Decomposition methods in optimization

Transcription

1 Decomposition methods in optimization I Approach I: I Partition problem constraints into two groups: explicit and implicit I Approach II: I Partition decision variables into two groups: primary and secondary I In Linear Programming, approaches I and II are essentially primal and dual views of the same method I Approach I: I known as Dantzig-Wolfe decomposition I leads to a reformulation amenable to column generation I typical application: guiding decisions of independently managed subsidiary I Approach II: I known as Benders decomposition I leads to a reformulation amenable to constraint generation (cutting plane) I typical application: multistage decision making, especially under uncertainty IOE 610: LP II, Fall 2013 Decomposition methods Page 271 Primal approach: focus on (P) (P) min c T x (D) max b T p + f T u s.t. Ax = b s.t. A T p + D T u apple c Dx = f x 0 where A us m 0 -by-n, D is m-by-n. I Think of (P) as (P) v? =min s.t. c T x Ax = b x 2 P, where P = {x 2< n : Dx = f, x 0} 6= ;. I Let x j, j 2 J be all the extreme points, and w K, k 2 K be all the extreme rays of P: 8 9 < X P = j x j + X = k w k : 0, e T =1, 0 : ; j2j k2k IOE 610: LP II, Fall 2013 Dantzig-Wolfe decomposition Page 272

2 Primal (full) master problem Using the above representation for x 2 P, (P)canbereformulated as X (MP) v? =min, (c T w k ) k s.t. (c T x j ) j + X Xj2J k2k (Aw k ) k (Ax j ) j + X Xj2J k2k j2j j 0, 0 = b =1 I Problem with m constraints, many variables I Amenable to column generation, a.k.a. Dantzig-Wolfe decomposition I In a column generation algorithm, (ResMP), with only a subset of the variables included, will be solved at every iteration IOE 610: LP II, Fall 2013 Dantzig-Wolfe decomposition Page 273 Dual (full) master problem The dual of the above (MP) is: (MD) v? =max p,r b T p +r s.t. (Ax j ) T p +r apple c T x j, j 2 J (Aw k ) T p apple c T w k, k 2 K I Problem with m variables, many constraints I Amenable to cutting plane approaches, a.k.a. Benders decomposition I In a cutting plane algorithm, (RelMD), with only a subset of the constraints included, will be solved at every iteration IOE 610: LP II, Fall 2013 Dantzig-Wolfe decomposition Page 274

3 The pricing problem I Let (, )/( p, r) be solutions found for (ResMP)/(RelMP) I Consider (recall: P 6= ;) Z( p) = min (c A T p) T x = min (c A T p) T x s.t. x 2 P s.t. Dx = f, x 0 I If Z( p) = 1, apple then(c A T p) T w k < 0 for some k 2 K; add I Aw k Column with obj. coef. c 0 T w k to (ResMP) I Constraint (Aw k ) T p apple c T w k to (RelMP) I If 1 < Z( p) < r, then(c A T p) T x j < r for some j 2 J; add apple I Ax j Column with obj. coef. c 1 T x j to (ResMP) I Constraint (Ax j ) T p + r apple c T x j to (RelMP) I If r apple Z( p), I ( p, r) X is an optimal solution to (MD) I x = (Aw k ) k is an optimal solution for (P) j2j (Ax j ) j + X k2k IOE 610: LP II, Fall 2013 Dantzig-Wolfe decomposition Page 275 Dual approach: focus on (D) (P) min c T x (D) max b T p + f T u s.t. Ax = b s.t. A T p + D T u apple c Dx = f x 0 where P = {x : Dx = f, x 0} has ex. points x j and ex. rays w k I Think of p as primary decisions, and u as secondary decisions. I.e., (D) max b T p + Z(p), p where v? = Z(p) =max f T u = min (c A T p) T x s.t. D T u apple c A T p s.t. x 2 P I Let u(p) denote an optimal solution of the appropriate representation of the pricing problem IOE 610: LP II, Fall 2013 Benders decomposition Page 276

4 Dual (full) master problem Z(p) =min x (c A T p) T x = max r r s.t. x 2 P s.t. (c A T p) T w k 0, k 2 K (c A T p) T x j r, j 2 J I By assumption, the problem is feasible I First group of constraints: Z(p) > 1 I Second group of constraints: Z(p) =min j (c Again, we derived: A T p) T x j (MD) v? =max p,r b T p +r s.t. (Ax j ) T p +r apple c T x j, j 2 J (Aw k ) T p apple c T w k, k 2 K IOE 610: LP II, Fall 2013 Benders decomposition Page 277 Cutting plane approach to solving (D) Analysis of iteration t I Let ( p, r) be the solution found by solving (RelMD t )andv t its value I Note: v? apple v t apple v t 1 (relaxation property) I Let Z( p) be the optimal value of the pricing problem I Proposition: If Z( p) > 1, v? v t +(Z( p) r) I Proof: ( p, Z( p)) is a feasible solution to (MD). Therefore, v? b T p + Z( p) =b T p + r +(Z( p) r) =v t +(Z( p) r) I In the algorithm described below, we allow for termination at a feasible solution of (D) with objective value v? + IOE 610: LP II, Fall 2013 Benders decomposition Page 278

5 Cutting plane approach to solving (D) I Initialization: Set v l = 1 and v u =+1, t =0 I Iteration t: Step 1 Solve (RelMD t )tofind( p, r) andv t. Update upper bound: v u v t Step 2 Solve subproblem Z( p) I If Z( p) = 1, addafeasibility cut to (RelMD) I Else, if 1 < Z( p) < r, addanoptimality cut to (RelMD) I Else, stop ( p, u( p)) is an optimal solution of (D) Step 3 Update lower bound and best solution: if v l < v t +(Z( p) r), v l v t +(Z( p) r) and best solution p If v u v l apple, terminate and output best solution (and corresponding u( )) as an approximate solution to (D). IOE 610: LP II, Fall 2013 Benders decomposition Page 279 Example: two-stage decision-making Application of Benders decomposition I There are two sets of decisions, one for each of the two consecutive stages. I The first-stage variables are p subject to constraints A T 0 p apple c 0. The direct contribution of the p decisions on the objective function is b T p. I The second-stage variables are u and are subject to constraints A T p + D T u apple c The direct contribution of the u decisions on the objective function is f T u I Recall: decomposition approach (D) max p : A T 0 papplec 0 b T p + Z(p), where Z(p) =max f T u s.t. D T u apple c A T p IOE 610: LP II, Fall 2013 Example of Benders decomposition Page 280

6 Two-stage linear optimization model under uncertainty I Often the data A, D, f, c are uncertain I We only learn the realized data values after we have made our first-stage decision p I Once the values of A, D, f, c are known, we then make our second-stage decisions u accordingly I Model: There are M possible future scenarios, with scenario! having a probability! of being realized, for! =1,...,M I The data A, D, f, c takes on values A!, D!, f!, c! with probability! for! =1,...,M I Let u! denote the second-stage decision under the condition that scenario! is realized, for! =1,...,M IOE 610: LP II, Fall 2013 Example of Benders decomposition Page 281 Two-stage linear optimization model under uncertainty Stochastic Programming The problem becomes: max b T p + f1 T u 1 + f2 T u fmu T M p, u 1,...,u M s.t. A T 0 p apple c 0 A T 1 p + D T 1 u 1 apple c 1 A T 2 p + D T 2 u 2 apple c A T Mp + D T Mu M apple c M f! =! f!, so objective function accounts for the expected value of the objective function contribution of the second period IOE 610: LP II, Fall 2013 Example of Benders decomposition Page 282

7 Block ladder constraint structure Stage-1 variables Stage-2 variables RHS Stage-1 cost Stage-2 expected cost Scenario 1 Scenario 2 Scenario 3 Scenario 4 IOE 610: LP II, Fall 2013 Example of Benders decomposition Page 283 Reformulation Reformulation: where, for! =1,...,M: max p b T p + M P!=1 s.t. A T 0 p apple c 0 z! (p) z! (p) = max u! f T! u! = min x! (c! A T! p) T x! s.t. D T! u! apple c! A T! p s.t. D! x! = f! x! 0 (separate second-stage cost calculation subproblem for each scenario) IOE 610: LP II, Fall 2013 Example of Benders decomposition Page 284

8 Reformulation I P = {x =(x 1,...,x M ):x! 2 P!,! =1,...,M}, where P! = {x! 0 : D! x = f! } has extr. points x j!, j 2 J! and extr. rays w k!, k 2 K! I Subproblem reformulation: z! (p) =min (c! A T!p) T x! = max r! r! s.t. x! 2 P! s.t. (c! A T!p) T x j! r!, j 2 J! (c! A T!p) T w! k 0, k 2 K! I Full master problem: (MD) max p,r b T p + M P r!!=1 s.t. A T 0 p apple c 0 (A! x j!) T p + r! apple c T!x j!, j 2 J!,! =1,...,M (A! w!) k T p apple c T!w!, k k 2 K!,! =1,...,M IOE 610: LP II, Fall 2013 Example of Benders decomposition Page 285 Algorithm outline At iteration t, I Solve (RelMD t ):: (RelMD t ) v t =max b T p + M P to find ( p, r 1,..., r M ).... to be continued r!!=1 p, z s.t. A T 0 p apple c 0 (A! x j!) T p + r! apple c T! x j!, for some j and! (A! w k!) T p apple c T! w k!, for some k and! IOE 610: LP II, Fall 2013 Example of Benders decomposition Page 286

9 Algorithm outline continued I For! =1,...,M: I Solve the corresponding second-stage subproblem: z! ( p) =min (c! A T! p)t x! s.t. x! 2 P! I If z! ( p) = 1, let w! be the extreme ray generated by the LP solver. Add the following constraint to (RelMD): (A! w! ) T p apple c T! w! I If 1 < z! ( p), let x! such that z! ( p) =(c! A T! p)t x! I If 1 < z! ( p) < r!, add the following constraint to (RelMD): I If for all! z! ( p) (A! x! ) T p + r! apple c T! x! r!,then p is optimal. Note: In this version of Benders decomposition, we might add as many as M new constraints per iteration. IOE 610: LP II, Fall 2013 Example of Benders decomposition Page 287 Example: planning in a multi-divison firm Application of Dantzig-Wolfe decomposition z? =min c T 1 x 1 + c T 2 x c T M x M s.t. A 1 x 1 + A 2 x A M x M = b D 1 x 1 = f 1 D 2 x 2 = f 2 I AfirmwithM divisions. D M x M = f M x 1, x 2, x M 0 I Each division makes n k decisions x k,subjecttoitsownm k constraints I The firm has m 0 coupling constraints, expressing, e.g., total budget, or performance goals I Two issues: (i) large-scale problem; (ii) the central planning unit may not be aware of individual divisions constraints IOE 610: LP II, Fall 2013 Example of Dantzig-Wolfe decomposition Page 288

10 Constraint structure illustrated Division-1 variables Division-2 variables Division-3 variables Division-4 variables RHS Division-1 cost Division-2 cost Division-3 cost Division-4 cost Coupling contrains Division 1 Division 2 Division 3 Division 4 IOE 610: LP II, Fall 2013 Example of Dantzig-Wolfe decomposition Page 289 Reformulation I P = {x =(x 1,...,x M ):x i 2 P i, i =1,...,M}, where P i = {x i 0 : D i x = f i } has extreme points x j i, j 2 J i and extreme rays w k i, k 2 K i I Full Master problem z? =min x,, s.t. M X i=1 X (c T i x j i ) j i + j2j i MX X (A i x j i ) j i + j2j i i=1 X j2j i I Dual variables: (p, r 1,...,r M ) MX X (c T i wi k ) i k i=1 k2k i MX X i=1 j i =1, i =1,...,M j i 0, k i 0, 8j, k, i k2k i (A i w k i ) k i = b IOE 610: LP II, Fall 2013 Example of Dantzig-Wolfe decomposition Page 290

11 Pricing subproblem I Z(p) = P M i=1 z i(p), where (SP i ) z i (p) =min(c T i p T A i ) x i s.t. x i 2 P i, i.e., each division solves its own version of the pricing subproblem, and generates its own columns/cuts: I If zi = 1, return w i k such that (c T i p T A i ) w i k < 0 corresponding i k has a negative reduced cost in the master problem I If 1 < zi < r i,return x j i such that (c T i p T A i ) x j i < r i corresponding j i has a negative reduced cost in the master problem I Otherwise (c T i p T A i )x j i r i 8j 2 J i and (c T i p T A i )wi k j 0 8k 2 K i all i s and k i s within this division have nonnegative reduced costs IOE 610: LP II, Fall 2013 Example of Dantzig-Wolfe decomposition Page 291 Economic interpretation (To keep is simple, assume P i s are bounded.) I Parent company with independent divisions I Each independent division would like to min c T i x i s.t. x i 2 P i I Parent company interests expressed through P i A ix i = b I D-W decomposition algorithm: indirect coordination of division activities by the central planner at the parent company: I Planner announces the value of p for each unit contribution towards common goal of b I If division i chooses xi, it experiences cost c T i x i, but gets rewarded in the amount (p T A i )x i by the company. Thus, division solves the corresponding pricing subproblem, and reports the solution as a proposal to the planner. I The planner (optimally) combines these proposals with previously obtained ones, and re-assesses p. I The process is repeated until the planner cannot elicit any improving proposals from any of the divisions. I Note: the central planner does not need to have any info on division s internal operations (i.e., c i, D i and P i ) IOE 610: LP II, Fall 2013 Example of Dantzig-Wolfe decomposition Page 292

12 Bounds on optimal cost Theorem 6.1 Let z? be the optimal cost of the (full) master problem, z be the cost of the restricted master problem at some iteration of D-W algorithm, r corresponding dual variables, and z i the optimal costs of (SP i ) in the same iteration. Then, assuming z i > 1 for all i, z + X (z i r i ) apple z? apple z. i Proof: I The upper bound is trivial. I For the lower bound, let ( p, r) solve (RelMD). Then ( p, z) isa feasible solution for (MD), so z? p T b + X i = p T b + X i z i r i + X i (z i r i )=z + X i IOE 610: LP II, Fall 2013 Example of Dantzig-Wolfe decomposition Page 293 (z i r i ).