Strong Price of Anarchy for Machine Load Balancing

20 Pages·2007·0.23 MB·English

Strong Price of Anarchy
for
Machine Load Balancing
Amos Fiat⋆, Haim Kaplan∗, Meital Levy⋆⋆, and Svetlana Olonetsky∗
Tel Aviv University
Abstract. As defined by Aumann in 1959, a strong equilibrium is a Nash equilibrium that is resilient
to deviations by coalitions. We give tight bounds on the strong price of anarchy for load balancing on
relatedmachines.Wealsogivetightboundsfork-strongequilibria,wherethesizeofadeviatingcoalition
is at most k, for unrelated machines.
Key words: Game theory,Strong Nash equilibria, Load balancing, Price of Anarchy
1 Introduction
Many concepts of game theory are now being studied in the context of computer science. This convergence of
different disciplines raises new and interesting questions not previously studied in either of the original areas
of study. Much of this interest in game theory within computer science is due to the seminal papers of Nisan
and Ronen [20] and Koutsoupias and Papadimitriou [17].
A Nash equilibrium ([19]) is a state in a noncooperative game that is stable in the sense that no agent can
gain from unilaterally switching strategies. There are many “solution concepts” used to study the behavior of
selfish agents in a non-cooperative game. Many of these are variants and extensions of the original ideas of
John Nash from 1951.
One immediate objection to Nash equilibria as a solution concept is that agents may in fact collude and
jointlychoosestrategiessoasto“profit”.Therearemanypossibleinterpretationsofthestatementthatasetof
agents“profit”fromcollusion.Onenaturalinterpretationofthisstatementisthenotionofastrongequilibrium
due to Aumann [5], where no coalition of players have any joint deviation such that every member strictly
benefits. Whereas mixed strategy Nash equilibria always exist for finite games [19], this is not in general true
for strong equilibria.
⋆
School of computer science, Tel Aviv University, Tel Aviv 69978, Israel. E-mail:
{fiat,haimk,levymeit,olonetsk}@post.tau.ac.il. Research partially supported by the Israel Science Foun-
dation and theGerman Israel Foundation.
⋆⋆
Theresearch has been supported by theEshkol Fellowship funded bythe Israeli Ministry of Science.
Holzman and Law-Yone [16] characterized the set of congestion games that admit strong equilibria. The
classofcongestiongamesstudiedwasextendedbyRozenfeldandTennenholtzin[21].[21]alsoconsideredmixed
strong equilibria andcorrelatedmixed strong equilibria under variousdeviation assumptions,pure, mixed and
correlated.Variantsofstrongequilibriaincludelimitingthesetofpossibledeviations(coalition-proofequililbria
[8]) and assuming static predefined coalitions ([15,14]).
The term price of anarchy was coined by Koutsoupias and Papadimitriou [17]. This is the ratio between
the cost of the worst-case Nash equilibria and the cost of the social optimum. A related notion is the price
of stability defined in [3], the ratio between the cost of the best Nash equilibria and the cost of the social
optimum. These concepts have been extensively studied in numerous settings, machine load balancing [17,18,
11,7,10], network routing [22,6,9], network design [4,12,1,3,13], etc.
Andelman et al. [2] initiated the study of the strong price of anarchy (SPoA), the ratio of the worst case
strongequilibriato the socialoptimum.Theauthorsalsodefine the notionofak-strongequilibrium,whereno
coalition of size up to k has any joint deviation where all strictly benefit. Analogous definitions can be made
for the k-strong price of anarchy.
One may argue that the strong price of anarchy (which is never worse than the price of anarchy) removes
the element of poor coordination and is entirely due to selfishness. Likewise, the k-strong price of anarchy
measures the cost of selfishness and restricted coordination (up to k agents at once).
Our work here is a direct continuation of the work of Andelman et al. [2], and addresses many of the open
problemscited there,inparticularinthe contextofaloadbalancinggame.Inthis setting agents(jobs)choose
a machine, and job j placed on machine i contributes w (i) to the load on machine i. Agents seek machines
j
withsmallload,andthesocialcostusuallyconsideredisthemakespan,i.e.,themaximalloadonanymachine.
Whereas [2] considered strong price of anarchy and k-strong price of anarchy for unrelated machines, herein
we primarily consider the strong price of anarchy for related machines (machines having an associatedspeed).
Our results.
1. Czumaj and Vocking [11] showed that the price of anarchy for load balancing on related machine is
Θ(logm/loglogm), we show that the strong price of anarchy for load balancing on related machine is
Θ logm/(loglogm)2 . This is our most technically challenging result.
(cid:0) (cid:1)
2. We also give tight results for the problems considered by [2]:
(a) In[2]thestrongpriceofanarchyforloadbalancingonmunrelatedmachineswasshowntoliebetween
m and 2m−1. We prove that the true value is always m.
(b) In [2], the k-strong price of anarchy for load balancing of n jobs on m unrelated machines is be-
tween O(nm2/k) and Ω(n/k). We prove that the k-strong price of anarchy falls in between and is
Θ(m(n−m+1)/(k−m+1)).
2
2 Preliminaries
AloadbalancinggameconsistsofasetM ={M ,...,M }ofmachines,asetN ={1,...,n}ofjobs (agents).
1 m
We use the terms machine i or machine M interchangeably. Each job j has a weight function w () such that
i j
w (i) is the running time of job j on machine M . When the machines are unrelated then w () is an arbitrary
j i j
positive realfunction. For relatedmachines,eachjobj has weight,denoted by w , andeachmachine M has a
j i
speed,denotedbyv(i).Therunningtimeofjobj onmachineiisw (i)=w /v(i)wherew istheweightofjob
j j j
j. In game theoretic terms, the set of strategies for job j is the set of machines M. A state S is an assignment
of jobs to machines. Let m (j) be the machine chosen by job j in state S. The load on machine M in state S
S i
is w (i).
j|Mi=mS(j) j
P
Givenastate S inwhichjobj isassignedto machine M ,we saythatthe load observed by job j is the load
i
on machine M in state S. The makespan of a state S is the maximum load of a machine in S. Jobs seek to
i
minimize their observedload.The state OPT(the socialoptimum)is a state withminimal makespan.We also
denote the makespan in state OPT by OPT, and the usage would be clear from the context.
A strong equilibrium is a state where no group of jobs can jointly choose an alternative set of strategies so
that every job in the grouphas a reducedobservedloadin the new state. In a k-strong equilibrium we restrict
such groups to include no more than k agents.The strong price of anarchy is the ratio between the makespan
of the worst strong equilibrium and OPT. The k-strong price of anarchy is the ratio between the makespan of
the worst k-strong equilibrium and OPT.
3 Related Machines
CzumajandVocking[11]showthatthepriceofanarchyofloadbalancingonrelatedmachinesisΘ(logm/loglogm).
Weshowthatthelowerboundconstructionof[11]isnotinstrongequilibrium.Wealsogiveasomewhatweaker
(and tight) lower bound of Ω(logm/(loglogm)2). We first present the lower bound of [11]and claim that it is
not resilient to deviation by coalitions.
The lower bound of [11]:
Consider the following instance in which machines are partitioned into ℓ+1 groups. Let these groups be
G , G , ..., G with m machines in group G . We define m = 1, and m = (ℓ−j)·m for j = 1,...,ℓ.
0 1 ℓ j j 0 j+1 j
Since the total number of machines m = ℓ m and m = ℓ!, it follows that ℓ ∼ logm/loglogm. Suppose
j=0 j ℓ
P
that all machines in G have speed 2(ℓ−j).
j
Consider the following Nash equilibrium, every machine in G receives ℓ−j jobs, each of weight 2(ℓ−j).
j
Eachsuchjobcontributes1totheloadofitsmachine.ThetotalloadoneverymachineinG isthereforeℓ−j.
j
Machines in group G have no jobs assigned to them.
ℓ
Themakespanisℓ,obtainedonmachinesofgroupG .Considersomejobassignedtoamachinefromgroup
0
G , this job has no incentive to migrate to a machine in a group of lower index since the load there is already
j
3
higher than the load it currently observes. If a job on a machine in G migrate to a machine in G then it
j j+1
would observe a load of ℓ−(j+1)+2(ℓ−j)/2(ℓ−(j+1)) =ℓ−j+1>ℓ−j and even higher loads on machines
of groups G ,...,G .
j+2 ℓ
For the minimal makespan, move all jobs on machines in G to machines in G (for j = 0,...,ℓ−1).
j j+1
There aresufficiently manymachinessothat nomachine getsmorethanone job,andthe loadonallmachines
is 2ℓ−j/2ℓ−(j+1) =2. The price of anarchy is therefore Ω(logm/loglogm). (This is also shown to be an upper
bound).
However, this is not a strong Nash equilibrium.
Lemma 1. The above Nash equilibrium is not in strong equilibrium for ℓ>8.
Proof. Consider the following scenario. We consider a deviation by a coalition consisting of all ℓ jobs on
some machine of group G (machine M) along with 3 jobs from each of ℓ different machines from group G
0 2
(machines N ,N ,...,N ). We describe the actions of the coalition in two stages, and argue that all members
1 2 ℓ
of the coalition benefit from this deviation.
All ℓ jobs located on machine M ∈G migrate to separate machines N ,N ,...,N in group G .
0 1 2 ℓ 2
Following this migration, the load on machines N is ℓ+2 (it was ℓ−2, we added a job from machine M
i
that contributedanextra 4to the load).The loadonmachineM has droppedto zero(alljobs wereremoved).
Now, remove 3 original jobs (with weight 2ℓ−2) from each of the N machines and place them on machine
i
M. The loadon machine N has dropped to ℓ−1, so the job that migratedfrom machine M to machine N is
i i
now experience lower load than before. The load on machine M is now 3ℓ·2ℓ−2/2ℓ =3ℓ/4<ℓ−2, for ℓ>8.
Thus, the jobs that migrated from machines in G to machine M also benefit from this coalition.
2
3.1 Lower bound on strong price of anarchy for related machines.
Theorem 1. The strong price of anarchy for m related machines and n jobs is Ω logm/(loglogm)2 .
(cid:0) (cid:1)
Proof. Considerthefollowinginstanceinwhichmachinesarepartitionedintoℓ+1groups.Letthesegroupsbe
G ,G ,...,G .WefurthersubdivideeachgroupG ,0≤i<ℓ,intologℓsubgroups,whereallmachineswithin
0 1 ℓ i
the samesubgrouphavethe samespeed,butmachinesfromdifferentsubgroupsofagroupdiffer inspeed.The
group G consists of a single subgroup F . In total, we have ℓlogℓ+1 subgroups F ,F ,...,F , where
ℓ ℓlogℓ 0 1 ℓlogℓ
subgroups F ,...,F area partitionof G , F ,...,F are the subgroupsof G , etc. The speed of
0 logℓ−1 0 logℓ 2logℓ−1 1
each machine in subgroup F is 2(ℓlogℓ−j).
j
Let m denote the number of machines in subgroup F , 0 ≤ j ≤ ℓlogℓ. Then m = 1, and for subgroup
j j 0
F such that F ⊂ G we define m = (ℓ−i)×m . It follows that the number of machines in subgroup
j+1 j i j+1 j
F is at least (ℓ!)logℓ and therefore m≥(ℓ!)logℓ and ℓ∼logm/(loglogm)2.
ℓlogℓ
4
Consider the following state, S. Each machine of group G is assigned ℓ−i jobs. Jobs that are assigned to
i
machines in subgroup F have weight 2(ℓlogℓ−j). As the speed of such machines is 2(ℓlogℓ−j), it follows that
j
each such job contributes one to the load of the machine it is assigned to. I.e., the load on all machines in G
i
is ℓ−i. Machines of F have no jobs assigned to them.
ℓlogℓ
The loadonthe machinesingroupG is ℓwhichisalsothe makespaninS.The minimalmakespan(OPT)
0
is attainedby movingthe jobs assignedto machines fromF eachto a separatemachine of subgroupF , for
j j+1
0≤j <ℓlogℓ. The load on all machines is now 2ℓlogℓ−j/2ℓlogℓ−(j+1) =2.
State S is a Nash equlibrium. A job assigned to a machine of subgroup F has no incentive to migrate to
j
a machine with a lower indexed subgroup since the current load there is equal or higher to the current load
it observes. There is no incentive to migrate to a higher indexed subgroup as it observes a load of at least
ℓ−j+1>ℓ−j.WenowarguethatstateS isnotonlyaNashEquilibriumbutalsoastrongNashequilibrium.
First, note that jobs residing on machines of group G , 0 ≤ i ≤ ℓ−2, have no incentive to migrate to
i
machines of group G , for j ≥i+2. This follows since the speed of each machine in group G is smaller by a
j j
factorofmorethan2logℓ =ℓfromthe speedofanymachineingroupG .Thus,evenifthe jobisaloneonsuch
i
a machine,the resultingloadis higher thanthe loadcurrentlyobservedby the job (currentloadis≤ℓ).Thus,
any deviating coalition has the property that jobs assigned to machines from group G may only migrate to
i
machines from groups G , for j ≤i+1.
j
Suppose that jobs that participate in a deviating coalition are from machines in groups G , G , ..., G ,
i i+1 j
1 ≤ i ≤ j ≤ ℓ. The load on machines from group G holding participating jobs must strictly decrease since
i
either jobs leave(andthe loadgoesdown)or jobs fromhigher orequalindexed groupsjoin(and then the load
muststrictlygodowntoo).IfmachinesfromgroupG havetheirloaddecrease,andalldeviatingjobsbelongto
i
groups i throughj, i<j, then there must be some machine M ∈G ,i<p≤j, with anincrease in load.Jobs
p
canmigratetomachineM eitherfromamachineingroupG ,orfromamachineingroupG forsomej ≥p.
p−1 j
If a deviating job migrates from a machine in G for some j ≥ p then this contradicts the increase in the
j
load on M. The contradiction arises as such jobs will only join the coalition if they become strictly better off,
and for this to happen the load on M should decrease.
However, this holds even if the deviating job migrates to M from a machine in G . The observed load
p−1
for this job prior to deviating wasℓ−(p−1) and it must strictly decrease.A job that migrates to machine M
from G increases the load by an integral value. A job that migrates away from machine M decreases the
p−1
loadbyanintegralvaluetoo.This implies thatthe new loadonM mustbe anintegersmallerthanℓ−(p−1),
which contradicts the increase in load on M. ⊓⊔
5
4 Upper bound on strong price of anarchy for related machines.
We assume that machines are indexed such that v(i) ≥ v(j) for i < j. We also assume that the speeds of the
machines are scaled so that OPT is 1. Let S be an arbitrary strong Nash equilibrium, and let ℓ be the
max
maximumloadofamachineinS.Ourgoalistogiveanupperboundonℓ .Whenrequired,wemayassume
max
thatℓ isasufficientlylargeconstant,sinceotherwiseanupperboundfollowstrivially.Recallthatmachines
max
are orderedsuch that v(1)≥v(2)≥···≥v(m)>0. Let ℓ(i) be the load on machine M , i.e., the total weight
i
of jobs assigned to machine M is ℓ(i)v(i).
i
4.1 Sketch of the proof
We prove that m = Ω(ℓ ℓmaxlogℓmax), which implies ℓ = O(logm/(loglogm)2). To show that m =
max max
Ω(ℓ ℓmaxlogℓmax) we partition the machines into consecutive disjoint “phases” (Definition 1), with the prop-
max
erty thatthe number ofmachinesinphasei is Ω(ℓ)times the number ofmachinesinphase i−1(Lemma 4.3),
where ℓ is the minimal load in phases 1 through i.
Fortechnicalreasonsweintroduceshiftedphases(s-phases,Definition2)whichareinone-to-onecorrespon-
dence to the phases. We focus on the s-phases of faster machines, so that the total drop in load amongst the
machines of these s-phases is about ℓ /2. We next partition the s-phases into consecutive blocks. Let δ be
max i
the loaddifference between slowestmachine in block i−1 and the slowestmachine in block i.By construction
we get that δ =Θ(ℓ ).
i max
P
We map s-phases to blocks such that each s-phase is mapped to at most one block as follows (Lemmas 7
and 8), see Fig. 1.
– If δ <1/logℓ ⇒ we map a single (1=⌈δ logℓ ⌉) s-phase to block i
i max i max
– If δ ≥1/logℓ ⇒ we map Ω(δ logℓ ) s-phases to block i
i max i max
Therefore the total number of s-phases is at least δ logℓ = Ω(ℓ logℓ ). Given the one-to-one
i max max max
P
mappingfroms-phasestophases,thisalsogivesusalowerboundofΩ(ℓ logℓ )onthenumberofphases.
max max
In Lemma 4.3 we prove that the number of machines in phase i is Ω(ℓ ) times the number of machines
max
in phase i−1. This allows us to conclude that the total number of machines m=Ω(ℓ ℓmaxlogℓmax), or that
max
ℓ =O(logm/(loglogm)2).
max
4.2 Excess Weight and Excess Jobs
Giventhatthe makespanofOPTis1,thetotalweightofalljobsassignedtomachineσ inOPTcannotexceed
v(σ),thespeedofmachineσ.Wedefinetheexcessweightonmachine1 ≤σ ≤mtobeX(σ)=(ℓ(σ)−1)v(σ).
(Note that excess weight can be positive or negative).
6
δ1>1/logℓ δ3≤1/logℓ δ4≤1/logℓ
B0 B1 B2
P(B1) P(B3) P(B4)
Fig.1.Themachinesaresortedinorderofdecreasingspeed(andincreasingindex),andpartitionedintos-phases.The
s-phases are furtherpartitioned into blocks Bi. The s-phases that are mapped to block i are marked P(Bi).
Given a set R⊂{1,...,m}, we define the excess weight on R to be
X(R)= X(σ). (1)
σX∈R
For clarity of exposition, we use intuitive shorthand notation for sets R forming consecutive subsequences of
1,...,m. In particular, we use the notation X(σ) for X({σ}), X(≤ w) for X({σ|1 ≤ σ ≤ w}), X(w...y) for
X({w≤σ ≤y}), etc.
Given that some set of machines R has excess weight X(R)>0, it follows that there must be some set of
jobsJ(R),oftotalweightatleastX(R),thatareassignedtomachinesinRbyS,butareassignedtomachines
in {1,...,m}−Rby OPT.Givensets ofmachines R andQ,letJ(R7→Q) be the setofjobs that areassigned
by S to machines in R but assigned by OPT to machines in Q, and let X(R 7→ Q) be the weight of the jobs
in J(R7→Q). Let R , and R be a partition of the set of machines R. Then we have
1 2
X(R)≤X(R7→{1,...,m}\R)=X(R7→R )+X(R7→R ) . (2)
1 2
In particular, using the shorthand notation above, we have that for 1≤y <σ ≤m,
X(≤y)≤X(≤y 7→>y)=X(≤y 7→y+1...σ)+X(≤y 7→σ+1...m) . (3)
Similarly, for 1≤σ <y ≤m we have
X(≤y)≤X(≤σ)+X(σ+1...y) . (4)
4.3 Partition into phases
Definition 1. We partition the machines 1,...,m into disjoint sets of consecutive machines called phases,
Φ , Φ , ..., where machines of Φ precede those of Φ . We define ρ =0 and ρ =max{j |j ∈Φ } for i≥1.
1 2 i i+1 0 i i
Thus, it follows that Φ = {ρ +1,...,ρ }. It also follows that machines in Φ are no slower than those of
i i−1 i i
Φ . Let n be number of machines in the ith phase, i.e., n =ρ −ρ , for i≥1.
i+1 i i i i−1
7
To determine Φ it suffices to know ρ and ρ . For i = 1 we define ρ = 1, as ρ = 0 it follows that
i i−1 i 1 0
Φ ={1}. We define ρ inductively using both ρ and ρ as follows.
1 i+1 i i−1
X(Φ )
i
ρ =argmin X(≤ρ 7→>σ)<X(≤ρ )+ . (5)
i+1 σ(cid:26) i i−1 2 (cid:27)
The phases have the following properties.
Lemma 2. Let ℓ be the minimal load of a machine in phases 1,...,i, (ℓ = min{ℓ(σ)|1 ≤ σ ≤ ρ }), then
i
n ≥n (ℓ−1)/2.
i+1 i
Proof. By the inductive definition of ρ above (Equation 5), we have that
i+1
X(Φ )
i
X(≤ρ 7→>ρ )<X(≤ρ )+ .
i i+1 i−1
2
Now, since X(≤ρ )≤X(≤ρ 7→Φ )+X(≤ρ 7→>ρ ), we have
i i i+1 i i+1
X(≤ρ 7→Φ )≥X(≤ρ )−X(≤ρ 7→>ρ ) (6)
i i+1 i i i+1
X(Φ )
i
>X(≤ρ )+X(Φ )− X(≤ρ )+ (7)
i−1 i i−1
(cid:18) 2 (cid:19)
X(Φ )
i
= ; (8)
2
Equation (6) follows by rewriting Equation (2). Equation (7) follows from the definition of ρ (Equation
i+1
(5)), and the rest is trivial manipulation.
Since the speed of any machine in Φ is no smaller than v(ρ ) (the lowest speed of any machine in phases
i i
1,...,i), and we have chosen ℓ to be the minimum load of any machine in the set ≤ ρ , for every machine
i
σ ∈Φ the excess weight X(σ)=(ℓ(σ)−1)v(σ)≥(ℓ−1)v(ρ ). Therefore by substituting this into (8) we get
i i
n (ℓ−1)v(ρ )
i i
X(≤ρ 7→Φ )> .
i i+1
2
In OPT, no machine can have a load greater than one. Therefore since the speed of any machine in Φ
i+1
is no larger than v(ρ ) at most v(ρ ) of the weight is on one machine in Φ , so there are at least (ℓ−1)n /2
i i i+1 i
machines in Φ . ⊓⊔
i+1
Lemma 3. Let j < i be two phases. If the minimal load of a machine ρ ≤ k ≤ ρ is at least 3 then
j−1 i
X(Φ )>X(Φ ).
i j
Proof. Clearly it suffices to prove that X(Φ )>X(Φ ) for every i>0.
i+1 i
Since in OPT the load of every machine is at most one we have that
X(≤ρ 7→Φ )≤ v(σ). (9)
i i+1
σ∈XΦi+1
8
This together with (8) gives that
X(Φ )
i
v(σ)> . (10)
2
σ∈XΦi+1
Let ℓ be the minimal load of a machine in Φ . From our definition follows that
i+1
X(Φ )≥(ℓ−1) v(σ) . (11)
i+1
σ∈XΦi+1
The lemma now follows by combining (10) and (11) together with the assumption that ℓ>3. ⊓⊔
Let ℓ be the minimal loadamong machines 1,...,ρ . Let Γ be the subset ofΦ that haveat least(ℓ−1)/2
i i i
of their load contributed by jobs of weight w ≤v(ρ ).
i+1
Lemma 4. For i>j, v(σ)≥v(ρ )n (ℓ−1)/(ℓ+3).
σ∈Γi j j
P
Proof. First we want to estimate X(Φ 7→≥ρ ). By rewriting Equation (2) we get that
i i+1
X(Φ 7→≥ρ )=X(≤ρ 7→≥ρ )−X(≤ρ 7→≥ρ ) .
i i+1 i i+1 i−1 i+1
Since X(≤ρ 7→≥ρ )≤X(≤ρ 7→>ρ ), we also have that
i−1 i+1 i−1 i
X(Φ 7→≥ρ )≥X(≤ρ 7→≥ρ )−X(≤ρ 7→>ρ ) . (12)
i i+1 i i+1 i−1 i
From the definition of ρ follows that
i+1
X(Φ )
i
X(≤ρ 7→≥ρ )≥X(≤ρ )+ , (13)
i i+1 i−1
2
Similarly, from the definition of ρ follows that
i
X(Φ )
i−1
X(≤ρ 7→>ρ )<X(≤ρ )+ . (14)
i−1 i i−2
2
Substituting Equations (13) and (14) into Equation (12) we get that
X(Φ ) X(Φ )
i i−1
X(Φ 7→≥ρ )≥X(≤ρ )+ − X(≤ρ )+
i i+1 i−1 i−2
2 (cid:18) 2 (cid:19)
X(Φ ) X(Φ )
i i−1
≥X(≤ρ )+X(Φ )+ − X(≤ρ )+
i−2 i−1 i−2
2 (cid:18) 2 (cid:19)
X(Φ ) X(Φ )
i−1 i
≥ + . (15)
2 2
Let A(σ) be the total weightof jobs j on machine σ with w ≤v(ρ ) and let A(Φ )= A(σ). Since
j i+1 i σ∈Φi
P
every job in J(Φ 7→≥ ρ ) has weight of at most v(ρ ), it follows that X(Φ 7→≥ ρ ) ≤ A(Φ ), and by
i i+1 i+1 i i+1 i
Equation (15)
X(Φ ) X(Φ )
i−1 i
A(Φ )≥ + . (16)
i
2 2
We claim that every machine σ with A(σ) >0 (i.e. the machine has at least one job j with w ≤v(ρ ))
j i+1
has load of at most ℓ+1. To prove the claim, let q ≤ρ be a machine that has load greater than ℓ+1 and
i+1
9
a job j with w ≤ v(ρ ), and let q′ be the machine among 1,...,ρ with load ℓ. This state is not a Nash
j i+1 i
equilibrium since if job j switches to machine q′ it would have a smaller cost. We get that
ℓ−1
A(Φ )≤ v(σ)(ℓ+1)+ v(σ)
i
2
σX∈Γi σ∈XΦi−Γi
ℓ+3 ℓ−1
= v(σ) + v(σ)
2 2
σX∈Γi σX∈Φi
ℓ+3 X(Φ )
i
≤ v(σ) + . (17)
2 2
σX∈Γi
Inequality (17) holds ℓ is the smallest load of a machine in 1,...,ρ . Combining (17) and (16) we get that
i
ℓ+3 X(Φ ) X(Φ ) X(Φ )
i i−1 i
v(σ) + ≥ + , (18)
2 2 2 2
σX∈Γi
and therefore
ℓ+3 X(Φ ) X(Φ ) ℓ−1
i−1 j
v(σ) ≥ ≥ ≥v(ρ )n . (19)
j j
2 2 2 2
σX∈Γi
Thefirstinequalityin(19)followsfrom(18),thesecondfollowsfromLemma3,andthethirdinequalityfollows
since ℓ is the smallest load of a machine in 1,...,ρ and since v(ρ ) is the smallest speed in Φ . From (19) the
i j j
lemma clearly follows. ⊓⊔
Recall that ℓ is the maximum load in S. Define k to be min{i|ℓ(i)<ℓ /2}. Let t be the phase such
max max
thatρ <k andρ ≥k.Considermachines1,...,ρ .Fromnowonℓwouldbe theminimalloadofamachine
t t+1 t
in this set of machines. Then, ℓ=Θ(ℓ ), and we may assume that ℓ is large enough.
max
Definition 2. We define another partition of the machines into shifted phases (s-phases) Ψ , Ψ , ... based on
1 2
the partition to phases Φ , Φ , ... as follows. We define ϕ =0. Let ϕ be the slowest machine in Φ such that
1 2 0 i i
at least (ℓ−1)/2 of its load is contributed by jobs with weight w ≤ v(ρ ) (there exists such a machine by
i+1
Lemma 4). We define Ψ ={ϕ +1,...,ϕ }.
i i−1 i
Note that there is a bijection between the s-phases Ψ ,Ψ ,... and the phases Φ ,Φ ,.... Furthermore, all
1 2 1 2
machines in Φ such that at least (ℓ−1)/2 of their load is contributed by jobs of weight ≤v(ρ ) are in Ψ .
i i+1 i
Lemma 5. The load difference between machines ϕ and ϕ , ℓ(ϕ )−ℓ(ϕ )>ℓ /4+4.
2 t 2 t max
Proof. According to the definition of Ψ , there is a job on machine ϕ with weight w ≤v(ρ )≤v(ϕ ) and
i i i+1 i+1
therefore it contributes load of at most 1 to machine ϕ . As S is a Nash equilibrium, the load difference
i+1
betweenmachinesϕ andϕ isatmost1.Theloadonthefastestmachineϕ isℓ(ϕ )>ℓ −1sinceeveryjob
i i+1 1 1 max
contributesaloadofatmost1toit.Thus,theloadonmachineϕ ∈Φ isatleastℓ(ϕ )≥ℓ(ϕ )−1≥ℓ −2.
2 2 2 1 max
The load on machine ϕ is ℓ(ϕ ) ≤ ℓ /2+1, since there is a job with weight of at most v(ϕ ) on ϕ
t t max t+1 s
and by the definition of ϕ there is a machine k ≤ϕ with load less than ℓ /2.
i t+1 max
10