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(    6dTf P f T Click to edit Master title style! !  0Qf  f RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0dQf `` f W*  0Qf `  f Y*  0$Rf `  f Y*H  0޽h ? ̙33 Default Design 0 @h.(  h h 0$Xf    x Y*  h 0DYf  @  x [* d h c $ ?HP  x h 0Xf  @0 x RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S h 6Xf   x Y*  h 6Wf  @ x [* H h 0jB ? ̙33, 00(  0  0 NOfkk s$  f u* J%%JJnn  0 NDVfkk  0$ f w* J%%JJnn 0 TUfkk s  f u* J%%JJnn 0 T$Ufkk  0 f w* J%%JJnnH 0 0jB ? ̙33G (    <Rf ]Branch-and-Price (Column Generation) Solving Integer Programs With a Huge Number of Variables^^$F  <Of S CP-AI-OR 02 School on Optimization Le Croisic, France March 2002AA$&H  0޽h ? ̙33 UMP (     <RfxY }EReasons for Considering Formulations with a Huge Number of VariablesFF  <DSfix P A compact formulation of a MIP may have a weak LP relaxation. Frequently the relaxation can be tightened by a reformulation that involves a huge number of variables. A compact formulation of a MIP may have symmetric structure that causes branch-and-bound to perform poorly because the problem barely changes after branching. A reformulation with a huge number of variables can eliminate this symmetry. Column generation provides a decomposition of the problem into master and sub problems. This decomposition may have a natural interpretation in the contextual setting allowing for the incorporation of additional important constraints. A formulation with a huge number of variables may be the only choice.XKbIKGH  0޽h ? ̙33% `e(    <dZf*/ q9Classic Example The Cutting Stock or Bin Packing Problem::H  <$Of 3BP/H@___PPT9" zInput: A set of n items of sizes a1, a2, & an, 2. A set of demands di for item i, i=1,& ,n 3. Bins of size L Output: A packing of the items into a minimum number of bins     ]Hs @`H  0޽h ? ̙332 2|2p[2(    <Zf7  >   <dWfn S  ?Example  <Yfys < ^ `  o 0#"     <Pf?`   K211 @`  <Tf?`  K395 @`  <4&x? K610 @`  <%x? J97 @`  <t%x? T  @`  <%x?`   J14 @`  <$x?`  J31 @`  <T$x? J36 @`  <#x? J45 @`   <#x? T  @`   <4#x?` `  I4 @`   <t"x?``  I3 @`   <"x?` I2 @`   <"x?` I1 @`  <T!x?` Ii @``B  0o ?```B  0o ?``B   0o ? ` ZB P s *o ?`B  0o ?`B Q 0o ?`B U 0o ?```B V 0o ?`B W 0o ?```B Y 0o ?`` ``B [ 0o ?` ` ``B ^ 0o ?  `B ` 0o ?`B h 0o ?  `B l 0o ?` `B n 0o ?`  x " <A ??p x B <A ??`0  p <!x 9L = 100 r < x0 BSolution with FFD  s 6&xP  >$ X t 0@ X u 00  X v 0  X w 0  X x 0  LB z c $DPLB { c $DP` ` LB | c $D@ LB } c $D@ LB ~ c $D@   < x)';# 445  <4 x  445  <x 4  445  <x  436  <T  414LB  c $D0  LB  c $D0 @@ LB  c $D0 0 0   < 4  436  <   436  <t 0  414  <ԁ < P  414LB  c $D  LB  c $D @@ LB  c $D  LB  c $D  LB  c $D   <4 4  436  <   436  < 4  436  <T   431  <   431  < 4  431  <t 0D  431  <Ԅ @T  431LB  c $D  LB  c $D  LB  c $D   <4F < @ 4x 48  <6 0  3x 1  <  3x 1  <T  5x 131  <&  5x 105  <   5x 199  <t g  5= 485  <4P s0 J EThere are 37 patternsx  <A ??P    <I_C ; Lower bound   <I C : rounded up   <t9X3 @= 416 in exampleH  0޽h ? ̙33 <(    <4{ KCompact Formulation  < /  ; and integer   <Ԭ0 Lp* zK is an upper bound on the number of bins The LP relaxation of this formulation is very bad since it only gives the trivial lower bound mentioned above. Solution won t change after branching because of symmetries. We need a better formulation for which the LP gives a good bound and the symmetries are eliminated.>>x  <A" N??pp NH  0޽h ? ̙33    M (    <4*q i1Formulation with an Exponential Number of Columns22x  <A ??M@   <7 zis the tth pattern*  x  <A ??0@@P   <0@ * &is the number of item i in tth pattern*'     <t@': ; In example     6ԯp 0 >$ LB   c $Dp0LB   c $Dp 0   <4f` 445  <f ` 445x  <A ??  s  x  <A ??l   <T@  $number of times tth pattern is used *%  < `  ; and integer   <t  Z*The difficulty is that T can be very large++x  <A Q?? P  QH  0޽h ? ̙33e   (     <4) # NLP bound for example is 452.25x   <A ??<   <7r  tDroundup gives 454 round down gives 450 and then 3 more for residualEEx   <A S??@ 80  S  <T 6   W'this is optimal because of the LP bound((H  0޽h ? ̙33     $L (  $ $ <u LLP Column Generation $ <t@zP/H@___PPT9" We solve the LP relaxation over a small subset of columns. We then need to find a column with negative reduced cost or prove that none exists implying the current solution is optional.,~D~D @`x $ <A ??`  $ <4/  Hreduced cost of column tx  $ <A "?? ] "  $ <Kz <   $ < 5wherex  $ <A $?? P $ $ <o  Kis an optimal dual solutionx $ <A &??  & $ <   7integer $ 0t@ ZThis is a knapsack problem that can be solved by dynamic programming or branch-and-bound. [[H $ 0޽h ? ̙33    (* (  ( ( <1 LLP Column GenerationX ( 0p@pPX ( 0@p ( <   ASolve Master LP ( <t0* CSolve Sub problemRB ( s *DP@ @ @RB  ( s *D@ @ LB  (@ c $D@LB  ( c $DRB  ( s *D@  ( <4Y wS = dual solution ( <yF s  7optimal ( <T  ECP may be useful hereRB (@ s *D ( < ; new columns r ( <t P  After solving the LP relaxation why can t we apply standard branch-and-bound To solve the IP. Difficulty: llx ( <A (?? P  ( ( <4 g   : fractional  ( <F P@ 7Branch: ( <v sCIt is difficult to enforce this constraint on the knapsack problem.DDx ( <A# U??  :  Ux ( <A% X??- @ XH ( 0޽h ? ̙33c  4(  4 4 < HRouting Problems 4 <`pZ <  4 0t   E Several Routes are needed Individual Routes are Constrained Master Problems: Puts together feasible routes at minimum cost Sub Problem: Generates routes based on minimum cost and dual variables Vehicle routing Crew scheduling Traveling tournament problemRB 4 s *D  H 4 0޽h ? ̙33 }8(  8 8 Z4gֳgֳ ?`@p MAirline Crew Schedulingw 8 Tgֳgֳ ? Input: A schedule of a fleet Objective: Find a set of crew itineraries (pairings) that partition all of the legs at minimum cost. :f f]|B 8 TD1?P@P|B 8 TD1? @ |B 8 TD1? @  8 T 1?Y AATL"  8 TT 1? M  7MIA   8 T 1?0 jP  AJAX"   8 Tt 1? } CMonday (duty 1)   8 T4 1? < _ NTuesday (duty 2)" B  8 ZD1?P`P B  8 ZD1?  B 8 ZD1?   8 TQ 1? p  D8:00$  8 TTR 1? \  E17:00$  8 TR 1?   E10:00$  8 TS 1? ;  E13:00$ B 8 ZD1? @` B 8 ZD1?P H 8 0޽h ? ̙33 RJ<(  < < ZSgֳgֳ ?@ >Pairings  < T4Tgֳgֳ ?P VRestrictions Connection times Duty elapsed time Flying time in a duty 8-in-24 rule Duty cost is the maximum of 3 quantities. Flying time, fraction of elapsed time, minimum guarantee Pairing cost is the maximum of 3 quantities. Sum of duty costs, elapsed time, minimum guarantee Daily problems 100 flights: 90,000 pairings 250 flights: 2.5 million pairings 1000 flights: more than a billion pairings F*9-3j F*9-3  jH < 0޽h ? ̙33 @+(  @ @ ZTgֳgֳ ?0 pp MAirline Crew Scheduling@ @ TTUgֳgֳ ?0P` A set partitioning problem: rows correspond to flights and columns to pairings aij=1 if flight i is in pairing j. xj=1 if pairing j is selected. n is very large, in billions.O`O   ,O ; @ c $A *??p x *dAfH @ 0޽h ? ̙33 =5D(  D D <V( QDelayed Column Generation D <tVz n/f^___PPT9@8 bWe solve the set partitioning problem over a subset of pairings. The dual variables give reduced costs for the pairings. We solve a constrained shortest path sub problem to generate new pairings or to show that the current LP solution is optimal. CP may be a good way to solve the sub problem because of the large number of constraints.Z=J=J   @`H D 0޽h ? ̙33U  H(  H H Z4Wgֳgֳ ?0`p ODelayed Column Generation| H TA jJ?@H H 0޽h ? ̙33  0 LJ(  L L <Wg JStandard Branching L <TX m; A pairing has a fractional value in an optimal LP solution<<x L <A ,??p  , L <Y@#:  < Two cases  x  L <A 3??0 Z  3x  L <A 4??   4  L <tY   l<is easy, fix the pairing and knock out the flights it covers==x L <A 6??   6 L <Y c  wGis hard, the pairing must be eliminated from the shortest path problemHHH L 0޽h ? ̙33~  . & @P (  P P <ZG KFollow-on Branching P <Zp VGiven a fractional solution to the LP there are fractional pairings r and s such that WWx P <A 8??` 8 P <T[[ 5andx P <A 9??` 9 P <\  ?are fractional.2  P <t\0  In r, flight k follows flight i In s, flight k appears but flight i doesn tLLx  P <A ;??0  ;  P <\ 5i k;  P <4] J   branch 1: flight k follows flight i in some pairing. Eliminate all pairings that contain i or k or both, but not consecutively. SA P <] y  rB[in the network delete all connections from i except the one to k]CCx P <A =?? `00  = P <0 '  i branch 2: k cannot follow i. Eliminate all pairings in which k immediately follows i. Mj P <   `0[in the network, delete the connector arc (i,k)]11x P <A ???  ?H P 0޽h ? ̙33< P T|(  T T <6e S0-1 Cutting Stock Branching T <I C a1A pattern contains are most one unit of an item, 22x T <A A??@@c A T <t(j vBranching is the same as crew pairing. A pattern with items r and s has a fractional value in an optimal LP solution.ww T <Y S |( branch 1: define a new item rs of size))x  T <A C?? @ C  T <y s  G(they must go together)  T < r  wE branch 2: add at most one of item r and s to the knapsack problemFFH T 0޽h ? ̙338 ` Xx(  X X < WGeneral Cutting Stock Branching   X <@W : f6A pattern may contain several units of different items77x X <A G??  G X < z L48.5 105.5 100.75 197.5 X 046  uC branch 1: patterns containing 2 or more units of item 2 207 DDx  X <A I??   I  X < @k  < branch 2:  x  X <A! K?? @  K X <T f*  b2These are simple constraints to add to the master.33H X 0޽h ? ̙33~ .&p\(  \ \ < IColumn Management \ 0tP@@  i We can add multiple problems from the sub problem to the master. It can be a good idea to sets of columns that yield feasible integer solutions to the master. Columns no longer active in the LP solution can be deleted from the master.XBlH \ 0޽h ? ̙33 ph`(  ` ` <4Z%T Y!Branch-and-Price Strategic Issues"" ` <z n/f^___PPT9@8 Column generation at a node can be terminated before proving LP optimality. If the LP value is above the upper bound, then we should continue column generation with the hope of fathoming the node. Strong branching may be helpful in difficult problems. 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