The Multi-shift Vehicle Routing Problem with Overtime

Transcription

1 The Mul-shf Vehcle Roung Problem wh Overme Yngao Ren, Maged Dessouy, and Fernando Ordóñez Danel J. Epsen Deparmen of Indusral and Sysems Engneerng Unversy of Souhern Calforna 3715 McClnoc Ave, Los Angeles, CA January 2010 Absrac: In hs paper, we sudy a new varan of he vehcle roung problem (VRP) wh me wndows, mul-shf, and overme. In hs problem, a lmed flee of vehcles s used repeaedly o serve demand over a plannng horzon of several days. The vehcles usually ae long rps and here are sgnfcan demands near shf changes. The problem s nspred by a roung problem n healhcare, where he vehcles connuously operae n shfs, and overme s allowed. We sudy wheher he radeoff beween overme and oher operaonal coss such as ravel cos, regular drver usage, and cos of unme demands can lead o a more effcen soluon. We develop a shf dependen (SD) heursc ha aes overme no accoun when consrucng roues. We show ha he SD algorhm has sgnfcan savngs n oal cos as well as he number of vehcles over consrucng he roues ndependenly n each shf, n parcular when demands are clusered or non-unform. Lower bounds are obaned by solvng he LP relaxaon of he MIP model wh specalzed cus. The soluon of he SD algorhm on he es problems s whn mes he opmal soluon dependng on he me wndow wdh, wh he smaller me wndows provdng he gher bounds. Keywords: Vehcle roung problem wh me wndows (VRPTW), mul-shf, overme, nseron heursc, abu search 1

2 1. Inroducon The basc Vehcle Roung Problem (VRP) s concerned wh fndng a se of roues o serve a gven number of cusomers, mnmzng he oal dsance raveled. In hs problem, he oal demand of all he cusomers on a roue mus no exceed he vehcle capacy. If n addon each cusomer specfes a me wndow whn whch he cusomer mus be vsed, he problem s nown as a VRP wh me wndows (VRPTW). There are many oher varans of he VRP. Classcal varans of he VRP am a plannng he roues of a flee of vehcles for a sngle plannng perod (a day or a shf). The above VRP s however unrepresenave of some suaons, n whch companes have o roue vehcles o sasfy demand for connuous operaons, n many cases roung and schedulng n a 24 hour perod dvded no shfs. In such cases, a soluon ha forces all vehcles o reurn o he depo before he end of each shf can perform subopmally, n parcular n suaons wh hgh demand near shf changes or wh long dsances. If he demands were he same n each shf we can smply fnd an opmal soluon (or a good soluon) and repea he same roues. These repeaed roues could sll be mproved wh overme f before reurnng o he depo a vehcle urns ou o be close o a demand of he nex shf. In pracce, however, he demand flucuaes causng repeaed roues o be neffcen. The objecve of hs paper s o nroduce a VRP model ha plans over several perods and allows roues o exceed shf lenghs a a ceran overme cos, f ha decson s economcal. Ths wor hen nvesgaes he mpac of he radeoff beween overme and oher operaonal coss (e.g. ravel cos, regular drver usage, and unme demands) on he effcency of he roung soluons. For example, f a cusomer n he nex shf happens o be on he reurn rp o he depo for a vehcle of he curren shf, he vehcle can serve ncurrng a small overme. Ths would reduce demands n he nex shf, possbly resulng n less oal ravel dsance and fewer drvers. We consder a mul-shf VRP wh overme. The problem s nspred by a roung problem n a leadng healhcare organzaon ha operaes abou 240 medcal offce buldngs n he Souhern Calforna regon. The healhcare provder connuously roues medcal samples, mals, x-rays, lab-specmens, documens, ec. beween varous medcal facles and a cenral lab for esng. 2

3 The medcal facles are locaed hroughou Souhern Calforna, causng ravel mes beween facles o be on he order of he enre shf lengh n he wors case. The organzaon has abou 70 vehcles o carry ou he delveres. Because of he random naure of cusomer demands n he healhcare ndusry, demands occur any me durng a day, even durng shf changes. In addon, mos medcal samples are pershable and mus be processed whn a ceran perod, so he demands have me wndows. There are no capacy consrans because he ems delvered are lgh and small. To serve hese demands, vehcles n hs problem ravel hrough mulple urban areas and are operaed on a 24/7 bass n hree consecuve shfs n each day. The cenral lab ncludes he vehcle depo where all roues n a shf sar and end. Overme s allowed a a hgher salary rae. Thrd pary vehcles, such as axs, are hred o serve he unme demand. There s lmed research on VRP ha plans mulple rps or over mulple perods. One VRP varan ha consders dependency beween roues s mul-rp VRP (MVRP) (Tallard e al., 1996; Brandão and Mercer, 1997, 1998; Pech and Salh, 2003; Campbell and Savelsbergh, 2004; Az e al., 2006; and Salh and Pech, 2007). In he MVRP, vehcles can be assgned o more han one roue whn a me perod. The MVRP s dfferen from our problem because he mulple rps sll occur n a sngle shf, and overme s consdered only for he las rp of a vehcle. Anoher VRP varan ha consders dependency beween perods s Perodc VRP (PVRP) (Angelell and Speranza, 2002; Francs and Smlowz, 2006; Hemmelmayr e al., 2009; and Alonso e al., 2008). The PVRP exends he classcal VRP o a plannng horzon of several days. Each cusomer requres a ceran number of vss whn hs me horzon whle here s some flexbly on he exac days of he vss. Hence, one has o choose he vs days for each cusomer and o solve a VRP for each day. In addon o decdng when a demand s servced, he PVRP s dfferen from our problem because he operaons consder only one regular shf n a perod (worday). In addon, overme s no consdered n he PVRP, and demands do no have me wndows. I s common n he producon plannng and schedulng leraure o use overme as an opon for shf schedulng (Lagodmos and Mhos, 2006; Merzfonluoğlu e al., 2007). For nsance, Lagodmos and Mhos (2006) show ha effecve use of overme leads o worforce reducons and mproved ulzaon n pacng shops. However, only a few pror wor has 3

4 consdered overme as an opon for vehcle roung and schedulng. In he 1970s, rans sysems generally used models o esmae he coss of bus sysems. In hese un cos models, he esmaed cos of a proposed meable for a rans sysem was smply he sum of wo cos facors: he number of buses and he oal mleage. These cos models for analyzng bus sysems were exended n Bodn e al. (1978, 1981) o nclude facors such as he number of bus drvers needed, overme, manenance coss, ec. Sneze and Bodn (2002) argue ha only consderng oal ravel me n he objecve funcon s no enough n evaluang VRP soluons, especally for non-homogeneous flees. Insead, hey deermne a Measure of Goodness, whch s a weghed lnear combnaon of many facors such as capal cos of a vehcle, salary cos of he drver, overme cos, mleage cos, and he ppng cos of a sanaon vehcle a he dsposal facly, o compare he varous soluons generaed. They use he cos models o generae and evaluae soluons o he Capacaed Arc Roung Problem wh Vehcle-Se Dependences (CARP-VSD). These cos models allow for he use of overme n generang roues and analyzng soluons. These models confrm a long-erm conjecure of he auhors ha usng overme can help o generae less expensve soluons o vehcle roung problems because of he savngs n capal cos of he vehcles. Recenly, Zäpfel and Bögla (2008) sudy a mul-perod vehcle roung and crew schedulng problem wh overme and ousourcng opons. The problem s dfferen from our problem because he operaon n her problem s no connuous. There are wo brea perods n each worday; second, he demands mus be served n he same perod as her occurrence; and hrd, he overme consran s mposed on a whole wee bass, no on each ndvdual shf. The res of hs paper s organzed as follows. In Secon 2 we presen a mxed neger programmng (MIP) formulaon of he problem. Secon 3 descrbes wo nseron heursc algorhms o solve he problem. Secon 4 descrbes an approach o oban a lower bound of he problem by solvng he relaxaon of he MIP model. Secon 5 presens he expermenal resuls, whch nclude comparson of he performance of he wo algorhms, and comparson of he bes soluon wh he lower bound. We conclude he paper and dscuss fuure research n Secon 6. 4

5 2 Problem Formulaon Assume we now n advance he demand for a plannng horzon of P days. We consder hree daly shfs of lengh L hours (e.g., he hree shfs are 0:00am-8:00am, 8:00am-16:00pm, and 16:00pm-24:00pm f L=8). Le Τ denoe he se of shfs, wh T = 3P and T={1, 2,, 3P}. For shf, T, he shf sar me s B = L( 1), and he shf end me s E = L. The se of demand nodes s denoed as D. A hard servce me wndow [ e, l ] s also assocaed wh each demand node D, where e and l represen he earles and laes servce sar mes, respecvely. Le n denoe he oal number of demand nodes wh n= D. We creae T +1 copes of he depo represened by nodes n+1, n+,, n+ T +1. Node n+1 represens he orgn depo of shf 1, node n+ represens he orgn depo of shf as well as he desnaon depo of shf -1, {2,,T}, and node n+ T +1 represens he desnaon depo of shf T. The problem can be defned on a dreced graph G=(V, E), where V = DU {n+1, n+,, n+ T +1}. Each arc (, j) A has an assocaed ravel me j, and he ravel cos s W per hour. Noe ha arcs (n+, n++1) are ncluded n he graph o model suaons n whch a vehcle s no used durng a shf. Le K denoe he se of vehcles. A vehcle can be reused by anoher drver n he nex shf afer reurns o he depo. A drver s ready o wor a he sar me of each shf. The worng me of a drver s from he shf sar me unl he/she reurns o he depo. A he end of each roue n a shf, all vehcles should reurn o he home depo. Because here are a lmed number of vehcles, he nex drver assgned o a vehcle has o wa unl he vehcle reurns o sar s roue. For example, f a vehcle from shf -1 reurns a B + 2, s earles sar me for he nex shf wll be B + 2. For a flee of 5 vehcles, f L=8 and he end mes of he roues for shf 1 are [5, 8, 9, 10, 6], hen he earles sar mes of he vehcles for shf 2 wll be [8, 8, 9, 10, 8]. We assume ha he overme lm s L hours. Tha s, he worng me of a drver canno be longer han L+ L hours. To deal wh he overme lm, we mpose a me wndow [ 1 E 1, E + L] on he depo node n+, {2,,T+1}. For node n+1, he me wndow s [0, 0]. I 5

6 s reasonable o assume L< L, snce overme canno be oo long. In hs case, he me wndows of node n+ and n++1 do no overlap. The regular salary rae s R per hour, and he overme salary rae s S per hour. In some suaons he use of overme s nevable. For example, f a demand occurs lae n shf -1 and e 0 > E 1, or a demand occurs early n a shf and + B + 0 > l, hen a vehcle n shf -1 has o use overme o mee such demands. Once a vehcle n a shf s used, s assumed ha he drver s pad for he enre shf even f he vehcle reurns o he depo early. The objecve s o deermne a se of roues and her schedulng o sasfy all demands and assocaed me wndows wh mnmum oal coss for he plannng horzon. Toal coss nclude ravel cos (he produc of oal ravel me and W), overme cos (he produc of oal overme and S), cos of drvers for regular me, and cos of unme demands. We assume ha each unme demand wll be served by a ax a a ransporaon cos of A per hour. The noaon s summarzed below: (1) Ses and problem sze parameers P : Number of days n he plannng horzon. T: The se of shfs n he plannng horzon, T={1, 2,, 3P}, and T = 3P. D: The se of demand nodes. V: DU {n+1, n+,, n+ T +1}. n: Toal number of demand nodes n he plannng horzon, n= D. K: The se of vehcles, defned a pror or deermned by he model. (2) Tme parameers j : Travel me from node o node j. L: Shf lengh (e.g. 8 hours). L : Overme lm for each shf mposed by company polcy. B : Begn me of shf, = L( 1), T. E : End me of shf, B E = L, T. 6

7 [ e, l ]: The servce me wndow of node, V. (3) Cos parameers W : Travelng cos per hour (e.g., gas). R : Regular salary rae for drvers. F: Drver cos of usng a vehcle n a shf, F=LR. S: Overme salary rae for drvers. A : Cos per hour for usng a ax o serve each unme demand. (4) Decson varables: x j =1 f vehcle ravels from node o node j, and 0 oherwse; y =1 f vehcle s used n shf, and 0 oherwse; u =1 f demand s served by ax, and 0 oherwse; w = he me a whch node s vsed by vehcle, w 0. The problem can be formulaed as he followng mxed neger program (MIP): Mnmze W j x + j S ( wn 1 E ) + F y + A + + u ( ) Subjec o j K (, j) E K (, j) E K T \{ T+ 1} K T x + u = 1, D (1) x 1, T, K (2) n+, = ( n+, ) E (, n+ ) E x, n+ = 1, {2,..., T + 1}, K (3) x ( j, ) E j x j (, j) E = 0, V \ { n+ 1, n+ T + 1}, K (4) D w + w + M (1 x ), V, j V, K (5) j j j e E w l, D, K (6) wn+ + 1 E + L, K, T (7) 7

8 y 1 xn+, n+ + 1 =, K, T (8) In he objecve funcon, he four cos erms are ravel, overme, regular drver usage, and ax respecvely. Consrans 1 ensure ha each demand s served exacly once, eher by nernal vehcles or by a ax. Consrans 2 and 3 ensure ha every vehcle wll vs all he depo nodes n+, { 1,..., T + 1} n he plannng horzon. Depo node n+ mus be vsed before n++1 because her me wndows are non-overlappng. The depo nodes dvde demands vsed by a vehcle no dfferen shfs. All demands vsed beween node n+ and n++1 are served n shf. If no demand s vsed beween hem for a vehcle, hen hs vehcle s no used and he roue for shf s empy. These consrans also ensure ha every roue n a shf sars from he orgn depo and ends a he desnaon depo. Consrans 4 ensure ha for all nodes excep n+1 and n+ T +1, he nflow equals o he ouflow. Consrans 5 force conssences of me varables, whch are also subour elmnaon consrans. M s an upper bound on l +, V, (, j) E. Consrans 6 are me wndow consrans for he demand nodes, and consrans 7 are me wndows consrans for he depo nodes. Consrans 8 calculae wheher vehcle s used n j shf or no. If xn, n =1, hen no demand s served by vehcle n shf. Therefore, he vehcle s no used and y =0. Oherwse, a leas one cusomer s served by vehcle, so y =1. 3. Heursc Algorhms In hs secon, we frs revew recen relevan research on heurscs for vehcle roung problems. We hen presen he buldng blocs of our heurscs, followed by wo heursc algorhms, SI and SD, for solvng our problem, and an mprovemen phase. In he soluon algorhm, we use an nseron-based heursc o generae he nal soluon and hen use a abu search algorhm o mprove. Inseron heurscs are effecve soluon mehods for VRP. Recen wors nclude Dana and Dessouy (2004), who presen a parallel regre nseron heursc o solve a large-scale dal-a-rde problem wh me wndows. The compuaonal resuls show he effecveness of hs approach n erms of radng-off soluon qualy and compuaonal mes. Lu and Dessouy (2006) presen an nseron-based consrucon heursc o solve he mul-vehcle pcup and delvery problem wh me wndows. 8

9 Ths heursc consders no only he classcal ncremenal dsance measure n he nseron evaluaon crera bu also he cos of reducng he me wndow slac due o he nseron. Tabu search has also been appled o major varans of VRP, e.g. VRP wh sof me wndows (Tallard e al. 1997), VRP wh spl delvery (Arche e al. 2006), and he pc-up and delvery problem (Banchess and Rghn 2006). I s shown ha abu search generally yelds very good resuls on a se of benchmar problems and some larger nsances (Gendreau e al. 2002). 3.1 The Buldng Blocs The Inseron Roune Frs we descrbe he Inseron Roune (Algorhm 1 n he Appendx), whch greedly nsers he cusomer wh he cheapes cos n he roues. There are four assocaed coss ( C1- C 4 ) n he algorhm. Smlar o Lu and Dessouy (2006), we consder he cos of reducng he me wndow slac due o he nseron, whch s denoed as C 1. The coss C 2, C 3, and C4 are respecvely he overme cos, he regular drver usage cos, and he cos of unme demands. The nseron algorhm res o nser as many cusomers as possble. The nseron cos of a new cusomer s he oal weghed ncrease of C 1, C 2, and C 3, whle C 4 s calculaed afer he algorhm s fnshed. In erms of he weghs of hese coss, we assume C 4 s he hghes; ha s, we ry o serve as many demands as possble wh he nernal vehcles. The weghs of C 1, C 2 and C 3 are W, S, and F respecvely. Whenever a new vehcle s used, a drver cos of F=LR s ncurred. Even f he vehcle leaves he depo afer he shf sar me, he salary s always F because he drver s avalable a he shf sar me. If he worng me s longer han L hours, here wll be an overme cos. Afer he roues are creaed, we calculae he cos of unme demand and add o he oal cos. We assume ha each unme demand needs a separae ax, so he cos for an unme demand s A + ). The oal unme demand cos s he sum of hose coss. ( 0 0 The nseron algorhm s a parallel nseron, n whch all roues are consruced smulaneously wh nal seeds used o begn he roue. If a par of nodes and j canno be served by he same vehcle,.e., e > l + j j and j j e + > l, hen he wo nodes are called an ncompable par. For 9

10 each shf, we consruc a compably graph by ang he nodes o be served n he shf, and creang an edge beween any ncompable par. Then we solve a mum clque problem approxmaely usng a greedy algorhm on he graph (Lu and Dessouy, 2006). The seeds are he nodes ha form he clque. Le Z denoe he se of seeds n shf. Snce each par n Z s ncompable, every node mus use a dfferen vehcle. Thus he mnmum number of vehcles used o serve he nodes n he shf s Z f all nodes are served by nernal vehcles. The role of he seeds s wo fold. Frs, helps o creae balanced roues. Because of he hgh cos of a new roue, whou seeds, he nseron algorhm ends o fufll a roue and hen naes anoher when he frs s full. Thus he las roue usually has few cusomers. Second, usng seeds o gude he consrucon of he roues avods he lae nseron of ncompable demand nodes, whch can lead o hgher coss or nfeasbles Inra-shf Tabu Search Algorhm We use an nra-shf abu search algorhm (Algorhm 2 n he Appendx) o mprove he roues for each shf. Ths mplemenaon of he abu search consders he neghborhoods obaned from he sandard 2-op exchange move (Ln, 1965) and he λ -nerchange move (Osman, 1993). The algorhm evaluaes soluons based on he objecve funcon,.e. he sum of he ravel cos, overme cos, cos of unme demands and drver cos. A each eraon he abu search generaes α λ -nerchange neghbors and β 2-op neghbors of he curren soluon. The abu search a each eraon moves o he bes neghbor, even f s worse han he curren bes soluon. The prevenon of cyclng s acheved by forbddng some moves for a gven number of eraons θ. In our mplemenaon, he number of abu eraons s randomly generaed n ( θ mn, θ ). For a λ -nerchange move, feasble moves of a soluon consder ha up o 2 nodes are exchanged beween wo roues of he soluon. The abu saus s overrdden f he new soluon s beer han he bes soluon so far and he algorhm ermnaes f here s no mprovemen n eraons Iner-shf Tabu Search Algorhm For algorhm SD, apar from he nra-shf abu search algorhm (Algorhm 2), an ner-shf abu search algorhm (Algorhm 3 n he Appendx) s mplemened. There are wo ypes of 10

11 moves ha can possbly reduce he oal cos. The frs ype s movng a node from he prevous shf o a non-empy roue n he curren shf. Because of he hgh fxed roue cos, he nseron algorhm (Algorhm 1) res o nser as many as possble demands n he exsng roues. As a resul, some demands may be served wh unnecessary hgh overme cos. In such a move, f he decrease n overme s larger han he ncrease n ravel cos, hen s made. The second ype of move s reschedulng a node from s curren shf o he prevous shf. The raonale s ha ncreasng overme mgh cause a larger savng n he ravel cos or new drver cos. 3.2 Heurscs Shf Independen (SI) Algorhm The smples way o acle he mul-shf VRP wh overme s o rea each shf ndependenly. Because of he lm n flee sze, we need o consder he avalably of vehcles. Tha s, we can only sar usng a vehcle n he curren shf afer reurns o he depo from he prevous shf. Specfcally, for vehcle n shf, he earles me s avalable s ( w n +, E 1 ). In he SI algorhm (Algorhm 4 n he Appendx), we avod usng overme as much as possble. If a demand can be served n eher he curren shf or prevous shf, s served n he curren shf o avod overme. To do hs, we frs classfy all demands D no dfferen shfs D, T, as shown n Fgure 1. In parcular, for demand ha occurs near shf changes, f B + 0 l and e + 0 B L (.e., can be served n eher shf or shf -1), hen we place he demand + n D. Oherwse f B > l D. + 0, hen can only be served n shf -1. So we place n 1 For example, suppose a demand has me wndow [ B, B +2]. If 0 2 and 0 L, hen can be served n shf or shf -1, and we place n D. On he oher hand, f 2, hen has o 0 > be served n shf -1 usng overme, and we place n D 1. 11

12 D 1 D 2 D 3 D 4 B 1 E 1 E 2 E 3 E 4 Fgure 1: The classfed demand nodes n SI algorhm Each D s scheduled ndependenly only consderng he reurn me of he vehcles of he prevous shf. As menoned before, we have seeds for each shf. There are a lo of opons o mach he seeds of he curren shf wh he roues of he prevous shf. One reasonable way s o mach seeds wh he roues ha have he earles avalable mes for he nex shf. In hs way, he roues naed by he seeds have more me avalable o serve he unscheduled demands, leadng o a hgher ulzaon level of he used vehcles. Ths s mplemened by frs ranng he roues n he prevous shf n ncreasng order of w +, and hen nserng he Z seeds n he frs Z roues. Because we consder each shf ndependenly, only he nra-shf abu search algorhm (Algorhm 2) s execued. n Shf Dependen (SD) Algorhm In he SD algorhm (Algorhm 5 n he Appendx), as opposed o he SI algorhm, a demand can be served n eher he curren shf or he prevous shf, as long as s feasble. Boh nrashf abu search algorhm (Algorhm 2) and ner-shf abu search algorhm (Algorhm 3) are execued. The mehod o mach seeds s also o place hem n he roues wh he earles avalable me. I also classfes he demands no each shf, and hen schedules he demands shf by shf sequenally. However, he classfcaon mehod s dfferen. In SD, he classfcaon dfferenaes demands ha can be served n boh shfs and hose ha can only be served n one shf. If wo nodes canno be served n he same shf, hey are called non-adjacen. To denfy all non-adjacen nodes, we classfy all demands nodes no 2T-1 dsjon ses. There are T ses called sngle shf (SS) demands,.e. SS conans nodes ha can only be served n shf, T. There are T-1 ses called ner-shf (IS) demands. Tha s, IS conans nodes ha can be served eher n shf or shf +1, { 1,..., T 1}. Sarng from shf 1, for node, f 12

13 E > l, hen can only be served n shf 1, and s placed n SS 1. If E l and e e 0 E L, hen can be served eher n shf 1 or shf 2, and s placed n IS 1. If > E L and E > l, hen can only be served n shf 2, and s pu n SS 2. In hs + 1+ way, we can classfy all demand nodes no SS or IS. Fgure 2 llusraes he classfed demand ses for T=4. We can see ha demands n SS, and hen schedules demands n IS -1. D = IS 1 SS, { 2,..., T }. For shf, SD frs schedules SS 1 IS 1 SS 2 IS 2 SS 3 IS 3 SS 4 B 1 E 1 E 2 E 3 E 4 Fgure 2: The classfed demand nodes n SD algorhm The Pos Improvemen (PI) Algorhm Ths Pos Improvemen (PI) algorhm can reduce overme by re-machng he roues n adjacen shfs whou changng he ravel me and sequence of nodes n each shf. The followng s a smple example. Le r denoe he roue of vehcle n shf. Suppose for roue r 1, e 1 = B +1, and for r, = + e B In addon, assume ha w e 1 for all r. Tha s, all demands of r can be served 1 hour earler. For roue l r 1, e = B, and for l 1 l r, e = B l In addon, assume l w 1 for all l l r. Tha s, all demands of l r can be served 1 hour laer. Now we have an overme of 2 hours for he four roues. Bu f we swap he second par of he roues of vehcles and l (ncludng all roues from o T ), we can oban an overme of only 1 hour for he four roues. The PI s mplemened by solvng a seres of mnmum cos newor flow problems, n parcular, assgnmen problems. Because overme cos n he curren shf depends on he roues of he prevous shfs, he algorhm s mplemened sequenally from shf 1 o T -1 (Algorhm 6 n he Appendx). The sub-problem for every wo consecuve shfs s a bpare assgnmen 13

14 problem, whch s solved o opmaly usng he Hungaran algorhm (Kuhn, 1955). In he subproblem of shfs and +1, he nodes are he roues from 1 o and he roues from o T of vehcle, K. The arc cos c l s he overme cos of machng he roues from 1 o of vehcle wh he roues from +1 o T of vehcle l,, l K. The graph consruced s bpare. We wan o fnd he machng ha has he mnmum oal overme cos. We need o solve he sub-problem T -1 mes. Afer SI and SD, PI s called o reduce he overme cos. Then, he fnal roues are oupued and he assocaed coss n ravel me, regular drver usage, overme, and unme demand are calculaed. 4. Lower Bound To esmae he performance of he SD algorhm relave o he opmal soluon, we presen a mehod o oban a lower bound of he objecve value. I s obaned by solvng he LP relaxaon of he MIP model of Secon 2.1 wh several ypes of cus, whch ae no accoun he specal srucure of he problem. Accordng o Cordeau a al. (2002), he LP relaxaon of he VRPTW model ofen provdes very loose lower bounds. To show hs, he auhors show a procedure o fnd near-opmal soluons n whch me consrans are nacve. We need o add some cus o ncrease he lower bound. In general he gher he me wndow, he gher he lower bound. Frs of all, we resrc he defnon of varables o only feasble ones. Recall ha all demands can be classfed no SS and IS (Fgure 2). Alhough he classfcaon s done only for SD, exss for he daa se, ndependenly of any algorhm. Varables x j are only defned for pars (, j) of adjacen nodes; oherwse here are one or more depo nodes beween hem, hus x = 0. In j addon, x j are only defned for edges (, j) for whch s feasble o vs j afer,.e. e l < 0. + j j 4.1 Mnmum umber of Requred Roues (M RR) In hs secon, we have four proposons o bound he mnmum number of requred roues. In each shf, he mnmum number of roues o be used s he sze of he clque n he compably graph nduced by SS f here s no unme demand. As before, he clque s obaned 14

15 by solvng a mum clque problem approxmaely usng a greedy algorhm (Lu and Dessouy, 2006). Recall ha Z s he se of nodes formng he clque obaned from SS. Proposon 1: The number of roues used for each shf sasfes he followng nequales: K D K D xn+ Z u, T, ; Z x, n+ Z u, {2... T + 1}. Proof: Snce every par of nodes n Z Z s ncompable, a leas Z vehcles have o be used n shf f hey are all served by he nernal vehcles. If any one of he demands s served by a ax, we need o decrease he sze of he clque by one. Therefore n shf a leas Z u vehcles have o ravel from he orgn depo node n+ o a demand node, and smlarly a leas Z vehcles have o reurn o he desnaon depo node n++1 from a demand node. u H H We exend he noon of ncompable par o ncompable rple. If hree nodes canno be served by one vehcle, hen hey are called an ncompable rple. We jus need o chec wheher all sx possble permuaons of he hree nodes are nfeasble o now wheher hey are ncompable or no. For example, for hree nodes, j and p, f e > l permuaon (, j, p) s nfeasble. Thus aes O( ncompable rples for shf. ( e +, e ) > l + j j or j j + jp p SS 3, hen he ) compuaon me o denfy all Proposon 2: Le K D K D x x IT be he se of all ncompable rples n n+, 2 u, T, P IT ; P, n+ 2 u, {2... T + 1}, P IT. P Proof: For an ncompable rple P n SS, we have SS, a leas wo vehcles are needed o serve he hree nodes n P f all hree nodes are served by nernal vehcles. The roue number s non-decreasng 15

16 f we serve more nodes by he wo vehcles. Consderng ha some nodes mgh be served by ax, we have Proposon 2. Proposon 3: In he compably graph, suppose here exss four dfferen nodes (, p, q, l) n SS, such ha node s ncden o nodes p, q, and l, and (p, q, l) s an ncompable rple (Fgure 3). Le K D K D x x TE be he se of all such erads n u TE n+, 3,, ; P u T, n+ 3, { },. P P T P TE l SS, we have: p q Fgure 3: The node paern for Proposon 3 Proof: Node needs one vehcle 1, and nodes p, q, and l, canno be served by 1 snce hey are ncden o node. Snce (p, q, l) s an ncompable rple, hey mus be served by a leas wo oher vehcles 2 and 3. Therefore, a leas hree vehcles are needed f hey are all served by nernal vehcles. Consderng he case of usng ax, we have proposon 3. Proposon 4: Suppose here exss fve dfferen nodes (, j, p, q, l) n SS, such ha s ncden o p and j, and j s ncden o q, and (l, p, j) and (l,, q) are boh ncompable rples (Fgure 4). Le K D K D PE be he se of all such penads n x x u PE n+, 3,, ; P T P SS, we have:, n+ 3 u, {2... T + 1}, P PE. P 16

17 l p q j Proof: Frs we assume ha all nodes n Fgure 4: The node paern for Proposon 4 SS are served by nernal vehcles. Snce nodes and j are ncden, a leas wo vehcles are needed. Suppose and j are served by vehcles 1 and 2 respecvely. Suppose wo vehcles can serve all 5 nodes, hen p mus be served by 2, q mus be served by 1, and l mus be served by 1 or 2. However, snce boh (l, p, j) and (l,, q) are boh ncompable rples, canno be served by eher 1 or 2. There s a conradcon. Thus here has o be a hrd vehcle 3 o serve all of hem. Consderng he case of usng ax, we have proposon 4. A sraghforward mplemenaon o fnd all penads ha sasfes proposon 4 s o chec every penad n SS. The compuaon me s O( SS 4 ) for shf. Smlarly, he compuaon me of proposon 5 s O( 5 SS ) for shf. 4.2 o Incompable odes Served by he Same Vehcle ( I SSV) Proposon 5 ( o Incompable Pars Served by he Same Vehcle): If nodes and j s an ncompable par, hen he followng nequales are vald: (, p) E x p + x jp 1, K. ( j, p) E 17

18 Proof: If nodes and j are ncompable, hen no vehcle can serve boh of hem n a feasble x p (, p) E soluon. Tha s, x jp ( j, p) E and canno be boh 1. Therefore x p + x jp 1. (, p) E ( j, p) E 4.3 o Two-node Cycles ( TC) Proposon 6 ( o Two-node Cycles): The followng nequales are vald: x j + x 1, (, j) E, K. j Proof: Snce here are no wo-node cycles n a feasble soluon, for any wo nodes and j, a mos one of x j and x can be 1. Thus x x 1. j j + j 4.4 Mnmum Overme Requred (MOR) Proposon 7 (Mnmum Overme Requred): The followng nequaly s vald: K w LK e E n+ + 1 ( 1) + ( + 0 )(1 ), {1... },. u Proof: For node n SS, { 2... T + 1}, he mnmum overme of he node s ( e + 0 E,0)(1 u ). Tha s, f u = 0, he earles me he vehcle servng node reurns o he depo s e + 0, wh a mnmum overme of ( e + 0 E,0). If u = 1, hen here s no overme assocaed wh hs node. For each node n SS, snce he earles sar me of depo node n+ s = L( 1), we have K w B LK( 1) + ( e + 0 E,0)(1 u ) LK( 1) + ( e + 0 T SS E )(1 u ). n+ + 1 Please noe ha o ge a lower bound, hese vald nequales are all ncluded from he begnnng of he formulaon. On he oher hand, for each proposon defnng he MNRR consrans, s convenen o nclude only one consran. The oher one s redundan because f vehcle n shf s used, mus go from he orgn depo node o a demand node and from a demand node o he = desnaon node,.e. x x 1. If vehcle s no used, hen D = D n+,, n+ + 1 = D x n+, x, n+ + 1 = 0. Thus, n+ = x D x,, n K D K D 18

19 5. Expermenal Resuls We randomly generae cusomers lyng on a 2-dmensonal Eucldean plane (a 200 mnue*200 mnue square) and hen consruc he ravel me marx. Problems generaed wh hs procedure wll have a symmerc ravel me marx, whch obeys he rangular nequaly. The depo s locaed a he cener of he square, (100, 100). We consder a me horzon of a wee (5 worng days), and each shf has 8 hours. There are a oal of 15 shfs. Noe ha we evaluae he resuls n a seady sae envronmen by solvng a larger nsance and removng several warm-up shfs and ermnaon shfs. In he seady sae, he cusomers ha occur early n he curren shf may be served by he vehcles of he prevous shf, and he cusomers ha occur lae n he shf may be served by he nex shf. The frs shf and he las shf do no mee boh requremens. I may ae several shfs o ener or leave he seady sae. Thus, we generae and solve an nsance of 21 shfs, bu only evaluae he mddle 15 shfs (shfs 4-18). The earles sar me ( e ) s generaed randomly n he nerval (0, 120), and he laes sar me ( l ) s e +2 snce we assume he base me wndow lengh s 2 hours. We also consder wo problem classes where me wndow lengh s 4 hours. We consder 4 problem classes: unform demands (TW=2 hours), clusered unform demands (TW=2 hours), clusered unform demands (TW=4 hours), and clusered non-unform demands (TW=4 hours). The problem classes are named accordng o he spaal (or geographcal) dsrbuon of cusomers and her arrval rae. Thus, unform demands means unform spaal dsrbuon and unform arrval rae; clusered unform demands means clusered spaal dsrbuon and unform arrval rae; clusered non-unform demands means clusered spaal dsrbuon and non unform arrval rae. For each problem class, we consder hree unform demand raes per shf: Q =30, 60 and 120. For each case, he number of vehcles s chosen such ha s jus enough o serve all he demands or almos all he demands for he SI algorhm. We generae 10 nsances n he same manner and repor he average resuls over he 10 nsances. Noe ha he number of vehcles does no ncrease lnearly wh he demand rae because of he aggregae effec,.e., each vehcle can serve more cusomers when he demand rae s hgh. 19

20 5.1 Parameer Seng and Tunng The values of he cos parameers are se o real world coss. W s obaned by consderng he fuel cos and fuel mleage of a sandard shule vehcle. R s based on $15/hour. Overme salary rae s 1.5 mes he regular salary rae. Tax cos rae s based on a ypcal fare n Los Angeles Couny. We do some expermens o une he parameers for abu search for he case of 30 demands/shf for unform demands. Smlar resuls are observed for oher cases, and hey are no repored for space consderaon. We frs show sensvy resuls for, α, and β n Table 1. In he able, he columns Travel, OT, Unme Demand and Roue are ravel cos, overme cos, unme demand cos, and roue cos, respecvely. Toal s he oal cos. CPU Tme s he CPU me n seconds. We noe ha as we ncrease he parameers, we observe a smaller mprovemen n soluon qualy, however soluon me ncreases seadly. The bes soluon s when =2000, α =1000, When =800, α =400, he CPU me s only abou 11 mnues. Hence, we use expermens. β =1000, however he CPU me s more han an hour. β =400, he soluon s whn 0.72% of he bes soluon, and =800, α =400, β =400 n laer Table 1: Parameer unng for abu search ( θ mn =10, θ =20) based on 10 nsances α β Travel OT Unme Demand Roue Toal CPU Tme

21 We show sensvy resuls for θ mn and θ n Table 2. We can see ha he changes n CPU me are no sgnfcan and he bes soluon s obaned when θ mn =10 and θ =20 so we use hese values n laer expermens. When he abu enure s hgh, he soluon converges faser, bu he soluon qualy deeroraes. Table 2: Sensvy analyss of θ mn and θ ( nsances θ mn θ Travel OT Unme =800, α =400, Demand Roue Toal CPU Tme β =400) based on 10 The parameer values n he problem nsances and algorhms are summarzed below. (1) Problem sze and me parameers P: 5 days T : 15 L: 8 hours L : 4 hours (2) Cos parameers W : $17.5/hour R : $15/hour F: $15*8=$120 S : $22.5/h (1.5 mes of he regular salary rae) A : $40/hour 21

22 (3) The parameer values n he abu search algorhms (Algorhms 2 and 3) : 800 α : 400 β : 400 θ mn : 10 θ : 20 All he expermens are performed on a Dell Precson 670 compuer wh a 3.2 GHz Inel Xeon Processor and 2 GB RAM runnng Red Ha Lnux 9.0. The larges nsance could be solved n abou an hour of CPU me. For a ypcal nsance of he problem class ha aes he longes runnng me (he hrd case n Table 5 solved by SD), he oal runnng me s 1 hour, he me used for consrucng he roues s 16 mnues, and he me used for abu search s 44 mnues. There are 45 abu searches because SD algorhms calls 3 abu searches n each eraon, so on average each abu search aes 1 mnue Unform Demands For unform demands, we consder hree cases: 30 demands/shf o be served by 7 vehcles, 60 demands/shf by 10 vehcles, and 120 demands/shf by 16 vehcles. The resuls are shown n Table 3, where he values are he raos of SD/SI for he four cos measures, he oal cos and he CPU me. In he able, Roue lengh s he rao of he average roue lengh, where roue lengh s defned as he sum of shf lengh L and overme of he roue. The rao of roue lengh r s an ndcaor of overme because for a problem nsance, r L+ OT SD =, where SD L+ OT SI OT s he average overme of he roues obaned by he SD algorhm, and OT SI s he average overme of he roues obaned by SI. Then we have L + OT = r L+ OT ),.e., OT = rot + ( r 1 L. I can be seen ha larger r corresponds o larger SD OT SD ( SI SD SI ). In he column CPU Tme, we gve CPU mes of boh SD and SI, where he frs value s for SD, and he second s for SI. Table 3: The rao of SD/SI for unform demand 22

23 Cases Travel Roue lengh Unme Demand Roue Toal CPU Tme SD/SI 450 cusomers, 7 vehcles / cusomers, 10 vehcles / cusomers, 16 vehcles /2133 From he resuls, we can see ha SD ouperforms SI n erms of oal cos wh a 7% savng. SD uses a lle more overme o ge savngs n ravel cos, unme demands, and cos of roues. The percenage savngs n unme demand s mos sgnfcan, meanng ha wh he same number of vehcles, SD can serve more cusomers han SI. The savngs n roue cos are also sgnfcan (a leas 18%). In erms of soluon me, SI s faser han SD. Bu he CPU me of SD s no a problem snce we are plannng for a wee. 5.3 Clusered Unform Demands We nvesgae he effec of geographcal dsrbuon of cusomers on he performance of he algorhms by consderng clusered unform demands. The mehod we use o generae clusered demands s smlar o Sungur e al. (2008). We randomly generae cusomer locaons n 5 dfferen clusers wh dencal radus of 25. We cener each cluser a a random locaon wh a dsance of 50 from he depo. The resuls are shown n Table 4. We consder he same demand quany as before, bu he flee sze s smaller because a vehcle can serve more cusomers for clusered demand. Table 4: The rao of SD/SI for clusered unform demand Cases Travel Roue lengh Unme Demand Roue Toal CPU Tme SD/SI 450 cusomers, 5 vehcles / cusomers, 8 vehcles / cusomers, 12 vehcles /2229 From Table 4 we can see ha he savngs n oal cos are larger han for unform demands. The savngs are abou 10%. The soluons n general use more overme han n he case of unform demand, and he savngs n roue cos are larger. The reason s ha n SD when he demands are clusered, he vehcles prefer o say longer n a cluser o serve more cusomers of he nex shf, 23

24 leadng o more overme cos and more reducon n he number of roues. However, he savngs n ravel cos are small. The reason s ha n SD, here are savngs from fewer rps from and o he depo; on he oher hand, snce here are fewer vehcles used, some demands are served wh longer ravel dsance. The wo effecs cancel ou and he oal ravel dsance s almos he same. Noe ha he rao of 1.00 of unme demand means ha here are no unme demands for boh algorhms. The soluon me s longer han he prevous problem class snce here are more feasble moves when he demands are clusered. 5.4 Clusered Unform Demands wh Relaxed Tme Wndows Now suppose ha he demands are unform and he me wndows are all 4 hours. The resuls are shown n Table 5. The savngs are even larger n hs case, a abou 22%. Compared wh he smaller me wndows, he roues use more overme. The savngs n roue cos and ravel cos are larger. The reason s ha n SD, wh longer me wndows, he vehcles can serve more requess on her reurn rps o he depo, ncurrng more overme cos bu larger savngs n ravel dsance and roue cos. The soluon me s longer han prevous problem classes snce here are more feasble moves wh wder me wndows. Table 5: The rao of SD/SI for clusered demand (me wndow 4 hours) Cases Travel Roue lengh Unme Demand Roue Toal CPU Tme SD/SI 450 cusomers, 5 vehcles / cusomers, 8 vehcles / cusomers, 12 vehcles / Clusered on-unform Demands wh Relaxed Tme Wndows In he real world, clusered non-unform demands are more lely o be observed. A new se of expermens s done for hs case. The me wndows are all 4 hours. Generally, we have he mos demand n shf 2, fewer n shf 3, and he leas n shf 1 for each day. Recall ha shf 1 corresponds o he ngh shf, shf 2 corresponds o he day shf, and shf 3 corresponds o he evenng shf. In parcular, we assume ha he demand rae s 0.25Q n shf 1, Q n shf 2, and 0.5Q n shf 3. For example, f he demand rae s 60/shf, we have 15, 60, and 30 demands n shfs 1, 2, and 3, respecvely. In he hree cases, he numbers of vehcles are 4, 7, and 11 24

25 respecvely. We use fewer vehcles snce he oal demand quany s less han n he prevous cases. The resuls are shown n Table 6. Table 6: The rao of SD/SI for non-unform clusered demand (me wndow 4 hours) Cases Travel Roue lengh Unme Demand Roue Toal CPU Tme SD/SI 260 cusomers, 4 vehcles / cusomers, 7 vehcles / cusomers, 11 vehcles /1576 From Table 6, we can see ha he savngs are even larger wh non-unform demands and relaxed me wndows. The reason s ha n SD, demands n he shf wh a lower demand rae end o be served by vehcles of he prevous shf usng overme, hus reducng subsanally he number of roues n he low demand shfs. Compared wh case 1, n bgger nsances (cases 2 and 3), more overme s used and larger savngs n every oher aspec are obaned. The CPU me s shorer han he prevous problem class because here are fewer cusomers. Noe ha PI s appled a he end of boh SD and SI. However, he mprovemen s very small (on average abou 0.01%) due o he effecveness of he abu search. I only aes 1~3 seconds, whch are ncluded n he CPU me of SD and SI Effec of he Algorhms on he umber of Vehcles In hs secon we sudy he effec of he algorhms on he mnmum number of vehcles requred o serve all cusomers. We only show he resuls for unform demands. Smlar resuls are obaned for oher problem classes. Table 7 shows he mnmum K requred o serve all cusomers for SD and SI. The daa shown s also he average resuls over 10 nsances. We can see ha SD can save on he number of used vehcles. The average savng s abou one vehcle. Wh a smaller flee sze, usng SD resuls n less fxed vehcle and manenance cos. Table 7: Mnmum K requred o serve all cusomers for unform demand Cases SI SD SD/SI 450 cusomers cusomers

26 1800 cusomers Comparson of he Lower Bound wh he Soluon of SD The CPLEX solver can only solve he case of 360 cusomers and 12 shfs for unform demands. Ths problem (wh cus) has varables and consrans afer presolve. For bgger nsances, he memory usage s more han 2G and exceeds he memory of he compuer. Usng he CPLEX solver, he LP relaxaon can be solved n a mos 30 mnues of CPU me. We compare he lower bound wh he soluon of SD. The resuls are shown n Table 8. Wh all he cus, he rao of he soluon of SD o he lower bound s beween 1.09~1.82. Whou addng he cus descrbed n Secon 4, he rao s 13.25~26.55, showng ha he cus are very effecve. From column 4, we can see ha he mos mporan cu s he Mnmum Number of Requred Roues (MNRR) as descrbed n Secon 4.1. From he resuls, we can see ha he gher he me wndow, he gher he lower bound. An mporan reason for he ncrease n he rao s he sze of he clque whch s que small wh wde me wndows. Hence n he lower bound, he cu mnmum roue number requred becomes loose, leadng o much less roue cos han n he algorhm. Table 8: Comparson of SD and LB Problem Classes Problem sze SD/LB (no cu) SD/LB (wh cu M RR) SD/LB (wh all cus) TW cusomers, 10 vehcles TW 60, non-unform 208 cusomers, 10 vehcles TW 60, clusered, non-unform 208 cusomers, 7 vehcles TW cusomers, 7 vehcles TW 120, non-unform 208 cusomers, 7 vehcles TW 120, clusered, non-unform 208 cusomers, 5 vehcles TW cusomers, 5 vehcles TW 240, non-unform 208 cusomers, 5 vehcles TW 240, clusered, non-unform 208 cusomers, 4 vehcles To show he performance of SD over he opmal soluon and nvesgae wheher a hgh rao s due o a loose lower bound, we solve a se of small nsances o opmaly and repor he raos 26

27 of he soluon of SD over he opmal soluon, and over he lower bound. The resuls are shown n Table 9. Noe ha he problem sze s he larges sze he CPLEX can solve. Table 9: Comparson of SD, opmal soluon, and LB for small nsances Problem classes Problem sze SD/OPT SD/LB TW cusomers, 4 vehcles, 3 shfs TW 60, non-unform 14 cusomers, 3 vehcles, 3 shfs TW 60, clusered, non-unform 14 cusomers, 3 vehcles, 3 shfs TW cusomers, 3 vehcles, 3 shfs TW 120, non-unform 14 cusomers, 3 vehcles, 3 shfs TW 120, clusered, non-unform 14 cusomers, 3 vehcles, 3 shfs TW cusomers, 2 vehcles, 2 shfs TW 240, non-unform 14 cusomers, 3 vehcles, 3 shfs TW 240, clusered, non-unform 14 cusomers, 3 vehcles, 3 shfs From Table 7, we can see ha he raos of SD/OPT are close o 1, meanng ha he soluon of SD s near opmal. On he oher hand, he raos of SD/LB are also hgh, meanng ha hgh rao s due o a loose lower bound. There are some dfferences n SD/LB beween small nsances and large nsances because n small nsances he MNRR cus are generally much gher. 6. Conclusons and fuure research Ths research provdes some praccal nsghs for mul-shf VRP wh overme. We provde wo algorhms: SI and SD. SI s based on schedulng each shf ndependenly, whle SD allows effecve use of overme. From he expermenal resuls, we can see ha SD can provde much beer soluons han SI n erms of oal cos. SD also uses fewer vehcles han SI o serve all cusomers. If we have geographcally clusered demands, he savngs are larger han n he case of unformly dsrbued demands. The savngs are even larger for wde me wndows and nonunform demand raes. We oban a lower bound o he problem by solvng he LP relaxaon problem wh cus, whch shows ha he soluon of SD s whn 1.09~1.82 mes he opmal soluon on he es problems. We also show ha he rao of he SD soluon o LB s hgh when he lower bound s loose. 27

28 The resuls can be generalzed o oher suaons where ner-shf dependences should be consdered. There are a lo of neresng exensons for hs problem. For example, n pracce some companes have mxed shf lenghs,.e., here are shorer shfs such as 4 or 5 hour shfs as well as 8 hour shfs. We can sudy he roung problem wh all hese shfs. I s also useful o deermne company polces, such as overme lm or overme salary rae o enhance operaonal effecveness or reduce oal cos. References: Alonso F., Alvarez M.J., and Beasley J.E., A abu search algorhm for he perodc vehcle roung problem wh mulple vehcle rps and accessbly resrcons. Journal of he Operaonal Research Socey, 59, , Angelell E. and Speranza M.G., The perodc vehcle roung problem wh nermedae facles. European Journal of Operaonal Research, 137, , Arche C., Herz A., and Speranza M., A abu search algorhm for he spl delvery vehcle roung problem. Transporaon Scence, 40, 64 73, Az N., Gendreau M., and Povn J., An exac algorhm for a sngle-vehcle roung problem wh me wndows and mulple roues. European Journal of Operaonal Research, 178, , Banchess N. and Rghn G., Heursc algorhms for he vehcle roung problem wh smulaneous pc-up and delvery. Compuers and Operaons Research, 34, , Bodn, L., Rosenfeld D., and Kydes A., UCOST, a mcro approach o he rans plannng problem. Journal of Urban Analyss, 15, 47 69, Bodn, L., Rosenfeld D., and Kydes A., Schedulng and esmaon echnques for ransporaon plannng. Compuers and Operaons Research, 8, 25 38, Brandão J. and Mercer A., A abu search algorhm for he mul-rp vehcle roung and schedulng problem. European Journal of Operaonal Research, 100, , Brandão J. and Mercer A., The Mul-Trp Vehcle Roung Problem. The Journal of he Operaonal Research Socey, 49, , Campbell A. and Savelsbergh M., Effcen nseron heurscs for vehcle roung and schedulng problems. Transporaon Scence, 38, ,

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