Cutting Planes in Mixed Integer Programming

Transcription

1 Cutting Planes in Mixed Integer Programming Santanu S. Dey School of Industrial and Systems Engineering, Georgia Institute of Technology. February, 20 (Many Thanks to Q. Louveaux for sharing some Images)

2 Outline Mixed Integer Programming (MIP) MIP are useful! MIP are difficult to solve How we go about solving MIPs Cutting Planes

3 Part 1 Mixed Integer Programming

4 Mixed Integer Programming: Linear Problem with Discrete Variables Mixed Integer Program (MIP) max c 1 x c nx n s.t. a 11 x a 1n x n b a m1 x a mnx n b m x 1, x 2,, x k Z + nonnegative integers I will typically use x", y" to represent decision variables and all the other letters to represent data. If k = n, then Integer Program. If k = 0, then Linear Program.

5 This seemingly simple class of optimization problems can be used to model a very large class of real world problems...

6 This seemingly simple class of optimization problems can be used to model a very large class of real world problems... I want to illustrate its expressive power with an elementary example

7 An Example: Production Planning 1. We have to draw the production plan for the next 5 months.

8 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units.

9 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units.

10 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6.

11 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8.

12 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8. x 1, x 2, x 3, x 4, x 5 : The amount I choose to produce in each month. y 1, y 2, y 3, y 4, y 5 : Whether I produce or not each month (0 = Do not Produce/ 1 = I decide to Produce).

13 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8. x 1 5

14 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8. x 1 5 x 1 + x

15 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8. x 1 5 x 1 + x x 1 + x 2 + x x 1 + x 2 + x 3 + x x 1 + x 2 + x 3 + x 4 + x

16 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8. x 1 20

17 An Example: Productions Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8. x 1 20y 1

18 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8. x 1 20y 1 x 2 20y 2 x 3 20y 3 x 4 20y 4 x 5 20y 5

19 An Example: Production Planning 1. We have to draw the production plan for the next 5 months. 2. The demands are known for the next 5 months: 5, 13, 7, 5, 12 units. 3. Suppose our production capacity each month is 20 units. 4. The cost of per unit production in the next 5 months is: $6, $7, $2, $3, $6. 5. If we choose to produce in any month, there is a extra startup cost of $ 8. min 6x 1 + 7x 2 + 2x 3 + 3x 4 + 6x 5 + 8y 1 + 8y 2 + 8y 3 + 8y 4 + 8y 5

20 An Example: Production Planning min 6x 1 + 7x 2 + 3x 3 + 6x 4 + 8y 1 + 8y 2 + 8y 3 + 8y 4 + 8y 5 x 1 5 x 1 + x x 1 + x 2 + x x 1 + x 2 + x 3 + x x 1 + x 2 + x 3 + x 4 + x x 1 20y 1 x 2 20y 2 x 3 20y 3 x 4 20y 4 x 5 20y 5 y 1 1 y x 1, x 2, x 3, x 4, x 5 Z + y 1, y 2, y 3, y 4, y 5 Z +

21 MIP is a Flexible Tool Logistics: Traveling Salesman problem, Vehicle Routing. Inventory (and Production) Planning. Facilities Location. Capacity planning: Matching, Assignment Problem. Data Mining: Classification, Regression.

22 MIP is a Flexible Tool Logistics: Traveling Salesman problem, Vehicle Routing. Inventory (and Production) Planning. Facilities Location. Capacity planning: Matching, Assignment Problem. Data Mining: Classification, Regression. Airline Industry: Schedule Planning, Fleet Assignment, Aircraft Rotation, Crew-pairing. Mining and Forestry Industry: Covering Models, partitioning models. National Security Planning. VLSI Chip Design. Computational Biology: Sequence Alignment, Genome Rearrangement. Health care: IMRT, Scheduling. Sports Scheduling, Timetabling.

23 Some "Slightly Dated" Applications of Integer Programming Papers from Interfaces. Company Year Type of Model Revenue Air New Zealand 20 Crew Scheduling NZ $15.6 million AT&T 2000 Network Restoration Hundreds of millions of dollars NBC 2002 Product Mix/ $200 million Commercials/Schedule Procter & Gamble 2006 Expressive Bidding $298.4 million Schindler Elevator 2003 Routing planning $1 million UPS 2004 Network Design $87 million Ford 20 Set-covering/ $250 million Product optimization Hewlett-Packard 2004 Inventory Optimization $130 million Merrill-Lynch 2002 Integrated Choice $80 million Strategy Motorola 2005 Bidding/ $200 million Supplier Negotiation

24 Part 2 MIPs are Difficult to Solve

25 A Natural Way to Solve MIPs 1. Enumerate all possible integer solution vectors.

26 A Natural Way to Solve MIPs 1. Enumerate all possible integer solution vectors. 2. For each potential solution vectors, check if it is feasible.

27 A Natural Way to Solve MIPs 1. Enumerate all possible integer solution vectors. 2. For each potential solution vectors, check if it is feasible. 3. If feasible, then compute objective function value.

28 A Natural Way to Solve MIPs 1. Enumerate all possible integer solution vectors. 2. For each potential solution vectors, check if it is feasible. 3. If feasible, then compute objective function value. 4. Pick the best solution.

29 Some Calculations Suppose we can evaluate 10 6 potential solution vectors in a second. Number of binary variables Number of vectors 2 n Time seconds

30 Some Calculations Suppose we can evaluate 10 6 potential solution vectors in a second. Number of binary variables Number of vectors 2 n Time seconds second minutes

31 Some Calculations Suppose we can evaluate 10 6 potential solution vectors in a second. Number of binary variables Number of vectors 2 n Time seconds second minutes days years

32 Some Calculations Suppose we can evaluate 10 6 potential solution vectors in a second. Number of binary variables Number of vectors 2 n Time seconds second minutes days years ,709 years ,709,791 years

33 Some Calculations Suppose we can evaluate 10 6 potential solution vectors in a second. Number of binary variables Number of vectors 2 n Time seconds second minutes days years ,709 years ,709,791 years This has got nothing" to do with speed of computers. Every time the speed is doubled It allows me to solve a problem with 1 extra variable.

34 Some Calculations Suppose we can evaluate 10 6 potential solution vectors in a second. Number of binary variables Number of vectors 2 n Time seconds second minutes days years ,709 years ,709,791 years This has got nothing" to do with speed of computers. Every time the speed is doubled It allows me to solve a problem with 1 extra variable. However, we routinely solve problems with thousands of integer variables.

35 Part 3 How we go about solving MIPs

36 Main Ideas Branch and Bound: Smart Enumerate Cutting Planes

37 A little bit of Linear Programming (LP) Geometry Polyhedron x >= x + 2y <= x + 2y <= 4 y>= x + 2y 6 x + 2y 4 x, y 0.

38 Using Linear Programming Techniques To Solve MIPs Convex Hull

39 Using Linear Programming Techniques To Solve MIPs Convex Hull

40 Cutting Planes (Cuts) A Simple Algorithm To Solve Integer Programs

49 How Do We Generate Cutting Planes? Cutting planes are difficult to obtain when there are a large number of variables! Two paradigms to generate cutting planes

50 How Do We Generate Cutting Planes? Cutting planes are difficult to obtain when there are a large number of variables! Two paradigms to generate cutting planes 1. Problem Specific Cutting Planes: These are useful for solving specific problems.

51 How Do We Generate Cutting Planes? Cutting planes are difficult to obtain when there are a large number of variables! Two paradigms to generate cutting planes 1. Problem Specific Cutting Planes: These are useful for solving specific problems. 2. Generic Cutting Planes: These are useful for solving any Mixed Integer Programs.

52 How Do We Generate Cutting Planes? Cutting planes are difficult to obtain when there are a large number of variables! Two paradigms to generate cutting planes 1. Problem Specific Cutting Planes: These are useful for solving specific problems. 2. Focus of the rest of the talk Generic Cutting Planes: These are useful for solving any Mixed Integer Programs.

53 Most Successful Cutting Planes for Generic Problems [Bixby, Rothberg (2007)] Disabled Cut Year Mean Performance Degradation Gomory Mixed Integer (GMIC) X Mixed Integer Rounding X Knapsack Cover X Flow Cover X Implied Bound X Flow path X Clique X GUB Cover X Disjunctive X

54 Chvátal-Gomory Cutting Planes x 1, x 2 0, x 1 + x 2 3, 5x 3y 3 x 1, x 2 Z Valid inequality for Continuous Relaxation: 4x 1 + 3x }{{} Z Black

55 Chvátal-Gomory Cutting Planes x 1, x 2 0, x 1 + x 2 3, 5x 3y 3 x 1, x 2 Z Valid inequality for Continuous Relaxation: 4x 1 + 3x }{{} Z This gives the following nontrivial valid inequality: 4x 1 + 3x

56 Cover Cuts 3x 1 + 5x 2 + 4x 3 + 2x 4 + 7x 5 8, 0 x 1, x 2, x 3, x 4, x 5 1, x 1, x 2, x 3, x 4, x 5 Z +

57 Cover Cuts 3x 1 + 5x 2 + 4x 3 + 2x 4 + 7x 5 8, 0 x 1, x 2, x 3, x 4, x 5 1, x 1, x 2, x 3, x 4, x 5 Z > 8 x 2 and x 3 cannot simultaneously be equal to 1

58 Cover Cuts 3x 1 + 5x 2 + 4x 3 + 2x 4 + 7x 5 8, 0 x 1, x 2, x 3, x 4, x 5 1, x 1, x 2, x 3, x 4, x 5 Z + x 2 + x 3 1

59 Cover Cuts 3x 1 + 5x 2 + 4x 3 + 2x 4 + 7x 5 8, 0 x 1, x 2, x 3, x 4, x 5 1, x 1, x 2, x 3, x 4, x 5 Z + x 2 + x 3 1 x 4 + x 5 1 x 1 + x 2 + x 3 2 x 1 + x 2 + x 4 2 x 1 + x 2 + x 5 2.

60 Basic Mixed-Integer Rounding Consider the basic set x y 2.5 x Z, y 0.

61 Basic Mixed-Integer Rounding Consider the basic set x y 2.5 x Z, y 0. MIR CUT It is called the Mixed-Integer-Rounding Inequality (MIR) x y 1 f, where f = is the fractional part of 2.5.

62 The Disjunctive Cuts Split Disjunctive cuts General family of cutting planes that include the MIR cuts. Based on a disjunction π T x π 0 or π T x π

63 The Disjunctive Cuts Split Disjunctive cuts General family of cutting planes that include the MIR cuts. Based on a disjunction π T x π 0 or π T x π SPLIT

64 The Disjunctive Cuts Split Disjunctive cuts General family of cutting planes that include the MIR cuts. Based on a disjunction π T x π 0 or π T x π

65 The Disjunctive Cuts Split Disjunctive cuts General family of cutting planes that include the MIR cuts. Based on a disjunction π T x π 0 or π T x π Split cut

66 Some recent trends...

67 Single Constraint Relaxation A simple example

70 Question: Can we expect any improvement by looking at multiple constraints simultaneously?

71 Single Constraint Vs Multiple Constraints Extra information by considering multiple constraints simultaneously!

73 Single Constraint Vs Multiple Constraints Extra information by considering multiple constraints simultaneously! Lets Drop This Constraint

74 Single Constraint Vs Multiple Constraints Extra information by considering multiple constraints simultaneously! Can we obtain this cut?

75 Single Constraint Vs Multiple Constraints Extra information by considering multiple constraints simultaneously! Problem Can we obtain this cut?

76 Single Constraint Vs Multiple Constraints Extra information by considering multiple constraints simultaneously! Problem Can we obtain this cut? NO!