Optimum and equilibrium in a transport problem with queue - - PowerPoint PPT Presentation

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Optimum and equilibrium in a transport problem with queue - - PowerPoint PPT Presentation

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Optimum and equilibrium in a transport problem with queue penalization effect. G. Crippa, (University of Parma (Italy)) C. Jimenez, (Universit de Bretagne Occidentale) A. Pratelli, (University of Pavia (Italy).) C. Jimenez, UBO JFCO 2008

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SLIDE 1

Optimum and equilibrium in a transport problem with queue penalization effect.

  • G. Crippa,

(University of Parma (Italy))

  • C. Jimenez,

(Université de Bretagne Occidentale)

  • A. Pratelli,

(University of Pavia (Italy).)

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 2

Presentation of the Problem

City: Ω ⊂ Rd bounded open set, k post-offices : x1, ..., xk ∈ Ω fixed, Population density: f dx a probability, Unknown: partition (Ai)i=1,...k of Ω: every person linving on Ai goes to xi. Time lost by a citizen living at x ∈ Ai: number of persons going to xi: ci =

  • Ai f(x) dx

Time = Displacement + queue = |x − xi|p + hi(ci) p ≥ 1. Total cost =

k

  • i=1
  • Ai

(|x − xi|p + hi(ci)) f(x)dx.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 3

Associated optimization problem

inf(Ai)i

  • k

i=1

  • Ai

|x − xi|p f(x)dx +

  • Ai

f(x)dx

  • hi
  • Ai

f(x)dx

  • (Ai)i=1,...k partition of Ω
  • = inf(ci)i
  • inf

(Ai)i partition of Ω

  • k
  • i=1
  • Ai

|x − xi|p f(x)dx :

  • Ai f(x)dx = ci, ci ≥ 0
  • +

i cihi(ci) : i ci = 1

  • = inf(ci)i
  • wp

p (fdx, k i=1 ciδxi) + k i=1 cihi(ci) : ci ≥ 0, i ci = 1

  • Wp(fdx, k

i=1 ciδxi) is the p-Wasserstein distance.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 4

Optimal Transportation

Wasserstein distance from fdx to k

i=1 ciδxi

Wp(fdx,

k

  • i=1

ciδxi) = inf

T

  • Ω

|x − Tx|pf(x)dx T(x) = xi ∀x ∈ Ai where (Ai)i=1,...k is a partition Ω such that:

  • Ai f(x)dx = ci.

Existence of a transport map It exists an optimal transport map T for Wp. p = 2 Brenier (87). p > 1 Rüschendorf (95), Gangbo, McCann (96). p = 1 Sudakov (79), Gangbo, McCann (96), Evans, Gangbo (99), Caffarelli, Feldmann, McCann (02), Ambrosio, Pratelli (03)...

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 5

Optimality transportation

Kantorovich duality Wp(fdx, k

i=1 ciδxi)

= supu, (αi)i

Ω

u(x) f(x)dx +

k

  • i=1

ciαi : u ∈ L1(Ω), u(x) − αi ≤ |x − xi|p a.e.x ∈ Ω

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 6

Optimal transportation

Primal-Dual Optimality conditions (Bouchitté) (u, (αi)i, T)

  • ptimal
  • ⇒

       T =

i xi1Ai

u(x) = infi |x − xi|p − αi =

i (|x − xi|p − αi) 1Ai

Ai = {x ∈ Ω : |x − xi|p − αi < |x − xi|p − αj} u(x) = infi |x − xi|p − αi =

i (|x − xi|p − αi) 1Ai

  • ⇒

T =

i xi1Aiand

u, (αi)i optimal. Consequence: the optimal partition is always unique.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 7

The optimization problem

inf(ci)i{W p

p (fdx, k i=1 ciδxi) + k i=1 cihi(ci) : ci ≥ 0, i ci = 1}

W p

p (fdx, k

  • i=1

ciδxi) = inf

(Ai)i

{

  • i
  • Ai

|x − xi| f(x)dx} under the constraints: (Ai)i is a partition of Ω, ci =

  • Ai

f(x)dx. Existence of an optimum If t → thi(t) is l.s.c. for any i = 1, ..k, it exists an optimal partition of Ω. Moreover if t → thi(t) is strictly convex, the

  • ptimal partition is unique.
  • C. Jimenez, UBO

JFCO 2008

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SLIDE 8

The optimization problem

Necessary and Sufficient Optimality Condition We assume hi is regular and t → thi(t) is convex for any i = 1, ..k, then a partition (Ai)i is optimal iff (up to negligible sets):    Ai = {x ∈ Ω : |x − xi|p + hi(ci) + cih′

i(ci)

< |x − xj|p + hj(cj) + cjh′

j(cj)}

ci =

  • Ai f(x)dx.

Moreover, there is only one partition which satisfies this condition.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 9

Example

Let Ω = [0, 1], f = 1, x1 = 0, x2 = 1, p = 1 and: h1(s) = 100, and h2(s) = for 0 ≤ s ≤ 0.999 1 for 0.999 < s ≤ 1. Optimum: A1 = [0, 0.001[, A2 =]0.001, 1]... Costumers in A1 may not be happy! A costumer living at x ∈ Ai will be happy if: |x − xi|p + hi(ci) = inf

j {|x − xj|p + hi(cj)}.

In the example: A1 = ∅, A2 = [0, 1].

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 10

The equilibrium problem

Nash Equilibrium A partition (Ai)i=1,..k is a Nash equilibrium if: Ai = {x ∈ Ω : |x − xi|p + hi(ci) < |x − xj|p + hj(cj)} ci =

  • Ai f(x)dx.

Existence Assume hi is continuous for any i = 1, ..k and gi is such that tg′

i(t) + gi(t) = hi(t).

Then, the minimizer of the following problem is an equilibrium: inf

(Ai)i partition ofΩ

  • i
  • Ai

|x − xi|p + gi

  • Ai

f(x)dx

  • f(x)dx
  • .

If, in addition, hi is non-decreasing, the equilibrium is unique.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 11

Example

Travellers game Traveller 1 2 ≤ t1 ≤ 100 t1 ∈ N Traveller 2 2 ≤ t2 ≤ 100 t2 ∈ N t1 = t2 t1 t2 t1 < t2 t1 + 2 t1 − 2 t1 > t2 t2 − 2 t2 + 2 Equilibrium: t1 = t2 = 2.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 12

Pareto Optimum

If (Ai)i is an equilibrium, is it possible to find another partition (Bi)i such that every citizen spends equal or less time ? Individual cost: C(x, (Bi)i) =

k

  • i=1
  • |x − xi|p + hi
  • Bi

f(x)dx

  • 1Bi(x).

Pareto Optimum A partition (Ai)i=1,..k is a Pareto optimum if there exists NO

  • ther partition (Bi)i of Ω such that:

C(x, (Bi)i) < C(x, (Ai)i) a.e. x ∈ Ω. Proposition Assume hi is strictly increasing. Then every equilibrium is a Pareto optimum.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 13

Sketch of proof

Important remark (Ai) is an equilibrium ⇔ C(x, (Ai)i) = infi=1,...k

  • |x − xi|p + hi
  • Ai f(x)dx
  • .

Remember the optimality condition for optimal transportation (slide 5)

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 14

Sketch of proof

Link between equilibrium and optimal transportation If (Ai)i is an equilibrium then: T =

i xi1Ai is optimal for Wp(f, i

  • Ai f(x)dx
  • δxi),

u(x) = C(x, (Ai)i) and (αi)i = −hi

  • Ai f(x)dx
  • are optimal

for the dual formulation of Wp. If (Ai)i is a partition such that u(x) = C(x, (Ai)i) and (αi)i = −hi

  • Ai f(x)dx
  • are optimal for the dual

formulation of Wp, then: (Ai)i is an equilibrium T =

i xi1Ai is optimal for Wp(f, i

  • Ai f(x)dx
  • δxi).
  • C. Jimenez, UBO

JFCO 2008

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SLIDE 15

Sketch of proof

Assume hi ր, (Ai)i is an equilibrium, (Bi)i is such that C(x, (Bi)i) ≤ C(x, (Ai)i)a.e. Let us show:

  • Ai f(x)dx =
  • Bi f(x)dx for every i.

Assume ∃j such that

  • Aj f(x)dx <
  • Bj f(x)dx. Then:

∀x ∈ Bj |x − xj|p + hj(

  • Bj

f(x)dx) = C(x, (Bi)i) ≤ C(x, (Ai)i) = min

i {|x − xi|p + hi(

  • Ai

f(x)dx)} ≤ |x − xj|p + hj(

  • Aj

f(x)dx). This is impossible because hj is strictly increasing.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 16

Sketch of proof

As (Ai)i is an equilibrium, T =

i xi1Ai is an optimal

transport map for Wp(f,

i

  • Ai f(x)dx
  • δxi), by

consequence:

  • i
  • Ai

|x − xi|p f(x)dx ≤

  • i
  • Bi

|x − xi|p f(x)dx. But as C(x, (Bi)i) ≤ C(x, (Ai)i) and

  • Ai f(x)dx =
  • Bi f(x)dx:
  • i
  • Ai

|x − xi|p f(x)dx ≥

  • i
  • Bi

|x − xi|p f(x)dx. By uniqueness of the optimal transport map, (Ai)i = (Bi)i up to negligible sets.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 17

Dynamic behaviour for 2 points

For simplicity, assume: Ω = [0, 1], x1 = 0, x2 = 1, f = 1, p = 1. At every day n, we set tn such that: A1 = [0, tn[, A2 =]tn, 1]. Day 1: the citizen has no information on the queue: t1 = 1/2, End of day: he knows h1(t1), h2(1 − t1). Day n: he knows h1(tn−1), h2(1 − tn−1), tn+1 + h1(tn) = (1 − tn+1) + h2(1 − tn) tn+1 := h2(1 − tn) − h1(tn) + 1 2 . Equilibrium: ¯ t such that: ¯ t = h2(1−¯

t)−h1(¯ t)+1 2

.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 18

Dynamic behaviour for 2 points

Convergence Let G(t) := h2(1−t)−h1(t)+1

2

. If G is a contraction mapping, then tn → ¯ t.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 19

Dynamic with memory

Assume the citizen "remembers" the K last days. The map G is the same as before but we set: tn+1 = K

i=1 G(tn−K+i)

K . Convergence Assume G is L-lipschitz with L < K, then tn → t.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 20

Sketch of proof for K = 2

Assume that for every k ≤ n : (i) |G(tk) − G(¯ t)| ≤ α (ii) |G(tk) − G(¯ t) + G(tk−1) − G(¯ t)| ≤ α Recall tn+1 = G(tn)+G(tn−1)

2

. |G(tn+1) − G(¯ t)| ≤ L|tn+1 − ¯ t| Lipschitz Property = L|tn+1 − G(¯ t)| ¯ t is a fixed point = L

  • G(tn) − G(¯

t) + G(tn−1) − G(¯ t) 2

  • ≤

Lα 2 by (ii) .

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 21

Sketch of proof for K = 2

Note that G ց.

(case 1) If (G(tn+1) − G(¯ t) and (G(tn) − G(¯ t) have different signs we have easily: |G(tn+1) − G(¯ t) + G(tn) − G(¯ t)| ≤ α. (case 2) Let us assume tn, tn+1 > ¯ t so that G(tn) − G(¯ t) ≤ 0. Then: tn+1−¯ t = tn+1 = G(tn) + G(tn−1) 2 −¯ t > 0 ⇒ G(tn−1)−G(¯ t) ≥ 0. Hence: |G(tn+1) − G(¯ t) + G(tn) − G(¯ t)| = |G(¯ t) − G(tn+1) + G(¯ t) − G(tn)| ≤ L(tn+1 − G(¯ t) + G(¯ t) − G(tn)) = L G(tn) + G(tn−1) 2 − G(¯ t)

  • + G(¯

t) − G(tn) ≤ (1 − L/2)α + L/2α = α.

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 22

Sketch of proof for K = 2

At time n + 1 we have:

  • (i)

|G(tn+1) − G(¯ t)| ≤ Lα

2

(ii) |G(tn+1) − G(¯ t) + G(tn) − G(¯ t)| ≤ α By repeating exactly the same arguments: (i) |G(tn+2) − G(¯ t)| ≤ Lα

2

(ii) |G(tn+2) − G(¯ t) + G(tn+1) − G(¯ t)| ≤ Lα

2

etc...

  • C. Jimenez, UBO

JFCO 2008

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SLIDE 23

Dynamic with memory

Convergence Assume G is strictly L-lipschitz with L ≤ K, then tn → t. Infinite memory: tn = n−1

i=1 G(ti)

n − 1 . Convergence Assume G is lipschitz, then tn → t.

  • C. Jimenez, UBO

JFCO 2008