An Optimal Lower Bound for Anonymous Scheduling

Mechanisms

Shahar Dobzinski

Joint work with Itai Ashlagi and Ron Lavi

Unrelated Machines Scheduling

• n jobs to be assigned to m selfish machines

• Each machine i needs tij time units to complete job j, and incurs a cost of tij.– Private Information.

• Goal: assign jobs to machines to minimize the maximal load (minimize the makespan). – We use payments to motivate truthfulness.

The Nisan-Ronen Conjecture• Nisan and Ronen: a simple mechanism gives

an upper bound of m, and there is a lower bound of 2.– Ignoring computational issues.– The ’99 paper that introduced Algorithmic

Mechanism Design!

• Conjecture [Nisan-Ronen]: The lower bound of m.

Previous Work• Many efforts to prove or disprove the conjecture.• Christodoulou, Koutsoupias, and Vidali ‘07: an

improved lower bound (about 2.41, and then 2.61).– A huge gap between the upper bound and the

lower bound.• Mu’alem and Schapira ’07 and Christodoulou et al

‘07 give a lower bound of 2 for randomized and fractional mechanisms.

• Dobzinski and Sundararajan ‘08, and Christodoulo, Koutsoupias, and Vidali ‘08 characterize the 2 machines case.

Previous Work – Special Cases• Lavi and Swamy ‘07 prove that in the two values

case (“low” and “high” jobs) there are constant approximation mechanisms.

• Dhangwatnotai, Dobzinski, Dughmi, and Roughgarden ‘08 show that if the machines are related there is a PTAS.– The problem was introduced by Archer and Tardos ‘01.

• Is the Nisan-Ronen conjecture true?

Anonymity

• We provide the first concrete evidence that the Nisan-Ronen conjecture is true.– A lower bound of m for anonymous mechanisms.

• A mechanism f is anonymous if the names of the machines do not matter.– Two machines that switch cost vectors also switch

their assignments.

• Very weak notion of anonymity.

Why Anonymity?

• That’s what we can prove • Very natural from an algorithmic perspective.• Powerful even from a mechanism design perspective.

– Related machines [Dhangwatnotai-Dobzinski-Dughmi-Roughgarden]

– 2 values [Lavi-Swamy]

• Recent interest in the AGT community. [Daskalakis-Papadimtriou]

• We will talk about fractional mechanisms later…• First evidence that the Nisan-Ronen conjecture is true.

– First lower bound for a large class that is super constant.– At least, the algorithm is “strange”.

• Still, for revenue maximization in digital goods naming the players helps! [Aggarwal-Fiat-Goldberg-Hartline-Immorlica-Sudan]: – But in a single parameter setting.

Weak Monotonicity• Definition: an allocation function f is weakly monotone if

for every ti, t’i, t-i: suppose that machine i is allocated S in f(ti, t-i), and that it is allocated T in f(t’i,t-i). Then,

ti(T) – t’i(T) ≥ ti(S) – t’i(S)

• Reminder: Fix t-i.Each bundle has an associated payment (independent of ti). The machine is allocated the bundle that maximizes its profit. Every truthful mechanism is weakly monotone.

• Interpretation of WMON: the profit from taking T must increase more than the profit from taking S.

• Easy corollary: If machine i is allocated S, and lowers its cost for all jobs in S while raising its cost for all jobs not in S, then machine i still receives S.

• We’ll also use WMON in more delicate ways.

The Main Result

• Theorem: Every anonymous mechanism that provides a finite approx ratio must allocate as follows:

• Thus it provides an approx ratio no better than m.

• Intuition: this is how VCG allocates the jobs.

J1 … Jm

M1 t1 … t1

M2 t2 … t2

.

.

.

Mm tm … tm

t1+ > tm > … > t2 > t1

Outline of the Proof

• Proof is by induction on the number of jobs.

• In this talk: only 3 machines and 3 jobs, hence a lower bound of 3.

• An easy base case, and 5 induction steps.• Steps are “modular”.

– More or less…

• Lots of omissions and shameless cheating, sometimes in technical details.

The Base Case: One Job, 3 Machines

J

M1 t1

M2 t2

M3 t3

Towardsa contradiction:

J

M1 t1

M2 t1

M3 t3

WMON

t3> t2 > t1A contradiction

to the anonymityof the mechanism!

J

M1 t2

M2 t1

M3 t3

WMON

Induction StepsJ1 J2 J3

M1 t1 t1 a

M2 t2 t2 a

M3 t3 t3 a

t3> t2 > t1 >> a >>

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 a

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2 t

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1 t1

M2 t2 t2 t2

M3 t3 t3 t3

Step 1

J1 J2 J3

M1 t1 t1 a

M2 t2 t2 a

M3 t3 t3 a

Informally: The cost of J3 is very small so we can ignore this job.

By the induction hypothesis it must allocate both “big” jobs to M1.More formally, we fix the costs of J3 and define a mechanism on J1 and J2. The induction hypothesis applies to this mechanism.

Induction StepsJ1 J2 J3

M1 t1 t1 a

M2 t2 t2 a

M3 t3 t3 a

t3> t2 > t1 >> a >>

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 a

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2 t

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1 t1

M2 t2 t2 t2

M3 t3 t3 t3

Step 2J1 J2 J3

M1 t1 t1 a

M2 t2 t2 a

M3 t3 t3 a

WMONJ1 J2 J3

M1 t1-a t1-a

M2 t2 t2 a

M3 t3 t3 a

Towardscontradiction

WMONJ1 J2 J3

M1

M2 t2 t2 a

M3 t3 t3 a

The mechanism does notprovide a finite approx ratio!

Induction StepsJ1 J2 J3

M1 t1 t1 a

M2 t2 t2 a

M3 t3 t3 a

t3> t2 > t1 >> a >>

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 a

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2 t

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1 t1

M2 t2 t2 t2

M3 t3 t3 t3

t3

Step 3

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

Step 3(a) Step 3(b)

Step 3(a)

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3- t3- a

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

WMON

Step 3

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

Step 3(a) Step 3(b)

Step 3(b)J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

Lemma 1:(M1 gets atleast one “big”job)

J1 J2 J3

M1 t1

M2 t2 t2 a

M3 t3 t3 t3

Lemma 2:(One big, 2small: M1 getseverything)

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

By Lemma 1, towards a contradiction

J1 J2 J3

M1 t1

M2 t2 t2 a

M3 t3 t3 t3

WMON

A contradiction to Lemma 2

Proof of3(b):

Given (no proof):

Induction StepsJ1 J2 J3

M1 t1 t1 a

M2 t2 t2 a

M3 t3 t3 a

t3> t2 > t1 >> a >>

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 a

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1 t1

M2 t2 t2 t2

M3 t3 t3 t3

t2

Step 4J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2 t

M3 t3 t3 t3

I.e., machine 2 gets nothing in such “ordered” instances.

J1 J2 J3

M1 t1 t1

M2 t2 t2 t

M3 t3 t3 t3Towards

contradiction

J1 J2 J3

M1 t1 t1

M2 t2 t2 t

M3 t2- t2- t2-

The 2nd machine got a job. A contradiction.

WMON

Induction StepsJ1 J2 J3

M1 t1 t1 a

M2 t2 t2 a

M3 t3 t3 a

t3> t2 > t1 >> a >>

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 a

J1 J2 J3

M1 t1 t1

M2 t2 t2 a

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2 t

M3 t3 t3 t3

J1 J2 J3

M1 t1 t1

M2 t2 t2 t2

M3 t3 t3 t3

t1

Step 5

J1 J2 J3

M1 t1 t1 t1

M2 t2 t2 t

M3 t3 t3 t3

Towards a contradiction

J1 J2 J3

M1 t1+ t1+

M2 t2 t2 t

M3 t3 t3 t3

A contradiction tothe previous step!

(M1 should get everything)

(A similar argument if M1 is allocated two jobs)

WMON

Summary

• We showed that anonymous mechanisms provide only a trivial approximation ratio.– First evidence that the Nisan-Ronen

conjecture is indeed correct

• Might help in proving a lower bound on all mechanisms: anonymity is without loss of generality for fractional mechanisms.

Tool: Induced Mechanisms

• Suppose f is a mechanism for n jobs and m machines.

• Define f’ (a mechanism for (n-1) jobs and m machines): fix identical costs for the n’th job. Allocate in f’ the first n-1 jobs as in f.

J1 J2 J3

M1 q w a

M2 r c a

M3 d e a

f:

J1 J2

M1 q w

M2 r c

M3 d e

f’:

Induced Mechanisms (cont.)

• Proposition: f is truthful f’ is truthful.

• Proof: – f satisfies weak monotonicity. To finish, we will

prove that f’ satisfies weak monotonicity too.

– Suppose that machine i gets S in f(ti,t-i), and T in f(t’i,t-i). f satisfies weak monotonicity:

• ti(T) - t’i(T) ≥ ti(S) - t’i(S)

– So in f’ we have that• ti(T \ {n}) - t’i(T \ {n}) ≥ ti(S \ {n}) - t’i(S \ {n})