Returning to Open Threads

Transcription

1 DM545 Linear and Integer Programming Lecture 14 Returning to Open Threads Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark

2 Outline

3 Assignments I corrected those from Assignment 0 and 30 from Assignment 1. Nice reviews, quite accurate grades although sometime a bit harsh with words like "stupid" Some copied from other peers. It is very easy to spot when it is wrong! Better to err with your own mind Often things are presented more complicated than they are. And lenghtier. (it was not meant to be a report) Some wrote results without explaining where they came from 3

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5 Today More on Branch and Bound More on Network Flows: Assignment and Transportation Problems More on More on Modeling 5

6 To come Next week: 2 exercise sessions. Possible to join groups? Rehearsal. is there space and will? Question time: Thursday 18th of June? Lecture notes will be extended 6

7 Exam Tilladt Håndscanner/digital pen og ordbogsprogrammet fra ordbogen.com Ikke tilladt at anvende digitalt kamera eller webcam o. lign. metoder for at digitalisere sin besvarelse Du afleverer efter fristen og kun en gang Exam Monitor er et lille program, som logger, hvilke programmer du afvikler på din computer under eksamen, samtidig med at din skærm optages. Internet Internet er ikke tilladt ved eksamener på NAT, men undtagelsesvis til denne eksamen er det tilladt, at benytte følgende webside http: // www. imada. sdu. dk/ ~marco/ DM545/ og siderne linket derfra. Det er ikke tilladt at benytte andre sider Vejledning og templates snart tilgænglig fra kurset web siden ved afsnittet Assessment Kom vel forberedet, bring noget at drikke og spise 7

8 Outline

9 rules Consider S = {x : a 0 x 0 + n j=1 a jx j b, l j x j u j, j = 0..n} Bounds on variables. If a 0 > 0 then: x 0 b a j l j a j u j /a 0 j:a j >0 j:a j >0 j:a j <0 and if a 0 < 0 then x 0 b a j l j a j u j /a 0 j:a j <0 Redundancy. The constraint n j=0 a jx j b is redundant if a j u j + a j l j b j:a j >0 j:a j <0 9

10 Infeasibility: S = if (swapping lower and upper bounds from previous case) a j l j + a j u j > b j:a j >0 j:a j <0 Variable fixing. For a max problem in the form max{c T x : Ax b, l x u} if i = 1..m : a ij 0, c j < 0 then fix x j = l j if i = 1..m : a ij < 0, c j > 0 then fix x j = u j Integer variables: l j x j u j Binary variables. Probing: add a constraint, eg, x 2 = 0 and check what happens 10

11 Example max 2x 1 + x 2 x 3 R1 : 5x 1 2x 2 + 8x 3 15 R2 : 8x 1 + 3x 2 x 3 9 R3 : x 1 + x 2 + x x x 2 1 x 3 1 {}}{ R1 :5x x 2 8x u 2 l 3 {}}{ 1 = 9 x 1 9/5 8x x 2 5x = 17 x 3 17/8 2x 2 5x 1 + 8x = 7 x 2 7/2, x 2 0 R2 :8x 1 9 3x 2 + x = 7 x 1 7/8 R1 :8x x 2 5x /8 = 101/8 x 3 101/64 R3 : x 1 + x 2 + x 3 9/ /64 < 6 Hence R3 is redundant 11

12 Example max 2x 1 + x 2 x 3 R1 : 5x 1 2x 2 + 8x 3 15 R2 : 8x 1 + 3x 2 x 3 9 7/8 x 1 9/5 0 x x 3 101/64 Increasing x 2 makes constraints satisfied x 2 = 1 Decreasing x 3 makes constraints satisfied x 3 = 1 We are left with: max{2x 1 : 7/8 x 1 9/5} 12

13 for Set Covering/Partitioning 1. if e T i A = 0 then the ith row can never be satisfied [ ] = if e T i A = e k then x k = 1 in every feasible solution [ ] = In SPP can remove all rows t with a tk = 1 and set x j = 0 (ie, remove cols) for all cols that cover t 13

14 3. if e T t A e T p A then we can remove row t, row p dominates row t (by covering p we cover t) t p 1 1 In SPP we can remove all cols j: a tj = 1, a pj = 0 4. if j S Ae j = Ae k and j S c j c k then we can cover the rows by Ae k more cheaply with S and set x k = 0 (Note, we cannot remove S if j S c j c k )

15 Outline

16 (see also lec. 10) Iterate: 1. define parameters 2. define variables 3. use variables to express objective function 4. use variables to express constraints a. problems with discrete input/output (knapsack, factory planning) b. problems with logical conditions c. combinatorial problems (sequencing, allocation, transport, assignment, partitioning) d. network problems 16

17 Variables discrete quantities decision variables indicator/auxiliary variables (for logical conditions) special ordered sets incidence vector of S Z n B n B n B n B n Assignment { } max c i,σ(i) σ : I J σ TSP min π i { } c i,π(i) π : {1..n} {1..n} and π is a circuit i COP min c j S F S N j S 17

18 Logical Conditions x y z binary integer continuous Linking constraints z R, x B if z = 0 then x = 0, if z > 0 then x = 1 z Mx 0 x = 1 = z > m z mx 0 Logical conditions and 0 1 variables X 1 X 2 x 1 + x 2 1 X 1 X 2 x 1 = 1, x 2 = 1 X 1 x 1 = 0 or (1 x 1 = 1) X 1 X 2 x 1 x 2 0 X 1 X 2 x 1 x 2 = 0 18

19 Examples (X A X B ) (X C X D X E ) x A + x B 1 x C + x D + x E 1 x A + x B 1 = x = 1 x = 1 = x C + x D + x E 1 x A + x B 2x 0 x C + x D + x E x Disjunctive constraints (encountered earlier) Constraint: x 1 x 2 = 0 1) replace x 1 x 2 by x 3 2) x 3 = 1 x 1 = 1, x 2 = 1 x 1 + x 3 0 x 2 + x 3 0 x 1 + x 2 x

20 z x, z R, x B 1) replace zx by z 1 2) impose: x = 0 z 1 = 0 x = 1 z 1 = z z 1 Mx 0 z + z 1 0 z z 1 + Mx M Special ordered sets of type 1/2 (for continuous or integer vars): SOS1: set of vars within which exactly one must be non-zero SOS2: set of vars within which at most two can be non-zero. The two variables must be adjacent in the ordering separable programming and piecewise linear functions (next 5 slides) 20

21 Separable Programming Separable functions: sum of functions of single variables: x x 2 + e x 3 YES x 1 x 2 + x 2 x x 3 NO (actually, some non-separable can also be made separable: 1. x 1 x 2 by y 2. relate y to x 1 and x 2 by: log y = log x 1 + log x 2 needs care if x 1 and x 2 close to zero.) non-linear separable functions can be approximated by piecewise linear functions (valid for both constraints and objective functions) 21

22 Convex Non-linear Functions We can model convex non-linear functions by piece-wise linear functions and LP min x1 2 4x 1 2x 2 x 1 + x 2 4 2x 1 + x 2 5 x 1 + 4x 2 2 x 1, x 2 0 x 2 x 2 1 LP Formulation a 1 a 2a 3 x 1 x = λ 0 a 0 + λ 1 a 1 + λ 2 a 2 + λ 3 a 3 y = λ 0 f (a 0 ) + λ 1 f (a 1 ) + λ 2 f (a 2 ) + λ 3 f (a 3 ) 3 i=0 λ i = 1 λ i 0 i = 0,..., 3 at most two adjacent λ i can be non zero (*) 22

23 To model (*) which are SOS2 we would need binary indicator variables and hence BIP as in next slide. However since the problem is convex, an optimal solution lies on the borders of the functions and hence we can skip introducing the binary variables and relax (*) 23

24 Non-convex Functions Piece-wise Linear Functions non-convex functions require indicator variables and IP formulation g(x) = j g j (x) g j non linear approximated by f (x) piecewise linear in the disjoint intervals [a i, b i ] convex hull formulation (convex combination of points) x = λ i a i + µ i b i y = λ i f (a i ) + µ i f (b i ) i I λ i + µ i = 1 λ i, µ i 0 Remember how we modeled disjunctive polyhedra... (cntd) 24

25 using indicator variables δs we obtain the BIP formulation: x = i I (λ ia i + µ i b i ) y = i I (λ if (a i ) + µ i f (b i )) λ i + µ i = δ i i I i I δ i = 1 λ i, µ i 0 i I δ i {0, 1} i I the δs are SOS1. 25

26 Good/Bad Models Number of variables: sometimes it may be advantageous increasing if they are used in search tree. 0 1 var have specialized algorithms for preprocessing and for branch and bound. Hence a large number solved efficiently. Good using. Binary expansion: 0 y u y = x 0 + 2x 1 + 4x 2 + 8x r x r r = log 2 u Making explicit good variables for branching: a j y j b j a j x j + u = b j u may be a good variable to branch (u is relaxed in LP but must be integer as well) 26

27 Symmetry breaking: Eg machine maintenance (in FPMM) y j Z vs x j B Difficulty of LP models depends on number of constraints: min t a t z t b t max t z t max t z + t z t z t a t z t b 1 z + t z t = a t z t b t z t b t a t z t more variables but less constraints With IP it might be instead better increasing the number of constraints. Make big M as small as possible in IP (reduces feasible region possibly fitting it to convex hull). 27

28 Practical Tips Units of measure: check them! all data should be scaled to stay in some software do this automatically Write few line of text describing what the equations express and which are the variables, give examples on the problem modeled. Try the model on small simple example that can be checked by hand. Be diffident of infeasibility and unboundedness, double check. Estimate the potential size. If IP problem large and no structure then it might be hard. If TUM then solvable with very large size If other structure, eg, packing, covering also solvable with large size Check the output of the solver and understand what is happening If all fail resort to heuristics 28

29 Conclusions Optimization Taxonomy (NEOS Server, University of Wisconsin) 29

30 Conclusions I hope this course has changed your way of thinking when you hear the word optimization I hope you gained enthusiasm and fascination for this subject. Where from here: DM841: Constraint Programming and Heuristics DM204: Scheduling Timetabling and Routing DM208/DM209: Combinatorial Optimization I/II DM817: Network Programming: Theory and Applications Bsc and Msc Thesis and ISA Instructor-Course assignment train timetabling quadratic programming nonlinear programming general problem solver practicum with sulum or come with your own problem! and apply for TA in this course! 30