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## Amos Fiat Tel Aviv University November 11, 2010

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**Some Research Problems in Algorithmic Game Theory:Incentive**compatible communicationsEnvy Free makespanGrad Student Research Seminar Amos Fiat Tel Aviv University November 11, 2010**Contention: Broadcast Channel**• n agents (with a packet each) at time 0 • No arrivals • Known number of agents time Slot #1 Slot #5 Slot #2 Slot #6 Slot #3 Slot #4**Broadcast Channel**time Slot #1 Slot #5 Slot #2 Slot #6 Slot #3 Slot #4 Transmission probability 1/n is not in equilibrium • Symmetric solution: every agent transmits with probability 1/n, the expected waiting time is O(n) slots. (Social optimum) • If all others transmit with probability 1/n, I am better off transmitting all the time, until success**Classical View versus AGT view**The classical view: • Find a “good” protocol • Assumes agents follow any protocol. Our view: • What would happen if agents are selfish • Agents can adjust their transmission probabilities • Rather than optimization consider equilibrium.**Related Work: Strategic MAC**• [Altman et al 04] • Incomplete information: number of agents • Stochastic arrival flow to each source • Restricted to a single retransmission probability • Shows the existence of an equilibrium • Numerical results • [MacKenzie & Wicker 03] • Multi-packet reception • Transmission cost [due to power loss] • Characterize the equilibrium and its stability • Also [Gang, Marbach & Yuen]**Equilibrium**Utility: Waiting time until success Strategy: Transmission probability is a function of the number of pending agents k and current waiting time t Equilibrium: Following the protocol is best response Protocol: Symmetric equilibrium**Broadcast Channel**Strategy: Always transmit! • Equilibrium • The channel is blocked anyway • Also in subgame perfect equilibrium • Remark: For at least 3 players • Not quite what we look for • Is this the only equilibrium? Slot #1 Slot #5 Slot #2 Slot #6 Slot #3 Slot #4**Summary of Results**• All protocols where transmission probabilities do not depend on the time have exponential latency • We give a “time-dependent” protocol where all agents are successful in linear time**Time-Independent Equilibrium**Theorem: There is a unique time-independent, symmetric, non-blocking protocol in equilibrium for latency cost with transmission probabilities: Very high “Price of Anarchy” • Expected Delay of the first transmitted packet: • Probability even one agent successful within polynomial time bound is negligible • Compare to social optimum: • All agents successful in linear time bound, with high probability**T**Deadline Main Intuition Effectively, no message gets through here Cost Time • Fight for every slot • Cooperation is more important when trying to prevent a large payment • How to create a large leap in cost function? • Using external payments • Agents go “crazy”: everyone continuously transmits • Time dependent • Analyze step cost function**Deadline Cost Function**Cost D (Deadline) Time • Deadline utility (scaled): • Success before deadline – cost 0 • Success after deadline – cost 1**Deadlines:**“Alright people, listen up. The harder you push, the faster we will all get out of here.” crowd in post office at tax filing deadline**2 agents 1 Slot before deadline**Suppose a non-blocking equilibrium exist: • Transmission probability: q < 1 Non-blocking equilibrium does not exists • Let Lisa play according to protocol • If Bart plays: • Quiescent: cost is 1 • Transmit: expected cost is q Transmit is dominant strategy Deadline Slot #17**Deadline Cost – Few slots**Theorem: In a symmetric equilibrium, whenever there are more agents than time slots until deadline,agents transmit (transmission probability 1) • Proof: By backward induction (on the time t) • At any time more agents than time slots • At times t’>t no successful transmission • “Fight” for the chance to succeed**Finite horizon Prisoners Dilemma**• Deadline reminds us of finite horizon prisoner’s dilemma • Defect the last game played • Inductively, no cooperation on any game Not our case: successful agents leave**Deadline Analysis: 2 Agents**• 2 time slots left Deadline Slot #16 Slot #17 • Bart plays quiescent • With probabilityqLisa will transmit and leave • Bart plays transmit • With probability1-qLisa will play quiescent q = 1-q ) q = ½**Deadline: non-blocking Equilibrium**Theorem: There exists a symmetric equilibrium, such that whenever there are at least as many time slots as agents, transmission probability is less than 1**Blocking**Solving with MATHEMATICA q20(t): Transmission probability when 20 agents are pending as a function of the time t, in equilibrium Transmission Probability 19 0.05 Time deadline**Efficiency of a linear deadline**Theorem: There exists a symmetric equilibrium for D-deadline cost function such that: ifthe deadline D > 20n then, the probability that not all agents succeed prior to the deadline is negligible (e-cD) If there is enough time for everyone, a “nice” equilibrium**Equilibrium Equations**Probability one of the other k-1 agents leaves = Quiescence Transmit (t+1) +(1- ) Ck,t+1 Ck-1,t+1+ (1 - ) Ck,t+1 = Probability the other k-1 agents are silent * Ck,t = expected cost of k agents at time t (t)= cost of leaving at timet**Equilibrium Equations**) k,t((t+1))+(1- k,t )Ck,t+1 = k,t Ck-1,t+1 + (1- k,t ) Ck,t+1 k,t((t+1)-Ck,t+1) = k,t(Ck,t+1-Ck-1,t+1) ) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1-Ck-1,t+1) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1- (t+1)+(t+1)-Ck-1,t+1) ) (1-qk,t)k-1(Fk,t+1) = (k-1)qk,t(1-qk,t)k-2(Fk,t+1-Fk-1,t+1) ) (1-qk,t)Fk,t+1 = (k-1)qk,t(Fk,t+1-Fk-1,t+1)**Equilibrium Equations**) k,t((t+1))+(1- k,t )Ck,t+1 = k,t Ck-1,t+1 + (1- k,t ) Ck,t+1 k,t((t+1)-Ck,t+1) = k,t(Ck,t+1-Ck-1,t+1) ) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1-Ck-1,t+1) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1- (t+1)+(t+1)-Ck-1,t+1) ) (1-qk,t)k-1(Fk,t+1) = (k-1)qk,t(1-qk,t)k-2(Fk,t+1-Fk-1,t+1) ) (1-qk,t)Fk,t+1 = (k-1)qk,t(Fk,t+1-Fk-1,t+1)**Equilibrium Equations**) k,t((t+1))+(1- k,t )Ck,t+1 = k,t Ck-1,t+1 + (1- k,t ) Ck,t+1 k,t((t+1)-Ck,t+1) = k,t(Ck,t+1-Ck-1,t+1) ) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1-Ck-1,t+1) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1- (t+1)+(t+1)-Ck-1,t+1) ) (1-qk,t)k-1(Fk,t+1) = (k-1)qk,t(1-qk,t)k-2(Fk,t+1-Fk-1,t+1) ) (1-qk,t)Fk,t+1 = (k-1)qk,t(Fk,t+1-Fk-1,t+1)**Equilibrium Equations**) k,t((t+1))+(1- k,t )Ck,t+1 = k,t Ck-1,t+1 + (1- k,t ) Ck,t+1 k,t((t+1)-Ck,t+1) = k,t(Ck,t+1-Ck-1,t+1) ) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1-Ck-1,t+1) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1- (t+1)+(t+1)-Ck-1,t+1) ) (1-qk,t)k-1(Fk,t+1) = (k-1)qk,t(1-qk,t)k-2(Fk,t+1-Fk-1,t+1) ) (1-qk,t)Fk,t+1 = (k-1)qk,t(Fk,t+1-Fk-1,t+1)**Equilibrium Equations**) k,t((t+1))+(1- k,t )Ck,t+1 = k,t Ck-1,t+1 + (1- k,t ) Ck,t+1 k,t((t+1)-Ck,t+1) = k,t(Ck,t+1-Ck-1,t+1) ) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1-Ck-1,t+1) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1- (t+1)+(t+1)-Ck-1,t+1) ) (1-qk,t)k-1(Fk,t+1) = (k-1)qk,t(1-qk,t)k-2(Fk,t+1-Fk-1,t+1) ) (1-qk,t)Fk,t+1 = (k-1)qk,t(Fk,t+1-Fk-1,t+1)**Equilibrium Equations**) k,t((t+1))+(1- k,t )Ck,t+1 = k,t Ck-1,t+1 + (1- k,t ) Ck,t+1 k,t((t+1)-Ck,t+1) = k,t(Ck,t+1-Ck-1,t+1) ) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1-Ck-1,t+1) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1- (t+1)+(t+1)-Ck-1,t+1) ) (1-qk,t)k-1(Fk,t+1) = (k-1)qk,t(1-qk,t)k-2(Fk,t+1-Fk-1,t+1) ) (1-qk,t)Fk,t+1 = (k-1)qk,t(Fk,t+1-Fk-1,t+1)**Equilibrium Equations**) k,t((t+1))+(1- k,t )Ck,t+1 = k,t Ck-1,t+1 + (1- k,t ) Ck,t+1 k,t((t+1)-Ck,t+1) = k,t(Ck,t+1-Ck-1,t+1) ) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1-Ck-1,t+1) ) (1-qk,t)k-1((t+1)-Ck,t+1) = (k-1)qk,t(1-qk,t)k-2(Ck,t+1- (t+1)+(t+1)-Ck-1,t+1) ) (1-qk,t)k-1(Fk,t+1) = (k-1)qk,t(1-qk,t)k-2(Fk,t+1-Fk-1,t+1) ) (1-qk,t)Fk,t+1 = (k-1)qk,t(Fk,t+1-Fk-1,t+1)**Transmission probability**when k players at time t Transmission Probability in Equilibrium Benefit from losing one agent Lemma (Manipulating equilibrium equations): 2/k > <1/2 1/k < > 1/2 >0 • Observation: • Either transmission probability in [1/k,2/k] • Or, limited benefit from losing one agent *Fk,t = Ck,t - (t) ; expected future cost Ck,t = expected cost of k agents at time t**Return to Deadline**We seek an upper bound onCn,0 = Fn,0 Fk,t = Fk-1,t+1 + (1- ) Fk,t+1 Recall: • Observation: • Either transmission probability in [1/k,2/k] • Or, limited benefit from loosing one agent Consider a tree of recursive computation for Fn,0**Fn,t**Fn,t+1 Fn,t Fn-1,t+1 Fn,t+1 Fn-1,t+1 Fn,t / Fn-1,t+1 Upper Bound on Cost One descendant Two descendants Fn,t+1 < 2 Fn-1,t+1 Transmission probability 2[1/n, 2/n ] (Fn,t+1 > 2 Fn-1,t+1 ) < 2 1- < 0.8 < 0.3 Fn,t = Fn-1,t+1 + (1-) Fn,t+1 Fn,t < Fn,t+1 < 2 Fn-1,t+1 Good edges Doubling edges**Fn,0**Fn,1 Fn-1,1 Fn-4,4 Fn-2,2 Fn-3,3 Fn-3,4 L1 cost=1 F1,D-9 = 0 cost=0 Upper Bound on Cost #Agents F17,D = 1 Deadline Time**1**cost=0 Upper Bound on Cost • The weight of such a path: • At least D-n good edges • Weight at most (1-β)D-n2n • Number of paths at most: Set D > 20n to get an upper bound of e-c n on cost**Protocol Design: from Deadline to Latency**Embed artificial deadline into “deadline” protocol • Deadline Protocol: • Before time 20n transmission probability as in equilibrium • If not transmitted until 20n: • Set transmission probability = 1 (blocking) • For exponential number of time slots • Equilibrium • Sub-game perfect equilibrium • Social optimum achieved with high probability**Summary**• Unique non-blocking equilibrium for Aloha like Protocols • Exponential latency • Deadlines: • If enough (linear) time, equilibrium is “efficient” • Protocol Design: • Make “ill behaved” latency cost act more “polite” • Using virtual deadlines • No monetary “bribes” or penalties**Open Problems I: Contention**• Prove the magical 4k threshold (!!!) • Extend to more general settings, multiple packets • Justify TCP/IP (Congestion vs. Contention)**Mechanism Design:Allocation problems**• Set Uof objects • m agents • [All] Objects to be allocated • Includes: • Combinatorial Auctions • Machine Scheduling • [Room / Paper] Assignment Problem • With / without capacity constraints • Payments/ Compensation**Allocation problems**• Possible Goals: • Social Welfare (sum of utilities) • Min makespan (min maximal disutility) • Revenue • Anything you can think of • Mechanism (M=<a,p>): receives agent valuations for object bundles as input • Returns: allocation a and payments p for the agents**Mechanisms for allocation problems**• nagents, m items • vi(S) – valuation of set of items S to agent i • Mechanism produces • allocation a = (a1,a2,…,am)and • prices (p1,p2,…,pm). • Utility of player i: vi(ai) - pi**Truthful mechanism**• Intuition: agent i whose valuation is vi would prefer “telling the truth” vi to the mechanism rather than any possible “lie” v’i • Mechanism is truthful (=incentive compatible): • Ifa = f(vi, v−i ) anda’= f (v’i, v-i ), • then vi(a) − pi (vi, v−i ) ≥ vi(a’) − pi(v’i, v−i ).**Envy Freeness**• Envy freeness: no one wants to switch places with another. • Envy freeness and Justice: • Rawls (A Theory of Justice - 2005), • Freud, Nietzsche (Forester - Justice, Envy and Psychoanalysis – 1997) • Aristole(322 BC), Mandeville (1730), etc. • The Envy Free Interpretation of Justice really means “no discrimination”**Envy Freeness**• We divide a cake amongst 3 children so that no one wants to switch with another. (Divisible Goods) • We divide household chores amongst 4 children so that no one wants to switch with another. • We assign rooms to faculty in a new building so that no one wants to switch with another. (Indivisible Goods)**Envy Freeness: Individual valuations**• A cake could be partly chocolate, partly vanilla, and has some cherries. Some people like chocolate more than vanilla, some like vanilla more than chocolate but hate cherries, etc. • Many different types of chores. Some kids hate washing dishes, others hate washing the dog, some like washing the dog. • Some rooms are larger, some have a view, some are closer to the grad student rooms. Some faculty like good views, others prefer larger rooms, etc.**Envy-free mechanism**• n agents, m items • vi(S) – valuation of agent ifor set S • Mechanism gives an allocation (a1,a2,…,am) and prices (p1,p2,…,pm). • Mechanism is envy-free if: vi(ai) – pi ≥ vi(ak) – pk**Ongoing Research Agenda**• Makespan minimization of unrelated machines: • Envy free mechanisms and lower bounds • Envy free and truthful mechanisms (?) • Combinatorial Auctions • Truthful and envy free (LOS is envy free). • Budgets ? • Assignment problems with capacities (the program committee problem): Truthful and envy free? • Lots and lots and lots of open problems**Nisan and Ronen 1999: Makespan Minimization for Unrelated**Machine Scheduling • There are m machines (or children), every machine (child) is an agent • There are n tasks (or household chores) • Every machine (child) says how long every task will take • The goal is to assign the jobs to the machines so as to well approximate the makespan. • This problem is APX but can be approximated.**Makespan minimization for unrelated machines**• Nisan and Ronen suggested the open problem of a truthful mechanism for (approximating) the minimal makespan for unrelated machine scheduling. This is still open. • The best known incentive compatible approximation is m and the lower bound is constant. • Hartline, Ieong, Mualem, Schapira and Zohar give an envy-free mechanism (not truthful) for approximating the minimal makespan for unrelated machine scheduling. • They give an envy free mechanism with an approximation factor upper bound of m/2 and a constant lower bound.**Our Results – Makespan Minimization**• We give an envy free mechanism that approximates the minimal makespan to within a factor of O(log m) • We show that no envy free mechanism can approximate the makespan to a factor better than Ω (log m / log log m) • Open problem: prove a better than O(1) lower bound for truthful and envy free mechanisms**Definitions**• Social welfare is sum of valuations : ∑ i vi(ai) • Allocation is locally efficient if the sum of valuations is maximized over all permutations of the assignments (forget payments) ∑ i vi(ai) ≥ ∑ i vi(aπ(i))