Outline. Integer LP Formulation. Time Windows. Resource Constrained Shortest Paths. s.t. Σ. x ji - x ij = -1, i=s j : (i,j) A. 0, i s,t 1, i=t.

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1 Reource Conrained Shore Pah Naahia Boland Irina Dumirecu Mahemaic and Saiic, Univeriy of Melbourne & HEC Monreal, Canada Reearch Chair in Diribuion Managemen Ouline Variou conrained hore pah model Single addiive reource conrained hore pah - Problem complexiy - Preproceing - A label eing algorihm - Lagrangian relaxaion - Uing Lagrangian informaion wihin label eing Elemenary conrained hore pah Time Window A pah in a nework G=(V,A) may repreen a roue for a delivery vehicle In hi cae, he ime ha he vehicle make each delivery may be conrained T ij = ime needed for vehicle o drive from node i o node j (may include unloading a i) [a i,b i ] = ime inerval mu arrive a node i in i = ime arrive a node i If (i,j) in pah, hen j = max{ i + T ij, a i } Require i b i for all i V Ineger LP Formulaion min x {0,} A,.. j : (j,i) A (i,j) A c ij x ij x ji - x ij = -, i= j : (i,j) A 0, i,, i= j i +T ij (b i +T ij -a j )(-x ij ), (i,j) A a i i b i, i V i V

2 Addiive Arc Reource A pah in a nework G=(V,A) may repreen a roue for a delivery vehicle The perol conumed by he vehicle may be conrained by i ank capaciy, Q D ij = perol conumed on arc (i,j) d i = oal perol conumed viiing all cuomer on he roue up o and including i If (i,j) in pah, hen d j = d i + D ij Require d i Q for all i V (or ju d Q) Special cae of ime window [0,Q] min x {0,} A.. j : (j,i) A ILP Formulaion (i,j) A c ij x ij x ji - x ij = -, i= j : (i,j) A 0, i,, i= (i,j) A D ij x ij Q i V Addiive Node Reource A pah in a nework G=(V,A) may repreen a roue for a delivery vehicle The quaniy ha can be carried on he vehicle may be conrained by i capaciy, Q D i = demand quaniy for node i d i = oal demand for all cuomer preceding on he roue, including i If (i,j) in pah, hen d j = d i + D j Require d i Q for all i V (or ju d Q) Special cae of addiive arc reource min x {0,} A.. j : (j,i) A ILP Formulaion (i,j) A c ij x ij x ji - x ij = -, i= j : (i,j) A 0, i,, i= (i,j) A D i x ij Q - D i V

3 Vecor Reource Arc (i,j) may ue r k ij uni of reource k Reource limi R k for each reource k = cumulaive uni of reource k ued on pah node from o i q k i If (i,j) in pah, hen q k j = qk i + rk ij Require q k Rk for all k (Alernaively, ome reource may be ime window ype, no ju addiive) min x {0,} A.. j : (j,i) A ILP Formulaion (i,j) A c ij x ij x ji - x ij = -, i= j : (i,j) A 0, i,, i= (i,j) A r k ij x ij Rk, reource k i V Renewable Reource A pah may repreen a equence of work for a crew in a job nework, perhap including everal duy period wih re in beween The reource can be replenihed a ome pecial node, R V If (i,j) in pah and j R, hen d j = d i + D j So if j R hen free o re-e d j = 0 Require d i Q for all i V Elemenary Pah Shore pah problem ofen arie a ubproblem in column generaion In hee cae here may be negaive lengh cycle a hore elemenary pah (no repeaed node) i ough Thi can be modelled by uing a vecor of addiive reource, one for each node i Reource r k i= if k=i, 0 oherwie i,k V Reource limi R k = for each k V

4 Oher Variaion Muli-crieria opimizaion Dynamic hore pah e.g. wih imedependen ravel ime, uch a due o peak-hour congeion Sube dijoin pah Many oher Single Reource The Reource Conrained Shore Pah Problem (RCSPP) Direced graph G=(V,A) Sar node, end node Co c a for arc a A Reource conumpion r a for a A Reource limi R Find he lea co pah from o conuming oal reource no more han R RCSPP: Noaion min c(p).. p a pah in G from o r(p) R c(p) = um of co of arc in p r(p) = um of reource conumed on arc in p Complexiy: RCSP and SPP The RCSPP can be ranformed ino an SPP on a new (acyclic!) graph Aumpion: he reource conumpion aociaed wih every arc i poiive The SPP i defined on he graph G =(V,A ) obained from he graph G=(V,A) Noaion for G: V={,,i 0,i,,i n } How do we obain G?

5 Reource conumpion : (Copy of he e of node) R- R- R Reource conumpion: (Copy of he e of node) R- R- R i 0 i 0 i 0 i 0 i 0 R- i 0 R- i 0 R i i i i i R- i R- i R i i i i i R- i R i R j j j j j R- j R- j R k k k k k R- k R- k R i n i n i n i n i n R- R- i R- n i R n R- R G Node, r k,c k k R ,,5 j i j r j,c j r ij,c ij 0 R- R G Node and arc RCSPP: Negaive Co Cycle Reource conumpion: (Copy of he e of node) Pah wih cycle: R- R- R i j, r i,c i i j i j i R- j R- R i,-,- j, i R j R- R- i R j R R G acyclic! RCSPP: Negaive Co Cycle Pah wih cycle: i j i j, r i,c i i j i 0,-,- j j 0, i i 5 i 6 i j 5 j 6 j (,i,j,): (,) (,i,j,i,j,): (5,-) (,i,j,i,j,i,j,): (7,-7)

6 RCSPP Complexiy Every reource-feaible pah in G from o (poibly no imple) correpond o a pah in G from o, of he ame co Every pah in G from o correpond o a reource-feaible pah in G from o (poibly no imple), of he ame co RCSPP in G equivalen o SPP in (acyclic) G SPP oluion ime polynomial in V =O( V R) Thu RCSPP i peudopolynomially olvable Previou approache Exac - kh-hore pah mehod: Eppein (997) - Dynamic programming/node labelling: Aneja e al (98), Deroier e al (98), Derocher & Soumi (988), Jaumard e al (996) - Lagrangian relaxaion: Handler & Zang (980), Bealey & Chriofide (989), Mehlhorn & Ziegelmann (000), Carlyle & Wood (00) Approximae - co caling: Hanen (979), Warburon (987), Hain (99), Lorenz & Raz (00) Algorihm Preproceing Preproceing no pah connecing and infeaible Reduced nework, U Lagrangian dual cheape pah from o i feaible opimal oluion Dual, parial pah, beer U pah of minimal reource conumpion from o i infeaible Label eing infeaible

7 Preproceing: Bound U = upper bound (e.g. co of lea reource pah) Lower bound on pah (w.r.. reource conumpion and co) i q q = reource conumed along he pah of minimal conumpion from o i i c c = co of he cheape pah from o i Preproceing U = upper bound (e.g. co of lea reource pah) i r, c q, b q = lower bound on reource needed for i pah b = lower bound on co needed for i pah r = lower bound on reource needed for i pah c = lower bound on co needed for i pah Eliminae node i if eiher q + r > R or b + c U Preproceing Upper Bound Updae i q, b (r ij,c ij ) j r, c q i r ij j r Eliminae arc (i,j) if eiher q + r ij + r > R or b + c ij + c U Updae upper bound if q + r ij + r R and co of he pah (, i, j, ) obained < U

8 Upper Bound Updae i b c ij j c Updae upper bound if b + c ij + c < U and reource of he pah (, i, j, ) obained R Repea preproceing unil no change!!! Label and Dominance Label (a a node): a pair (reource, co) aociaed wih a pah from o ha node,,5 reource co Pah: (,,), i correponding label a node : (,6) (reource,co ) dominae (reource,co ) if: reource <= reource, co <= co, label are no equal. Example: - (,) dominae (,5) and (,); - no dominance beween (,) and (,). Label Seing Algorihm (LSA) Treamen of a label: exend he pah along ougoing arc; check dominance; add he new label (if feaible and non-dominaed) p LSA (Derocher & Soumi, 998): Sar wih label (0,0) a ; i (r i,c i ) (reource(p), co(p)) (r ij,c ij ) (r i +r ij,c i +c ij ) The label no already viied are reaed in increaing order of heir reource conumpion; j (0,0),, LSA - Example,5,8,, 6, 5 reource co = =5 R=6

9 LSA - Example LSA - Example (0,0), (,),,5,8, = =5 R=6 (0,0), (,),,5,8, = =5 R=6, 6, 5 (,) reource co, 6, 5 (,) reource co LSA - Example LSA - Example (0,0), (,),,5,8 (,6), = =5 R=6 (0,0), (,),,5,8 (,6), = =5 R=6, 6, 5 (,) (,9) (,) reource co, 6, 5 (,) (,9) (,) reource co (7,)

10 LSA - Example LSA - Example (0,0), (,),,5,8 (,6), = =5 R=6 (0,0), (,),,5,8 (,6), = =5 R=6, 6, 5 (,) (,9) (,) reource co (7,), 6, 5 (,) (,9) (,) reource co (7,) (,8) LSA - Example Noe on Algorihm Ue of bucke yield O( V R ) algorihm (0,0),, (,),,5,8 (,6) 6, 5 (,) (,9) (,) reource co (7,), (,8) Opimal oluion = =5 R=6 Eaily exended o muliple reource: rea label in lexicographic reource vecor order Label correcing: rea all label on a node in one ep See Handbook in OR & MS Vol. 8, Chaper, Time- Conrained Rouing and Scheduling, Deroier, Duma, Solomon, Soumi Alo Mehlhorn & Ziegelmann for reamen in increaing co order, Carlyle & Wood for enumeraion

11 Modified LSA (MLSA) Exending a pah U = upper bound (r i +r ij,c i +c ij )? U = upper bound (r i,c i ) i (r ij,c ij ) j r, c (r i,c i ) i (r ij,c ij ) (r i +r ij,c i +c ij )? j r, c r = lower bound on reource needed for j pah c = lower bound on co needed for j pah Exend (r i,c i ) along arc (i,j) only if r i + r ij + r R and c i + c ij + c < U r = lea-reource j pah c = lea-co j pah Exend (r i,c i ) along arc (i,j) only if r i + r ij + r R and c i + c ij + c < U Exending a pah U = upper bound (r i,c i ) (r i +r ij,c i +c ij )? i j (r r, c ij,c ij ) r = lower bound on reource needed for j pah c = lower bound on co needed for j pah Can we do beer? Lagrangian Relaxaion (LR) z = z LR (λ) = min cx.. x in X Ax b hard conrain Lagrangian relaxaion: lif he hard conrain ino he objecive funcion; obain a problem ha i eaier o olve. min cx + λ(ax b).. x in X λ 0 Imporan propery: z LR (λ) z for any λ 0. Lagrangian dual problem = finding he be lower bound: LD: max z LR (λ), λ 0

12 LR for RCSPP P ij = e of pah from i o j min c(p) z RCSPP =.. P in P r(p) R Lagrangian relaxaion for λ 0: z LR (λ) = min c(p) + λ(r(p) R).. P in P = λr + min (c+λr)(p).. P in P Calculaing z LR (λ) mean calculaing a hore pah problem wih arc lengh given by c+λr, oluion P λ Lagrangian lower bound µ j Reource conumed o arrive o j Reource ill available: R- µ min c(p) z j (µ) =.. p in P j r(p) R - µ Lower bound obained for λ 0: min c(p) + λ(r(p) (R - µ)) z j LR (µ, λ) =.. p in P j = - λ(r - µ) + min c(p) + λr(p).. p in P j Parial RCSPP problem Finding hem Uing lengh c a + λr a for each a A, find he hore pah in G from j o, for all node j. z LR (λ) KCP Thi yield z j LR (µ, λ) for all j and any µ. Solve Langrangian dual of RCSPP for pah from o uing Kelley cuing plane mehod.

13 KCP KCP KCP KCP z LD

14 Kelley Cuing Plane Mehod Ued o olve he Lagrangian dual: keep wo acive value of λ: λ A and λ B uch ha r(p λ A ) R > 0 and r(p λa ) R < 0. define wo linear funcion: f A (λ)=(r(p λ A R)λ + c(p λa ) f B (λ)=(r(p λ B R)λ + c(p λb ) find λ a he inerecion of he wo line (i.e. f A (λ )=f B (λ )) STOP if r(p λ ) = R or z LR (λ ) = f A (λ ). if r(p λ ) < R hen λ B = λ, oherwie λ A = λ. Iniial λ value: λ A = 0, λ B = U c(cheape pah ->) = =6 R=0 Lagrangian - Example 6, 6,0 7, 7, reource co,0 5,, 5,, Min reource pah (,,6):, Min co pah (,,,6): 5,9 7,5 6 f A (λ)=9 + λ(5-0) = 9+5λ f B (λ)= + λ(-0) = -7λ f A (λ) = f B (λ) λ = / Find min c+ (/)r pah = =6 R=0 Lagrangian - Example 5 / 5 / / / 8 5 / / / Min reource pah (,,6):, Min co pah (,,,6): 5,9 7 / 6 f A (λ)=9 + λ(5-0) = 9+5λ f B (λ)= + λ(-0) = -7λ f A (λ) = f B (λ) λ = / f A (λ) = 0 / Find min c+ (/)r pah (,,,,6) of reduced co 7 z LR (λ) = 7 0 (/) = 0 / z LR (λ) < f A (λ) repea = =6 R=0 Lagrangian - Example 6, 6,0 7, 7,,0 5,, 5,, 7,5 z LD = 0 /8 RCSPP opimal pah i (,,,5,6), co Difference i dualiy gap Min red. co pah (,,,,6):,0 6 f A (λ)=0 + λ(-0) = 0+λ f B (λ)= + λ(-0) = -7λ f A (λ) = f B (λ) λ = /8 f A (λ) = 0 /8 Find min c+ (/8)r pah (,,,,6) of reduced co 7 7/8 z LR (λ) = 7 7/8 0 (/8) = 0 /8 z LR (λ) = f A (λ) finihed!

15 Finding hem Uing lengh c a + λr a for each a A, find he hore pah in G from j o for all node j. Thi yield z j LR (µ, λ) for all j and any µ. Solve Langrangian dual for pah from o uing Kelley cuing plane mehod. Save λ,, λ K generaed and lengh of hore pah from j o for all j, a each ieraion, L j,,lj K. z j (µ, LR λ h ) = - λ h (R- µ) + Lj h i a lower bound on he lea co j pah uing reource no more han R- µ. Exending a pah U = upper bound (r i,c i ) (r i +r ij, c i +c ij ) i j (r ij,c ij ) r, c r = lower bound on reource needed for j pah c = max z j LR (R-r i -r ij, λ h ) h=,,k Exend (r i,c i ) along arc (i,j) only if r i + r ij + r R and c ij + c ij + c < U? Te Problem Preproceing reul Cla : randomly generaed (Cherkaky e al.) Cla : grid nework wih random co, reource - ype and problem Cla 5: US road nework Cla 6: nework from digial elevaion model

16 Exended Pruning Wih Lagrangian Exra Upper Bound Updae (r i,c i ) i j r(sp j h ) r ij, c ij µ hore pah from j o, lengh: c+λ h r Exra Upper Bound Updae Check for all h (i.e. for λ,,λ K ), µ=r - (r i + r ij ). Be parial upper bound: c(µ) = min { c (SP j h ): r(spj h ) µ } h=,,k reource already conumed if r i +r ij +r(sp j h ) R hen c i +c ij +c(spj h ) i an upper bound Inide MLSA updae upper bound: U=min{U, c i + c ij + c(r - (r i + r ij ))}

17 Concluion Complee preproceing i very effecive Uing lower bound on pah compleion i very effecive Lagrangian lower bound can be efficienly calculaed Aggreively updaing upper bound i be A new approach Eliminae node Updae Lagrange muliplier Reduced nework, dual, pah,u Eliminae edge Updae Lagrange muliplier for an edge Reduced nework, dual, pah,beer U Enumeraion (cf Carlyle & Wood, 00) Lagrangian Node Eliminaion min c(p).. p a pah in G from o p pae hrough i r(p) R X i Iue in KCP Eliminaing node can change Φ (Can only move up) May need o recalculae SPT z i LR (λ) = min { (c+λ r)(p) - λr : p X i } Can ge lower bound for individual node by combining forward and revere hore pah ree Can eliminae node ha violae upper bound Eliminae node a each value λ,, λ K

18 Algorihm Eliminae node Updae Lagrange muliplier Reduced nework, dual, pah,u Eliminae edge Updae Lagrange muliplier for an edge Reduced nework, dual, pah,beer U Enumeraion Node Eliminaion Node can be grouped ino equivalence clae All node in he ame equivalence cla have ame lower bound Equivalence clae have a ree-like rucure wih hore pah from o a roo Node Equivalence Node Equivalence

19 Node Equivalence Theorem If node i and j are in he ame equivalence cla hen z i LR (λ) = zj LR (λ) Cla, ype Te Problem So if can eliminae one node in equivalence cla hen can eliminae hem all Equivalence clae hemelve form a ree i in cla higher in ree han cla of j mean z i LR (λ) zj LR (λ) So if can eliminae (cla of) j hen ame for i Performance: Small Nework Algorihm Eliminae node Updae Lagrange muliplier Reduced nework, dual, pah,u Eliminae edge Updae Lagrange muliplier for an edge Reduced nework, dual, pah,beer U Enumeraion

20 Aggreive Edge Eliminaion Performance: Large Nework Eliminae edge inead of node Opimize lower bound for every edge Updae edge concurrenly Search for upper bound a every ep Algorihm Eliminae node Updae Lagrange muliplier Reduced nework, dual, pah,u Eliminae edge Updae Lagrange muliplier for an edge Reduced nework, dual, pah,beer U Enumeraion Node Then Edge Eliminaing node allow u o find opimal Lagrange muliplier fa Eliminaing edge allow u o reduce inracable cae Enumeraion effecive a cloing gap and proving opimaliy Combine all hree ogeher

21 Performance: Large Nework Concluion Preproceing effecive Reduce ime o find opimal Lagrange muliplier Reduce number of inracable cae Finding opimal Lagrange muliplier dominae ime in average cae Enumeraion dominae in inracable cae Node/arc equivalence clae = peed-up? ERCSPP The Elemenary Reource Conrained Shore Pah Problem (ERCSPP) Direced graph G=(V,A) Sar node, end node Co c a (± ineger) for arc a A Reource conumpion r a (+ ineger) for a A Reource limi R The graph may have negaive co cycle Find he lea co elemenary pah from o conuming oal reource no more han R Previou approache -cycle eliminaion - Houck, Picard, Queyranne, Vemugani (980) - Chriofide, Mingozzi, Toh (98) RCSPP wih node reource - Idea decribed by Bealey & Chriofide (989) - Implemened by Gueguen, Dejax, Dror, Feille, Gendreau (000) wih rong dominance - Sae-pace relaxaion decribed by Kohl (995)

22 The ERCSPP min c(p).. p a pah in G from o r(p) R m i (p) for all node i V Mulipliciy of node i: m i (p) = number of ime i appear in pah p = number of arc leaving i in p A (ae-pace) relaxaion ERCSPP(S): min c(p).. p a pah in G from o r(p) R m i (p) for all node i S S V Node reource Each node in S a reource wih limi S = {,,7} (,5,0) (,,00) (,) 7 (-9,5) (,0,0) (6,) (0,0,000) 5 (,7) (0,,00) If Srong dominance + r(min reource pah) + r(min reource pah) > R (,) (0,0,000) S = {,,7} (,,00) 7 (6,) 5 (-9,5) (0,,00) (,0,) (,7) (,0,0)

23 The general ae-pace augmening algorihm S = olve ERCSPP(S) uing LSA le p* be opimal pah while (p* no elemenary) { augmen S olve ERCSPP(S) } Augmening S HMO: add ome mo repeaed node on p* HMO-All: add all mo repeaed node on p* MO-All: add all repeaed node on p* M-All: All: add all node repeaed on ome efficien pah S = V iniially V Problem Group Characeriic A Memory uage % neg. co arc # Label # Label Raio o HMO HMO HMO-All MO-All M-All All (6) (87) () () 66 ().0 (5).69 () 5.00 > 0 min ().7 (). ().5 (99) ().86 ().5 (6) ().66 ().9 (0) ().55 ().80 - Problem Group % neg. V A co arc CPU ime Time () Time Raio o HMO HMO HMO-All MO-All M-All All (6) (87) () ().05 () 8.75 (5). () 7. > 0 min ().9 ().8 ().9 (99) ().8 () 5.79 (6) ().98 () 6.6 (0) ().7 ().65 -

24 Bound Lower bound: value of ERCSPP(S) Upper bound: check all label on node afer olving ERCSPP(S) for feaible pah HMO: ypical profile S Time Max # Label Lower Bound Upper Bound The ERCSPP min c(p).. p a pah in G from o r(p) R m i (p) for all node i V Mulipliciy of node i: m i (p) = number of ime i appear in pah p = number of arc leaving i in p Sae-pace + Lagrangian relaxaion min c(p) + u i (m i (p) ) i V.. p a pah in G from o r(p) R m i (p) for all node i S Give a lower bound for S V, u 0 Can olve uing co c ij + u i on all arc (i,j) A

25 One approach S =, u = 0, iniialize UB olve ERCSPP(S,u), fix LB le p* be opimal pah while (p* no elemenary) { while (p* no elemenary) { updae u (coordinaewie) olve ERCSPP(S,u) updae LB, UB } augmen S, u = 0 olve ERCSPP(S,u) updae LB, UB } COULD BE VERY EFFICIENT Updaing u (0, -) (6,6) (,) (,-7) =7 Find malle co adjumen o ha la label on mo repeaed node i* from opimal pah would be dominaed by ome oher label Increae u i* by Which node? For HMO augmening of S, ofen many mo repeaed node: - Cloe o? - Cloe o? - Lowe index? For increaing u, ame ory! HMO Wih LR HMO wihou LR S Time () Label LB UB Hime () Time () Label LB UB

26 HMO wih LR HMO wihou LR S Time () Label LB UB Hime (Time () Label LB UB Concluion RCSPP - Uing lower bound on pah compleion i very effecive - Lagrangian lower bound can be efficienly calculaed - Aggreively updaing upper bound i be ERCSPP - Sae-pace relaxaion i much more efficien - Conervaive augmenaion i be - Lagrangian relaxaion look promiing Furher work Efficien re-opimizaion Alernaive Lagrangian opping condiion hould be eed Alernaive node elecion raegie hould be eed Alernaive Lagrangian approache hould be explored Branch-and-bound High co arc Low co arc

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