The Price of Routing Unsplittable Flow Yossi Azar Joint work with B. Awerbuch and A. Epstein

Outline Game Theory and Selfish Routing Price of Anarchy Network Model – Previous Results Network Model – Our Results

Selfish Routing Large networks Users are selfish Infeasible to maintain a central authority Users are selfish Each user tries to minimize its cost Each user is aware of network conditions Degradation of network performance

Nash Equilibrium Game Theory Nash Equilibrium Study and predict user behavior Nash Equilibrium Each agent minimizes its cost /maximizes its benefit No agent has an incentive change its behavior

Example-Prisoner’s Dilemma Nash Equilibrium (c,c) D C (-1,-1) (-4,0) (0,-4) (-3,-3)

Nash Equilibrium H T (1,-1) (-1,1) Every game has randomized Nash equilibrium In general a game may not have pure Nash equilibrium No Deterministic Nash Equilibrium Randomized Strategy pi,j= 0.5 is in N.E H T (1,-1) (-1,1)

Parallel Links (Machines) Model Two nodes m parallel (related) links n jobs User cost (delay) is proportional to link load Global cost (maximum delay) is the maximum link load

Nash Equilibrium - Example 2 identical links 4 jobs with weights : 1,2,3,4 2 Not a Nash Equilibrium 1 4 3 Nash Equilibrium m1 m2

Nash Equilibrium - Example 1 2 Optimal Solution 4 3 Also Nash Equilibrium m1 m2

Price of Anarchy Price of Anarchy (coordination ratio) : The worst possible ratio between: Global cost in Nash Equilibrium and Global cost in Optimum Global cost: maximum/total user’s cost

General Network Model A directed Graph G=(V,E) A load dependent latency function fe(.) for each edge e n users Bandwidth request (si, ti, wi) for user i Goal : route traffic to minimize total latency

Example Latency function f(x)=x 1 2 2 1 1 t s 2 2 2 2 Latency=2+1+2=5 Total latency =Σe fe(le)·le= Σe le· le =6·2·2+3·1·1=27

Example Traffic rate r=1 Nash total latency=1·0+1·1=1 f(x)=1 l=0 t s f(x)=x l=1

Example Traffic rate r=1 Optimal total latency=1·1/2+1/2·1/2=3/4 R=4/3 f(x)=1 l=1/2 t s f(x)=x l=1/2

Braess’s Paradox Traffic rate r=1 Optimal cost=Nash cost=2(1/2·1+1/2·1/2)=3/2 fl(x)=1 l=1/2 f(x)=x l=1/2 v t s f(x)=1 l=1/2 f(x)=x l=1/2 w

Braess’s Paradox Traffic rate r=1 Optimal cost did not change Nash cost=1·1+0·1+1·1=2 Adding edge negatively impact all agents fl(x)=1 l=0 f(x)=x l=1 v f(x)=0 l=1 t s f(x)=1 l=0 f(x)=x l=1 w

Related Work-General Network Roughgarden and Tardos (FOCS 2000) Assumption : each user controls a negligible fraction of the overall traffic Results : Linear latency functions - R=4/3 Continuous nondecreasing functions-bicriteria results Results hold also for nonnegligible splittable case (Roughgarden – SODA 2005) Without negligibility assumption : no general results

Our Results Unsplittable Flow, general demands Linear Latency Functions For weighted demands the price of anarchy is exactly 2.618 (pure and mixed) For unweighted demands the price of anarchy is exactly 2.5. Polynomial Latency Functions The price of anarchy - at most O(2ddd+1) (pure and mixed) The price of anarchy - at least Ω(dd/2)

Remarks Valid for congestion games Approximate computation (i.e approximate Nash) has limited affect

Routes in Nash Equilibrium Pure strategies – user j selects single path Q Qj Mixed strategies – user j selects a probability distribution {pQ,j} over all paths Q Qj

Routes in Nash Equilibrium Definition ( Pure Nash equilibrium): System S of pure strategies is in Nash equilibrium iff for every j {1,...,n}and Q’  {Qj} : , where Qj – path associated with request j

Example Latency function f(x)=x Path Q1 1 USER 1 : W1=1 2 2 1 1 Path Q CQ1,1 =2+1+2=5 CQ,1 =2+(1+1)+(1+1)+2=8

Routes in Nash Equilibrium Definition : The expected cost C(S) of system S of mixed strategies is (i.e. the expected total latency incurred by S)

Linear Latency Functions fe(x)=aex+be for each eE Theorem : For linear latency functions (pure strategies) and weighted demands R ≤ 2.618 Proof: For simplicity we assume f(x)=x Qj - the path of request j in system S -set of requests that are assigned to edge e - load of edge e For optimal routes : Qj* , J*(e) , le*

Weighted Sum of Nash Eq. According to the definition of Nash equilibrium: We multiply by wj and get We sum for all j, and get

Classification Classifying according to edges indices J(e) and J*(e), yields Using , we get

Transformation Using Cauchy Schwartz inequality, we obtain Define and divide by Then

Unweighted Demands Theorem : Proof : For linear latency functions, pure strategies and unweighted demands R ≤ 2.5. Proof :

Proof As in the previous proof Using , we get

Proof Applying properties Then

Linear Latency Functions Theorem : For linear latency functions and weighted demands R≥2.618. Proof: We consider a weighted network congestion game with four players

Linear Latency Functions v Player 1 : (u,v, φ) Player 2 : (u,w, φ) Player 3 : (v,w, 1) Player 4 : (w,v, 1) x u x x x w OPT=NASH1=2φ2 + 2·12 = 2φ+4

Linear Latency Functions v Player 1 : (u,v, φ) Player 2 : (u,w, φ) Player 3 : (v,w, 1) Player 4 : (w,v, 1) x u x x x w NASH2=2(φ+1)2 + 2·φ2 = 8 φ +6 R= φ+1=2.618

Linear Latency Functions Theorem : For linear latency functions and unweighted demands R≥2.5. Proof: The same example as in the weighted case with unit demands

Mixed Strategies Definition (Nash equilibrium): System S of mixed strategies is in Nash equilibrium iff for every j {1,...,n}and Q,Q’  {Qj}, with pQ,j>0 : cQ,j ≤ cQ’,j where XQ,j – indicates whether request j is assigned to path Q - load of edge e

Mixed Strategies Theorem : Proof : For linear latency functions (mixed strategies) and weighted demands R ≤ 2.618. Proof : Let {pQ,j} be the probability distribution of the system S. The expected latency of user j for assigning his request to path Q in S is

Step 1 According to the definition of Nash equilibrium for , hence We multiply by pQ,j·wj and get

Step 2 Sum over all paths and all users and classify according to the edges Augment to Obtain the same inequality as in the pure strategies case

General Latency Functions General functions-no bicriteria results Polynomial Latency Functions The price of anarchy - at most O(2ddd+1) (pure and mixed) The price of anarchy - at least Ω(dd/2)

Polynomial Latency Functions Theorem : For polynomials of degree d latency functions R = Ω(dd/2). Proof: We use the construction of Awerbuch et. al for the parallel links restricted assignment model.

Example l=3 OPT Group 1 Group 2 Group 3 m0 m0 m0 m0 m0 m0 m1 m1 m1 m1 NASH Group 2 Group 1 m0 m0 m0 m0 m0 m0 m1 m1 m1 m1 m1 m1 m2 m2 m2 m3

The Construction Total m=l! links each has a latency function f(x)=xd l+1 type of links For type k=0…l there are mk=T/k! links l types of tasks For type k=1…l there are k·mk jobs, each can be assigned to link from type k-1 or k OPT assigns jobs of type k to links of type k-1 one job per link.

System of Pure Strategies System S of pure strategies Jobs of type k are assigned to links of type k k jobs per link Lemma : The System S is in Nash Equilibrium.

The Coordination Ratio

Summary We showed results for general networks with unsplittable traffic and general demands For linear latency functions and weighted demands R=2.618 For linear latency functions and unweighted demands R=2.5 For Polynomial Latency functions of degree d , R=dӨ(d)

Related Work-Machines Model Main references Koutsoupias and Papadimitriou (STACS 99) Mavronicolas and Spirakis (STOC 2001) Czumaj and Vocking (SODA 2002) Awerbuch, Azar, Richter and Tsur (WAOA 2003) …

Related Work-Machines Model Main results (global cost – maximum user’s cost) For m identical links, identical jobs (pure) R=1 For m identical links (pure) R=2 For m identical links (mixed) R- Price of Anarchy

Mixed Strategies -Example Machines model (n=m) Pure strategy – Assign job i to link i maximum cost=1 Mixed strategy – assign jobs to links uniformly at random

Related Work (Cont’) Main results R- Price of Anarchy For 2 related links R=1.618 For m related links / restricted assignment (pure) : For m related links / restricted assignment (mixed) : R- Price of Anarchy