Load Balancing, Multicast routing, Price of Anarchy and Strong

Equilibrium

Computational game theorySpring 2008

Michal Feldman

Load Balancing Model: Unrelated Machines

• Set of machines M = {M1,…,Mm}• Set of jobs N = {1,…,n}• Unrelated machines model:

Job (player) i has load wij on machine j

• Strategy: select a machine

• Cost of a job = total load on selected machine

• Objective: minimize makespan (max load)

• Special cases: – Identical machines: wij=wij’ for all j,j’– Related machines: each machine j has a speed sj, and

each job i has load li, and wij=li/sj

M1M2

J157

J223

J341

5

4

3

L1(s)=9M1 M2

L2(s)=3

jobs

machines

(pure) equilibrium existence

• Potential function– Identical machines: sum of squares (why?)– Unrelated machines:

• Does sum of squares work? • No !

• Before migration: 10, after migration: 9, so cost decreased• Yet, sum of squares increased from 102+52 to 92+92

105

1 4

Lexicographic order

• Definition: a vector (l1,…lm) is smaller than (l1’,…,lm’) lexicographically if for some i, li < li’ and lk = lk’ for all k<I

• Definition: A joint action s is smaller than s’ lex. (ss’) if the vector of machine loads L(s), sorted in non-decreasing order, is smaller lex. than L(s’)

s

s’ss’

(Pure) NE Existence• Lemma: if a job i improves its cost by migration, then the

lexicographic order decreases • Proof sketch:

– a job migrating from blue machine to red machine– Only the load on these two machines change (blue decreases, red

increases)– But if the migrating job improves, red (in post-migration) must be

lower than blue (in pre-migration)– Thus after migration, both blue and red are lower than blue prior to

migration– Thus profile decreases lexicographically

• Conclusion 1: load balancing game admit a Nash equilibrium in pure strategies

• Conclusion 2: price of stability of any load balancing game is 1

Price of Anarchy for identical machines

• Theorem: in any load balancing game on identical machines, it holds that

• Proof: – Let s be a NE and let s* be OPT– Let i’ be a machine with highest cost in s, and let j’ be job

with lowest weight on machine i’– wlog, at least 2 jobs on machine i’ (why?), thus w j’≤ ½

cost(s)– Since s is a NE, for any machine i≠I’ (job j’ stays)

• li ≥ li’ – wj’ ≥ cost(s) – ½ cost(s) = ½ cost(s)

122

m

POA

mstm

m

mstst

m

lj

m

wst ji

i

2)(cos)1()1)((cos2

1)(cos*)(cos

122

12

*)(cos)(cos

mmm

stst

Convergence time of best response for identical machines

• Max-weight best response policy: – activate jobs, always activating job of max-weight

among unsatisfied jobs– activated job migrates to its best machines (i.e.,

performs a best-response)• Theorem: for any load balancing game on

identical machines, the max-weight best response policy converges to a NE, after each agent was activated at most once (from any initial profile)

Convergence time of best response for identical machines

• Proof sketch:– Claim: once a job was activated, it never gets unsatisfied again– Proof of claim is based on two observations (for identical

machines):• Job is satisfied IFF assigned to machine with minimal load (other than

itself)• Best response never decreases the minimal load among the machines

(why?)– Thus, a job can become unsatisfied only if another job migrated

to its own machine– Thus, sufficient to show that a migration of a job of lower

weight into one’s machine cannot make it unsatisfied– Proof in class..

• Note: order is crucial. Under “min-weight best response policy”, there may be instances with an exponential number of steps

Price of anarchy for unrelated machines

• POA for unrelated machines is unbounded

1

1

Job 1

Job 2

Machine 1 Machine 2

1 1

Machine 2Machine 1Machine 2

Machine 1

Social optimum Nash equilibrium

makespan= makespan=

PoA=1/

Allowing Coordination in Equilibrium

• Strong Equilibrium [Aumann’59]– No coalition can deviate and strictly improve the

utility of all of its members• very robust concept• may be a better prediction of rational behavior• most games do not admit Strong Eq.

– usually applied to pure Eq with pure deviations

Example 1: Prisoner’s Dilemma

0,5

5,0

cooperate

cooperate

defect

defect

Unique Nash Eq.

Strong Eq? .

Prisoner’s dilemma does not admit any Strong Eq .

Strong Price of Anarchy

• Determining SPoA requires two parts:– Proving existence of Strong Eq– Bounding the worst ratio

• SE NE SPoA ≤ PoA

Price of Anarchy (PoA) [KP00]:

optimum socialmequilibriuNash worst

PoA

Strong Price of Anarchy (SPoA):

optimum socialmequilibriu Strongworst

SPoA

k-Strong Equilibrium• A joint action sS is not resilient to a pure

deviation of a coalition if there is a pure action profile of such that ci(s-,)<ci(s) for any i – e.g., (defect,defect) in Prisoner’s dilemma

• A pure Nash Eq sS is resilient to pure deviation of coalitions of size k if there is no coalition of size at most k such that s is not resilient to a pure deviation by

• A k-Strong Equilibrium is a pure Nash Eq that is resilient to pure deviation of coalitions of size at most k

S=S1x…xSn

Strong Equilibrium Hierarchy

1-SE

2-SE

n-SE

=NE

=SE [Aumann]

Related Work

• Existence of Strong Equilibrium– monotone decreasing congestion games [Holzman+Lev-tov

1997, 2003]– monotone increasing congestion games + correlated SE

[Rosenfeld+Tennenholtz 2006]

• Related solution concepts– Coalition-proof Eq. [Bernheim 1987]– Group-strategyproof mechanisms

[Moulin+Shenker 2001]– Coalitions with transferable utilities

[Hayrapetyan et al 2006]

SECPE

NE

Existence of Strong Equilibrium in load balancing games

• Is every Nash Eq. on identical machines also a Strong Eq ?– NO ! (for m ≥ 3)

5

5

4 4

3 3

10 7 7

s

55

4 433

6 99

s’Coalition: 5,5,3,3

Strong Eq. Existence

• Theorem: in any load balancing game, the lex. minimal joint action s is a k-SE for any k

Recall Lexicographic Order

• Definition: a vector (l1,…lm) is smaller than (l1’,…,lm’) lexicographically if for some i, li < li’ and lk = lk’ for all k<I

• Definition: A joint action s is smaller than s’ lex. (ss’) if the vector of machine loads L(s), sorted in non-decreasing order, is smaller lex. than L(s’)

s

s’ss’

Proof of SE Existence• Suppose in contradiction that s (lex. minimal) is not a SE, and

let be the smallest coalition (deviating to s’).• Claim: the same set of machines are chosen by in s and in s’

(denote it M())– If a job migrates TO some

machine, another jobmigrates FROM it

• else contradicting s is NE – If a job migrates FROM some

machine, another jobmigrates TO it

• else contradicting minimality of • Since all jobs in must benefit, all loads of M() in s’ must be

smaller than max load of M() in s – Contradicting minimality of s

Price of Anarchy (PoA)• Recall: for unrelated machines, PoA may be unbounded

1

1

Job 1

Job 2

Machine 1 Machine 2

Objective: min makespan

Social optimumNash equilibrium Nash equilibrium

PoA=1/

1 1

M1

makespan=

M2

makespan=

M1 M2

Strong equilibrium Strong equilibrium

SPoA=1

Strong Price of Anarchy

• Theorem: for any job scheduling game with m unrelated machines and n jobs, SPoA ≤ m

Proof for SpoA ≤ m

• Claim 1: L1(s) ≤ OPT– else: coalition of all jobs to OPT

M1Mm Mi Mi-1 M1Mm Mi Mi-1

OPT

L1(s)

OPT

L1(s)

Proof for SpoA ≤ m

• Claim 1: L1(s) ≤ OPT– else: coalition of all jobs to OPT

• Claim 2: i Li(s)-Li-1(s) ≤ OPT – else: consider s’, where all jobs on machines i..m go to OPT. For all J

• cJ(s) > Li-1(s) + OPT• cJ(s’) ≤ Li-1(s) + OPT (since all J together add at most OPT)

M1Mm Mi Mi-1 M1Mm Mi Mi-1

>OPT OPT

Lm(s) ≤ m OPT

Li-1(s) L1(s)

Li(s)

Lower Bound (m machines)• Theorem: there exists a job scheduling game with m unrelated

machines for which SPoA ≥ m• Proof:

M1M2M3M4Mm

J111

J212

J313

J414

Jm1m

OPT = 1

makespan=mSE

Identical Machines

• Theorem: there exists a job scheduling game with m identical machines and n jobs, such that

m

SPoA 11

2

12m-1mJ1

Jm

Jm+1

J2m

1

1/m1 m-2 m-1m

OPT

SE

1+1/m

2

Mixed Deviations and Mixed Strong Eq

• Nash Eq – unilateral deviations– pure and mixed deviations are equivalent

• Strong Eq – coordinated deviation– Pure and mixed deviations are not equivalent– Given a mixed deviation, there might be no single pure

deviation which is good

J1

M1 M2

J3

J2Unique Nash Eq

J1

J2

¾ ¼

cJ1=cJ2=15/8

cJ1=cJ2=2

mixed deviation

J1

M1 M2

J3

J2

½½

Mixed Deviations and Mixed Equilibrium

• However, in many cases, allowing mixed deviations by a coalition eliminates all Nash Eq.

• Theorem: for m≥5 identical machines, and n>3m unit jobs, there is no 4-Strong Eq when mixed deviations are allowed– Based on a lemma that shows that the support of any two

“mixing” jobs must be disjoint

Strong equilibrium in multicast routing

Theorem: There exists a multicast routing game that does not posses a strong equilibrium.

Proof:

s

t1 t2

2

1 1

-½3ε-½3ε2+2ε

1-2ε

1+3εUnique NE: c1(S) = c2(S) = 2/2+1=2

deviation: ci(S) < 2

No SE in game

Strong Price of Anarchy

Theorem : The strong price of Anarchy of a multicast routing game with n players is at most H(n).

Proof:• Let S be a SE, and SΓ be the induced profile of players in Γ, and

let S* be OPT• For k=n,…,1, since S is SE, there exists a player

“k” k={1,…,k} that does not benefit from coal. deviation. i.e., ))(())(( )( ),()( 1

**** kkkkk SSScSScSc

kkk

))(()()(

1

SnHcj

cS eEe

eEe

Sn

j

ee

Potential function:

Proof (cont’d)

• We got for every k: • Summing over all players:

)()()(

))(()(

))(())(()(

*

*

**

**

OPTnHnHc

SnHcS

SSSc

See

eSe

e

nNi

i

))(())(()( 1**

kkk SSSc

Lower Bound• OPT: all users use indirect edge, c(OPT)=1+• Unique NE and SE: each user uses direct edge to ti,

c(NE)=c(SE)=H(n) PoA = SPoA = PoS = SPoS = H(n)

21

n

t1 tn-2t3t2 tn-1 tn

s

1 31

11

n n1 1+2

1