The common-line problem in congested transit networks

The common-line problem in congested transit networks R. Cominetti, J. Correa Abstract We analyze a general (Wardrop) equilibrium model for the common-line problem in transit networks under congestion conditions. Under very general conditions we establish the existence and uniqueness of an equilibrium. We characterize this equilibrium showing that, in contrast with the uncongested case, at some traffic levels no single set of common lines may be an equilibrium and there exist necessarily 2 sets of common-lines being used. In such ranges of flow we remark that an increment in passenger flow does not induce an increase in travel time. The general model is illustrated through some simple examples which show moreover that the equilibrium is not a social optimum and in some sense it is unstable. 1 Introduction In this paper we analyze a simple model for one aspect of the problem of passenger assignment in congested transit networks. Specifically we consider the common lines problem, that is to say the case of a single origin O connected to a single destination D by a set of n bus lines denoted l 1, l 2,..., l n. Each line l i is characterized by its in-vehicle travel time t i and its frequency f i. Aggregating those lines that have equal in-vehicle travel time and reordering them we can assume that t 1 < t 2 <... < t n. l 1 l 2 O D l n In a model that neglects the congestion effects for traveling from O to D, passengers choose a set s {1,..., n}, called the common lines, boarding the first bus in such a set to reach the destination. The chosen strategy s will be the one minimizing the mean total travel time, including waiting time and in-vehicle travel time (cf. [2, Chriqui & Robillard]). To estimate the expected waiting time we note that the total frequency corresponding to strategy s is i s f i, so that the waiting time can be estimated as 1 W s = i s f. i Universidad de Chile, Casilla 170/3 Correo 3, Santiago, Chile. Supported by FONDAP Matematicas Aplicadas. 1

On the other hand, for i s the probability of boarding a bus from the line l i is f i / j s f j, so that the mean in-vehicle travel time of the strategy s is given by T s = i s f i t i j s f j it follows that the optimal strategy should minimize the quantity T s = W s + T s = 1 + i s t if i i s f. i Intuitively, a passenger consenting to board line l i should also accept to board any of the faster lines l 1,..., l i 1, implying that an optimal strategy should be of the form s = {1,..., k }. Denoting k := {1,..., k} such kind of strategies, and T k the mean total travel time associated to it, it can be shown that the sequence T 1,..., T n is decreasing up to a certain k and strictly increasing from k + 1 on (it may happen that T k +1 = T k ). On the other hand, it is easy to see that T k+1 = αt k +(1 α)t k+1 with α (0, 1), from which it follows that T k+1 < T k t k+1 < T k. These remarks, whose proof can be found in [2] and also in Proposition 3.1, justifies the following algorithm for finding the common lines: k 1; while t k+1 < T k do k k + 1; end k k. The model described above neglects the congestion effects, since it assumes that every passenger can board the first bus in his strategy arriving to the bus stop. In a congested situation, due to the capacity restriction of the buses, it may happen that a passenger may not board the first bus that comes and might be forced to wait for the next one. A simple approach to model this situation is to assume that the frequency of each line l i is not fixed, but is a function of the flow v i boarding this line. More precisely we suppose that the line l i is characterized by an effective frequency f i : [0, v i ) [0, ) which is a continuous strictly decreasing function, with f i (v i ) 0 as v i v i. The constant v i > 0 (eventually v i = + ) is called the saturation flow of line l i. In this setting the travel time of the strategy s will be a function of the flows in the lines, T s (v) := 1 + i s t if i (v i ) i s f. i(v i ) and then the decision of each passenger depends on the decision of the other passengers. Hence the common lines problem may no longer be formulated as an optimization problem, but rather as an equilibrium problem. To describe the equilibrium model, we consider the flow x 0 of passengers going from O to D, and the set S := {s {1,..., n} : s φ} of possible strategies. The flow x is decomposed into flows y s 0 in the different possible strategies, so that x = s S y s. Using Wardrop s principle (1952), we say that a decomposition y = (y s) s S is an equilibrium if and only if all used strategies are of minimal time, that is (W ) y s > 0 T s (v ) = min s S T s (v ) where v = (v i )n is the vector of flows in the lines induced by the distribution y. 2

For computing the flow v induced by a given distribution y, we remark that a passenger using strategy s will board line l i, i s, with probability p i (s) = f i (v i )/ j s f j(v j ). Then the total flow in line l i can be computed as: (A) v i = s S:s i f i (v i ) y s j s f j(v j ) for all i = 1,..., n. In 2 we prove that, under natural conditions, these equations define an implicit function v = v(y). With this the travel time of strategy s becomes a function of the vector y, and the equilibrium condition (W ) may be stated as a variational inequality: (V I) Find y Ω(x) such that T (v(y )), y y 0 for all y Ω(x) where Ω(x) := {y 0 : x = s S y s} and T (v) := (T s (v)) s S. In 3 we describe the structure of the equilibria y, we prove their existence and uniqueness and we study their behavior as a function of the total flow x. At first sight one may think that the only effect of congestion, due to the reduction of frequencies and the corresponding increase of the waiting times, would be a change in the optimal strategy, which could be computed by using Chriqui & algorithm with the suitable frequencies f i (v i ) (eventually by an iterative process to adjust the flows v i ). This view, which assumes that every passenger uses the same strategy, is adopted for example in [1]. However we will prove that for the model (V I), and for wide ranges of flow, no pure strategy is a Wardrop equilibrium. In that case the flow x is decomposed into two consecutive strategies of the form k and k + 1. An interesting property is that, in such ranges, the total travel time at equilibrium remains constant, that is to say an increase in the flow x does not imply necessarily a decrease in the service quality, in terms of mean travel time. 2 Relation between strategy flows and line flows In this section we describe conditions under which the system (A) defines an implicit function v = v(y). This is necessary for the equilibrium problem to make sense. For that purpose we extend the function f i ( ) to IR by letting f i (x) = f i (0) x for x < 0, and we consider the change of variables z i := ln f i (v i ) defined for v i < v i. In these new variables system (A) is written as follows Defining q : IR n IR by where h i : IR + IR is given by 0 = f 1 i (e z i ) + y s s S:s i e z i j s ez j n q(z) := h i (e z i ) + y s ln e z j. s S j s u h i (u) := 1 f 1 i (τ) dτ, τ for all i = 1,..., n. (1) 3

system (1) can be expressed in the equivalent form q(z) = 0. It is easily proved that h i (u) is strictly convex for all i = 1,..., n, and so q(z) is also strictly convex. Hence the solutions of (1) are the minima of the function q. In what follows we give necessary and sufficient conditions for the existence of a minimum of q. The strict convexity of q implies uniqueness of this minimum. To this end we will use the following criteria (cf. [5, Theorem 27.1]): Argmin(q) φ iff q (d) > 0 for all d 0, where q ( ) denotes the recession function of q, that is, q q(x + td) q(x) (d) = lim. t t Proposition 2.1 The function q attains its minimum iff ( x ) n v i x i > s y s min j s x j, (2) where = {x IR n : x 0, n x i = 1} denotes the n dimensional simplex. Proof: Let us proceed to compute q ( ). Defining ψ i (z i ) := h i (e z i ) we obtain ψ i (d i ) = lim ψ t i(td i )d i = lim f 1 t i (e td i )d i = + if d i > 0 0 if d i = 0 v i d i if d i < 0. This equality, combined with the fact that for g s (y) = ln( i s ey i ) we have gs (d) = max i s d i (cf. [5]), implies that { q n v (d) = i d i + s S y s max j s d j if d 0, + if not. Hence q attains its minimum if and only if n v i d i + s S y s max j s d j > 0 d 0, d 0. Normalizing to vectors d such that n d i = 1 and multiplying by 1, we obtain the equivalence with condition (2). In what follows we give conditions equivalent to (2), which can be interpreted in more intuitive terms. To this end we note that the condition (2) is equivalent to L > 0 where: L := min x,λ { n v i x i + s S y s λ s : x, λ s + x j 0 s S, j s The previous expression is a bounded linear program, so we can use duality for writing it in the alternative way L = min x,λ max γ IR µ,β 0 n v i x i + n n y s λ s µ i x i γ( x i 1) β sj (λ s + x j ) s S s S,j s }. 4

= max γ IR µ,β 0 = max γ IR µ,β 0 = max γ IR β 0 n min x i ( v i µ i γ β si ) + λ s (y s β sj ) + γ x,λ s:s i s S j s γ : v i µ i γ β si = 0, i; β sj = y s, s S s i j s γ : β si v i γ, i; β sj = y s, s S. s i j s Therefore condition (2) is equivalent to There exists (y sj ) s S,j s such that y sj 0; y sj = y s, s S; j s y si < v i, i = 1,..., n. s i that is, the strategy flow vector y = (y s ) s S, may be decomposed into the lines without saturating any of them. The following theorem summarizes the preceding analysis. Theorem 2.1 Given y 0 the following conditions are equivalent: (a) For all x we have s S y s min j s x j < n v i x i. (b) There exists y si 0 such that i s y si = y s for all s S and s:s i y si < v i for i = 1,..., n. (c) For all s S we have s s y s < i s v i. These conditions are necessary and sufficient for the existence and uniqueness of the minimum of q, and so they are for the existence and uniqueness of the solution v of the system (A). Proof: We have already proved that (a) is necessary and sufficient for the existence and uniqueness of a minimum of q, as well as the equivalence between (a) and (b). (a) = (c): Let s S. Taking x i = 1/ s for i s, and x i = 0 for i s, condition (a) becomes now s s y s < i s v i. (c) = (b): In order to prove (b) we will use the maxflow-mincut theorem applied to the graph G defined by: the vertex sets V 1 = S, V 2 = {1,..., n}, and f, p two auxiliary vertex; and the arc sets E 1 = {(f, s) : s V 1 }, E 2 = {(i, p) : i V 2 } and E 3 = {(s, i) : s S, i s} as shown in the figure. V 1 V 2 f s i p We associate to each arc e a minimal capacity 0 and a maximal capacity given by y s if e = (f, s) E 1 c(e) = v i ɛ if e = (i, p) E 2 if e E 3 5

with ɛ > 0 chosen such that y s v i ɛ, s S. (3) s s i s Clearly for proving (b) it suffices to see that the maximum flow from f to p through the graph G is s S y s. By the maxflow-mincut theorem this is equivalent to proving that every (f, p) cut (W, W c ) has capacity c(w ) s S y s. To prove the latter let us observe that if there is an arc in E 3 going out from W, then c(w ) =. So it suffices to suppose that there are only arcs in E 1 and E 2 going out from W. Let A = W V 2. If A = φ it follows that c(w ) = s S y s. If not, we have c(w ) = ( v i ɛ) + ( y s y s ). i A s S s A so that, using (3) we obtain which completes the proof. c(w ) s S y s = ( v i ɛ) y s 0 i A s A To interpret condition (c) of the theorem let us remark that the amount s s y s is the total flow that must be carried through the set of lines s. In this way the condition may be interpreted by saying that every subset of lines s must have enough capacity to carry the total flow required for them. This condition is clearly necessary for the existence of a solution of the system (A). 3 Existence and uniqueness of equilibrium In this section we prove that given a total flow x for the system, there exists a unique Wardrop equilibrium y = (y s) s S, in the sense of the variational inequality (V I). Remember that the time T s associated to strategy s depends on the line flows as follows T s (v) := 1 + i s t if i (v i ) i s f, i(v i ) and implicitly is a function of the strategy flow vector y (through the solution of system (A)). The following elementary fact will be frequently used from now on. Lemma 3.1 Let v be a vector of flows in the lines, and s S a strategy. For every j s the time T s {j} (v) is a strict convex combination (scc) between T s (v) and t j, that is, there exists α (0, 1) such that T s {j} (v) = αt s (v) + (1 α)t j. Proof: It is enough to observe that ( i s T s {j} (v) = f ) ( ) i(v i ) f j (v j ) i s {j} f T s (v) + i(v i ) i s {j} f t j. i(v i ) 6

We start by proving that the only used strategies in an equilibrium situation are of the form k = {1,..., k}. Moreover there can be at most two strategies of this form being used at the same time. Proposition 3.1 Let v be a vector of flow in the lines. The strategies s that minimizes T s (v) are of the form s = k = {1,..., k}. More precisely there exists k = k(v) {1,..., n} such that T 1 (v) > T 2 (v) > > T k (v) T k+1 (v) < T k+2 (v) < < T n (v), so that the only minimal time strategies are k and eventually k + 1. Proof: Let s be a minimal time strategy. For j s the time T s {j} (v) is a scc between T s (v) and t j, and since T s {j} (v) T s (v), it follows that t j T s (v). Similarly, for i s the time T s (v) is a scc between T s\{i} and t i, and from T s\{i} T s (v), it follows t i T s (v). Hence, t i t j for all i s, j s, so that s is of the form k for some k. Let k be the smaller integer such that T k+1 (v) T k (v). Since T k+1 (v) is a scc between T k (v) and t k+1, it follows that T k+1 (v) t k+1 < t k+2. Moreover T k+2 (v) is also a scc between T k+1 (v) and t k+2. So that T k+2 (v) > T k+1 (v). By induction we conclude that T l (v) is strictly increasing from l = k + 1 on. To characterize the equilibria y and to determine the corresponding k = k(v(y )) in the previous proposition, we consider the situation in which the total flow x is carried only by strategy k, that is to say y k = x and y s = 0 for all s k. Let v k = v k (x) be the unique solution to the corresponding system (A), that is vi k f i (vi k = x ) kj=1 f j (vj k) i = 1,..., k ; vi k = 0 i = k + 1,..., n. (4) Lemma 3.2 The function x v k (x) is Lipschitz continuous. For each i = 1,..., k the component vi k( ) is strictly increasing and we have vk i (x) > vk+1 i (x) for all x > 0 and i = 1... k. Proof: We start proving the monotonicity of vi k( ). Let x > y 0 and suppose that vk j (y) vk j (x) for some j {1,..., k}, then we have v k j (y) f j (v k j (y)) vk j (x) f j (vj k (5) (x)). Due to equation (4), the quotient vi k( )/f i(vi k ( )) does not depends on i for i {1,..., k}, so that the inequality (5) is valid for all j = 1,..., k. From the fact that z z/f j (z) is strictly increasing we deduce that v j (y) v j (x) for j = 1,..., k. Adding these inequalities we get the contradiction y x. Let us prove the continuity. Let x, y IR and without loss of generality suppose that x y. Since v k (x) is strictly increasing on each component it follows that v k (x) v k (y) 1 = x y, and v k ( ) is Lipschitz continuous as claimed. Finally let us prove that v k i (x) > vk+1 i (x) for i = 1,..., k. Clearly vk+1 k+1 (x) > 0 = vk k+1 (x) and, since k k+1 x = vi k (x) = vi k+1 (x), 7

it follows that there exists j {1,..., k} such that v k j v k j (x) f j (v k j (x)) > (x) > vk+1 j (x), so that vk+1 j (x) f j (vj k+1 (x)). As mentioned before these quotients do not depend on the index j, and then the inequality holds for all j = 1,..., k. The monotonicity of z z/f j (z) implies that vj k (x) > vk+1 j (x) for all j = 1,..., k. In what follows we use the previous result in combination with the next property of strict monotonicity of the times T k ( ). Lemma 3.3 Let v, ˆv be such that v i > ˆv i for i = 1,..., k. If T k (ˆv) t k then T k (v) > T k (ˆv). Proof: The property is evident for T 1, so that it suffices to consider the case k > 1. Let w be any flow such that T k (w) t k. Observe that for each i = 1... k the time T k may be decomposed as T k (w) = t i + b(t s i (w) t i ) b + f i (w i ) ; b := j s i f j (w j ) (6) where s i denotes the strategy k\{i} whose associated time T si (w) does not depend on w i. From Lemma 3.1 it follows that T k (w) is a scc between T si (w) and t i, so that T si (w) > t i for i < k and T sk (w) t k. Decomposition (6) shows that in the region {w : T k (w) t k } the function T k ( ) is strictly increasing in the variable w i if i < k and non-decreasing in the variable w k. Using this property we can increase sequentially the variables w i from the level ˆv i to the level v i, to conclude that T k (v) > T k (ˆv). Let x k = k v i be the maximal capacity of strategy k. functions: For k {1,..., n 1} we consider the L k (x) := T k (v k (x)), U k (x) := T k (v k+1 (x)). defined over [0, x k ) and [0, x k+1 ) respectively. Combining the two previous Lemmas we obtain, Corollary 3.1 (a) If ˆx [0, x k ) is such that L k (ˆx) t k then L k ( ) is strictly increasing over [ˆx, x k ) with L k (x) as x x k. (b) If ˆx [0, x k+1 ) is such that U k (ˆx) t k then U k ( ) is strictly increasing over [ˆx, x k+1 ) with U k (x) as x x k+1. Proof: The monotonicity properties are immediate from the two previous Lemmas. To prove that L k (x) as x x k, it suffices to observe that in such situation we have vi k(x) v i for i {1,..., k}, and then the total effective frequency k f i (vi k(x)) tends to 0, so that L k(x) = T k (v k (x)) tends to. Similarly it can be shown that U k (x) as x x k+1. From this Corollary it results that for k {1,..., n 1}, the equations L k (x) = t k+1 and U k (x) = t k+1 have exactly one solution l k (0, x k ) and u k (0, x k+1 ) respectively, unless L k (0) t k+1 and/or U k (0) t k+1, in whose cases we define l k = 0 and/or u k = 0. With the convention u 0 = 0 and l n = x n we have, 8

Lemma 3.4 The solutions l k, u k are ordered as u 0 l 1 u 1 l k u k l k+1 l n 1 u n 1 < l n with strict inequality starting from the first positive solution. Proof: From Lemmas 3.2 and 3.3 it results U k (x) < max{t k, L k (x)}, so that if l k > 0 then U k (l k ) < t k+1 and then u k > l k. Similarly, given that L k+1 (x) is a scc between U k (x) and t k+1, if u k > 0 we will have that L k+1 (u k ) = t k+1 < t k+2 and consequently l k+1 > u k. With the preceding developments we are ready to state our main result. Theorem 3.1 For all flow x (0, x n ) there exists a unique Wardrop equilibrium y. In such equilibrium the only strategies s with y s > 0 are of the form s = k. More precisely, depending on the range where the flow x is found, the used strategies will be exactly: { k if x [uk 1, l k ], Proof: k, k + 1 if x (l k, u k ). Existence and characterization of the equilibrium: Case 1: x [u k 1, l k ]. Let us prove that y k = x and y s = 0 for s k, is a Wardrop equilibrium. In fact, the inequality x l k implies T k (v k (x)) t k+1, from where T k (v k (x)) T k+1 (v k (x)). Similarly, x u k 1 implies T k 1 (v k (x)) t k and then T k (v k (x)) T k 1 (v k (x)). Proposition 3.1 allows us to conclude that k is a minimal time strategy. Case 2: x (l k, u k ). In this case we have T k (v k+1 (x)) < t k+1 < T k (v k (x)). Defining v(u) := v k (u) + v k+1 (x u) it follows that T k (v(0)) < t k+1 < T k (v(x)) and, given that u T k (v(u)) is continuous, there exists u (0, x) such that T k (v(u )) = t k+1. From this we get T k (v(u )) = T k+1 (v(u )) = t k+1 and by Proposition 3.1 we conclude that k and k + 1 are the minimal time strategies associated to the flow v(u ). Then, choosing y k = u > 0, y k+1 = x u > 0, and y s = 0 for s k, k + 1, the vector y is a Wardrop equilibrium. Uniqueness of the equilibrium: Let y z be two Wardrop equilibria and let v, w be the corresponding flows in lines. From Proposition 3.1, there exists k and l such that y k > 0, y k+1 0, y s = 0 for s k, k + 1, z l > 0, z l+1 0, z s = 0 for s l, l + 1. Without loss of generality we suppose that l k. And in the case l = k we suppose z k+1 > y k+1. We start proving that w i < v i forall i = 1,..., k. Let us assume by contradiction that there exists 1 j k such that v j w j, so that v j /f j (v j ) w j /f j (w j ). From equation (A) we obtain and similarly v 1 f 1 (v 1 ) = = v k f k (v k ) = y k k f i (v i ) + y k+1 k+1 w 1 f 1 (w 1 ) = = f i(v i ) > y k+1 k+1 f i(v i ) = v k+1 f k+1 (v k+1 ) w l f l (w l ) > w l+1 f l+1 (w l+1 ), 9

So that v i /f i (v i ) w i /f i (w i ), and then v i w i, for i = 1,..., k. From here it follows that v k+1 < w k+1. Indeed, if l > k that comes from the inequality v k+1 f k+1 (v k+1 ) < v k f k (v k ) w k f k (w k ) = w k+1 f k+1 (w k+1 ), and in the case l = k this is a consequence of the assumption z k+1 > y k+1, together with the monotonicity of v k+1 ( ) and the fact that v = v k (y k ) + v k+1 (y k+1 ) and w = v k (z k ) + v k+1 (z k+1 ). From all this we conclude that x = k+1 v i < k+1 w i x, contradiction which establishes the claim: w i < v i for all i = 1,..., k. Let us see that these inequalities contradict the fact that z and y are Wardrop equilibria. Indeed, as T l (w) is minimal and is a scc between T l 1 (w) and t l, it follows that t l T l (w) T k (w). Then, from the inequalities w i < v i for i = 1,..., k and Proposition 3.3 it results T k (w) < T k (v). On the other hand the inequality T k (v) T k+1 (v) also implies T k (v) t k+1, so combining the previous we obtain t l < t k+1, from where l = k. Now as we suppose that in the case l = k we have z k+1 > y k+1, it follows that z k > 0, z k+1 > 0 and then T k (w) = T k+1 (w) = t k+1. As we have seen T k (v) > T k (w) then T k (v) > t k+1. So we obtain T k+1 (v) < T k (v) contradicting the optimality of T k (v). This contradiction completes the proof. Remark 3.1 When the total flow x lies between l k and u k for some k, the mean travel time given by the equilibrium is t k+1. However this is not the social optimum. In fact, if we force the flow x to use only the strategy k + 1 the flows v 1,..., v k will decrease, so that the time of the k-th strategy T k will decrease too. Then the time experimented by all the passengers will be T k+1 = αt k +(1 α)t k+1 < t k+1. This forced flow is not an equilibrium since in this case T k < T k+1. Moreover, this fenomena happens even before the flow x reaches the level l k. In fact it is easy to see that this happens as long as the flow x is greater than the smaller positive solution θ k of the equation T k (v k (x)) = T k+1 (v k+1 (x)) and the k-th strategy is being used. Notice that T k+1 (v k+1 (θ k )) < αt k (v k (θ k )) + (1 α)t k+1 so that we have θ k < l k. 4 Example In this section we provide a simple example illustrating the previous results. For simplicity we only consider two lines, which is enough to show that the predicted effects of the last Theorem are not marginal. Example: We consider a naïve point of view. We take the case where the buses have finite capacity K and each of the lines has a nominal frequency f i, i = 1, 2. Then, a passenger waiting at node O will see an effective frequency (in terms of number of seats by unit of time) of K f i v i for the line i. Hence we consider the frequency functions: f i (v i ) := f i v i /K. A simple calculation leads us to: l 1 = K( f 1 1 t 2 t 1 ) and u 1 = (1 + f 2 / f 1 )l 1, Moreover we can compute the equilibrium value of y 1 in terms of x: x if x l 1 f y 1 (x) = f 1 + 2 1 (t 2 t 1 )( f 1 + f if x (l 2 x/k) 1, u 1 ) 0 if x u 1 10

and obviously y 2 (x) = x y 1 (x). With all this we can explicitly determine x if x l 1 v 1 (x) = l 1 if x (l 1, u 1 ) x f 1 /( f 1 + f 2 ) if x u 1 v 2 (x) = 0 if x l 1 x l 1 if x (l 1, u 1 ) x f 2 /( f 1 + f 2 ) if x u 1. The equilibrium time turns out to be T 1 (x) if x l 1 T (x) = t 2 if x (l 1, u 1 ) T 2 (x) if x u 1. The following figure summarizes the situation: T t 2 l u 1 1 x y y 2 1 l u 1 1 x In the range x (0, l 1 ) the equilibrium strategy is 1, while in the range x (l 1, u 1 ) we have both 1 and 2 being used. Notice that when f 1 = f 2 the range (l 1, u 1 ) is as wide as (0, l 1 ) so that the case of equilibria with two strategies can not be considered as a marginal or degenerate case. 11

To illustrate what was said in the Remark let us suppose that K = 10; f1 = f 2 = 10; t 1 = 1; t 2 = 1.5. Then l 1 = 80 and u 1 = 160 so if x = 100 the time given by the equilibrium is 1.5. However if we force all the flow to use only the strategy 2 we will have that v 1 = v 2 = 50 and with this T 2 = 1.35 < 1.5. This will happen for flows larger than the first positive solution of the equation T 1 (v 1 (x)) = T 2 (v 2 (x)), which in our case is simply T 1 (x) = T 2 (x/2, x/2), and whose smaller positive solution is x = 69.377423..., and up to the flow level u 1 where strategy 1 is no longer used. 5 Extension Consider now the situation where the in-vehicle travel times t 1,..., t n are not ordered by strict inequality, as supposed before, but as t 1 t 2 t n. We will see that in this case in general there is existence of equilibrium but no uniqueness can be assured. We illustrate this statement by a simple example as follows. Let us consider three bus lines with on board times t 1 = 1; t 2 = 1.5; t 3 = 1.5, nominal frequencies f 1 = f 2 = f 3 = 10, capacity K = 10 and frequency functions given by f i = 10 v i 10 for i = 1, 2, 3. We consider a total flow x = 100. For u [0, 20] let (v 1, v 2, v 3 ) be the vector of flow with components v 1 = 80, v 2 = u, v 3 = 20 u, which can be obtained from the flows y in many ways, one possibility is to take y 12 = (20 u)(100 + u) u(120 u) ; y 13 = 80 + u 100 u ; y 1 = 100 y 12 y 13 ; y s = 0 s {1}, {12}, {13}. The vector y described above (or any other that generates the same line flow) is a Wardrop equilibrium. In fact we have that T 1 (v 1 ) = 1.5; T 12 (v 1, v 2 ) = T 13 (v 1, v 3 ) = T 123 (v 1, v 2, v 3 ) = 1.5 (which comes straightforward from Lemma 3.1 and does not depend on the frequency functions) and T 23 = 1. 5. With this the only strategies that can be used are such of minimal time, that is {1}; {1, 2}; {1, 3}; {1, 2, 3}. For that reason a flow as the previously described is a Wardrop equilibrium. The question comes immediately: Is there a most natural equilibrium? The answer we can give is based only on intuitive arguments: the only strategies that should be used are {1} and {1, 2, 3}. Indeed, suppose that a passenger only consents to board the lines 1 and 2, and suddenly arrives a bus from line 3. Will he discard the bus without boarding it knowing that the next bus could be from line 2, which has the same in-vehicle travel time as line 3? Since there is no way to distinguish a bus from lines 2 and 3 once it arrives at the bus stop, it seems natural aggregate lines 2 and 3 as a single line and then the only used strategies will be {1} and {1, 2, 3}. In general if we have t 1 t 2 t k < t k+1 = = t k+i t n we should not consider strategies of the form {1,..., k} J with J {k + 1,..., k + i} except J = φ and J = {k + 1,..., k + i}. References [1] De Cea J., Fernández E., Transit Assignment for Congested Public Transport Systems: An Equilibrium Model, Transportation Science Vol 27, No 2, pp 133-147. [2] Chriqui, Robillard, Common bus lines, Transportation Science 9, 115-121, 1975. [3] Hao Wu, Florian Marcotte, Transit equilibrium assignment: a model solution algorithms, Transportation Science Vol 28, No 3, August 1994. 12

[4] Nguyen, Pallotino, Equilibrium traffic assignment in large scale transit networks, European Journal of Operational Research 37, 1988. [5] R.T. Rockafeller, Convex Analysis, Princeton University Press, 1970. [6] Spiess, Florian, Optimal strategies: a new assignment model for transit networks, Transportation Research, Vol 23B, No 2, pp 83-102, 1989. 13