Static Dial-a-Ride Problem with Money as an Incentive : Study of the Cost Constraint

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1 Statc Dal-a-Rde Problem wth Money as an Incentve : Study of the Cost Constrant Alan Faye, Dmtr Watel To cte ths verson: Alan Faye, Dmtr Watel. Statc Dal-a-Rde Problem wth Money as an Incentve : Study of the Cost Constrant <hal > HAL Id: hal Submtted on 3 Apr 207 HAL s a mult-dscplnary open access archve for the depost and dssemnaton of scentfc research documents, whether they are publshed or not. The documents may come from teachng and research nsttutons n France or abroad, or from publc or prvate research centers. L archve ouverte plurdscplnare HAL, est destnée au dépôt et à la dffuson de documents scentfques de nveau recherche, publés ou non, émanant des établssements d ensegnement et de recherche franças ou étrangers, des laboratores publcs ou prvés.

2 Statc Dal-a-Rde Problem wth Money as an Incentve : Study of the Cost Constrant Alan Faye, Dmtr Watel Laboratore CEDRIC, ENSIIE, square de la Résstance 9025 Evry, FRANCE Abstract In ths paper, we study a tax sharng problem. Gven a set of clents wth dfferent transportaton requests, the problem s to defne a set of taxs that wll be shared by the clents n order to reduce ther bll by a gven factor α <. To acheve ths, each clent shares the cost of the rde wth other passengers. More precsely, the fragments of the rde n whch the clent s alone s fully pad by ths clent. On the contrary, for each fragment n whch the clent shares the tax wth other passengers, the cost s equally dvded between the passengers. Frstly, we show that the problem consstng n searchng for a unque tax satsfyng ths cost constrant s NP-Complete, even f we do not mpose ths tax to pck up every clent. In addton, we show that, even f the factor α and the ctes of the taxs are fxed, the problem of satsfyng all the requests s NP-Complete. Secondly, we propose a nonlnear MIP model. Nonlnearty s due to the computaton of the shared prce of the clents. We propose dfferent lnearzatons. Then we establsh some necessary condtons that are used to reduce the sze of the problem. The end of the paper s dedcated to a numercal study of the model n order to evaluate ts performances. Keywords: Dal-a-Rde Problem, Complexty, Nonlnear Mxed Integer Program, Lnearzaton.. Introducton The Dal-a-Rde problem (DARP) conssts n the search for an optmal route for many vehcles n order to drve people from ther respectve orgn to ther respectve destnaton. Ths model s used, for example, to determne an optmzed route for taxs n order to pck up more passengers. We focus n ths artcle on a verson of ths tax-sharng problem n whch the prce pad by each passenger s shared. Ths problem, called Dal-a-rde problem wth money as an ncentve, was prevously ntroduced and studed n [20, 2]. Rde-sharng, ncludng Tax-sharng, has been massvely studed for the last ffteen years due to the economcal mpact and the ecologcal mpact of such a research. Indeed, optmzatons reducng the number of vehcles or the number of travels s an obvous way to reduce the costs and the greenhouse gas emssons. DARP can be seen as a subproblem of the general pckup and delvery problem (GPDP) descrbed n [22] n whch the goal s to transport a resource from dfferent pckup locatons to drop off locatons. In DARP, we consder a human resource (the clents) and each pckup or drop off locaton s assocated wth exactly one clent. The consequence of ths specfc resource s that one must be aware of the user nconvenence... Related work DARP can hardly be defned as a unque problem. The feasble and optmal solutons of a Dal-a-rde problem depend on the measure, the fleet parameters and the clents constrants. Thus, the varety of studes about DARP s not surprsng. Consderng the measure, one may optmze the vehcle travel cost, see for example [3, 8, 9], the total travel tme [0] or the proft [7]. Another opton s to maxmze the number of satsfed requests or a combnaton of all those parameters [20, 2, 23]. Emal addresses: alan.faye@ense.fr (Alan Faye), dmtr.watel@ense.fr (Dmtr Watel) Preprnt submtted to Elsever Aprl 27, 206

3 Some constrants modelze the user convenence. A usual opton s to search for a feasble soluton consderng tme wndows [4, 0, 9, 23] as t has been done for the more general pckup and devvery problem [8]. Ths last problem s solved wth a column generaton scheme where columns defne admssble routes. In [0, 9], the authors develop a smlar approach mergng a branch-and-cut algorthm wth column generaton. In [4, 23], the problem s solved usng a Tabu search heurstc. Another opton to modelze the user convenence s to tend to mnmze the excess rde tme [2,, 4]. Some models contan custom constrant n order to deal wth the restrctons and partculartes of the applcaton. For nstance, n [7], heterogeneous vehcles and customers are consdered. Each vehcle has ts own cty and s not able to drve all the customers. They developped a framework ncludng a column generaton algorthm and a neghborhood search algorthm. In [2], the authors consder electrc vehcles. In [3], the authors merge DARP and GPDP. Each tax drves customers and carres some parcels. A reduced verson of ths problem s proposed. Parcels are nserted n a gven route handlng people requests. MILP formulatons are gven for these problems and solved by a standard solver. Fnally one can consder ether the statc problem n whch all the requests are known n advance or the dynamc verson n whch the requests may occur at any tme [, 6,, 20, 2], ths problem s usually solved usng a local search heurstc. A recent revew about the Dal-a-rde problem and some of ts generalzatons may be found n [5]. We refer the reader to [5, 0] for a more specfc revew about DARP. Our verson may be classfed wth the lst of problems where the goal s to fnd a feasble soluton satsfyng a clent cost constrant. Few paper focused on that constrant. In [20, 2], the authors study the dynamc verson of the problem n whch each clent, travelng by tax, shares the cost of the rde wth other passengers. More precsely, the fragments of the rde n whch the clent s alone s fully pad by ths clent. On the contrary, for each fragment n whch the clent shares the tax wth other passengers, the cost s equally dvded between the passengers. The problem conssts n the search for a rde n whch every clent do not pay more than the cost he would pay alone n a tax travelng drectly from hs orgn to hs destnaton. Ths problem s called Dal-a-Rde problem wth Money as Incentve and s denoted by DARP-M. The authors used a GRASP heurstc to solve the dynamc verson of ths problem. A dfferent cost constrant may be found n [4], n whch the tax fare s reduced. Consequently, taxs are able to devate from the shortest path n order to pckup more clents wthout ncreasng the prce of the rde. Moreover, a hgh devaton s counterbalanced by a new reducton of the tax fare..2. Our contrbutons We study a dfferent verson of DARP-M. As n the orgnal problem, each tax starts at a depot and ends at a depot, each tax has the same cty and the clents must satsfy tme wndows. Contrary to [20, 2], we focus on the statc verson of the problem. Moreover, we generalze the cost constrant as we ask for a rde n whch each clent do not pay more than α tmes the cost he would pay alone, where α < s a constant real. As any feasble soluton would be acceptable for all the clents, the optmzaton measure we use benefts to the taxs company: we consder the total travel dstance of all the taxs. In ths paper, our frst contrbuton s a study about the complexty of the problem. In addton to the obvous NP-Completeness of the problem, due to the tme wndows constrant, whch was already proved n [20, 2], we determne how each constrant mpacts the complexty of the whole problem and, partcularly, how hard s ths cost constrant. Our second contrbuton s the modelzaton of ths cost constrant wth a lnear programmng model. We frst gve a complete mathematcal model of DARP-M. We secondly study the lnearzaton of the cost constrant. We then gve some reducton rules concernng ths constrant. We fnally experment the model wth a numercal study. Ths paper s organzed as follow: n the second secton, we formally defne the problem and study ts complexty; the thrd secton s dedcated to the mathematcal programmng modelzaton and the fourth secton contans the lnearzaton of the problem. In Sectons 5 and 6 we respectvely gve some reducton rules and the results of our numercal study. 2

4 2. Defnton and complexty of DARP-M In ths secton, we ntroduce formally DARP-M and prove t s NP-Complete. We partcularly focus on the hardness of fndng a tax satsfyng the cost constrant. 2.. Notatons and defnton of the problem We work n a complete drected graph G = (V, A). We are gven a set of n clents arbtrarly numbered n ; n. The -th clent s attached to three parameters : two nodes v and v, whch are respectvely the orgn and the destnaton of the clent and an nteger τ whch s ts departure tme. We respectvely defne V c, V c and T as {v, n}, {v, n} and {τ, n}. In addton, we are gven m taxs arbtrarly numbered n ; m. There s a unque depot d from where all the taxs start. Note that, frstly, the graph does not contan a node whch s not a pckup of a clent, a delvery of a clent or the depot of the taxs, and, secondly, that G contans one node for each pckup and each delvery, even f two of those nodes correspond to the same place. Consequently, V = V c V c {d} * and V = 2 n +. We search for at most m taxs, defned by the route (the path) they follow n the graph, such that each clent travels on tme and reduces ts cost by a gven factor. We now defne the four constrants that a tax must satsfy. Precedence constrant. The route P of a tax must start and end at the depot. For each clent, P contans the org f and only f t contans the destnato. In that case, v appears n P before v. We say that clent travels n that tax. Tme constrant. Each arc a s weghted wth a non-negatve nteger t(a) called the duraton of a. Ths value s the tme needed to travel through the arc wth a tax. Ths weght satsfes the trangular nequalty. We are also gven two ntegers t 0 and t 0. The frst one defnes the tme wndow durng whch the tax can pck up a clent. The latter one s the maxmum duraton the tax can delay a clent because of the other clents. More formally, the moment h d when a tax leaves the depot d s any non-negatve nteger; the moment when a tax reaches the j-th node u j of ts route P s h d + t(d, u ) + j the moment h v when a tax reaches the org of clent s after τ t 0 and before τ ; k= t(u k, u k+ ); the moment h v when a tax reaches the destnato of clent s before h v + t(v, v ) + t 0. Note that, there may be no non-negatve nteger h d satsfyng those constrants (for example, f for each arc a leavng the depot t(a) =, and for each clent τ = 0). We may fx ths f we allow a negatve h d. On the other hand, we may assume that each parameter τ s hgh enough such that, whatever the frst org of the route s, there s a non-negatve h d satsfyng h d + t(d, v ) = τ t 0. In a usual mplementaton, ths assumpton s true as τ s a tmestamp (countng the number of seconds snce 970). Nevertheless, f ths assumpton s false, we can translate each tme τ by max (t(d, v j) (τ j t 0 )). In j ;n that case, any tax startng at a non-negatve tme may reach any pckup node between tmes τ t 0 and τ. Capacty constrant. We consder that each tax has the same number of seats. Ths number s defned as the cty of the taxs. Ths cty s no more than n as there s no more than n clents and s at least 2. For each arc a = (u, v) n the route of a tax, we defne n a = n u as the number of clents travelng n the tax mmedately after u. The tax must satsfy n u. * The symbol s used to denote a dsjont unon : A B equals the unon A B, and ponts out that A and B are dsjont. Ths notaton cannot be used f A B. 3

5 Cost constrant. In addton to the weght t(a), each arc a = (u, v) s weghted wth a non-negatve nteger ω(a), correspondng to the cost that a clent would pay alone n a tax drvng from u to v. Ths weght satsfes the trangular nequalty too. We defne the desred gan α < as the mnmum factor reducng the bll of each clent. The cost pad s dvded between the passengers travelng on the same arc : for each clent, f a, a 2,..., a j are the arcs through whch the tax drves from v to v, the cost pad by that clent s ω = j k= ω(a k ) n ak. The route of the tax must reduce at least by a factor α the cost ω(v, v ) that the clent would pay alone. In other words, ths cost must satsfy ω α ω(v, v ). Note that there would not be any feasble soluton f the cty of the tax s. Ths s why ths case s forbdden. We can now defne the problem DARP-M. Problem. Dal-a-rde problem wth money as an ncentve (DARP-M). Gven n clents, m taxs, the correspondng complete drected graph G = (V = V c V c {d}, A) wth two non-negatve weghts t and ω over the arcs, the correspondng departure tmes T, two ntegers t 0 and t 0, a cty n of the taxs and a desred gan α <, the DARP-M problem conssts n fndng at most m m taxs satsfyng the precedence constrant, the cty constrant, the tme constrant and the cost constrant such that for each clent, there s a unque tax n whch that clent travels Complexty The Travelng salesman problem (TSP) conssts, gven a drected complete graph H and non-negatve weghts over the arcs, n fndng a mnmum cost Hamltonan path (the usual defnton asks for a cycle, but the two problems reduce to each other n lnear tme [6]). TSP s NP-Hard, even f the weghts satsfy the trangular nequalty [6]. There s a reducton from TSP to DARP-M. The reducton s based on the tme constrants of the clents. The proof can be found n [20, 2]. Consderng ths reducton, DARP-M can be seen as a generalzaton of TSP n whch we add the cost constrant. As a consequence, DARP-M s NP-Complete. The problem of ths reducton s that t does not reflect the hardness of the cost constrant. In TSP, the weght constrant s not the reason why the problem s hard. If we relax the hamltonan constrant, we would then have to fnd a path of mnmum weght n H, whch s obvously the empty path. On the contrary, the sole cost constrant of DARP-M s enough to make the problem hard. Theorem. The problem consstng n searchng for a unque tax satsfyng the precedence constrant and the cost constrant s NP-Complete, even f we do not mpose the tax to pck up every clent. Remark 2.. We show, n the followng proof, that we can force the tax to pck up every clents usng the sole cost constrant. Proof. The problem s n NP, as, gven a tax, one can easly check the constrants t must satsfy. We reduce the hamltonan path problem to ths problem. Gven a drected graph H, and two nodes u and v, the problem of fndng an hamltonan path n H from u to v s NP-Hard [9]. Let I = (H, u, v) be an nstance of the hamltonan path problem. We now buld an nstance J = (V c, V c, T, d, ω, t, t 0, t 0,, α) of our problem. Ths reducton s llustrated on Fgure. Let Ω be an nteger whch value wll be defned later. We defne n = n H + clents where n H s the number of nodes n H and m = tax. Thus, the graph G contans 2 (n H + ) + nodes. For each node w of H, G contans the node w tself and an addtonal node w. In addton, G contans three other nodes u, v + and the depot d. We defne the costs ω as followed : ω(a) = f a H, ω(d, u ) = 0, ω(u, u) = 0, ω(v, v + ) = Ω, ω(u, v + ) = Ω, and, for each node w of H, ω(w, w ) = Ω. In addton, we choose any path P startng at v + and gong through all the nodes w, for w H. Each arc of P s weghted wth 0. Fnally, we recall that n an nstance of that problem, G s a complete drected graph and that ω satsfes the trangular nequalty. In order to complete the graph, we frst compute n the current graph, for every par of nodes (u, v), the cost of the shortest path from u to v, and then add an arc (u, v) weghted wth that cost. If there s no path from u to v, the arc s weghted wth + that s a suffcently bg nteger B. The frst clent goes from u to v +. For each node w n H, we add a clent gong from w to w. 4

6 x u H y v G Ω Ω d x y 0 u 0 Ω u v v + Ω Ω Ω x y u v Fgure : Example of reducton from the hamltonan path problem to DARP-M. In ths example, we search for an hamltonan path n H from u to v. Note that G should be a complete drected graph, but we do not represent all the arcs for readablty. Every other arc s weghted such that ω satsfes the trangular nequalty. For each arc a of G, t(a) = 0. For each clent, the departure tme τ s 0. We set the cty to n H +. The desred gan α s n + Ω H n H + Ω. Each clent must pay Ω n a prvate rde. Consequently, n a feasble soluton, he has to pay at most α Ω = n H + Ω n H +. If an hamltonan path from u to v exsts n H, then, there s a feasble soluton for J consstng n a tax startng at u pckng up the frst clent, usng the arc (u, u), gong through each node of the hamltonan path, pckng up each clent of H, usng exstng arcs n H, whch weghts are, and gong to v + usng the arc (v, v + ) wth n H + clents and fnally gong through all the nodes w for w n H usng the path P of weght 0. Consequently, the cost pad by the frst clent s n H + Ω n H +. The cost pad by every other clent s less than ths value. Consequently, the precedence and the cost constrants are satsfed. If there s no hamltonan path, then we show that there s no tax satsfyng the cost constrant, f the value of Ω s correctly chosen. Ether the tax pcks up every clents or not. If t does, then t may go through an arc of nfnte weght and thus does not satsfy the cost constrant; t may go through an arc whch does not exst n H or goes through the same node twce, and, consequently, the frst clent should pay at least α Ω + n H, thus the cost constrant s not satsfed. If the tax does not pck every clents, then, as the weght of every path gong from an orgn to a destnaton of a clent s at least Ω, the clents n the tax should pay at least Ω n H. We now fx the value of Ω such that Ω n H > α Ω, whch s equvalent to Ω > n H (n H + ) ( n H ). Note that α < snce n H > α. As a consequence, I has a feasble soluton f and only f there s, n J, a tax satsfyng the precedence and the cost constrants. Ths complete the proof of the theorem. Remark 2.2. In the prevous reducton, the cty of the tax s gnored. In other words, we consder the tax to have a suffcently bg cty n order to pck up every clents at the same tme. Note that t s not known f the problem of fndng a unque tax satsfyng the precedence constrant, the cost constrant, and the cty constrant s NP-Complete when the cty of the tax s fxed and when we do not mpose the tax to pck up every clent. Fnally, note that ths problem cannot be solved by enumeratng all the routes satsfyng at most clents. Indeed, even f the cty of the tax s, t may satsfy more than requests. 5

7 2.3. Parameterzed complexty We fnally consder the parameterzed complexty of DARP-M wth respect to the parameters α and. Theorem 2. Unless P = NP, DARP-M parameterzed wth α and s not n XP, even f the tme constrant s relaxed. Remark 2.3. The relaxaton of the tme constrant means, for example, that t 0 and t 0 are large enough to let the tax choose any route n the graph ; or that t(a) = 0 for each arc a and τ = 0 for each clent. Proof. Let and α be two fxed numbers, satsfyng α > α. We now descrbe a polynomal reducton from the hamltonan path problem. Let I = (H, u, v) be an nstance of the hamltonan path problem. We now buld an nstance J = (V c, V c, T, d, ω, t, t 0, t 0,, α) of our problem. An example s gven on Fgure 2. x u H y v G d u 0 x 0 x 0 y 0 y 0 x Φ 0 u x x x u Φ y y Φ v y y v 0 u 0 0 v 0 u u u Φ v Ω v v 0 0 v + v + v + Ω v + v + v + u 0 Fgure 2: Example of reducton from the hamltonan path problem to DARP-M. In ths example, we search for an hamltonan path n H from u to v. In ths nstance of DARP-M, α = and = 4. In G, all the arc of the cycles of 3.5 nodes wth the same name are weghted wth 0. Note that G should be a complete drected graph, but we do not represent all the arcs for readablty. Every other arc s weghted such that ω satsfes the trangular nequalty. Let Ω and Φ be two ntegers whch value wll be defned later. We defne n = + (n H + ) ( α ) clents where n H s the number of nodes n H and m = tax. Thus, the graph G contans 2 ( + (n H + ) ( α )) + nodes. For each node w of H, G contans α copes of w and α copes of a new node w. In addton, G contans two new nodes u, u, the depot d, and α copes of a new node v+ and α copes of a new node v+. We defne the weghts ω as followed : the cost of an arc lnkng two copes of a same node s 0, ω(w, w 2 ) = Φ f (w, w 2 ) H, ω(w, w ) = f w H, ω(d, u ) = 0, ω(u, u) = 0, ω(v, v + ) = 0, ω(v +, v + ) = Ω, ω(u, u ) = Ω, ω(v +, u ) = ω(u, v + ) = 0. Fnally, we recall that n an nstance of that problem, G s a complete drected graph and that ω satsfes the trangular nequalty. In order to complete the graph, we frst compute n the current graph, for every par of nodes (u, v), the cost of the shortest path from u to v, and then add an arc (u, v) weghted wth that cost. If there s no path from u to v, the arc s weghted wth + that s a suffcently bg nteger B. The frst clent goes from u to u. Let C H be the set of n H ( α ) clents, t contans α clents per node w n H, gong from w to w. Fnally, let C v be the set of α clents gong from v+ to v +. The tmes t 0 and t 0 are 0. For each clent, τ = 0. For each arc a, t(a) = 0. The class XP contans every couples (Π, k) of parameterzed problems such that, when k s fxed, Π becomes polynomal. More formally, there exsts an algorthm for Π runnng n tme O(n f(k) ) for some functon f where n s the sze of the nstance. 6

8 The frst clent and all the clents of C v must pay Ω n a prvate rde. Consequently, n a feasble soluton, they have to pay less than α Ω. The clent of C H must pay less than α n a feasble soluton. We wll frstly prove the followng clams. Clam 2.. In a feasble soluton, any clent of C v cannot move to a node whch s not v +, v + or u. Proof of Clam 2.. The cost of every arc from v +, v + or u to another node of G s nfnte. Thus, f a clent of C v uses one of those arcs, he cannot satsfy the cost constrant. Clam 2.2. In a feasble soluton, f Φ > α, any clent of C H, gong from a node w to w cannot move to a node whch s nether w or w. Proof of Clam 2.2. The cost of every arc from w or w to another node of the graph s at least Φ, Φ and thus, such a clent of C H would have to pay at least > α. As a consequence, the cost constrant s not satsfed. Clam 2.3. In a feasble soluton, f Φ > α, the frst clent must go through every node of H. Proof of Clam 2.3. There are, for each node w n H, α clents gong from w to w. A tax drvng those clents from w to w must contan an addtonal clent to satsfy the cost constrant. By Clams 2. and 2.2, ths addtonal clent s nether a clent of C v nor a clent of C H not gong from w to w. Thus ths addtonal clent must be the frst clent. If an hamltonan path from u to v exsts n H, then, there s a feasble soluton for J consstng n a tax startng at d, gong to u and then u, pckng up the frst clent, gong through an hamltonan path n H from u to v, pckng up and drvng all the clents of C H, and, fnally, drvng the frst clent and the clents of C v through v +, v + and u. The clents of C H pay α α. The clents of C v pay Ω α α Ω. The cost pad by the frst clent s (n H+Ω) α + Φ (n H ). The cost constrant s then satsfed for every clent f (n H+Ω) + Φ (n α H ) α Ω. We now assume there s no hamltonan path, then we show that there s no tax satsfyng the cost constrant, f the values of Ω and Φ are correctly chosen. Φ By Clam 2.3, f > α, n a feasble soluton, the frst clent must go through each node w and w of H otherwse the cost constrant s not satsfed. As there s no hamltonan path n H, t must go through n H arcs of cost Φ. By Clams 2. and 2.2, he travels alone on those arcs. Moreover, by Clams 2. and 2.2, there cannot be more than α clents that can go through an arc (w, w ) or an arc (v +, v + ). Consequently, f there s no hamltonan path, the frst clent must pay at least (n H+Ω) + Φ (n α H ). Thus, f Ω also satsfy (n H+Ω) + Φ (n α H ) > α Ω, there s no feasble soluton n J. Consequently, f Φ > α and α Φ(n H )+n H Ω < α α Φn H+n H, the nstance I contans a α α α feasble soluton f and only f the nstance J contans a feasble soluton. We then fx Φ = α + and Ω = α Φ(n H 2 )+n H α α. Ths last value s well defned as we assumed that α > α. DARP-M s then NP-Hard, even f α and are fxed, even f the tme constrant s relaxed. As ths problem s n NP, t s NP-Complete. Consequently, f P NP, ths problem cannot be XP wth respect to the parameters α and. Remark 2.4. Ths proof focus on the case α > α. If < α, then there s no feasble soluton as a tax cannot carry enough clents so that they can share the cost of the rde. The problem s then obvously polynomal. The cases = α and > α = α are open. In ths secton, we studed how each constrant affects the complexty of DARP-M. Partcularly, we focused on the cost constrant, the tme constrant and the constrant of pckng up every clents. The latter constrant seems to be the strongest one as we can easly relax the tme constrant or the cost constrant wthout reducng ts complexty. Especally, Theorem 2 shows that, although the tme constrant s relaxed and the parameters α and are fxed, the problems remans NP-Complete. However, Theorem shows that, even f we may pck up only a subset of clents, the problems remans hard, due to the cost constrant leadng, n some cases, to pck up all the clents. Fnally, note that fndng a tax only pckng up a subset of clent and satsfyng the precedence constrant, the cty 7

9 constrant and the tme constrant s a polynomal problem as one can exhbt a tax pckng up only one clent. Consequently, the cost constrant s somehow stronger than the tme constrant. In the next secton, we consder an optmzatoerson of DARP-M and propose a mathematcal model. 3. Modelzaton of the problem DARP-M For the mathematcal model, we slghtly modfy the graph G = (V, A) defned n Secton 2.. Frst, we replace the depot node d by nodes d j for j =,..., m. The physcal locatons attached to these nodes are dentcal to the locaton of the depot. Each node d j s connected by an arc (d j, v ) to the departure node of each clent and each delvery node v s connected to each d j by an arc (v, d j). These arcs are weghted n the same manner than the arcs gong from the depot d to the departure of a clent or from the delvery nodes to the depot d. Two dstnct nodes d j and d j are not connected and for each d j we add a loop (d j, d j ). Ths loop allows a tax j to stay at the depot and thus not to be used. Addtonally, we remove the arc (v, v ) (destnaton toward orgn) for each clent. We denote the set of depot nodes by D = {d j : j =,..., m}. s We recall that DARP-M s a decson problem. We are here nterestng n fndng a soluton, f t exsts, that mnmzes the total dstance covered by the taxs. For each arc a = (v, w) of G we defne a length c(a) whch s the dstance for gong from v to w. Let us, now, ntroduce the varables of the problem. Each node v receves a number u v correspondng to the sequencng order: u v < u w means that node v s deserved before node w. h v s the tme a tax reaches node v. One can always assume that h v 0 as outlned n Secton 2.. s the number of clents n a tax after hs passage on node v. Each node v receves a prce p v. Ths prce s the prce that would have to pay a fctve clent travelng from the frst node encountered by a tax to the node v. The last varables are the decsoarables: x vw s a bnary varable that equals one f and only f arc (v, w) s ncluded n the route of a tax. The followng problem (P tax ), on Page 9, modelzes the selecton of the arcs n A that wll consttute routes of taxs and schedule of clents. The objectve s to buld a set of routes whose total length s mnmum and satsfyng, for each tax, the precedence constrant, the tme constrant, the cty constrant and the cost constrant. Note that we do not use varables x ndexed by taxs as, for example, n [3, 0, 3, 7, 9]. Ths allows us to reduce the number of varables. Ths s possble because the taxs have the same cty. We buld a cycle n G ncludng all customer nodes and the number of necessary tax depot nodes. So, the routes of the selected taxs are peces of ths cycle. In a route, one arc, exactly, s leavng a node (.a) and exactly one arc s enterng a node (.b). Constrants (2.a, 2.b) gve an ncreasng number to the pckup and drop-off nodes followng ther order n the tax route. If x vw =, then u w = u v +. M num s an upper bound of the order number whch guarantes that the constrants are vald when x vw = 0. Constrants (2.c) defne the numbers allocated to the depots. A node n poston k n the tax j wll be numbered (j ) 4n+k. Wth ths conventon, two nodes v and w n dfferent taxs, satsfes u w u v 2n +. A tax goes to the pckup node of a clent before the delvery node (2.d). The pckup and the delvery of a clent must be n the same tax. So u v u v 2n snce a tax can deserved at most 2n ponts. On the contrary, f v and v were n two dfferent taxs then u v u v 2n +. So, the constrants (2.e) are satsfed f and only f the delvery and the pckup of the same clent are n the same tax. Constrants (3.a, 3.b) compute the tme a tax reaches a clent node w when t comes from node v. If x vw = then h w = h v +t(v, w). M tme s an upper bound of the moment allocated to a clent node whch guarantes that the constrants are vald when x vw = 0. Constrants (3.c, 3.d) are the tme constrants of each clent. 8

10 Constrants (4.a,4.b) ncrease by one the number of passengers when the tax s arrvng on a pckup node. Constrants (4.c,4.d) decrease by one the number of passengers when the tax s arrvng on a drop-off node. The number of passengers must not be greater than the cty of the tax (4.e). There s no passenger on a depot (4.f). mn c(v, w)x vw s.t. (v,w) A w:(v,w) A v:(v,w) A x vw = x vw = v V w V (.a) (.b) Precedence constrants u v + M num x vw M num + u w v V, w V \ D, (v, w) A u v + (M num ) ( x vw ) + u w v V, w V \ D, (v, w) A u dj = (j ) 4n j ; m u v + u v ; n u v u v 2n ; n (2.a) (2.b) (2.c) (2.d) (2.e) (P tax ) Tme constrants h v + (M tme + t(v, w)) x vw M tme h w v V, w V \ D, (v, w) A h v + (t(v, w) M tme ) x vw + M tme h w v V, w V \ D, (v, w) A τ t 0 h v τ ; n h v h v t(v, v ) + t 0 ; n Capacty constrants + x vv + ; n, v V \ D, (v, v ) A + ( ) x vv + ; n, v V \ D, (v, v ) A + ( ) x vv ; n, v V \ D, (v, v ) A + ( ) x vv + ; n, v V \ D, (v, v ) A v V \ D n dj = 0 j ; m (3.a) (3.b) (3.c) (3.d) (4.a) (4.b) (4.c) (4.d) (4.e) (4.f) Cost constrants ( { p v + ω(v, w) mn, } + M cost ) x vw M cost p w v, w V \ D, (v, w) A p v ( x djv )M cost ; n, j ; m, (d j, v ) A p v p v α ω(v, v ) ; n (5.a) (5.b) (5.c) x vw {0, } u v, N p v 0 h v 0 (v, w) A v V v V \ D v V (6.a) (6.b) (6.c) (6.d) 9

11 The prce for gong through the arc (v, w) s calculated by Constrants (5.a). If x vw =, then p w p v s no less than the prce for a passenger alone dvded by the number of clents whch are n the tax. If there s no clent n the tax ( = 0), the prce for gong through the arc s set to zero, {, {, } thus mn = 0. In the other case,, thus mn =. M cost s an upper bound of ths prce whch guarantes the valdty of the constrants (5.a) when x vw = 0. The prce of the frst pckup node encountered by a tax (after departure from the depot) s set to zero (5.b). By Constrants (5.c), the prce that has to pay a clent s no more than the prce he would have to pay n travelng alone multpled by the desred gan α. Constrants (5.a) only set that p w p v + ω vw mn } {, the prce of node w when x vw =, but t s not necessary to add constrants n order to have equalty because the prce pad by clent s at most p v p v, and by Constrants (5.c), n any feasble soluton, ths prce s lower than α ω(v, v ). Fnally, Constrants (6.a, 6.b, 6.c, 6.d) defne the type of the varables. It can be easly proved that the ntegralty condton of varables u v, can be relaxed wthout relaxng the problem. One can consder a varant of ths problem where an empty tax returns to the depot. There are, n a practcal pont of vew, many clents and, consequently, a tax s not often empty. Thus, a soluton of (P tax ) usually satsfes ths constrant. Moreover, t s a way to lmt the duraton of the route of a tax. For ths varant, we add the followng constrants. + x v d ; n (7) d D (v,d) A Indeed, f there s no clent after a delvery node v ( = 0), then the tax must go through an arc from v to a depot node. We shall see, n Sectons 4, 5 and 6, that addng ths constrant brngs some help to solve the problem. We now defne the upper bounds M num, M tme and M cost. Because of Constrants (2.c), the maxmum value for u dj s (m ) 4n. There are at most 2n nodes n a route of a tax. Moreover, for each couple of successve nodes v and w n that route, u w = u v + by Constrants (2.a) and (2.b). Thus, the maxmum value for u v, v V, s (m ) 4n + 2n, and, then, an acceptable value for M num s (m ) 4n + 2n. M tme = max {τ + t(v, v )} + ;n t 0 s the mnmum upper bound of the delvery tmes, and t gves a vald upper bound for the pckup tmes and departures tmes of the taxs. n Fnally, ω(v, v ) s a trval upper bound for the prces allocated to the clent nodes. Ths value = } for s acceptable for M cost, however, we are gong to gve a lower one. The varables p v are ncreasng along the tax route. So, n order to evaluate the worst prce, we assume that there s only one tax that drves all the clents. We need some defntons. Let A route A be the set of arcs of the route of that tax from the frst pckup node to the last delvery node (the depot nodes are excluded). Let A κ A route be the set of arcs of the route whch contans κ clents (the arcs (v, w) where = κ). The sets A κ, for κ ;, gve a partton of A route. The prce of the last node of the route s the sum of the prces of the arcs (v, w) of the route dvded by. Ths gves: ω(v, w) = (v,w) A route ω(v, w) κ (v,w) A κ (8) Fnally, let A κ A κ be the set of arcs whch contans κ clents and such that clent s one of these κ clents. n Lemma. Let κ ;, ω(a) = κ ω(a) = a A κ a A κ Proof. We have A κ = n = A κ and for each a A κ, appears n exactly κ subsets A κ for, n. n {a} A κ = κ. Ths means that the arc a = 0

12 For concson, we ntroduce the followng notatons l κ = ( Proposton. M cost = α + α L κ = a A κ ω(a) ω(a) a A κ ) n ω(v, v ) s an upper bound of the prce p v allocated to a node v. = Proof. A κ s the set of arcs from the pckup node to the delvery node of clent n the route of the tax. So, the prce pad by that clent s gven by l κ κ. So, by Constrants (5.c), we have: l κ κ α ω(v, v ) (9) In the route of a tax, the path for gong from pckup node to delvery node of clent may nvolve more than one arc. So, by trangle nequalty of ω, the prce of ths path s larger than the prce of the drect path, and we have: By summng (9) for = ; n and by Lemma, we obtan : n = l κ ω(v, v ) (0) l κ κ = n L κ α ω(v, v ) () In the same way, by summng (0) for = ; n and by Lemma, we obtan : n = l κ = Multplyng ths last nequalty by, we obtan : κl κ = n ω(v, v ) (2) = n κl κ ω(v, v ) (3) Now, we make the lnear combnaton we obtan: By (8), the hgher prce s equal to κ Let us defne λ = The functon f(κ) = κ ( + = ) () + (3) of these two last nequaltes and ( + κ ) ( L κ α + α ) n ω(v, v ) = L κ κ. Now, we have to prove that + κ κ for κ ;.. Note that λ s well defned as >. We have 0 λ for κ ;. s convex for κ > 0 hence, a fortor, for κ. Thus, = f (( λ) + λ ) κ ( λ)f() + λf() = ( λ) + λ = + κ

13 { The problem (P tax ) s a mxed nteger nonlnear problem snce the terms x vw mn, }, nvolvng varables x vw and, appear n the constrants (5.a). In the next secton, we propose some lnearzatons. 4. Lnearzaton of the problem 4.. Frst lnearzaton We have to lnearze the term x vw mn { }, for gong from node v to node w. Frst, we lnearze mn n (5.a), the constrants that compute the prce { }, by ntroducng new bnary varables y vκ, κ ;. Let us denote by ˆη v the varable that wll take place for mn followng constrants. = ˆη v = κy vκ y vκ κ y vκ y vκ {0, } v V v V v V v V, κ ; {, }. We use the (4.a) (4.b) (4.c) (4.d) Constrants (4.a,4.b) lnk varables and y vκ : = κ f and only f y vκ = wth κ and = 0 f and only f y vκ = 0 for all κ. Constrants (4.c) defne ˆη { v, wth a ratonal value between 0 and : = 0 ˆη v = 0 and ˆη v = and fnally ˆη v = mn, }. The varable y vκ must be bnary (4.d). Now, we lnearze the product x vw ˆη v. We ntroduce the varable π vw that takes place for x vw ˆη v and state the followng constrants. { 0 πvw x vw v V \ D, w V \ D, (v, w) A ˆη v ( x vw ) π vw ˆη v v V \ D, w V \ D, (v, w) A (4.e) (4.f) Wth constrants (4.e), x vw = 0 π vw = 0, and x vw = 0 π vw whch s true snce, for x vw =, we have π vw = ˆη v. Wth constrants (4.f), x vw = π vw = ˆη v, and x vw = 0 ˆη v π vw ˆη v whch s true snce ˆη v 0 and, for x vw = 0, we have π vw = Improvement of the lnearzaton When > 0 the lnearzaton can be mproved. The followng lnearzaton can be used n the varant of the problem (P tax ) where Constrants (7) are added (an empty tax returns to the depot). In ths case, > 0 for any pckup and delvery clent node v except f v s the last clent node of the tax. But n ths case, the tax returns to the depot and there s no clent prce to calculate. In that case, ; and then. Consequently, we can wrte { mn, } = = + ( κ ) y vκ We ntroduce a new varable ˆη v and defne t wth the followng constrants. ˆη v = ( κ ) y vκ v V \ D (5.a) Note that ˆη v s stll a non-negatve varable and that hs maxmum value s the maxmum value of ˆη v prevously used n Secton 4.. whch s less than 2

14 The product x vw s now equal to x vw + x vw ˆη v. There appear a lnear term and a new product that we have to lnearze, but ths product s smaller than the prevous one. We ntroduce the varable π vw that takes place for x vw ˆη v and state the followng constrants. 0 π vw x vw ( ˆη v ( x vw ) ( ) ) π vw ˆη v v V \ D, w V \ D, (v, w) A v V \ D, w V \ D, (v, w) A (5.b) (5.c) These lnearzaton constrants are smlar to (4.e, 4.f). But observe that the coeffcent of x vw and x vw s now <. Another way to reduce the product x vw s to wrte Then we set ˆη v = κ=2 = κ=2 ( κ) yvκ. ( ) y vκ v V \ D (6.a) κ and we have 0 ˆη v. The product x vw s now equal to x vw x vw ˆη v. Then, we lnearze the product x vw ˆη v n a smlar way than prevously. 5. Reducng the sze of the problem The number of decsoarables of (P tax ) s equal to the number of arcs of the graph G = (V, A). An arc (v, w) modelzes a potental transton from node v toward node w. But some of these transtons may be forbdden by the cost constrants (5.c) whch say that the prce pad by a clent must be at least reduced by α. In the followng, we are gong to gve condtons that enable to remove arcs from A and consequently to reduce the sze of (P tax ). We recall that the costs ω satsfy the trangle nequalty. 5.. Frst set of reducng condtons We gve a frst set of condtons that are based on two clents and j. Frst, we consder the arcs (v, v j ) and (v j, v ) gong from a pckup node to a delvery node. Proposton 2. Let and j be two clents. If ω(v,v j ) be removed from A. + ω(v j,v ) > α ω(v, v ) then the arc (v, v j ) can Proof. Suppose that the arc (v, v j ) s ncluded n the route of a tax. Then, the clent s travelng on ths arc (v s ts pckup node) and has to pay at least ω(v,v j ) snce s the maxmum number of clents n a tax. Next, he has to go from v j to hs delvery pont v. Then, he has to pay at least ω(v j,v ) snce the costs ω satsfy the trangle nequalty. Thus, n total, he has to pay at least ω(v,v j ) + ω(v j,v ). If ths cost exceeds α ω(v, v ), the constrants (5.c) are volated for clent whch s mpossble. So, the arc (v, v j ) cannot be ncluded n a route of a tax and can be removed from A. Proposton 3. Let and j be two clents. If ω(v,vj) be removed from A. + ω(vj,v ) > α ω(v, v ) then the arc (v j, v ) can The proof of Proposton 3 s smlar to the proof of Proposton 2. Next, we consder the arcs (v, v j ) lnkng two pckup nodes. Proposton 4. Let and j be two clents. If ω(v,vj) + ω(vj,v ) > α ω(v, v ) then the arc (v, v j ) can be removed from A. Moreover, f ω(vj,v ) + ω(v,v j ) replaced by ω(v,vj) + ω(vj,v j )+ω(v j,v ) > α ω(v, v ). > α ω(v j, v j ) then the prevous condton can be 3

15 Proof. Suppose that the arc (v, v j ) s ncluded n the route of a tax. Then, on ths arc, there wll be at most clents snce v j s a pckup node. The cost pad by clent for gong from v to v j wll be at least ω(v,vj). After v j, the route of the tax must go through the delvery node v. The end of the route costs at least ω(vj,v ) snce the costs ω satsfy the trangle nequalty and a tax contans at most passengers. So, n total, clent wll pay at least ω(v,vj). If ths prce s greater than + ω(vj,v ) α ω(v, v ) then the constrants (5.c) are volated by clent. So, the arc (v, v j ) cannot be ncluded n the route of a tax and then, can be removed from A. Suppose the tax route goes through v before v j. Then, clent j would have to pay at least ω(vj,v ) + ω(v,v j ). If ω(vj,v ) + ω(v,v j ) > α ω(v j, v j ), then the constrants (5.c) would be volated by clent j. In ths case, a tax must deserve v j before v, and the lower bound of the cost for gong from v j to v can be mproved by ω(vj,v j )+ω(v j,v ). At last, we consder the arcs (v j, v ) lnkng two delvery nodes. Proposton 5. Let and j be two clents. If ω(v,v j ) + ω(v j,v ) > α ω(v, v ) then the arc (v j, v ) can be removed from A. Moreover, f ω(vj,v) + ω(v,v j ) replaced by ω(v,vj)+ω(vj,v j ) + ω(v j,v ) > α ω(v, v ). The proof of Proposton 5 s smlar to the proof of Proposton 4. > α ω(v j, v j ) then the prevous condton can be 5.2. Second set of reducng condtons In the prevous subsecton, we have consdered the arcs of the form pckup-delvery, pckup-pckup and delvery-delvery nodes. We want now to obtan results concernng the delvery-pckup nodes. To acheve ths, we have to consder the varant of the problem (P tax ) where Constrants (7) are added (an empty tax returns to the depot). Proposton 6. Let and j be two clents. For all clents k dfferent from and j, let ŵ k = ω(v k, v ) f ω(v, v k ) + ω(v k, v ) α ω(v, v ) ŵ k = ω(v k, v ) + ω(v, v ) otherwse ŵ jk = ω(v j, v k ) f ω(v j, v k ) + ω(v k, v j ) α ω(v j, v j ) ŵ jk = ω(v j, v j ) + ω(v j, v k ) otherwse Suppose that the clents k, j are renumbered from to n 2 and n such a way that β k = ŵk ŵ jk α ω(v k, v k ) are n ncreasng order. If, ω(v, v j) k then the arc (v, v j) can be removed from A. + + β k > 0 k =,..., (7) Proof. Suppose there are p clents n a tax gong through arc (v, v j). Frst, note that p snce v j s a pckup node. Let k < k 2 < < k p be the numbers of the p clents on arc (v, v j). The number k p s not strctly lower than p because, n the worst case, we have k =, k 2 = 2,..., k p = p. By (7), we have ω(v,vj) p + β p > 0. As k p p, we have β kp β p and hence, ω(v,vj) p + β kp > 0. Replacng β kp by ts defnton, we obtan ω(v,vj) p + ŵkp + ŵjk p > α ω(v k p, v k p ). The trp of clent k p s dvded nto three parts. On the part from v kp to v, the prce wll be at least snce there are at most passengers n a tax and the costs ω satsfy the trangle nequalty. ω(v kp,v ) Moreover, f ω(v,v kp )+ω(v kp,v ) > α ω(v, v ) the k p cannot be on the route from v to v. Indeed, otherwse, clent would volate the prce constrants (5.c). So, n ths case, v s on the route from v kp to v and the prce from v k p to v wll be at least ω(v kp,v)+ω(v,v ). The same holds for the trp from v j to v k p. For the trp on the arc (v, v j), clent k p has to pay exactly ω(v,vj) p. So, n total, clent k p has 4

16 to pay at least ω(v,vj) p + ŵkp + ŵjk p whch s greater than α ω(v k p, v k p ). Then, clent k p volates the prce constrants (5.c). Hence, k p cannot be n a tax gong through arc (v, v j). There s a contradcton and the hypothess p s false. Hence, there s no clent on the arc (v, v j) whch s mpossble because, due to Constrants (7), an empty tax goes back to the depot. So, f the condtons (7) are satsfed, the arc (v, v j) s never used and t can be removed from A. To compute the condtons (7) t s not necessary to sort the β k for k ; n wth k and k j. We have just to sort the smaller β k. Ths can be done n O(( )n) whch s lnear f the cty of a tax s fxed. 6. Numercal results Numercal tests have been performed on the lnearzed versons of (P tax ) wth dfferent nstances. The nstances are generated n the followng way. We have extracted from Google Maps fourteen locatons n Pars wth travel tmes (n mnutes) and dstances (n meters) between them. We set the cost of a trp from a locaton to a locaton j equal to the tme for gong from to j. In order to generate an nstance, frst, we randomly select locatons that can be vewed as reference locatons. The number of reference locatons depends on the number of clents. We assgn departures and arrvals of the clents to the reference locatons n such a way that (P tax ) admt at least one feasble soluton for α = 0.5. Next, n order to have dstnct locatons, we randomly move the departures and arrvals n the neghbor of ther reference locatons. To do ths, we choose an α > 0.5 and the ampltude of the perturbaton, that s the maxmum travel tme from the new locaton to the reference locaton, s computed n such a way that (P tax ) stll admt at least one feasble soluton for the chosen α. The Manathan dstance s used to measure the travel tme and the dstance from the new departure and arrval locatons from ther reference locatons. The locaton of the tax depot has been arbtrarly chosen at Chatelet, a locaton at the centre of Pars. The tests concentrate on the cost constrant whch has been shown n Secton 2.2 to ntroduce most of the complexty (see Theorem ). So, we assume that tme wndows of the clents are large enough to ensure that the tme constrant s satsfed n any schedulng. Ths can be acheved by takng t 0 and t 0 suffcently large. Note that, due to Theorem 2, ths assumpton does not make the problem easer to solve. We test the problem (P tax ) and ts verson n whch an empty tax returns to the depot, modelzed by Constrants (7). We denote by (P tax ) ths second verson. We mplemented the models n AMPL language and solved them usng the CPLEX software package (CPLEX 2.6), on a quad-core processor Intel(R) Core(TM) CPU (3.2 GHz) wth 4 GB of RAM. The results are summarzed n the followng tables. Each lne of these tables s obtaned by averagng over 5 nstances. The columns of the tables are the followng: n s the number of clents, m s the number of taxs, α s the desred gan of the clents, graph reducton s the percentage of suppressed varables (edges of the graph) usng the propostons gven n Secton 5, max CPU tme s the maxmum CPU tme n seconds allowed to solve the problem, gap s the relatve gap between the value of the best soluton found wthn the tme lmt and the best lower bound of the optmal soluton, BB nodes s the number of branch-and-bound nodes of the search tree, unsolved nstances s the number of nstances for whch no soluton has been found wthn the tme lmt, used CPU tme s the tme used to solve the problem; when ths tme s smaller than max CPU tme ths means that the solver has found an optmal soluton wthn the allowed tme lmt. For all the nstances, the cty of a tax s equal to 4. Note that, agan, due to Theorem 2, the values chosen for the cty and α do not make the problem easer to solve. Table s devoted to problem (P tax ). The problem s lnearzed followng the method descrbed n Secton 4.. The used reducton rules are the ones descrbed n Secton 5.. The problem s dffcult to solve but we can obtan solutons up to n = 8. We can exactly solve some nstances for n = 8 and 0 (the used CPU tme s strctly less than the max CPU tme). For n 2 we obtan only suboptmal solutons whose gap s ncreasng wth the number of clents. Table 2 s devoted to the same problem but n ths case the graph reducton s not used. Comparson of the results n Tables and 2 shows the gan obtaned by the graph reducton. Ths can be observed on three ponts: the used CPU tme for n = 8 and n = 0, the gap, and, for n 2, the number of unsolved nstances. These entres are greater f no reducton s done. 5

17 Problem (P tax ) n m α graph max CPU unsolved used CPU gap % BB nodes reducton % tme s. nstances tme s Table : Results for (P tax ) Problem (P tax ) wthout graph reducton n m α graph max CPU unsolved used CPU gap % BB nodes reducton % tme s. nstances tme s Table 2: Results for (P tax ) wthout graph reducton 6

18 The dffculty to solve the problem s due to the poor LP relaxaton (contnuous relaxaton). But ths s not the unque reason. The cost constrant nduces numercal troubles. To see ths, we solved the problem wthout the cost constrant,.e. wth Constrants (5.a-5.c) removed, and so wthout graph reducton. We can solve ths relaxed problem more easly even f ts LP relaxaton s poor too. The results are reported n Table 3. The column LP gap gves the relatve gap between the best soluton found and the LP relaxaton of the problem. The computatons have been done wthn the same tme lmts used above. The number of unsolved nstances s sgnfcantly reduced when the cost constrant s removed. (P tax ) (P tax ) (P tax ) wth reducton wthout reducton cost constrant relaxed unsolved LP n m α gap % unsolved LP gap % unsolved LP gap % nstances gap % nstances gap % nstances gap % Table 3: Comparson of problems wth and wthout cost constrant Table 4 s devoted to problem (P tax ) where an empty tax returns to the depot. In ths table, the results for the three lnearzatons descrbed n Secton 4 are reported. The frst lnearzaton s the one descrbed n Secton 4.. The second and thrd lnearzatons are descrbed n Secton 4.2 and defned respectvely by (5.a-5.c) and (6.a). For a gven trplet (n, m, α), the results of the frst, second and thrd lnearzaton are reported n ths order. The graph reducton s more mportant for ths knd of problem snce one can use the addtonal reductons gven n Secton 5.2. Ths provdes a lttle bt more easness to solve the problem. Ths can be observed especally by the number of unsolved nstances whch s less mportant than n Table. The second and thrd lnearzatons mprove the frst lnearzaton. Ths can be observed by a decreasng gap and more especally by a decreasng number of unsolved nstances. As the tme wndows are large enough, optmal solutons of (P tax ) use only one tax. On the contrary, optmal solutons of (P tax ) may use several taxs. However, such a soluton can be transformed nto a soluton of (P tax ) wth a lower cost by lnkng the routes of two successve taxs. The routes of taxs k and k +, for k ; m, are merged by gong drectly from the last clent of k to the frst clent of k +. The return to the depot s removed. Table 5 contans the relatve gaps between the heurstc solutons bult from the solutons of (P tax ) and the solutons of (P tax). The gaps are very tght (a negatve value ndcates an mprovement of the soluton of (P tax )). Ths s an nterestng results snce (P tax ) s more easly solved than (P tax) and t provdes good heurstc solutons for ths last one. 7. Concluson We have studed a tax sharng problem n whch the prce of a trp s evenly shared between the passengers of the trp. The bll of the passengers must be reduced by a gven factor. Partcularly, ths last constrant strongly affects the complexty of the problem. We have shown that determnng a feasble soluton s an NP-complete problem even f we relax the tme wndows constrant and f we do not ask the taxs to drve all the clents to ther destnatons. However, t s not known f ths result remans true f the cty of the taxs s fxed. Ths open queston s worth to be nvestgated. We have proposed a mxed nteger model that takes nto account the prce reducton constrants. These constrants are nonlnear. We have proposed some lnearzaton approaches. Next, we have proposed some condtons, based on the cost constrants, that provde sgnfcant reductons of the sze of the problem. Nevertheless, for now, the model can only solve small nstances. Some mprovements 7