load balancing, multicast routing, price of anarchy and strong equilibrium
DESCRIPTION
Load Balancing, Multicast routing, Price of Anarchy and Strong Equilibrium. Computational game theory Spring 2008 Michal Feldman. Load Balancing Model: Unrelated Machines. machines. Set of machines M = {M 1 ,…,M m } Set of jobs N = {1,…,n} - PowerPoint PPT PresentationTRANSCRIPT
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