14 Reapplying fathoming test 1 (bound) to remaining branch of Subproblem 3: Bound = 13 Z * = 14 (fathomed). João Miguel da Costa Sousa / Alexandra Moutinho 305

47 Solution tree after Iteration 4 João Miguel da Costa Sousa / Alexandra Moutinho 306

48 Other options in Branch and Bound Branching can be done e.g. from the best bound rather than from the most recently created subproblem. Bounding is done by solving a relaxation. Another possible one is e.g. the Lagrangian g relaxation. Fathoming criteria can be generally stated as: Crit. 1: feasible solutions of subproblem must have Z Z *, Crit. 2: the subproblem has no feasible solutions, or Crit. 3: an optimal solution of subproblem has been found. Some adjustments necessary for Branch and bound to find multiple optimal solutions. João Miguel da Costa Sousa / Alexandra Moutinho 307

49 Branch and bound for MIP General form of the problem: Maximize Z cjxj n j= 1 n = j= 1 subject to ax b, for i= 1,2,, m, and ij j i x 0, for j= 1,2,, n j x is integer, for j= 1,2,, I; I n. j João Miguel da Costa Sousa / Alexandra Moutinho 308

50 Branch and bound for MIP Similar to BIP algorithm. Solving LP relaxations are the basis for bounding and fathoming. 4 changes are needed: 1. Choice of branching variable. Only integer variables that have a noninteger value in the optimal solution for the LP relaxation can be chosen. João Miguel da Costa Sousa / Alexandra Moutinho 309

51 Branch and bound for MIP 2. As integer variables can have a large number of possible values, create just two new subproblems: x * j* : noninteger value of optimal solution for LP relaxation. [x j* ] = greatest integer x j*. Range of variables for two new subproblems: bl x j* [x j* ] and x j* [x j* ] + 1. Each inequality becomes an additional constraint. Example: x j* = 3.5, then: x j* 3 and x j* 4. When changes 1. and 2. are combined a recurring branching variable can occur, see figure. João Miguel da Costa Sousa / Alexandra Moutinho 310

52 Recurring branching variable João Miguel da Costa Sousa / Alexandra Moutinho 311

53 Branch and bound for MIP Changes needed: 3. Bounding step: value of Z was rounded down in BIP algorithm. Now some variables are not integerrestricted so bound is value of Z without rounding. 4. Fathoming test 3: optimal solution for the subproblem s s LP relaxation at must only ybe integer for integer restricted variables. João Miguel da Costa Sousa / Alexandra Moutinho 312

54 MIP branch and bound algorithm Initialization: Set Z * =. Apply bounding, fathoming and optimization steps described below to the whole problem. If not fathomed, perform iteration. Steps for each iteration: 1. Branching: Among the remaining subproblems, bl select the one created most recently. From integer variables that have a noninteger value in the optimal solution for the LP relaxation choose the first one. Let x * j be this variable and x j its value. Branch from this creating two subproblems by adding the respective constraints: x * * * * j [x j ] and x j [x j ] + 1. João Miguel da Costa Sousa / Alexandra Moutinho 313

55 MIP branch and bound algorithm 2. Bounding: For each new subproblem, obtain its bound by applying its LP relaxation. Use Z without rounding for resulting optimal solution. 3. Fathoming: For each new subproblem, apply the three fathoming tests, and discard subproblems that are fathomed by the tests. Test 1: Its bound Z *, where Z * is value of Z for current incumbent. Test 2: Its LP relaxation has no feasible solutions. João Miguel da Costa Sousa / Alexandra Moutinho 314

56 MIP branch and bound algorithm 3. Fathoming (cont.): Test 3: Optimal solution for its LP relaxation has integer values for integer restrictedrestricted variables. (If this solution is better it becomes new incumbent, and test 1 is reapplied for all unfathomed subproblems). Optimality test: Stop when there are no remaining subproblems. The current incumbent is optimal. Otherwise, perform another iteration. See MIP examples in PL#7 and in page 518 of Hillier s book. João Miguel da Costa Sousa / Alexandra Moutinho 315

57 Branch and cut approach to BIP Branch and bound was develop and refined in the 60 s and early 70 s. Can solve problems up to 100 variables. Branch and cut approach was introduced in the mid 80 s, and can solve problems with thousands of variables. Only solve large problems if they are sparse (less than 5 or even 1% of nonzero values in functional constraints). Uses a combination of automatic problem processing, generation of cutting planes and B&B techniques. João Miguel da Costa Sousa / Alexandra Moutinho 316

58 Automatic problem processing for BIP Computer inspection of IP formulation to spot reformulations that make the problem quicker to solve: Fixing variables: identify variables that can be fixed at 0 or 1, because other value cannot lead to feasible and optimal solution. Eliminating redundant constraints: identify and eliminate constraints that are automatically satisfied by solutions that satisfy all other constraints. Tightening constraints: tighten constraints in a way that reduces feasible region of LP relaxation without eliminating any feasible solutions for the BIP problem. João Miguel da Costa Sousa / Alexandra Moutinho 317

59 Tightening gconstraints LP relaxation including LP relaxation after tightening feasible region. constraint. João Miguel da Costa Sousa / Alexandra Moutinho 318

60 Generating cutting planes for BIP Cutting plane (or cut) is a new functional constraint that reduces feasible region for LP relaxation without eliminating any feasible solutions of IP problem. Procedure for generating cutting planes: 1. Consider functional constraint in form with only nonnegative coefficients. 2. Find a group of N variables such that a) Constraint is violated if every variable in group = 1 and all other variables = 0. b) It is satisfied if value of any variables ibl changes from 1 to Resulting cutting plane: sum of variables ibl in group N 1. João Miguel da Costa Sousa / Alexandra Moutinho 319

61 Constraint Programming g Combination of artificial intelligence with computer programming languages in the mid 80 s. Flexibility in stating (nonlinear) constraints: 1. Mathematical constraints, e.g., x + y < z. 2. Disjunctive constraints, e.g., times of certain tasks cannot overlap. 3. Relational constraints, e.g., at least three tasks should be assigned to a certain ti machine. João Miguel da Costa Sousa / Alexandra Moutinho 320

62 Stating constraints 4. Explicit constraints, e.g., x and y have same domain {1,2,3,4,5}, but (x, y) must be (1, 1), (2, 3) or (4, 5). 5. Unary constraints, e.g. z is integer between 5 and Logical constraints, t e.g., if x = 5, then y [6, 8]. Allows use of standard logical functions such as IF, AND, OR, NOT. Constraint programming gapplies domain reduction and constraint propagation. The process creates a tree search. João Miguel da Costa Sousa / Alexandra Moutinho 321

63 Example Consider: x 1 {1,2}, x 2 {1,2}, x 3 {1,2,3}, x 4 {1,2,3,4,5} Constraints: 1. All variables must have different values; 2. x 1 +x 3 = 4 Apply domain reduction and constraint propagation to obtain feasible solutions: x 1 {1}, x 2 {2}, x 3 {3}, x 4 {4,5}. João Miguel da Costa Sousa / Alexandra Moutinho 322

64 Constraint Programming g Steps in Constraint Programming: 1. Formulate a compact model for the problem by using a variety it of constraint t types (most not of IP type). 2. Efficiently find feasible solutions that satisfy all these constraints. 3. Search among feasible solutions for an optimal one. Strength of constraint programming gis in first two steps, whereas the main strength of IP is in step 3. Current research: integrate CP and IP! João Miguel da Costa Sousa / Alexandra Moutinho 323