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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 t ij time units to complete job j, and incurs a cost of t ij .
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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 mselfish 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 monotoneif 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. t1+e > 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 Towardsa contradiction: WMON WMON t3> t2 > t1 A contradictionto the anonymityof the mechanism!
Induction Steps t3> t2 > t1 >> a >> d
Step 1 • 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 Steps d t3> t2 > t1 >> a >> d
Step 2 WMON Towards contradiction WMON The mechanism does notprovide a finite approx ratio!
Induction Steps t3 t3> t2 > t1 >> a >> d
Step 3 Step 3(a) Step 3(b)
Step 3(a) WMON
Step 3 Step 3(a) Step 3(b)
Step 3(b) Given (no proof): Lemma 1: (M1 gets atleast one “big”job) Lemma 2: (One big, 2small: M1 getseverything) Proof of 3(b): WMON By Lemma 1, towards a contradiction A contradiction to Lemma 2
Induction Steps t2 t3> t2 > t1 >> a >> d
Step 4 I.e., machine 2 gets nothing in such “ordered” instances. The 2nd machine got a job. A contradiction. WMON Towardscontradiction
Induction Steps t1 t3> t2 > t1 >> a >> d
Step 5 WMON A contradiction tothe previous step! (M1 should get everything) Towards a contradiction (A similar argument if M1 is allocated two jobs)
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. f’: 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})