Deux problèmes sur les réseaux

Deux problèmes sur les réseaux

1er octobre 2008

Ecole des Ponts, France

Affectation dynamique du trafic : il y a un équilibre ! avec Nicolas Wagner.

Traffic assignment

Traffic assignment

Traffic assignment

Traffic assignment

Static traffic assignment

Supply A directed graphG = (V,A), acost function c_a :R+ →R+for eacha∈A.Assumptions : c_a isnon-decreasingandcontinuousandGis strongly connected.

Demand Ademand functionb :V ×V →R+.

Realization Aflowx : R →R+ whereR = (R_o,d)_(o,d)∈V×V, such that X

r∈R_o,d

x(r) = b(o,d), withR_o,d being the set of allo–d route ofG.

Equilibrium Wheneverr,r⁰ ∈R_o,d, we have

x(r)>0⇒X

a∈r

c_a(x_a)≤ X

a∈r⁰

c_a(x_a),

Static traffic assignment

Supply A directed graphG = (V,A), acost function c_a :R+ →R+for eacha∈A.Assumptions : c_a isnon-decreasingandcontinuousandGis strongly connected.

Demand Ademand functionb :V ×V →R+.

Realization Aflowx : R →R+ whereR = (R_o,d)_(o,d)∈V×V, such that X

r∈R_o,d

x(r) = b(o,d), withR_o,d being the set of allo–d route ofG.

Equilibrium Wheneverr,r⁰ ∈R_o,d, we have x(r)>0⇒X

a∈r

c_a(x_a)≤ X

a∈r⁰

c_a(x_a),

wherex_a :=P

r∈R:a∈rx(r)(Wardrop ).

Static traffic assignment

Supply A directed graphG = (V,A), acost function c_a :R+ →R+for eacha∈A.Assumptions : c_a isnon-decreasingandcontinuousandGis strongly connected.

Demand Ademand functionb :V ×V →R+.

Realization Aflowx : R →R+ whereR = (R_o,d)_(o,d)∈V×V, such that X

r∈R_o,d

x(r) = b(o,d), withR_o,d being the set of allo–d route ofG.

Equilibrium Wheneverr,r⁰ ∈R_o,d, we have x(r)>0⇒X

a∈r

c_a(x_a)≤ X

a∈r⁰

c_a(x_a),

Static traffic assignment

Supply A directed graphG = (V,A), acost function c_a :R+ →R+for eacha∈A.Assumptions : c_a isnon-decreasingandcontinuousandGis strongly connected.

Demand Ademand functionb :V ×V →R+.

Realization Aflowx : R →R+ whereR = (R_o,d)_(o,d)∈V×V, such that X

r∈R_o,d

x(r) = b(o,d), withR_o,d being the set of allo–d route ofG.

Equilibrium Wheneverr,r⁰ ∈R_o,d, we have x(r)>0⇒X

a∈r

c_a(x_a)≤ X

a∈r⁰

c_a(x_a),

wherex_a :=P

r∈R:a∈rx(r)(Wardrop ).

Static traffic assignment : results

Theorem

There is always an equilibrium. Moreover, the values c_a P

r∈R:a∈rx(r)

at equilibrium are unique.

Formulation as a convex program→computable.

Static traffic assignment : limitation

This model has alimited significance:

the time needed to travel along the route is not modelled.Each user occupies the whole route continuously.

→need ofdynamic traffic assignment model.

Static traffic assignment : limitation

This model has alimited significance:

the time needed to travel along the route is not modelled.Each user occupies the whole route continuously.

→need ofdynamic traffic assignment model.

Dynamic traffic assignment

Since the 70’s, several models has been proposed, for instance :

Vickrey (1969)

Merchant and Nemhauser (1978) Friez and al. (1989)

90’s : Leurent (LADTA), Bellei, Gentile and Papola, Akamatsu and Kuwahara

Roughly speaking, the models are such that

Dynamic traffic assignment model

Time intervalI = [0,H].

Users aredynamic flows:x :R×I. The quantityx(r,h)is the number ofusers choosing the router at timeh.

y_a :I →R+. The quantityy_a(h)is the number ofusers entering the arcaat timeh.

Useful notation : X_r(h) :=

Z h h⁰=0

x(r,h⁰) and Y_a(h) :=

Z h h⁰=0

y_a(h⁰)

(cumulated flow).

Dynamic traffic assignment model

Supply A strongly connected directed graphG = (V,A), aarc travel time functiont_afor eacha∈Asuch thatt_a(Y_a)is aI →R+ continuous map,

depending continuously of the cumulated flowYa. Demand Ademand functionb :V ×V ×I →R+.

Realization X

r∈Ro,d

x(r,h) =b(o,d,h)for all(o,d)∈V ×V andh∈I.

Equilibrium Wheneverr,r⁰ ∈Ro,d, we have

x(r,h)>0⇒t_r(X~)(h)≤t_r⁰(X~)(h),

where forr =a₁. . .an, we havetr(X~)(h) := Pn

i=1ta_i(Ya_i)(h_i) withY_ai :=φ_ai(X)and

h₁:=handh_i+1:=h_i +t_ai(Y_ai)(h_i)fori =1, . . . ,n−1.

Dynamic traffic assignment model

Supply A strongly connected directed graphG = (V,A), aarc travel time functiont_afor eacha∈Asuch thatt_a(Y_a)is aI →R+ continuous map,

depending continuously of the cumulated flowYa. Demand Ademand functionb :V ×V ×I →R+.

Realization X

r∈Ro,d

x(r,h) =b(o,d,h)for all(o,d)∈V ×V andh∈I.

Equilibrium Wheneverr,r⁰ ∈Ro,d, we have

x(r,h)>0⇒t_r(X~)(h)≤t_r⁰(X~)(h),

where forr =a . . .a , we havet (X~)(h) := Pn

t (Y )(h)

Dynamic traffic assignment model

Supply A strongly connected directed graphG = (V,A), aarc travel time functiont_afor eacha∈Asuch thatt_a(Y_a)is aI →R+ continuous map,

depending continuously of the cumulated flowYa. Demand Ademand functionb :V ×V ×I →R+.

Realization X

r∈Ro,d

x(r,h) =b(o,d,h)for all(o,d)∈V ×V andh∈I.

Equilibrium Wheneverr,r⁰ ∈Ro,d, we have

x(r,h)>0⇒t_r(X~)(h)≤t_r⁰(X~)(h),

where forr =a₁. . .an, we havetr(X~)(h) := Pn

i=1ta_i(Ya_i)(h_i) withY_ai :=φ_ai(X)and

h₁:=handh_i+1:=h_i +t_ai(Y_ai)(h_i)fori =1, . . . ,n−1.

Dynamic traffic assignment model

Supply A strongly connected directed graphG = (V,A), aarc travel time functiont_afor eacha∈Asuch thatt_a(Y_a)is aI →R+ continuous map,

depending continuously of the cumulated flowYa. Demand Ademand functionb :V ×V ×I →R+.

Realization X

r∈Ro,d

x(r,h) =b(o,d,h)for all(o,d)∈V ×V andh∈I.

Equilibrium Wheneverr,r⁰ ∈Ro,d, we have

x(r,h)>0⇒t_r(X~)(h)≤t_r⁰(X~)(h),

where forr =a . . .a , we havet (X~)(h) := Pn

t (Y )(h)

Dynamic traffic assignment

TheY_acan be computed from thet_aknowing allX_r (under some assumptions).

Existence of an equilibrium ? In general, it is an open question.

Zhu and Marcotte prove an equilibrium for Friez’s model (with a stronger assumptions on travel time).

Ladta : Unknown.

Lindsey + Leurent, Wagner studied time departure choice on one arc (Vickrey model) : their is an equilibrium.

Dynamic traffic assignment

TheY_acan be computed from thet_aknowing allX_r (under some assumptions).

Existence of an equilibrium ? In general, it is an open question.

Zhu and Marcotte prove an equilibrium for Friez’s model (with a stronger assumptions on travel time).

Ladta : Unknown.

Lindsey + Leurent, Wagner studied time departure choice on one arc (Vickrey model) : their is an equilibrium.

Results

Our result

• a general model with minimal assumption :continuity, fifoness (in a weak form), causality, no infinite speed.

• existence of an equilibrium that containsallprevious results.

Traffic assignement as a continuous game

R: set ofroutes,I := [0,H]time interval.

M(R×I): set of measures onR×I (choice of time departure allowed).

T_r : set of continuous mapsM(R×I)→ C(R,R)(t_r ∈ T_r is such thattr(~X)(h)gives the time needed to traverse the route r).

user: characterized by a functionu : S

r∈RT_r

×R×I →R¯ (that is upper semicontinuous inr andh, and continuous in the travel times, (theutility-function) :u(~t,r,h): a choice(r,h)7→

the payoff, when route travel times are~t := t_r1,t_r2, . . . ,.

U: space of usersu.

Atraffic gameis a measureU onU.Number of users:

Traffic equilibrium

Arealizationof the traffic game is a measure onU ×R×I.

Denotetr₁,tr₂, . . .by~t.

Given a traffic gameU, a realizationDis aCournot-Nash equilibriumif we haveD_U =U and

D {(u,r,h) : for each(r⁰,h⁰)∈R×I, u(~t(D_R×I),r,h)≥u(~t(D_R×I),r⁰,h⁰) o

=N

D_R×I is exactly the cumulated flowsX. Set X_r(J) := D_R×I({r} ×J).

Khan-Mas-Colell’s theorem ?

Khan-Mas-Colell’s theorem tells us thatthere is a Nash equilibrium, provided that the utility function are continuous in the realization of the game (=here the travel times).

For traffic assignement :DR×I 7→u(~t(DR×I),·,·)mustbe continuous.

Reformulation : Our purpose : prove that~X 7→~t(~X)is continuous.

Khan-Mas-Colell’s theorem ?

Khan-Mas-Colell’s theorem tells us thatthere is a Nash equilibrium, provided that the utility function are continuous in the realization of the game (=here the travel times).

For traffic assignement :DR×I 7→u(~t(DR×I),·,·)mustbe continuous.

Reformulation : Our purpose : prove that~X 7→~t(~X)is continuous.

Khan-Mas-Colell’s theorem ?

Khan-Mas-Colell’s theorem tells us thatthere is a Nash equilibrium, provided that the utility function are continuous in the realization of the game (=here the travel times).

For traffic assignement :DR×I 7→u(~t(DR×I),·,·)mustbe continuous.

Reformulation : Our purpose : prove that~X 7→~t(~X)is continuous.

Arc exit time

Givent_a, thearc exit timefunction is

Ha(Y)(h) := h+ta(Y)(h) forY ∈ M(R)andh∈R

Assumptions on travel time

Continuity Ha :M(R)→ C(R)is continuous.

Limited speed there existst_min >0such that for all Y ∈ M(R)and allh∈R, we have H_a(Y)(h)>h+t_min.

Fifo Leth₁ <h₂inRand letY ∈ M(R). Whenever Y[h₁,h₂]6=0, we haveH_a(Y)(h₁)<H_a(Y)(h₂).

Causality For allh ∈RandY ∈ M(R), we have Ha(Y|_h)(h) = Ha(Y)(h).

Assumptions on travel time

Continuity Ha :M(R)→ C(R)is continuous.

Limited speed there existst_min >0such that for all Y ∈ M(R)and allh∈R, we have H_a(Y)(h)>h+t_min.

Fifo Leth₁ <h₂inRand letY ∈ M(R). Whenever Y[h₁,h₂]6=0, we haveH_a(Y)(h₁)<H_a(Y)(h₂).

Causality For allh ∈RandY ∈ M(R), we have Ha(Y|_h)(h) = Ha(Y)(h).

Assumptions on travel time

Continuity Ha :M(R)→ C(R)is continuous.

Limited speed there existst_min >0such that for all Y ∈ M(R)and allh∈R, we have H_a(Y)(h)>h+t_min.

Fifo Leth₁ <h₂inRand letY ∈ M(R). Whenever Y[h₁,h₂]6=0, we haveH_a(Y)(h₁)<H_a(Y)(h₂).

Causality For allh ∈RandY ∈ M(R), we have Ha(Y|_h)(h) = Ha(Y)(h).

Assumptions on travel time

Continuity Ha :M(R)→ C(R)is continuous.

Limited speed there existst_min >0such that for all Y ∈ M(R)and allh∈R, we have H_a(Y)(h)>h+t_min.

Fifo Leth₁ <h₂inRand letY ∈ M(R). Whenever Y[h₁,h₂]6=0, we haveH_a(Y)(h₁)<H_a(Y)(h₂).

Causality For allh ∈RandY ∈ M(R), we have Ha(Y|_h)(h) = Ha(Y)(h).

Arc flowing function

Let~Y_abe inM(R×R); the flow of users following router and leaving the arcaon the time subsetJis :

ψ_a^r(~Y_a)(J) :=ψa(Y~_a)({r}×J) :=

Y_a^r(H_a(Y_a)⁻¹(J)) ifa ∈r

0 if not.

(1)

Flows on routes → flow on arcs

AnoutflowofX~ is a measureYa :=P

rY_a^r onRsuch that for everyr =a₁a₂..a_n:

1 Y_a^r

1 =X_r

2 Y_a^r

i =ψ_r^ai−1(Y~a_i−1)fori =2..n

3 Y_a^r =0ifa ∈/ r

Uniqueness and continuity of the outflow

Proposition

GivenX~, there exists a unique outflowY_a. Moreover, the map φa :X~ 7→Ya is continuous.

Continuity of t

is proved !

Continuity oftr is proved.

Indeed

t_r(X~)(h) :=

n

X

i=1

t_ai(Y_ai)(h_i)

withh₁ :=handh_i+1 :=h_i +t_ai(Y_ai)(h_i)fori = 1, . . . ,n−1 can be rewritten :

t_r(~X)(h) = H_an

φ_an(~X)

◦. . .◦H_a1

φ_a1(~X)

(h)−h for allh ∈I.

Continuity of t

is proved !

Continuity oftr is proved.

Indeed

t_r(X~)(h) :=

n

X

i=1

t_ai(Y_ai)(h_i)

withh₁ :=handh_i+1 :=h_i +t_ai(Y_ai)(h_i)fori = 1, . . . ,n−1 can be rewritten :

t_r(~X)(h) = H_an

φ_an(~X)

◦. . .◦H_a1

φ_a1(~X)

(h)−h for allh ∈I.

There is an equilibrium

Continuity of thet_r ⇒continuity of the utility functionu(t_r,·,·)

⇒we can apply the theorem.

Theorem

Given a directed graphG= (V,A)with arc travel time

functions(ta)a∈Asatisfying assumptions of causality, fifoness, limited speed and continuity and given a measureUon the set of possible users (identified with their utility function), there is a Nash equilibrium.

There is an equilibrium

Continuity of thet_r ⇒continuity of the utility functionu(t_r,·,·)

⇒we can apply the theorem.

Theorem

Given a directed graphG= (V,A)with arc travel time

functions(ta)a∈Asatisfying assumptions of causality, fifoness, limited speed and continuity and given a measureUon the set of possible users (identified with their utility function), there is a Nash equilibrium.

A versatile model

The model is reallyversatileand contains previous models.

Playing with the mapu(~t(X~),r,h),

1 we cover also the case when there is no possible choice of time departure.

2 we can put arbitrarily taxes on some routes or time departure.

They will always be a Nash equilibrium : contains previous results of equilibrium.

A versatile model

The model is reallyversatileand contains previous models.

Playing with the mapu(~t(X~),r,h),

1 we cover also the case when there is no possible choice of time departure.

2 we can put arbitrarily taxes on some routes or time departure.

They will always be a Nash equilibrium : contains previous results of equilibrium.