Price of Anarchy Bounds Price of Anarchy Convergence Based on Slides by Amir Epstein and by Svetlana Olonetsky Modified/Corrupted by Michal Feldman and Amos Fiat

Equal Machine Load Balancing = Parallel Links 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

Price of Anarchy Price of Anarchy: The worst possible ratio between: -Objective function in Nash Equilibrium and -Optimal Objective function Objective function: total user cost, total user utility, maximal/minimal cost, utility, etc., etc.

Identical machines Main results (objective function – maximum load) -For m identical links, identical jobs (pure) R=1 -For m identical links (pure) R=2-1/(m+1) -For m identical links (mixed) Lower bound – easy : uniformly choose machine with prob. 1/m Upper bound – assume opt = 1, opt = max expected ≤ 2 in NE (otherwise not NE, NE = expected max ≤ log m / loglog m due to Hoeffding concentration inequality

Related Work (Cont’) Main results -For 2 related links R= For m related links (pure) -For m related links (mixed) -For m links restricted assignment (pure) -For m links restricted assignment (mixed)

m (=3) machines n (=4) jobs v i – speed of machine i w j – weight of job j v 1 = 4 v 2 = 2 v 3 = 1 1 (4) 2 (4) 2 (2) 1 (2) L 1 = 1 L 2 = 3 L 3 = 2 L i – load on machine i

Price of Anarchy: Lower Bound k! / (k-i)! GiGi k-i k! 1 k GkGk k k-1 k(k-1) k-2 G0G0 G1G1 G2G2 v=2 k-i v=1 v=2 k w=2 k-i w=2 k v=2 w=2 v=2 k-1 w=2 k-1

Price of Anarchy: Lower Bound GiGi k-i k! 1 k GkGk k k-1 k(k-1) k-2 G0G0 G1G1 G2G2 k! ~ m k ~ log m / log log m k! / (k-i)! v=2 k-i v=1 v=2 k w=2 k-i w=2 k v=2 w=2 v=2 k-1 w=2 k-1

1 1 Its a Nash Equilibrium GiGi k-i k! 1 k GkGk k k-1 k(k-1) k-2 G0G0 G1G1 G2G2 k! / (k-i)! 2 v=2 k-i v=1 v=2 k w=2 k-i w=2 k v=2 w=2 v=2 k-1 w=2 k-1

1 Its a Nash Equilibrium GiGi k-i k! 1 k GkGk k k-1 k(k-1) k-2 G0G0 G1G1 G2G2 k! / (k-i)! 2 4 v=2 k-i v=1 v=2 k w=2 k-i w=2 k v=2 w=2 v=2 k-1 w=2 k-1

1 The social optimum k! / (k-i)! GiGi k-i k! 1 k GkGk k k-1 k(k-1) k-2 G0G0 G1G1 G2G2 2 1 v=2 k-i v=1 v=2 k w=2 k-i w=2 k v=2 w=2 v=2 k-1 w=2 k-1

The social optimum k! / (k-i)! GiGi k-i k! 1 k k k-1 k(k-1) k-2 G0G0 G1G1 G2G2 2 v=2 k-i v=1 v=2 k w=2 k-i w=2 k v=2 w=2 v=2 k-1 w=2 k-1 1 GkGk

1 2 The social optimum k! / (k-i)! GiGi k-i k! 1 k k k-1 k(k-1) k-2 G0G0 G1G1 G2G v=2 k-i v=1 v=2 k w=2 k-i w=2 k v=2 w=2 v=2 k-1 w=2 k-1 GkGk

Related Machines: Price of Anarchy upper bound Normalize so that Opt = 1 Sort machines by speed The fastest machine (#1) has load Z, no machine has load greater than Z+1 (otherwise some job would jump to machine #1) We want to give an upper bound on Z

Related Machines: Price of Anarchy upper bound Normalize so that Opt = 1 The fastest machine (#1) has load Z, but Opt is 1, consider all the machines that Opt uses to run these jobs. These machines must have load ≥ Z-1 (otherwise job would jump from #1 to this machine) There must be at least Z such machines, as they need to do work ≥ Z

Related Machines: Price of Anarchy upper bound Take the set of all machines up to the last machine that opt uses to service the jobs on machine #1. The jobs on this set of machines have to use Z(Z-1) other machines under opt. Continue, the bottom line is that n ≥ Z!, or that Z ≤ log m / log log m

Restricted Assignment to Machines m0m0 m0m0 m0m0 m0m0 m0m0 m1m1 m0m0 m1m1 m1m1 m1m1 m1m1 m1m1 m2m2 m2m2 m2m2 m3m3 NASH Group 1 m0m0 m0m0 m0m0 m0m0 m0m0 m1m1 m0m0 m1m1 m1m1 m1m1 m1m1 m1m1 m2m2 m2m2 m2m2 m3m3 Group 2Group 3 Group 1 Group 2 Group 3 OPT l=3

Network models (Many models) Symmetric (all players go from s to t) –No weights on the players (all bandwidth requests are one) –Arbitrary monotonic increasing link delay function –Polynomial time –How bad a solution can this be?

Network models (Many models) Asymmetric with weights –Negligible load (one car out of 100,000 cars traveling from Tel Aviv to Jerusalem) Famouse as Waldrop equilibrium –Atomic Splitable (the cars are all controlled by one agent, but the agent can split the routes taken by the cars) –Atomic Unsplitable (all cars / oil / communications must flow through the same path.

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

Example s t          Latency=2+1+2=5 Latency= =8 Latency function f(x)=x Total latency =Σ e f e (l e )·le= Σ e le· le  =6·2·  ·  · 

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

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 s t w v f(x)=x l=1 f(x)=x l=1 f(x)=1 l=0 fl(x)=1 l=0 f(x)=0 l=1

Negligible networks - POA Roughgarden and Tardos (FOCS 2000) Assumption : each user controls a negligible fraction of the overall traffic Results : -Linear latency functions - POA=4/3 -Continuous nondecreasing functions-bicriteria results Without negligibility assumption : no general results

Azar, Epstein, Awerbuch Unsplittable Flow, general demands Linear Latency Functions -For weighted demands the price of anarchy is exactly (pure and mixed) - For unweighted demands the price of anarchy is exactly 2.5. Polynomial Latency Functions -The price of anarchy - at most O(2 d d d+1 ) (pure and mixed) -The price of anarchy - at least Ω(d d/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  Q j Mixed strategies – user j selects a probability distribution {p Q,j } over all paths Q  Q j

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

Example s t          C Q1,1 =2+1+2=5 Latency function f(x)=x Path Q 1 USER 1 : W 1 =1 C Q,1 =2+(1+1)+(1+1)+2=8 Path Q

Routes in Nash Equilibrium Definition (Nash equilibrium): System S of mixed strategies is in Nash equilibrium iff for every j ,...,n}and Q,Q’  Q j }, with p Q,j >0 : c Q,j ≤ c Q’,j where X Q,j – indicates whether request j is assigned to path Q - load of edge e

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 f e (x)=a e x+b e for each e  E Theorem : For linear latency functions (pure strategies) and weighted demands R ≤ Proof: For simplicity assume f(x)=x Q j - the path of request j in system S -set of requests that are assigned to edge e - load of edge e For optimal routes : Q j *, J * (e), l e *

Weighted Sum of Nash Eq. According to the definition of Nash equilibrium: We multiply by w j 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

Mixed Strategies Theorem : For linear latency functions (mixed strategies) and weighted demands R ≤ Proof : Let {p Q,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 p Q,j ·w j 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

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

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

Proof Applying inequalities Then

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

Linear Latency Functions u v w x x 0 0 xx OPT=NASH1=2φ 2 + 2·1 2 = 2φ+4 Player 1 : (u,v, φ) Player 2 : (u,w, φ) Player 3 : (v,w, 1) Player 4 : (w,v, 1)

Linear Latency Functions u v w x x 0 0 xx NASH2=2(φ+1) 2 + 2·φ 2 = 8 φ +6 R= φ+1=2.618 Player 1 : (u,v, φ) Player 2 : (u,w, φ) Player 3 : (v,w, 1) Player 4 : (w,v, 1)

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

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

General Latency Functions Polynomial Latency Functions -The price of anarchy - at most O(2 d d d+1 ) (pure and mixed) -The price of anarchy - at least Ω(d d/2 )

The Construction Total m=l! links each has a latency function f(x)=x l+1 type of links For type k=0…l there are m k =T/k! links l types of tasks For type k=1…l there are k·m k 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

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

Summary We showed results for general networks with unsplittable traffic and general demands -For linear latency functions R≤ For Polynomial Latency functions of degree d, R=d Ө(d)