19. Logic constraints, integer variables

Transcription

1 CS/ECE/ISyE 524 Introduction to Optimization Spring Logic constraints, integer variables If-then constraints Generalized assignment problems Logic constraints Modeling a restricted set of values Sudoku! Laurent Lessard (

2 If-then constraints A single simple trick (with suitable adjustments) can help us model a great variety of if-then constraints The trick We d like to model the constraint: if z = 0 then a T x b. Let M be an upper bound for a T x b. Write: a T x b Mz If z = 0, then a T x b 0 as required. Otherwise, we get a T x b M, which is always true. 19-2

3 If-then constraints Slight change: if z = 1 then a T x b Again, let M be an upper bound for a T x b Write: a T x b M(1 z) Reversed inequality: if z = 0 then a T x b Write constraint as a T x + b 0 Let m be an upper bound on a T x + b Write: a T x + b mz. Same as: a T x b mz Note: m is a lower bound on a T x b. 19-3

4 If-then constraints The converse: if a T x b then z = 1 Equivalent to: if z = 0 then a T x > b (contrapositive). The strict inequality is not really enforceable. Instead, write: if z = 0 then a T x b + ε where ε is small. Let m be a lower bound for a T x b and we obtain the equivalent constraint: a T x b mz + ε(1 z) If z = 0, we get a T x b + ε, as required. Otherwise, we get: a T x b m, which is always true. Note: If a, x, b are integer-valued, we may set ε =

5 If-then constraints (summary) Logic statement Constraint if z = 0 then a T x b a T x b Mz if z = 0 then a T x b a T x b mz if z = 1 then a T x b a T x b M(1 z) if z = 1 then a T x b a T x b m(1 z) if a T x b then z = 1 a T x b mz + ε(1 z) if a T x b then z = 1 a T x b Mz ε(1 z) if a T x b then z = 0 a T x b m(1 z) + εz if a T x b then z = 0 a T x b M(1 z) εz Where M and m are upper and lower bounds on a T x b. 19-5

6 Return to fixed costs and lower bounds Modeling a fixed cost: if x > 0 then z = 1. Use the contrapositive: if z = 0 then x 0. Apply the 1 st rule on Slide Modeling a lower bound: either x = 0 or x m. Equivalent to: if x > 0 then x m. Equivalent to the following two logical constraints: if x > 0 then z = 1, and if z = 1 then x m. The first one is a fixed cost (see above) The second one is the 4 th rule on Slide

7 Generalized assignment problems (GAP) Set of machines: M = {1, 2,..., m} that can perform jobs. (think of these as the facilities in the facility problem) Machine i has a fixed cost of h i if we use it at all. Machine i has a capacity of b i units of work (this is new!) Set of jobs: N = {1, 2,..., n} that must be performed. (think of these as the customers in the facility problem) Job j requires a ij units of work to be completed if it is completed on machine i. Job j will cost c ij if it is completed on machine i. Each job must be assigned to exactly one machine. 19-7

8 GAP model minimize x,z subject to: c ij x ij (fixed cost + assignment cost) h i z i + i M i M j N x ij = 1 j N (one machine per job) i M a ij x ij b i i M (work budget) j N x ij z i i M, j N (if x ij > 0 then z i = 1) x ij, z i {0, 1} i M, j N (all binary!) z i = 1 if machine i is used, and x ij = 1 if job j is performed by machine i. Note: many choices possible for M i and aggregations. 19-8

9 New constraints Let s make GAP more interesting If you use k or more machines, you must pay a penalty of λ. 2. If you operate either machine 1 or machine 2, you may not operate both machines 3 and 4 at the same time. 3. If you operate both machines 1 and 2, then machine 3 must be operated at 40% of its capacity. 4. Each job j N has a duration d j. Minimize the time we have to wait before all jobs are completed. (this is called the makespan). 19-9

10 GAP 1 If you use k or more machines, you must pay a penalty of λ. Using k or more machines is equivalent to saying that z 1 + z z m k Let δ 1 = 1 if we incur the penalty. We now have the if-then constraint: if i M z i k then δ 1 = 1. Use the 6 th rule on Slide 19-5 and obtain: i M z i mδ 1 + (k 1)(1 δ 1 ) add λδ 1 to the cost function

11 GAP 2 If you operate either machine 1 or machine 2, you may not operate both machines 3 and 4 at the same time. Operating machine 1 or machine 2: z 1 + z 2 1. Not operating machines 3 and 4: z 3 + z 4 1 We must model z 1 + z 2 1 = z 3 + z 4 1 Same trick as before: model this in two steps: z 1 + z 2 1 = δ 2 = 1 and δ 2 = 1 = z 3 + z 4 1 First follows from 6 th rule on Slide 19-5 Second follows from 3 rd rule on Slide 19-5 Result: z 1 + z 2 2δ 2 and z 3 + z 4 + δ

12 GAP 2 (cont d) If you operate either machine 1 or machine 2, you may not operate both machines 3 and 4 at the same time. We didn t do anything to ensure that when z i = 1, the machines are actually operating! (we didn t explicitly disallow paying the fixed cost without using the machine). To force the converse as well, include the constraint: if z i = 1 then j N x ij 1 Use the 4 th rule on Slide Result: j N x ij z i (for i = 1, 2, 3, 4) 19-12

13 GAP 3 If you operate both machines 1 and 2, then machine 3 must be operated at 40% of its capacity. Operate both machines 1 and 2: z 1 + z 2 2 Capacity of machine 3 drops: b 3 becomes 0.4b 3. Two parts to the implementation: z1 + z 2 2 = δ 3 = 1. (6 th rule on Slide 19-5) δ3 = 1 = j N a 3jx 3j 0.4b 3. (3 rd rule on Slide 19-5) Equivalently, just replace b 3 by: b 3 (1 δ 3 ) + 0.4b 3 δ

14 GAP 4 Each job j N has a duration d j. Minimize the time we have to wait before all jobs are completed. (the makespan) Machine i completes all its jobs in time: j N x ijd j Minimax problem (no integer variables needed!) Let t be the makespan; t = max i M ( j N x ijd j ) Model: minimize t subject to: t j N x ij d j for all i M 19-14

15 Logic constraints A proposition is a statement that evaluates to true or false. One example we ve seen: a linear constraint a T x b. We ll use binary variables { δ i to represent propositions P i : 1 if proposition P i is true δ i = 0 if proposition P i is false The term for this is that δ i is an indicator variable. How can we turn logical statements about the P i s into algebraic statements involving the δ i s? Some standard notation: means or = means implies means and means if and only if means not means exclusive or 19-15

16 Boolean algebra Basic definitions: P Q P Q P Q P Q Useful relationships: (P 1 P k ) = P 1 P k (P 1 P k ) = P 1 P k P (Q R) = (P Q) (P R) P (Q R) = (P Q) (P R) P Q = (P Q) ( P Q) 19-16

17 Logic to algebra Statement Constraint P 1 δ 1 = 0 P 1 P 2 δ 1 + δ 2 1 P 1 P 2 δ 1 + δ 2 = 1 P 1 P 2 δ 1 = 1, δ 2 = 1 (P 1 P 2 ) δ 1 = 0, δ 2 = 0 P 1 = P 2 δ 1 δ 2 (equivalent to: ( P 1 ) P 2 ) P 1 = ( P 2 ) δ 1 + δ 2 1 (equivalent to: (P 1 P 2 )) P 1 P 2 δ 1 = δ 2 P 1 = (P 2 P 3 ) δ 1 δ 2, δ 1 δ 3 P 1 = (P 2 P 3 ) δ 1 δ 2 + δ 3 (P 1 P 2 ) = P 3 δ 1 + δ δ 3 (P 1 P 2 ) = P 3 δ 1 δ 3, δ 2 δ 3 P 1 (P 2 P 3 ) δ 1 = 1, δ 2 + δ 3 1 P 1 (P 2 P 3 ) δ 1 + δ 2 1, δ 1 + δ

18 More logic to algebra Statement P 1 P 2 P k (P 1 P k ) = (P k+1 P n) at least k out of n are true exactly k out of n are true at most k out of n are true P n (P 1 P k ) Constraint k δ i 1 i=1 k (1 δ i ) + n δ i 1 i=1 i=k+1 n δ i k i=1 n δ i = k i=1 n δ i k i=1 k δ i δ n, δ n δ j, j = 1,..., k i=1 k P n (P 1 P k ) δ n + k 1 + δ i, δ j δ n, j = 1,..., k i=

19 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 binary variables y 1,..., y m and the constraints x = m a j y j, j=1 m y j = 1, y j {0, 1} j=1 y i serves to select which a i will be selected. The set of variables {y 1, y 2,..., y m } is called a special ordered set (SOS) of variables

20 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 associated cost are shown below: Size Cost Warehouse sizes and costs 19-20

21 Example: building a warehouse 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 =

22 What about integers? What if x is an integer, i.e. x {1, 2,..., 10} First option: use 10 separate variables: 10 x = k y k, k=1 10 k=1 y k = 1, y k {0, 1} Another option: use 4 binary variables (less symmetry): x = y 1 + 2y 2 + 4y 3 + 8y 4, 1 x 10, y k {0, 1} Performance is solver-dependent. If the solver allows integer constraints directly, that s often the right choice

23 Example: Sudoku fill grid with numbers {1, 2,..., 9} each row and each column contains distinct numbers each 3 3 cluster contains distinct numbers 19-23

24 Example: Sudoku Decision variables: X {0, 1} (729 binary variables) { 1 if (i, j) entry is a k X ijk = 0 otherwise Can fill in known entries right away. Basic constraints: (324 in total) 9 k=1 X ijk = 1 i, j (SOS constraint) 9 i=1 X ijk = 1 j, k (column j contains exactly one k) 9 j=1 X ijk = 1 i, k (row i contains exactly one k) (i,j) C X ijk = 1 C, k (cluster C contains exactly one k) Much trickier to model using other integer representations! Julia code: Sudoku.ipynb 19-24