Worst-case Equilibria Elias Koutsoupias and Christos Papadimitriou Presenter: Yishay Mansour Tight Bounds for Worst-case Equilibria Artur Czumaj and Berthold Vocking

Outline Motivation Model Unit speed links Weighted speed links

Motivation Internet users: –very selfish and spontaneous behavior, –No one is thinking to achieve the “social optimum”. Game theory as an analysis tool: –rational behavior and Nash Equilibrium. Nash equilibrium: –no optimization of overall system performance. –design mechanisms that encourage behaviors close to the social optimum.

Motivation Nash Equilibrium versus global optimum Many cases: best Nash Equilibrium is global (social) optimal Worse case analysis –Compare worse Nash to optimum –How bad can things get

Current Work Coordination ratio - the ratio between –the worst possible Nash equilibrium and – social (global) optimum This works: –Very simple network model. –Derive upper and lower bounds. –Evaluate the price due to lack of coordination.

Model Simple routing model: –Two nodes –m parallel links with speeds s i –n jobs/connection weights w j Load model: –The delay of a connection is proportional to load on link

Cost Measure Each job selects a link Jobs(j) jobs assigned to link j Cost of jobs assigned to link j –L j =  j in Jobs(i) w j /s j Total cost of a configuration –Max j {L j } Social optimum –Min Max j {L j }

Nash Equilibria Each job i assigns a probability p(i,j) to link j –Support(i) = { j : p(i,j) > 0} –Deterministic: one p(i,j) =1 other p(i,j’)=0 Expected link j load –E[L j ] =  i p(i,j) w i / s j Job i view of link j: –Cost(i,j) = w i /s j +  k  i p(k,j) w k / s j = E[L j ] + (1-p(i,j))w i –Cost after job i moves to link j

Nash Equilibria For every job i Min_cost(i) = MIN j cost(i,j) For every link j: –IF cost(i,j) > min_cost(i) THEN p(i,j)=0

Example Two links, unit speed: –s 1 = s 2 =1 Social optimum is hard: –Problem is NP-complete –Partition Two trivial lower bounds: –Max weight job: w max = MAX i {w i } –Average over machines:  i w i /m

Example I Deterministic Example –2 jobs of weight 2 –2 jobs of weight one Optimum = 3 Nash = 4 Coordination ratio  4/3

Example Stochastic Example –2 jobs of weight 2 Optimum = 2 Nash: –P(i,j)= ½ –Expected Cost = 3 Coordination ratio  3/2

Upper bound: Deterministic Load L 1 and L 2 ; L 1 > L 2 Difference at most w max ; L 1 – L 2 = v  w max Nash_Cost = L 1 –IF L 2 > v/2 THEN OPT_cost  L 2 + v/2 Nash cost = L 2 + v Coordination ratio  3/2 –Otherwise opt_cost  w max & L 1  (3/2 )w max Coordination ratio  3/2

Upper Bound: Stochastic Contribution probability q i of job i: –Probability that it is in the unique max load link (assume tie breaker) –Cost =  i q i w i Collision probability t(i,k) of jobs i and k –Probability they select the same link –Both contribute to social cost only if they collide: q i + q k  1+t(i,k)

Upper bound proof Lemma:  i  k t(i,k) w k = min_cost(i) – w i Claim: Theorem: The coordination ratio for two unit speed links is 3/2

Unit speed: many links – DET. L max = MAX L j ; L min = MIN L j L max – L min  w max IF L min  w max THEN –OPT cost  w max & L max  2 w max OTHERWISE: –OPT cost  L min & L max  2 L min Coordination ratio  2

Unit speed: many links – STOCH. Lower bound: –m links m jobs –p(i,j) =1/m –m balls in to m buckets. –Probability of k balls approx. 1/ k k –Need probability of 1/m –Max load  ( log m / log log m)

Unit speed: many links – STOCH. Upper bound: –Nash load  2 OPT –Large deviation bound. –bound α by log m / log log m

Multiple speeds: Each link i has speed s i Assume s 1 ≥... ≥ s m

Multiple speeds: Lower bound Let K = log m /log log m K+1 groups of links –N j links in group j N k =  m N j = (j+1) N j+1 N 0 = K!  m Group k has speed 2 k Assignment: –Each Link in group k has k jobs of weight 2 k

Multiple speeds: Lower bound Configuration load = K OPT load < 2 System in Nash Lower bound for deterministic NASH

Multiple speeds: Upper bound c = MAX E[L j ] LEMMA:

Multiple speeds: Upper bound C = E[ MAX{L j }] LEMMA:

Expected Load I Let J k =r if the least index link with load less than k*OPT is r+1 Every link j  J k has load at least k*OPT Link J k +1 has load less than k*OPT Let c* =  (c-OPT)/OPT  Target: show that J 1 > c*! Since J 1  m then a [log m /log log m] bound.

Expected Load I Claim: E[L 1 ]  c –OPT Proof: By contradiction –consider the most loaded link –Any job J from it can move to link 1 –Its running time of link 1 is at most OPT –Job J improves its load. Corollary: J c*  1

Expected Load I Lemma: J k  (k+1) J k+1 Proof: T are jobs in links 1 to J k+1 –Claim: OPT can not allocate job from T to link r>J k Jobs in T observe load at least (k+1)*OPT Link J k +1 has load less than k*OPT. No job from T wants to move to link J k +1=u Minimum weight in T at least s u *OPT On any link r>u any job from T will run more than OPT

Expected Load I –Claim: IF OPT allocates jobs from T to links 1 to J k THEN J k  (k+1) J k+1 W sum of weights of jobs in T W   j s j E[L j ]  (k+1) OPT  j  J(k+1) s j Since OPT allocate jobs in T in links 1 to J k W  OPT  j  J(k) s j  j  J(k) s j  (k+1)  j  J(k+1) s j Since link speeds are decreasing claim follows.

Expected Load II c=O( log (s 1 / s m ) ) –CLAIM: for 1  k  c-3 –Corollary: s m  2 -(c-5)/2 s 1 –Or: c  2 log (s 1 /s m ) + O(1)

Proof OPT schedule some job i: –Nash in j in {1.. J k+2 } cost(i,j)  (k+2)*OPT –OPT in j’ in {J k+2 +1,... m} w i  S J(k+2)+1 OPT –cost(i,J k +1)  k*OPT + w i / s J(k)+1 Nash implies: –cost(i,j)  cost(i,J k +1)

Expected Maximum Load Large deviation result Each link near its expectation. Separates small and large jobs Large jobs: contribution proportional to weight. Small jobs: use Hoeffding relative bound.