IE418 Integer Programming

Transcription

1 IE418: Integer Programming Department of Industrial and Systems Engineering Lehigh University 24th January 2005 Name two reasons why anyone would want to program with integers? What is the knapsack problem? (MIP): max{c T x + h T y Ax + Gy b, x Z n +, y R p + } What is Mixed 0-1 Programming? What is Pure Integer Programming What is Binary Programming? What is 0-1 Programming? What is Q n +? What is my son s name?

2 Outline Set {Packing, Covering, Partitioning} TSP Models of Choice Pick one! Piecewise Linear Functions Algorithimic Modeling A couple examples Set {Pack,Cover,Partition}ing Traveling Salesman Selecting from a Set We can use constraints of the form j T x j 1 to represent that at least one item should be chosen from a set T. Similarly, we can also model that at most one or exactly one item should be chosen. Example: Set covering problem If A in a 0-1 matrix, then a set covering problem is any problem of the form Set Packing: Ax e min c T x s.t. Ax e 1 x j {0, 1} Set Partitioning: Ax = e 1 It is common to denote the vector of 1 s as e j

3 Set {Pack,Cover,Partition}ing Traveling Salesman COP Set{Cover, Pack, Partition}ing A combinatorial optimization problem CP = (N, F) consists of A finite ground set N, A set F 2 N of feasible solutions, and A cost function c Each row of A represents an item from N Each column A j represents a subset N j F of the items. Each variable x j represents selecting subset N j. The constraints (for covering and partitioning) say that {j xj =1}N j = N. In other words, each item must appear in at least, (or exactly one selected subset. Set {Pack,Cover,Partition}ing Traveling Salesman Vehicle Routing x 1 x 2 x 3... Customer 1 : Customer 2 : Customer 3 : Customer 4 : Customer 5 : = 1. = 1. = 1. = 1. = 1 This is a very flexible modeling trick You can list all feasible routes, allowing you to handle weird constraints like time windows, strange precedence rules, nonlinear cost functions, etc.

4 Set {Pack,Cover,Partition}ing Traveling Salesman The Farmer s Daughter This is The Most Famous Problem in Combinatorial Optimization! A traveling salesman must visit all his cities at minimum cost. Given directed (complete) graph with node set N. (G = (N, N N)) Given costs c ij of traveling from city i to city j Find a minimum cost Hamiltonian Cycle in G Variables: x ij = 1 if and only if salesman goes from city i to city j Set {Pack,Cover,Partition}ing Traveling Salesman TSP (cont.) min c ij x ij i N j N x ij = 1 j N Enter Each City i N x ij = 1 i N Leave Each City j N x ij {0, 1} i N, j N Subtour elimination constraint: x ij 1 S N, 2 S N 2 i S j S Alternatively: x ij S 1 S N, 2 S N 2 i S j S ( No Beaming ) ( No Beaming )

5 Set {Pack,Cover,Partition}ing Traveling Salesman TSP Trivia Time! What is This Number? Is this... a) The number of times an undergraduate student asked me where room 355 was this week? b) The number of subatomic particles in the universe? c) The number of subtour elimination constraints when N = 299. d) All of the above? e) None of the above? Set {Pack,Cover,Partition}ing Traveling Salesman Answer Time The answer is (e). (a) (c) are all too small (as far as I know) :-). (It is (c), for N = 300). Exponential is really big. Yet people have solved TSP s with N > 16, 000! You will learn how to solve these problems too! The trick is to only add the subset of constraints that are necessary to prove optimality. This is a trick known as branch-and-cut, and you will learn lots about branch-and-cut in this course.

6 SOS1 SOS2 Modeling a Restricted Set of Values We may want variable x to only take on values in the set {a 1,..., a m }. We introduce m binary variables y j, j = 1,..., m and the constraints x = m a j y j, j=1 m y j = 1, j=1 y j {0, 1} The set of variables {y 1, y 2,... y m } is called a special ordered set (SOS) of variables. SOS1 SOS2 Example Building a warehouse Suppose we are modeling a facility location problem in which we must decide on the size of a warehouse to build. The choices of sizes and their associated cost are shown below: Size Cost Warehouse sizes and costs

7 SOS1 SOS2 Warehouse Modeling Using binary decision variables x 1, x 2,..., x 5, we can model the cost of building the warehouse as COST 100x x x x x 5. The warehouse will have size SIZE 10x x x x x 5, and we have the SOS constraint x 1 + x 2 + x 3 + x 4 + x 5 = 1. SOS1 SOS2 Piecewise Linear Cost Functions We can use binary variables to model arbitrary piecewise linear functions. The function is specified by ordered pairs (a i, f(a i )) We have a binary variable y i, which indicates whether a i x a i+1. f(a 6 ) f(a 5 ) f(a 5 ) f(x) f(a 3 ) f(a 2 ) f(a 1 ) a 1 a 2 a 3 a 4 x a 5 a 6

8 SOS1 SOS2 Minimizing Piecewise Linear Cost Functions To evaluate the function, we will take linear combinations k i=1 λ if(a i ) of the given functions values. This only works if the only two nonzero λ is are the ones corresponding to the endpoints of the interval in which x lies. The Key Idea! If y j = 1, then λ i = 0, i j, j+1. kx min λ i f(a i ) i=1 kx s.t. λ i = 1, i=1 λ 1 y 1, λ i y i 1 + y i, i = 2,..., k 1, λ k y k 1, Xk 1 y i = 1, i=1 λ i 0, y i {0, 1}. SOS1 SOS2 SOS2 A better formulation involves the use of special ordered sets of type 2 SOS2 A set of variables of which at most two can be positive. If two are positive, they must be adjacent in the set. k min λ i f(a i ) i=1 k s.t. λ i = 1 i=1 λ i 0 {λ 1, λ 2,..., λ k } SOS2 The adjacency conditions of SOS2 are enforced by the solution algorithm (All) commercial solvers allow you to specify SOS2

9 Modeling Disjunctive Constraints We are given two constraints a T x b and c T x d with nonnegative coefficients. Instead of insisting both constraints be satisfied, we want at least one of the two constraints to be satisfied. To model this, we define a binary variable y and impose a T x yb, c T x (1 y)d, y {0, 1}. More generally, we can impose that at least k out of m constraints be satisfied with (a i )T x b i y i, i {1, 2,... m} m i=1 y i k, y i {0, 1} There are lots of things you can model with binary variables: Indicator variables (Positivity of variables) Limiting the Number of Positive Variables Fixed Charge problems Minimum production level Indicator variables (Validity of constraints) Either-or If-then k out of n

10 Nonconvex regions Economies of Scale Discrete Capacity Extensions Maximax or Minimin The problem is that sometimes to see the modeling trick is difficult. For example... Use a 0-1 variable δ to indicate whether or not the constraint 2x 1 + 3x 2 1 is satisfied. x 1, x 2 are nonnegative continuous variables that are 1 δ = 1 2x 1 + 3x 2 1 The Slide of Tricks. Indicator Variables... δ = 1 j N a jx j b j N a jx j + Mδ M + b j N a jx j b δ = 1 j N a jx j (m ɛ)δ b + ɛ δ = 1 j N a jx j b j N a jx j + mδ m + b j N a jx j b δ = 1 j N a jx j (M + ɛ)δ b ɛ Definitions δ: Indicator variable (δ {0, 1}). M: Upper bound on j N a jx j b m: Lower bound on j N a jx j b ɛ: Small tolerance beyond which we regard the constraint as haven been broken. If a j Z, x j Z, then we can take ɛ = 1.

11 Modeling Trick #1 Indicating Constraint (Non)violation Suppose we wish to indicate whether or not an inequality j N a jx j b holds by means of an indicator variable δ. Implication We Wish to Model δ = 1 j N a j x j b This can be represented by the constraint j N a jx j + Mδ M + b Trick #1... The Logic δ = 1 j N a jx j b j N a jx j + Mδ M + b (Thinking)... δ = 1 j N a jx j b 0 1 δ = 0 j N a jx j b 0 j N a jx j b M(1 δ) Does it work? δ = 0 j N a jx j b M (true by definition of M) δ = 1 j N a jx j b 0

12 Modeling Trick #2 j N a jx j b δ = 1 δ = 0 j N a jx j b δ = 0 j N a jx j > b δ = 0 j N a jx j b + ɛ If a j, x j are integer, we can choose ɛ = 1 Model as j N a jx j (m ɛ)δ b + ɛ m is a lower bound for the expression j N a jx j b Some Last Modeling Tricks δ = 1 j N a jx j b Model as j N a jx j + mδ m + b j N a jx j b δ = 1 Model as j N a jx j (M + ɛ)δ b ɛ You can obtain these by just transforming the constraints to form and using the first two tricks.

13 Back To Our Example... Use a 0-1 variable δ to indicate whether or not the constraint 2x 1 + 3x 2 1 is satisfied. x 1, x 2 are nonnegative continuous variables that cannot exceed 1. δ = 1 2x 1 + 3x 2 1 M : Upper Bound on 2x 1 + 3x works m : Lower Bound on 2x 1 + 3x works. ɛ : 0.1 Example, Cont. ( ) Recall the trick. δ = 1 j N a jx j b j N a jx j + Mδ M + b 2x 1 + 3x 2 + 4δ 5 ( ). Recall the trick. j N a jx j b δ = 1 j N a jx j (m ɛ)δ b + ɛ 2x 1 + 3x δ 1.1 2x 1 + 3x 2 + 4δ 5 2x 1 + 3x δ 1.1

14 A More Realistic Example PPP Production Planning Problem. (A simple linear program). An engineering plant can produce five types of products: p 1, p 2,... p 5 by using two production processes: grinding and drilling. Each product requires the following number of hours of each process, and contributes the following amount (in hundreds of dollars) to the net total profit. p 1 p 2 p 3 p 4 p 5 Grinding Drilling Profit PPP More Info Each unit of each product take 20 manhours for final assembly. The factory has three grinding machines and two drilling machines. The factory works a six day week with two shifts of 8 hours/day. Eight workers are employed in assembly, each working one shift per day.

15 PPP maximize 55x x x x x 5 (Profit/week) subject to 12x x 2 + 0x x x (Grinding) 10x 1 + 8x x 3 + 0x 4 + 0x (Drilling) 20x x x x x Final Assembly x i 0 i = 1, 2,... 5 Another PPP Modeling Example Let s model the following situation. If we manufacture P 1 or P 2 (or both), then at least one of P 3, P 4, P 5 must also be manufactured. We first need an indicator variable z j that indicate when each of the x j > 0. How do we model x j > 0 z j = 1?. Hint: This is equivalent to z j = 0 x j = 0

16 Modeling the Logic Answer: x j Mz j Given that we have included the constraints x j Mz j, we d like to model the following implication: z 1 + z 2 1 z 3 + z 4 + z 5 1 Can you just see the answer? I can t. So let s try our formulaic approach. Think of it in two steps z 1 + z 2 1 δ = 1 δ = 1 z 3 + z 4 + z 5 1. Look up the Tricks First we model the following: z 1 + z 2 1 δ = 1 The formula from the bag o tricks j N a jx j b δ = 1 j N a jx j (M + ɛ)δ b ɛ M : Upper Bound on j N a jz j b M = 1 in this case. (z 1 1, z 2 1, b = 1). ɛ : Tolerance Level indicating the minimum about by which the constraint can be violated. ɛ = 1 in this case! If the constraint is going to be violated, then it will be violated by at least one.

17 Modeling z 1 + z 2 1 δ = 1, Cont. Just plug in the formula j N a jx j (M + ɛ)δ b ɛ z 1 + z 2 2δ 0 Does this do what we want? z 1 z 2 δ /2( = 1) 1 0 1/2( = 1) Second Part Want to model the following: δ = 1 z 3 + z 4 + z 5 1. The formula from the bag o tricks δ = 1 j N a jx j b j N a jx j + mδ m + b m : lower bound on j N a jx j b. m = 1. (z 1 0, z 2 0, b = 1). Plug in the formula: z 3 + z 4 + z 5 δ 0 It works! (Check for δ = 0, δ = 1).

18 PPP, Make 1 or 2 make 3, 4, or 5 maximize 55x x x x x 5 (Profit/week) subject to 12x x 2 + 0x x x x 1 + 8x x 3 + 0x 4 + 0x x x x x x x i M i z i i = 1, 2,... 5 z 1 + z 2 2δ 0 z 3 + z 4 + z 5 δ 0 x i 0 i = 1, 2,... 5 z i {0, 1} i = 1, 2,... 5 δ {0, 1} Cool Things You Can Now Do Either constraint 1 or constraint 2 must hold Create indicators δ 1, δ 2, then δ 1 + δ 2 1 At least one constraint of all the constraints in M should hold i M δ i 1 At least k of the constraints in M must hold i M δ i k If x, then y δ y δ x

19 That s All Folks! That s it for modeling. You should have read N&W I.1.1 by now. Those of you who want to get a jump start on the homeworks can consider doing problems??????