Tutorial: Operations Research and Constraint Programming. Why Integrate OR and CP? Computational Advantage of Integrating CP and OR

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1 Computatonal Advantage of Integratng CP and OR Usng CP + relaaton from MILP Tutoral: Operatons Research and Constrant Programmng Focacc, Lod, Mlano (999) Problem Lesson tmetablng Speedup to tmes faster than CP John Hooer Carnege Mellon Unversty June 8 Refalo (999) Hooer & Osoro (999) Thorstensson & Ottosson () Pecewse lnear costs Flow shop schedulng, etc. Product confguraton to tmes faster than MILP to tmes faster than MILP. to tmes faster than CP, MILP Slde Slde Computatonal Advantage of Integratng CP and MILP Usng CP + relaaton from MILP Problem Speedup Why Integrate OR and CP? Sellmann & Fahle () Van Hoeve () Automatc recordng Stable set problem to tmes faster than CP, MILP etter than CP n less tme Slde Complementary strengths Computatonal advantages Outlne of the Tutoral ollapragada, Ghattas & Hooer () ec & Refalo () Slde Structural desgn (nonlnear) Schedulng wth earlness & tardness costs Up to 6 tmes faster than MILP. problems: <6 mn vs > hrs for MILP Solved 67 of 9, CP solved only Complementary Strengths Computatonal Advantage of Integratng CP and MILP Usng CP-based ranch and Prce CP: Inference methods Modelng Eplots local structure OR: Relaaton methods Dualty theory Eplots global structure Let s brng them together! Yunes, Moura & de Souza (999) Easton, Nemhauser & Trc () Problem Urban transt crew schedulng Travelng tournament schedulng Speedup Optmal schedule for trps, vs. for tradtonal branch and prce Frst to solve 8-team nstance Slde Slde 6

2 Jan & Grossmann () Thorstensson () Tmpe () Computatonal Advantage of Integratng CP and MILP Usng CP/MILP enders methods Problem Mn-cost plannng & schedung Mn-cost plannng & schedulng Polypropylene batch schedulng at ASF Speedup to tmes faster than CP, MILP tmes faster than Jan & Grossmann Solved prevously nsoluble problem n mn Detaled Outlne Why Integrate OR and CP? Complementary strengths Computatonal advantages Outlne of the tutoral A Glmpse at CP Early successes Advantages and dsadvantages Intal Eample: Integrated Methods Freght Transfer ounds Propagaton Cuttng Planes ranch-nfer-and-rela Tree Slde 7 Slde enost, Gaudn, Rottembourg () Hooer () Hooer () Slde 8 Computatonal Advantage of Integratng CP and MILP Usng CP/MILP enders methods Problem Call center schedulng Mn-cost, mn-maespan plannng & cumulatve schedulng Mn tardness plannng & cumulatve schedulng Speedup Solved twce as many nstances as tradtonal enders - tmes faster than CP, MILP - tmes faster than CP, MILP Detaled Outlne CP Concepts Consstency Hyperarc Consstency Modelng Eamples CP Flterng Algorthms Element Alldff Dsunctve Schedulng Cumulatve Schedulng Lnear Relaaton and CP Why rela? Algebrac Analyss of LP Lnear Programmng Dualty LP-ased Doman Flterng Eample: Sngle-Vehcle Routng Dsunctons of Lnear Systems Slde Outlne of the Tutoral Why Integrate OR and CP? A Glmpse at CP Intal Eample: Integrated Methods CP Concepts CP Flterng Algorthms Lnear Relaaton and CP Med Integer/Lnear Modelng Cuttng Planes Lagrangean Relaaton and CP Dynamc Programmng n CP CP-based ranch and Prce CP-based enders Decomposton Detaled Outlne Med Integer/Lnear Modelng MILP Representablty. Dsunctve Modelng. Knapsac Modelng Cuttng Planes - Knapsac Cuts Gomory Cuts Med Integer Roundng Cuts Eample: Product Confguraton Lagrangean Relaaton and CP Lagrangean Dualty Propertes of the Lagrangean Dual Eample: Fast Lnear Programmng Doman Flterng Eample: Contnuous Global Optmzaton Slde 9 Slde

3 Detaled Outlne Dynamc Programmng n CP Eample: Captal udgetng Doman Flterng Recursve Optmzaton CP-based ranch and Prce asc Idea Eample: Arlne Crew Schedulng CP-based enders Decomposton enders Decomposton n the Abstract Classcal enders Decomposton Eample: Machne Schedulng What s constrant programmng? It s a relatvely new technology developed n the computer scence and artfcal ntellgence communtes. It has found an mportant role n schedulng, logstcs and supply chan management. Slde Slde 6 acground Readng Early commercal successes Crcut desgn (Semens) Contaner port schedulng (Hong Kong and Sngapore) Ths tutoral s based on: J. N. Hooer, Integrated Methods for Optmzaton, Sprnger (7). Contans 9 eercses. J. N. Hooer, Operatons research methods n constrant programmng, n F. Ross, P. van ee and T. Walsh, eds., Handboo of Constrant Programmng, Elsever (6), pp Real-tme control (Semens, Xero) Slde Slde 7 Applcatons A Glmpse at Constrant Programmng Early Successes Advantages and Dsadvantages Job shop schedulng Assembly lne smoothng and balancng Cellular frequency assgnment Nurse schedulng Shft plannng Mantenance plannng Arlne crew rosterng and schedulng Arport gate allocaton and stand plannng Slde Slde 8

4 Applcatons CP vs. MP Producton schedulng chemcals avaton ol refnng steel lumber photographc plates tres Transport schedulng (food, nuclear fuel) Warehouse management Course tmetablng In mathematcal programmng, equatons (constrants) descrbe the problem but don t tell how to solve t. In constrant programmng, each constrant nvoes a procedure that screens out unacceptable solutons. Much as each lne of a computer program nvoes an operaton. Slde 9 Slde Advantages and Dsadvantages Advantages of CP CP vs. Mathematcal Programmng Relaaton ranchng Slde MP Numercal calculaton Atomstc modelng (lnear nequaltes) Independence of model and algorthm Logc processng Inference (flterng, constrant propagaton) Hgh-level modelng (global constrants) ranchng CP Constrant-based processng etter at sequencng and schedulng Slde where MP methods have wea relaatons. Addng messy constrants maes the problem easer. The more constrants, the better. More powerful modelng language. Global constrants lead to succnct models. Constrants convey problem structure to the solver. etter at hghly-constraned problems Msleadng better when constrants propagate well, or when constrants have few varables. Programmng programmng Dsdvantages of CP In constrant programmng: programmng = a form of computer programmng (constrant-based processng) In mathematcal programmng: programmng = logstcs plannng (hstorcally) Slde Weaer for contnuous varables. Due to lac of numercal technques May fal when constrants contan many varables. These constrants don t propagate well. Often not good for fundng optmal solutons. Due to lac of relaaton technology. May not scale up Dscrete combnatoral methods Software s not robust Younger feld Slde

5 Obvous soluton Integrate CP and MP. More on ths later. Eample: Freght Transfer Transport tons of freght usng 8 trucs, whch come n szes Truc sze Number avalable Capacty (tons) 7 Cost per truc 9 6 Slde Slde 8 Trends Number of trucs of type CP s better nown n contnental Europe, Asa. Less nown n North Amerca, seen as threat to OR. CP/MP ntegraton s growng Eclpse, Mozart, OPL Studo, SIMPL, SCIP, ARON Heurstc methods ncreasngly mportant n CP Dscrete combnatoral methods MP/CP/heurstcs may become a sngle technology. Slde 6 Knapsac pacng constrant Slde 9 mn {,,,} Truc type Number avalable Capacty (tons) 7 Cost per truc 9 6 Knapsac coverng constrant ounds propagaton Intal Eample: Integrated Methods mn {,,,} Freght Transfer ounds Propagaton Cuttng Planes ranch-nfer-and-rela Tree = 7 Slde 7 Slde

6 Reduced doman ounds propagaton mn {,,},,, {,,,} = 7 Cuttng Planes egn wth contnuous relaaton mn , Replace domans wth bounds Ths s a lnear programmng problem, whch s easy to solve. Its optmal value provdes a lower bound on optmal value of orgnal problem. Slde Slde ounds consstency Cuttng planes (vald nequaltes) Let {L,, U } be the doman of A constrant set s bounds consstent f for each : Slde = L n some feasble soluton and = U n some feasble soluton. ounds consstency we wll not set to any nfeasble values durng branchng. ounds propagaton acheves bounds consstency for a sngle nequalty s bounds consstent when the domans are {,,} and,, {,,,}. ut not necessarly for a set of nequaltes. Slde mn , We can create a tghter relaaton (larger mnmum value) wth the addton of cuttng planes. ounds consstency Cuttng planes (vald nequaltes) ounds propagaton may not acheve bounds consstency for a set of constrants. Consder set of nequaltes + wth domans, {,}, solutons (, ) = (,), (,). mn , Cuttng plane Contnuous relaaton ounds propagaton has no effect on the domans. ut constrant set s not bounds consstent because = n no feasble soluton. Slde Slde 6 All feasble solutons of the orgnal problem satsfy a cuttng plane (.e., t s vald). ut a cuttng plane may eclude ( cut off ) solutons of the contnuous relaaton. Feasble solutons 6

7 Cuttng planes (vald nequaltes) mn , Cuttng planes (vald nequaltes) mn , {,} s a pacng because 7 + alone cannot satsfy the nequalty, even wth = =. Mamal Pacngs {,} {,} {,} Knapsac cuts Slde 7 Slde Knapsac cuts correspondng to nonmamal pacngs can be nonredundant. Cuttng planes (vald nequaltes) Contnuous relaaton wth cuts mn , {,} s a pacng So, + (7 + ) Knapsac cut mn , Knapsac cuts Slde 8 whch mples (7 + ) + = ma{,} Optmal value of. s a lower bound on optmal value of orgnal problem. Slde Cuttng planes (vald nequaltes) Let have doman [L,U ] and let a. In general, a pacng P for a a satsfes P a a a U P ranchnfer-andrela tree Propagate bounds and solve relaaton of orgnal problem. { } {} {} {} = (⅓,,⅔,) value = ⅓ and generates a napsac cut a au P P ma{ a } P Slde 9 Slde 7

8 ranch-nferand-rela tree ranch on a varable wth nonntegral value n the relaaton. { } {} {} {} = (⅓,,⅔,) value = ⅓ {,} = ranch-nferand-rela tree ranch agan. { } { } { } { } nfeasble relaaton { } {} {} {} = (⅓,,⅔,) value = ⅓ {,} = { } { } { } {} = (,,¾,) value = 7½ { } {} {} {} = (,.6,,) value = 6 = {,,} = {,} Slde Slde 6 ranch-nferand-rela tree Propagate bounds and solve relaaton. Snce relaaton s nfeasble, bactrac. Slde { } { } { } { } nfeasble relaaton { } {} {} {} = (⅓,,⅔,) value = ⅓ = {,} ranch-nferand-rela tree Soluton of relaaton s ntegral and therefore feasble n the orgnal problem. Ths becomes the ncumbent soluton. Slde 7 { } { } { } { } nfeasble relaaton { } {} {} {} = (⅓,,⅔,) value = ⅓ = {,} { } { } { } {} = (,,¾,) value = 7½ { } {,} { } { } { } = (,,,) value = feasble soluton = { } {} {} {} = (,.6,,) value = 6 = {,,} ranch-nferand-rela tree Propagate bounds and solve relaaton. ranch on nonntegral varable. { } { } { } { } nfeasble relaaton { } {} {} {} = (⅓,,⅔,) value = ⅓ = {,} { } {} {} {} = (,.6,,) value = 6 = {,,} ranch-nferand-rela tree Soluton s nonntegral, but we can bactrac because value of relaaton s no better than ncumbent soluton. { } { } { } { } nfeasble relaaton { } {} {} {} = (⅓,,⅔,) value = ⅓ = {,} { } { } { } {} = (,,¾,) value = 7½ { } {} {} {} = (,.6,,) value = 6 = {,,} Slde Slde 8 { } {,} { } { } { } = (,,,) value = feasble soluton = { } { } { } { } = (,½,,½) value = bactrac due to bound 8

9 ranch-nferand-rela tree Another feasble soluton found. No better than ncumbent soluton, whch s optmal because search has fnshed. Slde 9 { } { } { } { } nfeasble relaaton { } {} {} {} = (⅓,,⅔,) value = ⅓ {,} = { } { } { } {} = (,,¾,) value = 7½ { } {,} { } { } { } = (,,,) value = feasble soluton = { } {} {} {} = (,.6,,) value = 6 = {,,} { } { } { } { } = (,½,,½) value = bactrac due to bound { } { } { } { } = (,,,) value = feasble soluton Consstency A constrant set s consstent f every partal assgnment to the varables that volates no constrant s feasble..e., can be etended to a feasble soluton. Consstency feasblty Consstency means that any nfeasble partal assgnment s eplctly ruled out by a constrant. Fully consstent constrant sets can be solved wthout bactracng. Slde Two optmal solutons = (,,,) = (,,,) Consstency Consder the constrant set + {,} It s not consstent, because = volates no constrant and yet s nfeasble (no soluton has = ). Addng the constrant = maes the set consstent. Slde Slde + other constrants {,} = = Constrant Programmng Concepts Consstency Hyperarc Consstency Modelng Eamples subtree wth 99 nodes but no feasble soluton y addng the constrant =, the left subtree s elmnated Slde Slde 9

10 Hyperarc Consstency Graph colorng problem that can be solved by arc consstency mantenance alone. Color nodes wth red, green, blue wth no two adacent nodes havng the same color. Also nown as generalzed arc consstency. A constrant set s hyperarc consstent f every value n every varable doman s part of some feasble soluton. That s, the domans are reduced as much as possble. If all constrants are bnary (contan varables), hyperarc consstent = arc consstent. Doman reducton s CP s bggest engne. Slde Slde 8 Graph colorng problem that can be solved by arc consstency mantenance alone. Color nodes wth red, green, blue wth no two adacent nodes havng the same color. Graph colorng problem that can be solved by arc consstency mantenance alone. Color nodes wth red, green, blue wth no two adacent nodes havng the same color. Slde 6 Slde 9 Graph colorng problem that can be solved by arc consstency mantenance alone. Color nodes wth red, green, blue wth no two adacent nodes havng the same color. Graph colorng problem that can be solved by arc consstency mantenance alone. Color nodes wth red, green, blue wth no two adacent nodes havng the same color. Slde 7 Slde 6

11 Graph colorng problem that can be solved by arc consstency mantenance alone. Color nodes wth red, green, blue wth no two adacent nodes havng the same color. Popular - model Let = f cty mmedately precedes cty, otherwse mn s.t. =, all =, all V W c {,} { }, all dsont V, W,, n Subtour elmnaton constrants Slde 6 Slde 6 Graph colorng problem that can be solved by arc consstency mantenance alone. Color nodes wth red, green, blue wth no two adacent nodes havng the same color. A CP model Let y = the th cty vsted. The model would be wrtten n a specfc constrant programmng language but would essentally say: Varable ndces mn y y + s.t. alldff( y,, y ) c y {,, n} n Global constrant Slde 6 Slde 6 Modelng Eamples wth Global Constrants Travelng Salesman Travelng salesman problem: Let c = dstance from cty to cty. Fnd the shortest route that vsts each of n ctes eactly once. An alternate CP model Let y = the cty vsted after cty. mn y s.t. crcut( y,, y ) c y {,, n} n Hamltonan crcut constrant Slde 6 Slde 66

12 Element constrant The constrant c y can be mplemented: z ( y c c z) element,(,, ), The constrant y can be mplemented z (ths s a slghtly dfferent constrant) ( y z) element,(,, ), n n Assgn z the yth value n the lst Add the constrant z = y CP model Mnmze holdng and setup costs mn qyt y + h t s t t Inventory balance s.t. s + = d + s, all, t, t t t t Producton capacty t C, st, all, t ( yt ) ( t = ), all, t Slde 67 Slde 7 Modelng eample: Lot szng and schedulng Day: CP model Varable ndces Mnmze holdng and setup costs Slde 68 A A Product At most one product manufactured on each day. Demands for each product on each day. Mnmze setup + holdng cost. mn qyt y + h t s t t Inventory balance s.t. s + = d + s, all, t, t t t t t C, st, all, t ( yt ) ( t = ), all, t Producton capacty Slde 7 Producton level of product n perod t Product manufactured n perod t Integer programmng model (Wolsey) Slde 69 mn htst + qδt t, t s.t. s + = d + s, all, t, t t t t z y y, all, t t t, t z y, all, t t t z y, all, t t, t δ y + y, all,, t t, t t δ y t, t, all,, t δ y, all,, t t t t Cy, all, t t y, z, δ {,} t t t t, s t y =, all t t Many varables Cumulatve schedulng constrant Used for resource-constraned schedulng. Total resources consumed by obs at any one tme must not eceed L. Job start tmes (varables) Slde 7 ( t t p p c c L) cumulatve (,, ),(,, ),(,, ), n n n Job processng tmes Job resource requrements

13 Cumulatve schedulng constrant Precedence constrants Mnmze maespan (no deadlnes, all release tmes = ): L resources mn Slde 7 z ( t t ) s.t. cumulatve (,, ),(,,,,),(,,,,),7 z t + z t + Mn maespan = 8 tme Resources used Processng tmes Job start tmes L,, Slde 76, , 9 6,,, Modelng eample: Shp loadng Wll use ILOG s OPL Studo modelng language. Eample s from OPL manual. The problem Load tems on the shp n mnmum tme (mn maespan) Each tem requres a certan tme and certan number of worers. Total of 8 worers avalable. Use the cumulatve schedulng constrant. mn z s.t. z t +, z t +, etc. ( t t ) cumulatve (,, ),(,,,),(,,,),8 t t +, t t +, etc. Slde 7 Slde 77 Item Labor Item Duraton Duraton Labor OPL model Problem data nt capacty = 8; nt nbtass = ; range Tass..nbTass; nt duraton[tass] = [,,,6,,]; nt totalduraton = sum(t n Tass) duraton[t]; nt demand[tass] = [,,,,,]; struct Precedences { nt before; nt after; } {Precedences} setofprecedences = { <,>, <,>,, <,> }; 7 6 Slde 7 Slde 78

14 schedulehorzon = totalduraton; Actvty a[t n Tass](duraton[t]); DscreteResource res(8); Actvty maespan(); mnmze maespan.end subect to forall(t n Tass) a[t] precedes maespan; forall(p n setofprecedences) a[p.before] precedes a[p.after]; forall(t n Tass) a[t] requres(demand[t]) res; }; Slde 79 mn s.t. T Slde 8 b T u +, all s t R, all ( t v e m) cumulatve,,, b v = u + t, all s s b + su C, all r n cumulatve u,,,,, u t Maespan b b e p s s n e = (,,) Job release tme m storage tans Job duraton Tan capacty p pacng unts Modelng eample: Producton schedulng wth ntermedate storage Manufacturng Unt Storage Tans Capacty C Capacty C Pacng Unts Modelng eample: Employee schedulng Schedule four nurses n 8-hour shfts. A nurse wors at most one shft a day, at least days a wee. Same schedule every wee. No shft staffed by more than two dfferent nurses n a wee. A nurse cannot wor dfferent shfts on two consecutve days. A nurse who wors shft or must do so at least two days n a row. Slde 8 Capacty C Slde 8 Fllng of storage tan Two ways to vew the problem Level Need to enforce capacty constrant here only Assgn nurses to shfts Sun Mon Tue Wed Thu Fr Sat Shft Shft Shft A C D C D A C D A D A C A C A D Assgn shfts to nurses Fllng starts Pacng starts Slde 8 t u t + (b/r) u + (b/s) Manufacturng rate atch sze Fllng ends Pacng ends Pacng rate Nurse A Nurse Nurse C Nurse D Slde 8 Sun Mon Tue = day off Wed Thu Fr Sat

15 Use both formulatons n the same model! Frst, assgn nurses to shfts. Let w sd = nurse assgned to shft s on day d Remanng constrants are not easly epressed n ths notaton. So, assgn shfts to nurses. Let y d = shft assgned to nurse on day d alldff( w d, w d, w d ), all d The varables w d, w d, w d tae dfferent values Slde 8 That s, schedule dfferent nurses on each day alldff Slde 88 (,, ) y y y, all d d d d Assgn a dfferent nurse to each shft on each day. Ths constrant s redundant of prevous constrants, but redundant constrants speed soluton. Use both formulatons n the same model! Frst, assgn nurses to shfts. Let w sd = nurse assgned to shft s on day d alldff( w, w, w ), all d d d d ( w A C D ) cardnalty (,,, ),(,,,),(6,6,6,6) A occurs at least and at most 6 tmes n the array w, and smlarly for, C, D. That s, each nurse wors at least and at most 6 days a wee Remanng constrants are not easly epressed n ths notaton. So, assgn shfts to nurses. Let y d = shft assgned to nurse on day d alldff ( y d, yd, yd ), all d ( y,sun y,sat P ) stretch,, (,),(,),(6,6),, all Every stretch of s has length between and 6. Every stretch of s has length between and 6. So a nurse who wors shft or must do so at least two days n a row. Slde 86 Slde 89 Use both formulatons n the same model! Frst, assgn nurses to shfts. Let w sd = nurse assgned to shft s on day d ( w d wd wd ) d ( w A C D ) ( ws,sun ws,sat ) s alldff,,, all cardnalty (,,, ),(,,,),(6,6,6,6) nvalues,...,,, all Remanng constrants are not easly epressed n ths notaton. So, assgn shfts to nurses. Let y d = shft assgned to nurse on day d alldff ( y d, yd, yd ), all d ( y,sun y,sat P ) stretch,, (,),(,),(6,6),, all Slde 87 The varables w s,sun,, w s,sat tae at least and at most dfferent values. That s, at least and at most nurses wor any gven shft. Here P = {(s,),(,s) s =,,} Whenever a stretch of a s mmedately precedes a stretch of b s, (a,b) must be one of the pars n P. So a nurse cannot swtch shfts wthout tang at least one day off. Slde 9

16 Now we must connect the w sd varables to the y d varables. Use channelng constrants: w y y dd wsdd =, all, d = s, all s, d Channelng constrants ncrease propagaton and mae the problem easer to solve. Flterng for element Varable domans can be easly fltered to mantan hvperarc consstency. Doman of z ( y z) element,(,, ), D D D z z Dy n { } D D D D y y z { } Dz f Dy = D D otherwse Slde 9 Slde 9 The complete model s: w y y dd wsdd ( w d wd wd ) d ( w A C D ) ( ws,sun ws,sat ) s alldff,,, all cardnalty (,,, ),(,,,),(6,6,6,6) nvalues,...,,, all alldff ( y d, yd, yd ), all d ( y,sun y,sat P) stretch,, (,),(,),(6,6),, all =, all, d = s, all s, d Flterng for element Eample... ( y z) element,(,,, ), The ntal domans are: The reduced domans are: D = D D D D D z y = {,,6,8,9 } {,, } {,} {,} {,,8,9 } {,,7} = = = = D = D D D D D z y = { 8,9} { } {,} {,} { 8,9} {,,7 } = = = = Slde 9 Slde 9 Flterng for alldff ( y y ) alldff,, n CP Flterng Algorthms Domans can be fltered wth an algorthm based on mamum cardnalty bpartte matchng and a theorem of erge. It s a specal case of optmalty condtons for ma flow. Element Alldff Dsunctve Schedulng Cumulatve Schedulng Slde 9 Slde 96 6

17 Flterng for alldff Indcate domans wth edges Consder the domans y y y y y { } {,,} {,,, } {, } {,,,,,6 } y y y y Fnd mamum cardnalty bpartte matchng. y 6 Slde 97 Slde Indcate domans wth edges Indcate domans wth edges y y Fnd mamum cardnalty bpartte matchng. y y y y Mar edges n alternatng paths that start at an uncovered verte. y y y 6 y 6 Slde 98 Slde Indcate domans wth edges Indcate domans wth edges y Fnd mamum cardnalty bpartte matchng. y Fnd mamum cardnalty bpartte matchng. y y y y Mar edges n alternatng paths that start at an uncovered verte. y y y 6 y 6 Slde 99 Slde 7

18 Indcate domans wth edges Flterng for alldff y y y y Fnd mamum cardnalty bpartte matchng. Mar edges n alternatng paths that start at an uncovered verte. Mar edges n alternatng cycles. Domans have been fltered: y y y y y { } {,, } {,,, } {, } {,,,,,6 } y y y y y { } {,} {,} { } {,6} y 6 Hyperarc consstency acheved. Slde Slde 6 Indcate domans wth edges Dsunctve schedulng y Fnd mamum cardnalty bpartte matchng. Consder a dsunctve schedulng constrant: ( s s s s p p p p ) dsunctve (,,, ),(,,, ) y y Mar edges n alternatng paths that start at an uncovered verte. Start tme varables y Mar edges n alternatng cycles. y 6 Remove unmared edges not n matchng. Slde Slde 7 Indcate domans wth edges Edge fndng for dsunctve schedulng y Fnd mamum cardnalty bpartte matchng. Consder a dsunctve schedulng constrant: ( s s s s p p p p ) dsunctve (,,, ),(,,, ) y y Mar edges n alternatng paths that start at an uncovered verte. Processng tmes y Mar edges n alternatng cycles. y 6 Remove unmared edges not n matchng. Slde Slde 8 8

19 Edge fndng for dsunctve schedulng Consder a dsunctve schedulng constrant: ( s s s s p p p p ) dsunctve (,,, ),(,,, ) Varable domans defned by tme wndows and processng tmes s [, ] s [, ] s [,7 ] s [,7 ] Edge fndng for dsunctve schedulng ut let s reduce of the deadlnes to 9: We wll use edge fndng to prove that there s no feasble schedule. Slde 9 Slde Edge fndng for dsunctve schedulng Consder a dsunctve schedulng constrant: dsunctve (( s, s, s, s ),( p, p, p, p )) A feasble (mn maespan) soluton: Edge fndng for dsunctve schedulng We can deduce that ob must precede obs and : {,} ecause f ob s not frst, there s not enough tme for all obs wthn the tme wndows: L{,,} E{,} < p{,,} Slde Tme wndow Slde E {,} 7<++ L {,,} Edge fndng for dsunctve schedulng ut let s reduce of the deadlnes to 9: Edge fndng for dsunctve schedulng We can deduce that ob must precede obs and : {,} ecause f ob s not frst, there s not enough tme for all obs wthn the tme wndows: L{,,} E{,} < p{,,} Latest deadlne Slde Slde E {,} 7<++ L {,,} 9

20 Edge fndng for dsunctve schedulng We can deduce that ob must precede obs and : {,} ecause f ob s not frst, there s not enough tme for all obs wthn the tme wndows: L{,,} E{,} < p{,,} Edge fndng for dsunctve schedulng In general, we can deduce that ob must precede all the obs n set J: J If there s not enough tme for all the obs after the earlest release tme of the obs n J LJ { } EJ < pj { } L{,,} E{,} < p{,,} Earlest release tme Slde E {,} 7<++ L {,,} Slde 8 Edge fndng for dsunctve schedulng We can deduce that ob must precede obs and : {,} ecause f ob s not frst, there s not enough tme for all obs wthn the tme wndows: L{,,} E{,} < p{,,} Total processng tme Edge fndng for dsunctve schedulng In general, we can deduce that ob must precede all the obs n set J: J If there s not enough tme for all the obs after the earlest release tme of the obs n J LJ { } EJ < pj { } L{,,} E{,} < p{,,} Now we can tghten the deadlne for ob to: { L p } mn J J L{,} p{,} = J J Slde 6 E {,} 7<++ L {,,} Slde 9 Edge fndng for dsunctve schedulng We can deduce that ob must precede obs and : {,} So we can tghten deadlne of ob to mnmum of L {} p{} = L {} p{} = L{,} p{,} = Snce tme wndow of ob s now too narrow, there s no feasble schedule. Edge fndng for dsunctve schedulng There s a symmetrc rule: J If there s not enough tme for all the obs before the latest deadlne of the obs n J: LJ EJ { } < pj { } Now we can tghten the release date for ob to: { E + p } ma J J J J Slde 7 E {,} 7<++ L {,,} Slde

21 Edge fndng for dsunctve schedulng Edge fndng for dsunctve schedulng Problem: how can we avod enumeratng all subsets J of obs to fnd edges? LJ { } EJ < pj { } and all subsets J of J to tghten the bounds? { L p } mn J J J J Key result: We only have to consder sets J whose tme wndows le wthn some nterval. { L p } mn J J J J Note: Edge fndng does not acheve bounds consstency, whch s an NP-hard problem. e.g., J = {,} Slde Slde Edge fndng for dsunctve schedulng Edge fndng for dsunctve schedulng One O(n ) algorthm s based on the Jacson pre-emptve schedule (JPS). Usng a dfferent eample, the JPS s: Key result: We only have to consder sets J whose tme wndows le wthn some nterval. { L p } mn J J J J e.g., J = {,} Slde Slde Edge fndng for dsunctve schedulng Edge fndng for dsunctve schedulng One O(n ) algorthm s based on the Jacson pre-emptve schedule (JPS). Usng a dfferent eample, the JPS s: Key result: We only have to consder sets J whose tme wndows le wthn some nterval. Removng a ob from those wthn an nterval only weaens the test LJ { } EJ < pj { } Slde { L p } mn J J J J There are a polynomal number of ntervals defned by release tmes and deadlnes. e.g., J = {,} For each ob Scan obs J n decreasng order of L Select frst for whch L E < p + p Conclude that J Update E to JPS(, ) Slde 6 Jobs unfnshed at tme E n JPS J Jobs n J wth L L Latest completon tme n JPS of obs n J

22 Not-frst/not-last rules We can deduce that ob cannot precede obs and : {,} ( ) ecause f ob s frst, there s too lttle tme to complete the obs before the later deadlne of obs and : L{,} E < p + p + p Not-frst/not-last rules In general, we can deduce that ob cannot precede all the obs n J: J ( ) f there s too lttle tme after release tme of ob to complete all obs before the latest deadlne n J: LJ E < pj Now we can update E to mn E + p J { } Slde 7 E 6<++ L {,} There s a symmetrc not-last rule. The rules can be appled n polynomal tme, although an effcent algorthm s qute complcated. Slde Not-frst/not-last rules We can deduce that ob cannot precede obs and : {,} Now we can tghten the release tme of ob to mnmum of: E E + p = + p = ( ) Cumulatve schedulng Consder a cumulatve schedulng constrant: ( s s s p p p c c c C) cumulatve (,, ),(,, ),(,, ), A feasble soluton: E 6<++ L {,} Slde 8 Slde Not-frst/not-last rules In general, we can deduce that ob cannot precede all the obs n J: J ( ) f there s too lttle tme after release tme of ob to complete all obs before the latest deadlne n J: LJ E < pj Now we can update E to mn E + p J { } Edge fndng for cumulatve schedulng We can deduce that ob must fnsh after the others fnsh: > {, } ecause the total energy requred eceeds the area between the earlest release tme and the later deadlne of obs,: ( ) e + e > C L E {,} {,} {,,} Slde 9 Slde

23 Edge fndng for cumulatve schedulng Edge fndng for cumulatve schedulng We can deduce that ob must fnsh after the others fnsh: > {, } ecause the total energy requred eceeds the area between the earlest release tme and the later deadlne of obs,: Total energy requred = Slde ( ) e + e > C L E {,} {,} {,,} 9 8 We can deduce that ob must fnsh after the others fnsh: > {, } We can update the release tme of ob to E Slde 6 {,} Energy avalable for obs, f space s left for ob to start anytme = Ecess energy requred by obs, = ej ( C c )( L{,} E{,} ) + c Edge fndng for cumulatve schedulng Edge fndng for cumulatve schedulng We can deduce that ob must fnsh after the others fnsh: > {, } ecause the total energy requred eceeds the area between the earlest release tme and the later deadlne of obs,: Total energy requred = Area avalable = Slde ( ) e + e > C L E {,} {,} {,,} 9 8 We can deduce that ob must fnsh after the others fnsh: > {, } We can update the release tme of ob to E Slde 7 {,} Energy avalable for obs, f space s left for ob to start anytme = Ecess energy requred by obs, = ej ( C c )( L{,} E{,} ) + c Move up ob release tme / = unts beyond E {,} E Edge fndng for cumulatve schedulng We can deduce that ob must fnsh after the others fnsh: > {, } We can update the release tme of ob to E {,} Energy avalable for obs, f space s left for ob to start anytme = ej ( C c )( L{,} E{,} ) + c Edge fndng for cumulatve schedulng In general, f ej { } > C ( LJ EJ { } ) then > J, and update E to ma J J ej C c LJ E ( )( ) > J e ( )( ) J C c L E J J EJ + c In general, f ej { } > C ( LJ { } EJ ) then < J, and update L to mn J J e C c L E ( )( ) > J J J e ( )( ) J C c L E J J LJ c Slde Slde 8

24 Edge fndng for cumulatve schedulng There s an O(n ) algorthm that fnds all applcatons of the edge fndng rules. Why Rela? Solvng a relaaton of a problem can: Tghten varable bounds. Possbly solve orgnal problem. Gude the search n a promsng drecton. Flter domans usng reduced costs or Lagrange multplers. Prune the search tree usng a bound on the optmal value. Provde a more global vew, because a sngle OR relaaton can pool relaatons of several constrants. Slde 9 Slde Other propagaton rules for cumulatve schedulng Etended edge fndng. Tmetablng. Not-frst/not-last rules. Energetc reasonng. Some OR models that can provde relaatons: Lnear programmng (LP). Med nteger lnear programmng (MILP) Can tself be relaed as an LP. LP relaaton can be strengthened wth cuttng planes. Lagrangean relaaton. Specalzed relaatons. For partcular problem classes. For global constrants. Slde Slde Motvaton Slde Lnear Relaaton Why Rela? Algebrac Analyss of LP Lnear Programmng Dualty LP-ased Doman Flterng Eample: Sngle-Vehcle Routng Dsunctons of Lnear Systems Lnear programmng s remarably versatle for representng real-world problems. LP s by far the most wdely used tool for relaaton. LP relaatons can be strengthened by cuttng planes. - ased on polyhedral analyss. LP has an elegant and powerful dualty theory. - Useful for doman flterng, and much else. The LP problem s etremely well solved. Slde

25 Algebrac Analyss of LP Algebrac analyss of LP An eample mn , + 7 = + Optmal soluton = (,) Wrte mn c A = b as mn c + c + N = b, N N Solve constrant equaton for : = b NN All solutons can be obtaned by settng N to some value. The soluton s basc f N =. N N where A = [ N] + 6 It s a basc feasble soluton f N = and. Slde Slde 8 Algebrac Analyss of LP Rewrte mn + 7 Slde , as In general an LP has the form mn = 6 + =,,, mn c A = b Eample mn = 6 + =,,, Slde 9, basc, basc, basc +, basc, basc = basc feasble soluton + 6, basc Algebrac analyss of LP Algebrac analyss of LP Wrte m n matr mn c A = b Slde 7 as asc varables mn c + c + N = b, N N N N Nonbasc varables where A = [ N] Any set of m lnearly ndependent columns of A. These form a bass for the space spanned by the columns. Wrte mn c A = b Slde as mn c + c + N = b, N N N Solve constrant equaton for : = b NN Epress cost n terms of nonbasc varables: c b ( c c N) N N Vector of reduced costs N where A = [ N] Snce N, basc soluton (,) s optmal f reduced costs are nonnegatve.

26 Eample mn = 6 + =,,, Consder ths basc feasble soluton Eample c c N N [ ] + [ ] mn = NN b, asc soluton s = b N = b N / 6 = = =, basc, basc Slde Slde Eample Wrte as c c N N [ ] + [ ] mn + 7 mn 7 + = = + =,,, NN b, Slde Eample c c N N [ ] + [ ] mn = NN, Slde asc soluton s = b N = b Reduced costs are c N N / 6 = = = c N / = [ 7 ] [ ] = [ ] [ ] Soluton s optmal Eample Lnear Programmng Dualty An LP can be vewed as an nference problem c c N N [ ] + [ ] mn = NN b, mn c A b = ma v A b c v mples Dual problem: Fnd the tghtest lower bound on the obectve functon that s mpled by the constrants. Slde Slde 6 6

27 Eample An LP can be vewed as an nference problem mn c A b Slde 7 = ma v A b c v From Faras Lemma: If A b, s feasble, That s, some surrogate (nonnegatve lnear combnaton) of A b domnates c v λa λb domnates c v A b c v ff for some λ λa c and λb v Prmal mn + 7 = + 6 ( λ ) +, Slde 6 ( λ ) Dual ma 6λ + λ = λ + λ λ + λ 7 λ, λ ( ) ( ) A dual soluton s (λ,λ ) = (,) + 6 ( λ = ) ( λ = ) domnates + 7 Dual multplers Surrogate Tghtest bound on cost Wea Dualty An LP can be vewed as an nference problem mn c A b = ma v A b c v From Faras Lemma: If A b, s feasble, = ma λb Ths s the λa c λ classcal LP dual λa λb domnates c v A b c v ff for some λ If * s feasble n the prmal problem mn c A b and λ* s feasble n the dual problem ma λb λa c λ then c* λ*b. Ths s because c* λ*a* λ*b λ* s dual feasble and * * s prmal feasble and λ* Slde 8 λa c and λb v Slde 6 Dual multplers as margnal costs Ths equalty s called strong dualty. mn c A b = ma λb λa c λ If A b, s feasble Ths s the classcal LP dual Note that the dual of the dual s the prmal (.e., the orgnal LP). Suppose we perturb the RHS of an LP (.e., change the requrement levels): The dual of the perturbed LP has the same constrants at the orgnal LP: mn c A b + b ma λ( b + b) λa c λ So an optmal soluton λ* of the orgnal dual s feasble n the perturbed dual. Slde 9 Slde 6 7

28 Dual multplers as margnal costs Suppose we perturb the RHS of an LP (.e., change the requrement levels): Slde 6 mn c A b + b y wea dualty, the optmal value of the perturbed LP s at least λ*(b + b) = λ*b + λ* b. Optmal value of orgnal LP, by strong dualty. So λ * s a lower bound on the margnal cost of ncreasng the -th requrement by one unt ( b = ). If λ * >, the -th constrant must be tght (complementary slacness). Dual of an LP n equalty form Prmal mn c + c + N = b, N N N N Slde 66 ( λ) Dual ma λb λ c λn c λ unrestrcted ( ) ( ) Recall that reduced cost vector s cn c N = cn λn λ Chec: λ = c = c ths solves the dual λn = c N cn f (,) solves the prmal N ecause reduced cost s nonnegatve at optmal soluton (,). Dual of an LP n equalty form Dual of an LP n equalty form Prmal mn c + c + N = b, N N N N ( λ) Dual ma λb λ c λn c N λ unrestrcted ( ) ( ) Prmal mn c + c + N = b, N N N N ( λ) Dual ma λb λ c λn c N λ unrestrcted ( ) ( ) Slde 6 Recall that reduced cost vector s cn c N = cn λn λ ths solves the dual f (,) solves the prmal In the eample, / λ = c = [ ] = [ ] Slde 67 Dual of an LP n equalty form Dual of an LP n equalty form Prmal mn c + c + N = b, N N N N ( λ) Dual ma λb λ c λn c N λ unrestrcted ( ) ( ) Prmal mn c + c + N = b, N N N N ( λ) Dual ma λb λ c λn c N λ unrestrcted ( ) ( ) Recall that reduced cost vector s cn c N = cn λn λ ths solves the dual f (,) solves the prmal Recall that reduced cost vector s cn c N = cn λn λ Note that the reduced cost of an ndvdual varable s r = c λa Column of A Slde 6 Slde 68 8

29 LP-based Doman Flterng Let mn c A b be an LP relaaton of a CP problem. One way to flter the doman of s to mnmze and mamze subect to A b,. - Ths s tme consumng. A faster method s to use dual multplers to derve vald nequaltes. - A specal case of ths method uses reduced costs to bound or f varables. - Reduced-cost varable fng s a wdely used technque n OR. Supposng mn c A b has optmal soluton *, optmal value v*, and optmal dual soluton λ*: We have found: a change n that changes A by b ncreases the optmal value of LP at least λ * b. Snce optmal value of the LP optmal value of the CP U, we have λ * b U v*, or * U v b * λ Slde 69 Slde 7 Suppose: mn c A b Slde 7 has optmal soluton *, optmal value v*, and optmal dual soluton λ*. and λ * >, whch means the -th constrant s tght (complementary slacness); and the LP s a relaaton of a CP problem; and we have a feasble soluton of the CP problem wth value U, so that U s an upper bound on the optmal value. Supposng mn c A b has optmal soluton *, optmal value v*, and optmal dual soluton λ*: We have found: a change n that changes A by b ncreases the optmal value of LP at least λ * b. Snce optmal value of the LP optmal value of the CP U, we have λ * b U v*, or * U v b * λ Snce b = A A * = A b, ths mples the nequalty * U v A b + * λ whch can be propagated. Slde 7 Supposng mn c A b has optmal soluton *, optmal value v*, and optmal dual soluton λ*: If were to change to a value other than *, the LHS of -th constrant A b would change by some amount b. Snce the constrant s tght, ths would ncrease the optmal value as much as changng the constrant to A b + b. So t would ncrease the optmal value at least λ * b. Eample mn ( λ = ) + ( λ = ), Suppose we have a feasble soluton of the orgnal CP wth value U =. Snce the frst constrant s tght, we can propagate the nequalty * U v A b + * λ or = 6. Slde 7 Slde 7 9

30 Reduced-cost doman flterng Suppose * =, whch means the constrant s tght. The nequalty A * U v b + becomes λ * * U v r Assgnment Relaaton mn = =, all {, }, all, c = f stop mmedately precedes stop Stop s preceded and followed by eactly one stop. The dual multpler for s the reduced cost r of, because ncreasng (currently ) by ncreases optmal cost by r. Smlar reasonng can bound a varable below when t s at ts upper bound. Slde 7 Slde 78 Eample mn ( λ = ) + ( λ = ), Snce * =, we have or =. Suppose we have a feasble soluton of the orgnal CP wth value U =. * U v r If s requred to be nteger, we can f t to zero. Ths s reduced-cost varable fng. Assgnment Relaaton mn = =, all, all, c = f stop mmedately precedes stop Stop s preceded and followed by eactly one stop. ecause ths problem s totally unmodular, t can be solved as an LP. The relaaton provdes a very wea lower bound on the optmal value. ut reduced-cost varable fng can be very useful n a CP contet. Slde 76 Slde 79 Eample: Sngle-Vehcle Routng A vehcle must mae several stops and return home, perhaps subect to tme wndows. The obectve s to fnd the order of stops that mnmzes travel tme. Ths s also nown as the travelng salesman problem (wth tme wndows). Stop Dsunctons of lnear systems Dsunctons of lnear systems often occur naturally n problems and can be gven a conve hull relaaton. A dsuncton of lnear systems represents a unon of polyhedra. mn c ( A b ) Travel tme c Stop Slde 77 Slde 8

31 Relang a dsuncton of lnear systems Dsunctons of lnear systems often occur naturally n problems and can be gven a conve hull relaaton. A dsuncton of lnear systems represents a unon of polyhedra. We want a conve hull relaaton (tghtest lnear relaaton). mn c ( A b ) Why? To derve conve hull relaaton of a dsuncton Wrte each soluton as a conve combnaton of ponts n the polyhedron mn c A b, all = y = y y Change of varable = y mn c A b y, all y = y = Conve hull relaaton (tghtest lnear relaaton) Slde 8 Slde 8 Relang a dsuncton of lnear systems Dsunctons of lnear systems often occur naturally n problems and can be gven a conve hull relaaton. The closure of the conve hull of Slde 8 s descrbed by mn c ( A b ) mn c A b y, all = y = y Slde 8 Med Integer/Lnear Modelng MILP Representablty Dsunctve Modelng Knapsac Modelng Why? To derve conve hull relaaton of a dsuncton Wrte each soluton as a conve combnaton of ponts n the polyhedron Slde 8 mn c A b, all = y = y y Conve hull relaaton (tghtest lnear relaaton) Motvaton A med nteger/lnear programmng (MILP) problem has the form We can rela a CP problem by modelng some constrants wth an MILP. If desred, we can then rela the MILP by droppng the ntegralty constrant, to obtan an LP. The LP relaaton can be strengthened wth cuttng planes. The frst step s to learn how to wrte MILP models. Slde 86 mn c + dy A + by b, y y nteger

32 MILP Representablty R n A subset S of s MILP representable f t s the proecton onto of some MILP constrant set of the form A + u + Dy b, y n m R, u R, y {,} Eample Mnmze a fed charge functon: Feasble set mn f = f + c f > Slde 87 Slde 9 MILP Representablty R n A subset S of s MILP representable f t s the proecton onto of some MILP constrant set of the form A + u + Dy b, y n m R, u R, y {,} Recesson cone of polyhedron Eample Mnmze a fed charge functon: mn f = f + c f > n Theorem. S R s MILP representable f and only f S s the unon of fntely many polyhedra havng the same recesson cone. Polyhedron Unon of two polyhedra P, P P Slde 88 Slde 9 Eample: Fed charge functon mn Mnmze a fed charge functon: f = f + c f > Eample Mnmze a fed charge functon: mn f = f + c f > Unon of two polyhedra P, P P P Slde 89 Slde 9

33 Eample Eample Mnmze a fed charge functon: mn f = f + c f > Start wth a dsuncton of lnear systems to represent the unon of polyhedra mn = M f + c The polyhedra have dfferent recesson cones. P P P P Slde 9 P recesson cone P recesson cone Slde 96 M Eample Eample Mnmze a fed charge functon: Add an upper bound on The polyhedra have the same recesson cone. P P mn f = f + c f > M Start wth a dsuncton of lnear systems to represent the unon of polyhedra Introduce a - varable y that s when s n polyhedron. Dsaggregate to create an for each. mn = M f + c mn c =, My, c + fy y + y =, y {,} = + Slde 9 M P recesson cone P recesson cone Slde 97 Modelng a unon of polyhedra Start wth a dsuncton of lnear systems to represent the unon of polyhedra. The th polyhedron s { A b} Introduce a - varable y that s when s n polyhedron. Dsaggregate to create an for each. Slde 9 mn c ( A b ) mn c A b y, all = y = y {,} Eample To smplfy: Replace wth. Replace wth. My, c + fy Replace y wth y. y + y =, y {, } = + Ths yelds Slde 98 mn My fy + c y {,} mn or =, mn fy + c My y {,} g M

34 Dsunctve Modelng Uncapactated faclty locaton Dsunctons often occur naturally n problems and can be gven an MILP model. Recall that a dsuncton of lnear systems (representng polyhedra wth the same recesson cone) ( A b ) Slde 99 has the MILP model mn c mn c A b y, all = y {,} y = MILP formulaton: mn f y + c y, all, y {,} Slde Dsunctve model: mn z + c =, all, all, all z z f = =, all No factory at locaton Factory at locaton Eample: Uncapactated faclty locaton m possble factory locatons n marets Locate factores to serve marets so as to mnmze total fed cost and transport cost. No lmt on producton capacty of each factory. Uncapactated faclty locaton MILP formulaton: mn f y + c y, all, y {,} egnner s model: mn f y + c y {,} ny, all, Mamum output from locaton Fed cost f Slde c Transport cost Slde ased on capactated locaton model. It has a weaer contnuous relaaton (obtaned by replacng y {,} wth y ). Ths begnner s mstae can be avoded by startng wth dsunctve formulaton. Uncapactated faclty locaton m possble factory locatons n marets Dsunctve model: mn z + c =, all, all, all z z f = =, all Fracton of maret s demand satsfed from locaton Knapsac Modelng Knapsac models consst of napsac coverng and napsac pacng constrants. The freght transfer model presented earler s an eample. We wll consder a smlar eample that combnes dsunctve and napsac modelng. Most OR professonals are unlely to wrte a model as good as the one presented here. Fed cost f c Transport cost No factory at locaton Factory at locaton Slde Slde

35 Note on tghtness of napsac models The contnuous relaaton of a napsac model s not n general a conve hull relaaton. - A dsunctve formulaton would provde a conve hull relaaton, but there are eponentally many dsuncts. Knapsac cuts can sgnfcantly tghten the relaaton. Eample: Pacage transport MILP model mn y, all,, y {,} Q y a ; =, all a Q y, all c y Modelng trc; unobvous wthout dsunctve approach Most OR professonals would omt ths constrant, snce t s the sum over of the net constrant. ut t generates very effectve napsac cuts. Slde Slde 8 Eample: Pacage transport Each pacage has sze a Each truc has capacty Q and costs c to operate Truc used Dsunctve model mn Q y a ; =, all y = z = c y = a z, all Q = =, all, y {,} z Knapsac constrants Truc not used Cuttng Planes - Knapsac Cuts Gomory Cuts Med Integer Roundng Cuts Eample: Product Confguraton Slde 6 f truc carres pacage f truc s used Slde 9 Eample: Pacage transport To revew MILP model mn Q y a ; =, all a Q y, all y, all,, y {,} c y Dsunctve model mn Q y a ; =, all y = z = c y = a z, all Q = =, all, y {,} z A cuttng plane (cut, vald nequalty) for an MILP model: s vald - It s satsfed by all feasble solutons of the model. cuts off solutons of the contnuous relaaton. - Ths maes the relaaton tghter. Cuttng plane Feasble solutons Contnuous relaaton Slde 7 Slde

36 Motvaton Cuttng planes (cuts) tghten the contnuous relaaton of an MILP model. Knapsac cuts - Generated for ndvdual napsac constrants. - We saw general nteger napsac cuts earler. - - napsac cuts and lftng technques are well studed and wdely used. Roundng cuts - Generated for the entre MILP, they are wdely used. - Gomory cuts for nteger varables only. - Med nteger roundng cuts for any MILP. Slde Eample J = {,,,} s a cover for Inde set J s a cover f a > a J Slde Ths gves rse to the cover nequalty The cover nequalty J s a - napsac cut for a a J Only mnmal covers need be consdered. - Knapsac Cuts - napsac cuts are desgned for napsac constrants wth - varables. The analyss s dfferent from that of general napsac constrants, to eplot the specal structure of - nequaltes. Sequental lftng A cover nequalty can often be strengthened by lftng t nto a hgher dmensonal space. That s, by addng varables. Sequental lftng adds one varable at a tme. Sequence-ndependent lftng adds several varables at once. Slde Slde - Knapsac Cuts - napsac cuts are desgned for napsac constrants wth - varables. The analyss s dfferent from that of general napsac constrants, to eplot the specal structure of - nequaltes. Consder a - napsac pacng constrant a a. (Knapsac coverng constrants are smlarly analyzed.) Inde set J s a cover f a > a J The cover nequalty J s a - napsac cut for a a J Sequental lftng To lft a cover nequalty J J add a term to the left-hand sde + π J J where π s the largest coeffcent for whch the nequalty s stll vald. So, π = J ma {, } a a a J J for J Ths can be done repeatedly (by dynamc programmng). Slde Only mnmal covers need be consdered. Slde 6 6

37 Eample Gven 6 To lft add a term to the left-hand sde where π = ma Ths yelds { } {,} for {,,,} Further lftng leaves the cut unchanged π ut f the varables are added n the order 6,, the result s dfferent: Eample Gven 6 To lft Add terms ρ(8) + ρ() 6 where ρ(u) s gven by Ths yelds the lfted cut ( / ) + (/ ) 6 Slde 7 Slde Sequence-ndependent lftng Gomory Cuts Sequence-ndependent lftng usually yelds a weaer cut than sequental lftng. ut t adds all the varables at once and s much faster. Commonly used n commercal MILP solvers. When an nteger programmng problem has a nonntegral soluton, we can generate at least one Gomory cut to cut off that soluton. - Ths s a specal case of a separatng cut, because t separates the current soluton of the relaaton from the feasble set. Separatng cut Soluton of contnuous relaaton Gomory cuts are wdely used and very effectve n MILP solvers. Feasble solutons Slde 8 Slde Sequence-ndependent lftng To lft a cover nequalty J J add terms to the left-hand sde + ρ( a ) J where wth J J f A u A + and {,, p } ρ( u) = + ( u A ) / f A u < A and {,, p } p + ( u Ap ) / f Ap u = J a a J = {,, p} A =a = A = Gomory cuts Gven an nteger programmng problem mn c A = b and ntegral Let (,) be an optmal soluton of the contnuous relaaton, where ˆ ˆ = b NN ˆ ˆ b = b, N = N Then f s nonntegral n ths soluton, the followng Gomory cut s volated by (,): ˆ ˆ + N N b Slde 9 Slde 7

38 Eample Med Integer Roundng Cuts mn , and ntegral or mn + + = + = 6 and ntegral Optmal soluton of the contnuous relaaton has = = / / / Nˆ = / 9 / 9 ˆ b = / Med nteger roundng (MIR) cuts can be generated for solutons of any relaed MILP n whch one or more nteger varables has a fractonal value. Le Gomory cuts, they are separatng cuts. MIR cuts are wdely used n commercal solvers. Slde Slde 6 Eample mn , and ntegral or mn + + = + = 6 and ntegral The Gomory cut ˆ ˆ + N N b s + [ /9 / 9] / Optmal soluton of the contnuous relaaton has = = / / / Nˆ = / 9 / 9 ˆ b = / or In, space ths s + MIR cuts Gven an MILP problem mn c + dy A + Dy = b, y and y ntegral In an optmal soluton of the contnuous relaaton, let J = { y s nonbasc} K = { s nonbasc} N = nonbasc cols of [A D] Then f y s nonntegral n ths soluton, the followng MIR cut s volated by the soluton of the relaaton: frac( Nˆ ) ˆ ˆ ˆ + y ˆ ˆ + N y + N + N N b frac( ˆ + ) frac( ˆ J ) J b b K where J { J frac( Nˆ ) frac( ˆ b )} = J = J \ J Slde Slde 7 Eample Eample mn , and ntegral or mn + + = + = 6 and ntegral Optmal soluton of the contnuous relaaton has = = / / / Nˆ = / 9 / 9 ˆ b = / + 6y y = + y y =, y, y nteger Tae basc soluton (,y ) = (8/,7/). Then / / Nˆ = / 8 / 8 / ˆ b = 7 / J = {}, K = {}, J =, J = {} / The MIR cut s y + / + y + (/ ) + 8 / / / y + (/ ) y + or bˆ = / Slde Gomory cut + Gomory cut after re-solvng LP wth prevous cut. Slde 8 8

39 Eample: Product Confguraton Ths eample llustrates: Combnaton of propagaton and relaaton. Processng of varable ndces. Contnuous relaaton of element constrant. To solve t: ranch on domans of t and q. Propagate element constrants and bounds on v. Varable nde s converted to specally structured element constrant. Vald napsac cuts are derved and propagated. Use lnear contnuous relaatons. Specal purpose MILP relaaton for element. Slde 9 Slde The problem Propagaton Choose what type of each component, and how many Memory Memory Memory Memory Memory Memory Personal computer Power supply Ds drve Ds drve Ds drve Ds drve Ds drve mn c v v = q A, all t L v U, all Ths s propagated n the usual way Power supply Power supply Power supply Slde Slde Model of the problem Unt cost of producng attrbute Amount of attrbute produced (< f consumed): memory, heat, power, weght, etc. mn c v v = q A, all t L v U, all Quantty of component nstalled Amount of attrbute produced by type t of component t s a varable nde Propagaton mn c v v = q A, all t L v U, all v z, all = ( n ) element t,( q, A,, q A ), z, all, Ths s rewrtten as Ths s propagated n the usual way Slde Slde 9

40 Propagaton v z, all = ( n ) element t,( q, A,, q A ), z, all, Relaaton v z, all = ( n ) element t,( q, A,, q A ), z, all, Ths can be propagated by (a) usng specalzed flters for element constrants of ths form mn c v v = q A, all t L v U, all Ths s relaed by relang ths and addng the napsac cuts. Ths s relaed as v v v Slde Slde 8 Propagaton v z, all = ( n ) element t,( q, A,, q A ), z, all, Relaaton v z, all = ( n ) element t,( q, A,, q A ), z, all, Ths s propagated by (a) usng specalzed flters for element constrants of ths form, (b) addng napsac cuts for the vald nequaltes: Slde 6 { } ma A q v, all Dt { } mn A q v, all Dt and (c) propagatng the napsac cuts. [ v, v ] s current doman of v Slde 9 Ths s relaed by replacng each element constrant wth a dsunctve conve hull relaaton: z = A q, q = q Dt D t Relaaton Relaaton mn v = q A, all t L v U, all Slde 7 c v Ths s relaed as v v v So the followng LP relaaton s solved at each node of the search tree to obtan a lower bound: Slde mn v = A q, all Dt q = q, all Dt v v v, all q q q, all { } napsac cuts for ma A q v, all Dt { } napsac cuts for mn A q v, all c v q, all, Dt

41 Computatonal Results Lagrangean Dualty Seconds 8 6 CPLEX CLP Hybrd Consder an nequalty-constraned problem mn f( ) g( ) S Hard constrants Easy constrants. The obect s to get rd of (dualze) the hard constrants by movng them nto the obectve functon.. Problem Slde Slde Lagrangean Dualty Slde Lagrangean Relaaton Lagrangean Dualty Propertes of the Lagrangean Dual Eample: Fast Lnear Programmng Doman Flterng Eample: Contnuous Global Optmzaton Consder an nequalty-constraned problem mn f( ) g( ) S Slde It s related to an nference problem ma v s S g( ) b f ( ) v mples Lagrangean Dual problem: Fnd the tghtest lower bound on the obectve functon that s mpled by the constrants. Motvaton Lagrangean relaaton can provde better bounds than LP relaaton. The Lagrangean dual generalzes LP dualty. It provdes doman flterng analogous to that based on LP dualty. - Ths s a ey technque n contnuous global optmzaton. Lagrangean relaaton gets rd of troublesome constrants by dualzng them. - That s, movng them nto the obectve functon. - The Lagrangean relaaton may decouple. Prmal Dual mn f ( ) ma v g( ) s S g( ) b f ( ) v S Surrogate Let us say that S g( ) f ( ) v ff λg( ) domnates f ( ) v for some λ λg() f() v for all S That s, v f() λg() for all S Slde Slde 6

42 Prmal Dual mn f ( ) ma v g( ) s S g( ) b f ( ) v S Surrogate Let us say that S g( ) f ( ) v ff λg( ) domnates f ( ) v for some λ λg() f() v for all S That s, v f() λg() for all S Or mn { ( ) λ g ( )} S Eample mn + + +, {,,, } Strongest surrogate The Lagrangean relaaton s θ ( λ, λ ) = mn { + λ ( + ) λ ( + ) } {,,} = mn {( + λ λ ) + ( λ λ ) + λ} {,,} The Lagrangean relaaton s easy to solve for any gven λ, λ : f + λ λ = otherwse f λ λ = otherwse Optmal soluton (,) Slde 7 Slde Prmal mn f ( ) g( ) S Let us say that Surrogate Dual ma v s S g( ) b f ( ) v S λg( ) domnates f ( ) v g( ) f ( ) v ff for some λ λg() f() v for all S That s, v f() λg() for all S Or v mn { f ( ) λ g ( )} So the dual becomes ma v v mn { f ( ) λg( ) } for some λ S Slde 8 S Eample mn + + +, {,,, } Optmal soluton (,) Value = Slde θ(λ,λ ) s pecewse lnear and concave. λ θ(λ)=7. θ(λ)= θ(λ)=9 /7 θ(λ)= θ(λ)= λ Soluton of Lagrangean dual: (λ,λ ) = (/7, /7), θ(λ) = 9 /7 Note dualty gap between and 9 /7 (no strong dualty). Now we have Eample Prmal mn f ( ) g( ) S These constrants are dualzed Dual ma v v mn { f ( ) λg( ) } for some λ or ma θ( λ) λ S where θ ( λ) = mn f ( ) λg( ) S { } mn + + +, {,,, } Note: n ths eample, the Lagrangean dual provdes the same bound (9 /7) as the contnuous relaaton of the IP. Ths s because the Lagrangean relaaton can be solved as an LP: θ( λ, λ ) = mn {( + λ λ ) + ( λ λ ) + λ } {,,} = mn {( + λ λ ) + ( λ λ ) + λ } Lagrangean relaaton Vector of Lagrange multplers Lagrangean dualty s useful when the Lagrangean relaaton s tghter than an LP but nonetheless easy to solve. Slde 9 The Lagrangean dual can be vewed as the problem of fndng the Lagrangean relaaton that gves the tghtest bound. Slde

43 Propertes of the Lagrangean dual Wea dualty: For any feasble * and any λ*, f(*) θ(λ*). In partcular, mn f ( ) g( ) S ma θ( λ) λ Concavty: θ(λ) s concave. It can therefore be mamzed by local search methods. At root node, solve Dualze Specal structure, e.g. varable bounds mn c A b D d ( λ) The (partal) LP dual soluton λ* solves the Lagrangean dual n whch θ( λ) = mn c λ( A b) D d { } Complementary slacness: If * and λ* are optmal, and there s no dualty gap, then λ*g(*) =. Slde Slde 6 Solvng the Lagrangean dual Let λ be the th terate, and let λ + = λ + α ξ Subgradent of θ(λ) at λ = λ If solves the Lagrangean relaaton for λ = λ, then ξ = g( ). Ths s because θ(λ) = f( ) + λg( ) at λ = λ. The stepsze α must be adusted so that the sequence converges but not before reachng a mamum. At root node, solve Dualze Specal structure, e.g. varable bounds mn c A b D d ( λ) The (partal) LP dual soluton λ* solves the Lagrangean dual n whch θ( λ) = mn c λ( A b) D d At another node, the LP s Here θ(λ*) s stll a lower bound on the optmal value of the LP and can be qucly calculated by solvng a specally structured LP. { } mn c A b D d H h ( λ) ranchng constrants, etc. Slde Slde 7 Eample: Fast Lnear Programmng In CP contets, t s best to process each node of the search tree very rapdly. Lagrangean relaaton may allow very fast calculaton of a lower bound on the optmal value of the LP relaaton at each node. The dea s to solve the Lagrangean dual at the root node (whch s an LP) and use the same Lagrange multplers to get an LP bound at other nodes. Doman Flterng Suppose: mn f ( ) g( ) S has optmal soluton *, optmal value v*, and optmal Lagrangean dual soluton λ*. and λ * >, whch means the -th constrant s tght (complementary slacness); and the problem s a relaaton of a CP problem; and we have a feasble soluton of the CP problem wth value U, so that U s an upper bound on the optmal value. Slde Slde 8

44 mn f ( ) has optmal soluton *, optmal value v*, and Supposng g( ) optmal Lagrangean dual soluton λ*: S If were to change to a value other than *, the LHS of -th constrant g () would change by some amount. Snce the constrant s tght, ths would ncrease the optmal value as much as changng the constrant to g (). So t would ncrease the optmal value at least λ *. (It s easly shown that Lagrange multplers are margnal costs. Dual multplers for LP are a specal case of Lagrange multplers.) Eample: Contnuous Global Optmzaton Some of the best contnuous global solvers (e.g., ARON) combne OR-style relaaton wth CP-style nterval arthmetc and doman flterng. The use of Lagrange multplers for doman flterng s a ey technque n these solvers. Slde 9 Slde 6 mn f ( ) has optmal soluton *, optmal value v*, and Supposng g( ) optmal Lagrangean dual soluton λ*: S We have found: a change n that changes g () by ncreases the optmal value at least λ *. Snce optmal value of ths problem optmal value of the CP U, we have λ * U v*, or * U v * λ Contnuous Global Optmzaton ma + = + [,], [,] Feasble set Global optmum Local optmum Slde 6 Slde 6 mn f ( ) has optmal soluton *, optmal value v*, and Supposng g( ) optmal Lagrangean dual soluton λ*: S We have found: a change n that changes g () by ncreases the optmal value at least λ *. Snce optmal value of ths problem optmal value of the CP U, we have λ * U v*, or * U v * λ Snce = g () g (*) = g (), ths mples the nequalty * U v g ( ) * λ whch can be propagated. Slde 6 To solve t: Search: splt nterval domans of,. Each node of search tree s a problem restrcton. Propagaton: Interval propagaton, doman flterng. Use Lagrange multplers to nfer vald nequalty for propagaton. Reduced-cost varable fng s a specal case. Relaaton: Use functon factorzaton to obtan lnear contnuous relaaton. Slde 6

45 Interval propagaton Relaaton (functon factorzaton) Propagate ntervals [,], [,] through constrants to obtan [/8,7/8], [/,7/] The lnear relaaton becomes: mn + y = + + y + + y +, =, Slde 6 Slde 68 Relaaton (functon factorzaton) Relaaton (functon factorzaton) Factor comple functons nto elementary functons that have nown lnear relaatons. Wrte = as y = where y =. Ths factors nto lnear functon y and blnear functon. Lnear functon y s ts own lnear relaaton. Solve lnear relaaton. Slde 66 Slde 69 Relaaton (functon factorzaton) Relaaton (functon factorzaton) Factor comple functons nto elementary functons that have nown lnear relaatons. Wrte = as y = where y =. Ths factors nto lnear functon y and blnear functon. Lnear functon y s ts own lnear relaaton. lnear functon y = has relaaton: + y + + y + where doman of s [, ] Solve lnear relaaton. Snce soluton s nfeasble, splt an nterval and branch. [,.7] [.,] Slde 67 Slde 7

46 [,.7] [.,] Relaaton (functon factorzaton) mn + y = + + y + + y +, =, Assocated Lagrange multpler n soluton of relaaton s λ =. Slde 7 Slde 7 [,.7] [.,] Relaaton (functon factorzaton) Soluton of relaaton s feasble, value =. Ths becomes ncumbent soluton mn + y = + Assocated Lagrange multpler n soluton of relaaton s λ =. + y + + y +, =, Slde 7 Ths yelds a vald nequalty for propagaton:.8. + =.. Value of relaaton Slde 7 Lagrange multpler Value of ncumbent soluton [,.7] [.,] Soluton of relaaton s feasble, value =. Ths becomes ncumbent soluton Soluton of relaaton s not qute feasble, value =.8 Also use Lagrange multplers for doman flterng Dynamc Programmng n CP Eample: Captal udgetng Doman Flterng Recursve Optmzaton Slde 7 Slde 76 6

47 Motvaton Dynamc programmng (DP) s a hghly versatle technque that can eplot recursve structure n a problem. Doman flterng s straghtforward for problems modeled as a DP. DP s also mportant n desgnng flters for some global constrants, such as the stretch constrant (employee schedulng). Nonseral DP s related to bucet elmnaton n CP and eplots the structure of the prmal graph. DP modelng s the art of eepng the state space small whle mantanng a Marovan property. Eample: Captal udgetng + + {, } In general the recurson for a b s f ( s ) = ma { f ( s + a )} D + oundary condton: f f sn+ b ( s ) = otherwse n+ n+ We wll eamne only one smple eample of seral DP. f (s ) for each state s Slde 77 Slde 8 Eample: Captal udgetng We wsh to bult power plants wth a total cost of at most mllon Euros. There are three types of plants, costng, or mllon Euros each. We must buld one or two of each type. The problem has a smple napsac pacng model: Number of factores of type + + {, } Eample: Captal udgetng + + {, } The problem s feasble. Each path to s a feasble soluton. Path : = (,,) Path : = (,,) Path : = (,,) Possble costs are 9,,. f (s ) for each state s Slde 78 Slde 8 Eample: Captal udgetng + + {, } State s Stage Doman Flterng + + {, } In general the recurson for a b s f ( s ) = ma { f ( s + a )} D + = f there s a path from state s to a feasble soluton, otherwse Slde 79 State s sum of frst terms of a = = f (8) = ma{f (8+ ), f (8+ )} = ma{,} = f ()= f ()= To flter domans: observe what values of occur on feasble paths. D = {, } D = {, } D = { } Slde 8 = = = = = = 7

48 Recursve Optmzaton ma {, } Mamze revenue The recurson ncludes arc values: f ( s ) = ma { c + f ( s + a )} D + CP-based ranch and Prce = value on ma value path from s to fnal stage (value to go) Slde 8 Arc value f (8) = ma{ +f (8+ ), +f (8+ )} = ma{, } = f ()= f ()= Slde 86 asc Idea Eample: Arlne Crew Schedulng f Recursve optmzaton ma {, } oundary condton: ( s ) n+ n+ The recurson ncludes arc values: f ( s ) = ma { c + f ( s + a )} + f sn+ b = otherwse 9 f (s ) for each state s Motvaton ranch and prce allows soluton of nteger programmng problems wth a huge number of varables. The problem s solved by a branch-and-rela method. The dfference les n how the LP relaaton s solved. Varables are added to the LP relaaton only as needed. Varables are prced to fnd whch ones should be added. CP s useful for solvng the prcng problem, partcularly when constrants are comple. CP-based branch and prce has been successfully appled to arlne crew schedulng, transt schedulng, and other transportaton-related problems. Slde 8 Slde 87 Recursve optmzaton ma {, } 9 asc Idea Suppose the LP relaaton of an nteger programmng problem has a huge number of varables: mn c A = b The mamum revenue s 9. The optmal path s easy to retrace. (,, ) = (,,) We wll solve a restrcted master problem, whch has a small subset of the varables: Column of A mn J J c A = b ( λ) f (s ) for each state s Addng to the problem would mprove the soluton f has a negatve reduced cost: r = c λa < Slde 8 Slde 88 8

49 asc Idea Addng to the problem would mprove the soluton f has a negatve reduced cost: r = c λa < Computng the reduced cost of s nown as prcng. So we solve the prcng problem: mn cy Cost of column y λy y s a column of A If the soluton y* satsfes c y* λy* <, then we can add column y to the restrcted master problem. Arlne Crew Schedulng There are crew members, and the possble rosters are: (,,), (,,6), (,,), (,,6) The LP relaaton of the problem s: Cost of assgnng crew member to roster = f we assgn crew member to roster, = otherwse. Each crew member s assgned to eactly roster. Each flght s assgned at least crew member. Slde 89 Slde 9 asc Idea The prcng problem ma λy y s a column of A need not be solved to optmalty, so long as we fnd a column wth negatve reduced cost. However, when we can no longer fnd an mprovng column, we solved the prcng problem to optmalty to mae sure we have the optmal soluton of the LP. If we can state constrants that the columns of A must satsfy, CP may be a good way to solve the prcng problem. Arlne Crew Schedulng There are crew members, and the possble rosters are: (,,), (,,6), (,,), (,,6) The LP relaaton of the problem s: Cost of assgnng crew member to roster = f we assgn crew member to roster, = otherwse. Each crew member s assgned to eactly roster. Each flght s assgned at least crew member. Rosters that cover flght. Slde 9 Slde 9 Eample: Arlne Crew Schedulng We want to assgn crew members to flghts to mnmze cost whle coverng the flghts and observng comple wor rules. Flght data Start tme Fnsh tme Slde 9 A roster s the sequence of flghts assgned to a sngle crew member. The gap between two consecutve flghts n a roster must be from to hours. Total flght tme for a roster must be between 6 and hours. For eample, flght cannot mmedately precede 6 flght cannot mmedately precede. The possble rosters are: (,,), (,,6), (,,), (,,6) Arlne Crew Schedulng There are crew members, and the possble rosters are: (,,), (,,6), (,,), (,,6) The LP relaaton of the problem s: Slde 9 Cost of assgnng crew member to roster Rosters that cover flght. = f we assgn crew member to roster, = otherwse. Each crew member s assgned to eactly roster. Each flght s assgned at least crew member. 9

50 Arlne Crew Schedulng There are crew members, and the possble rosters are: (,,), (,,6), (,,), (,,6) The LP relaaton of the problem s: Slde 9 Cost of assgnng crew member to roster Rosters that cover flght. = f we assgn crew member to roster, = otherwse. Each crew member s assgned to eactly roster. Each flght s assgned at least crew member. Arlne Crew Schedulng There are crew members, and the possble rosters are: (,,), (,,6), (,,), (,,6) The LP relaaton of the problem s: Slde 98 Cost of assgnng crew member to roster Rosters that cover flght 6. = f we assgn crew member to roster, = otherwse. Each crew member s assgned to eactly roster. Each flght s assgned at least crew member. Arlne Crew Schedulng There are crew members, and the possble rosters are: (,,), (,,6), (,,), (,,6) The LP relaaton of the problem s: Cost of assgnng crew member to roster = f we assgn crew member to roster, = otherwse. Each crew member s assgned to eactly roster. Each flght s assgned at least crew member. Arlne Crew Schedulng There are crew members, and the possble rosters are: (,,), (,,6), (,,), (,,6) The LP relaaton of the problem s: Cost c of assgnng crew member to roster = f we assgn crew member to roster, = otherwse. Each crew member s assgned to eactly roster. Each flght s assgned at least crew member. Slde 96 Rosters that cover flght. Slde 99 In a real problem, there can be mllons of rosters. Arlne Crew Schedulng There are crew members, and the possble rosters are: (,,), (,,6), (,,), (,,6) The LP relaaton of the problem s: Cost of assgnng crew member to roster = f we assgn crew member to roster, = otherwse. Each crew member s assgned to eactly roster. Each flght s assgned at least crew member. Arlne Crew Schedulng We start by solvng the problem wth a subset of the columns: Optmal dual soluton u u v v v v v v 6 Rosters that cover flght. Slde 97 Slde

51 Arlne Crew Schedulng We start by solvng the problem wth a subset of the columns: Dual varables u u v v v v v v 6 Prcng problem Crew member Crew member Each s-t path corresponds to a roster, provded the flght tme s wthn bounds. Slde Slde Arlne Crew Schedulng We start by solvng the problem wth a subset of the columns: Dual varables u u v v v v v v 6 The reduced cost of an ecluded roster for crew member s c u v n roster We wll formulate the prcng problem as a shortest path problem. Prcng problem Crew member Crew member Cost of flght f t mmedately follows flght, offset by dual multpler for flght Slde Slde Prcng problem Prcng problem Cost of transferrng from home to flght, offset by dual multpler for crew member Crew member Crew member Dual multpler omtted to brea symmetry Crew member Crew member Slde Slde 6

52 Prcng problem Crew member Length of a path s reduced cost of the correspondng roster. Prcng problem The shortest path problem cannot be solved by tradtonal shortest path algorthms, due to the bounds on total path length. It can be solved by CP: Set of flghts assgned to crew member Path length Graph Crew member Path global constrant Setsum global constrant Path( X, z, G), all flghts mn ( ) X T f s T X { flghts }, z <, all Duraton of flght ma Slde 7 Slde Prcng problem Crew member Arc lengths usng dual soluton of LP relaaton 6 Crew member -9 6 Slde 8 Slde Prcng problem Crew member Reduced cost = Add to problem. Crew member Reduced cost = Add to problem. Slde 9-9 Soluton of shortest path problems 6 6 After and are added to the problem, no remanng varable has negatve reduced cost. CP-based enders Decomposton Slde enders Decomposton n the Abstract Classcal enders Decomposton Eample: Machne Schedulng