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Efficient Contention Resolution Protocols for Selfish Agents Amos Fiat, Joint work with Yishay Mansour and Uri Nadav Tel-Aviv University, Israel Workshop on Algorithmic Game Theory, University of Warwick, UK

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Alright people, listen up. The harder you push, the faster we will all get out of here. Tax deadline Deadlines:

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Deadline Analysis: 2 Symmetric Agents / 2 Time slots / Service takes 1 time slot Both agents are aggressive with prob. q, and polite with prob. 1-q Deadline Bart is polite: With probability q Lisa will get service and depart Bart is aggressive: With probability 1-q Lisa will be polite and Bart will be successful Slot #16Slot #17

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2 agents 1 Slot before deadline And Samson said, "Let me die with the Philistines!" Judges 16:30 Deadline Let Lisa be polite with prob. q If Bart is: polite - cost is 1 aggressive - expected cost is q Aggression is dominant strategy Slot #17

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Solving with MATHEMATICA q 20 ( t ) : Prob. of aggression when 20 agents are pending as a function of the time t, in equilibrium Time Aggression Probability deadline Blocking no one gets served

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Solving with MATHEMATICA q k (4 k ) : Aggression prob. when k agents are pending before deadline in 4 k time slots (Deadline: when lunch trays are removed at U. Warwick, CS department) # agents

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Deadline Cost – Few slots Theorem: In a symmetric equilibrium, whenever there aremore agents than time slots until deadline, agents transmit (transmission probability 1)

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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

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Efficiency of a linear deadline Theorem: There exists a symmetric equilibrium for D -deadline cost function such that: if the 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

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Switch Subject: Broadcast Channel / Latency Slot #1 Slot #2 Slot #3 Slot #4 Slot #5 Slot #6 time n agents (with a packet each) at time 0 No arrivals Known number of agents

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Broadcast Channel Slot #1 Slot #2 Slot #3 Slot #4 Slot #5 Slot #6 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, agent is better off transmitting all the time and has constant latency time Transmission probability 1/n is not in equilibrium

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Classical Results Maximizing the throughput Aloha (fixed probability) 0.37 More advanced algorithms 0.48 [MoH85] Impossibility result 0.56 [TsL88]

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Well established research. –Mostly in the 80s To learn more

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Related Work: Strategic MAC (Multiple Access Channel) [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]

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Protocol in Equilibrium Agent utility: Minimize latency Protocol in equilibrium: No incentive not to follow protocol Agent strategy: Transmission probability is a function of the number of pending agents k and current waiting time t Symmetry: All agents are symmetric

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Broadcast Channel Slot #1 Slot #2 Slot #3 Slot #4 Slot #5 Slot #6 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?

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Summary of (Latency) Results 1.All protocols where transmission probabilities do not depend on the time have exponential latency 2.We give a time-dependent protocol where all agents are successful in linear time

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Time-Independent Equilibrium Theorem: There is a unique time-independent, symmetric, non- blocking protocol in equilibrium for latency cost with transmission probabilities: 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 Very high Price of Anarchy

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Latency Equilibrium Proof idea (assuming q = q k q k-1 ) For the other k-1 agents: – α k-1 = Pr[all silent] = (1-q) k-1 – β k-1 = Pr[success] = q(k-1)(1-q) k-2 Consider always Transmit: –Expected Cost: 1/α k-1 Consider Quiescence and then Transmit –Expected cost: 1/β k-1 +1/α k-2

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Latency Equilibrium Proof idea (assuming q = q k q k-1 ) Equilibrium Equation: 1/α k-1 = 1/β k-1 +1/α k-2 Simplifying: 1-q-(k-1)q 2 =0 –Solution q 1/k A major simplification: q k q k-1

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Fight for every slot Cooperation is more important when trying to avoid a large payment (deadline) How can one create a sudden jump in cost? –Using external payments Agents go crazy: everyone continuously transmits –Time dependence Analyze step cost function (Deadline) Translate Latency Minimization to Deadline Cost Time Deadline Effectively, no message gets through here T

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Deadline Cost Function Deadline utility (scaled): Success before deadline – cost 0 Success after deadline – cost 1 Cost Time D (Deadline)

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Deadline vs. Repeated Prisoners Dilemma For finite horizon prisoners dilemma: Defect on last game. Inductively, no cooperation on any game Not our case: successful agents leave

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(t+1) +(1- ) C k,t+1 C k-1,t+1 + (1 - ) C k,t+1 Equilibrium Equations (Deadline, Latency, etc.) * C k,t = expected cost of k agents at time t (t) = cost of leaving at time t = Quiescence Transmit Probability one of the other k-1 agents leaves Probability the other k-1 agents are silent =

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Equilibrium Equations k,t ( (t+1)-C k,t+1 ) = k,t (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 - (t+1)+ (t+1)-C k,t+1 ) (1-q k,t ) k-1 (F k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (F k,t+1 -F k-1,t+1 ) (1-q k,t ) F k,t+1 = (k-1)q k,t (F k,t+1 -F k-1,t+1 ) k,t ( (t+1))+(1- k,t )C k,t+1 = k,t C k-1,t+1 + (1- k,t ) C k,t+1

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Equilibrium Equations k,t ( (t+1)-C k,t+1 ) = k,t (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 - (t+1)+ (t+1)-C k,t+1 ) (1-q k,t ) k-1 (F k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (F k,t+1 -F k-1,t+1 ) (1-q k,t ) F k,t+1 = (k-1)q k,t (F k,t+1 -F k-1,t+1 ) k,t ( (t+1))+(1- k,t )C k,t+1 = k,t C k-1,t+1 + (1- k,t ) C k,t+1

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Equilibrium Equations k,t ( (t+1)-C k,t+1 ) = k,t (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 - (t+1)+ (t+1)-C k,t+1 ) (1-q k,t ) k-1 (F k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (F k,t+1 -F k-1,t+1 ) (1-q k,t ) F k,t+1 = (k-1)q k,t (F k,t+1 -F k-1,t+1 ) k,t ( (t+1))+(1- k,t )C k,t+1 = k,t C k-1,t+1 + (1- k,t ) C k,t+1

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Equilibrium Equations k,t ( (t+1)-C k,t+1 ) = k,t (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 - (t+1)+ (t+1)-C k,t+1 ) (1-q k,t ) k-1 (F k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (F k,t+1 -F k-1,t+1 ) (1-q k,t ) F k,t+1 = (k-1)q k,t (F k,t+1 -F k-1,t+1 ) k,t ( (t+1))+(1- k,t )C k,t+1 = k,t C k-1,t+1 + (1- k,t ) C k,t+1

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Equilibrium Equations k,t ( (t+1)-C k,t+1 ) = k,t (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 - (t+1)+ (t+1)-C k,t+1 ) (1-q k,t ) k-1 (F k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (F k,t+1 -F k-1,t+1 ) (1-q k,t ) F k,t+1 = (k-1)q k,t (F k,t+1 -F k-1,t+1 ) k,t ( (t+1))+(1- k,t )C k,t+1 = k,t C k-1,t+1 + (1- k,t ) C k,t+1

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Equilibrium Equations k,t ( (t+1)-C k,t+1 ) = k,t (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 - (t+1)+ (t+1)-C k,t+1 ) (1-q k,t ) k-1 (F k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (F k,t+1 -F k-1,t+1 ) (1-q k,t ) F k,t+1 = (k-1)q k,t (F k,t+1 -F k-1,t+1 ) k,t ( (t+1))+(1- k,t )C k,t+1 = k,t C k-1,t+1 + (1- k,t ) C k,t+1

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Equilibrium Equations k,t ( (t+1)-C k,t+1 ) = k,t (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 - (t+1)+ (t+1)-C k,t+1 ) (1-q k,t ) k-1 (F k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (F k,t+1 –F k-1,t+1 ) (1-q k,t ) F k,t+1 = (k-1)q k,t (F k,t+1 -F k-1,t+1 ) k,t ( (t+1))+(1- k,t )C k,t+1 = k,t C k-1,t+1 + (1- k,t ) C k,t+1

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Equilibrium Equations k,t ( (t+1)-C k,t+1 ) = k,t (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 -C k,t+1 ) (1-q k,t ) k-1 ( (t+1)-C k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (C k-1,t+1 - (t+1)+ (t+1)-C k,t+1 ) (1-q k,t ) k-1 (F k,t+1 ) = (k-1)q k,t (1-q k,t ) k-2 (F k,t+1 -F k-1,t+1 ) (1-q k,t ) F k,t+1 = (k-1)q k,t (F k,t+1 -F k-1,t+1 ) k,t ( (t+1))+(1- k,t )C k,t+1 = k,t C k-1,t+1 + (1- k,t ) C k,t+1

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> 1/2 Transmission Probability in Equilibrium Lemma (Manipulating equilibrium equations): >0 1/k < Benefit from losing one agent * F k,t = C k,t - (t) ; expected future cost C k,t = expected cost of k agents at time t Transmission probability when k players at time t Observation: –Either transmission probability in [ 1/k,2/k ] –Or, limited benefit from loosing one agent <1/2 2/ k >

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Analysis of Deadline utility F k,t = F k-1,t+1 + (1- ) F k,t+1 We seek an upper bound for C n,0 = F n,0 Recall: Observation: –Either transmission probability in [ 1/k,2/k ] –Or, limited benefit from getting rid of one agent Consider a tree of recursive computation for F n,0

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F n,t F n,t+1 F n-1,t+1 Upper Bound on Cost Two descendantsOne descendant (F n,t+1 > 2 F n-1,t+1 ) F n,t+1 < 2 F n-1,t+1 1- F n,t = F n-1,t+1 + (1- ) F n,t+1 F n,t < F n,t+1 < 2 F n-1,t+1 < 2 Good edgesDoubling edges F n,t+1 F n-1,t+1 F n,t F n,t / F n-1,t+1 < 0.8 < 0.3 Transmission probability

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Upper Bound on Cost # Agents TimeDeadline Fn,0Fn,0 F n,1 F 17,D = 1 F n-3,4 F n-1,1 F n-2,2 F n-3,3 F n-4,4 F 1,D-9 = 0 cost=0 L 1 cost=1

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Upper Bound on Cost The weight of such a path: –At least D-n good edges –Weight at most (1-β) D-n 2 n Number of paths at most: cost=0 1 Set D > 20n to get an upper bound of e -c n on cost

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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 Sub-game perfect equilibrium Social optimum achieved with high probability Equilibrium

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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

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Future Research General cost functions Does the time-independent equilibrium induces an optimal expected latency? Protocol in equilibrium for an arrival process Arrival times / duration in general congestion games: –Atomic traffic flow: dont leave home until 9:00 AM and get to work earlier

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Two users : Equilibrium Always transmit: Best response is to Quiescence agents: Eq. q = ½, minimizes time to first success Notation: C k,t = expected latency with k agents at time t F k,t = C k,t - t

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Two users : Equilibrium Always transmit: Best response is to be quiescent agents: Eq. q = ½, minimizes time to first success

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In a strictly mixed Equilibrium, individual is indifferent between Transmit and Quiescent Equilibrium Equations + C k-1,t+1 (1 - ) C k,t+1 (t+1) (1- ) C k,t+1 + = * C k,t = expected cost of k agents at time t (t) = cost of leaving at time t = Quiescent Transmit Probability one of the other k-1 agents leaves Probability the other k-1 agents are silent

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Broadcast Channel Slot #1 Slot #2 Slot #3 Slot #4 Slot #5 Slot #6 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 Transmission probability 1/n is not in equilibrium time

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Upper Bound on Cost The weight of such a path: –At least D-n good edges –Weight at most (1-β) D-n 2 n Number of paths at most: cost=0 1 Set D > 20n to get an upper bound of e -c n on cost Let D = αn, then total weight at most 2(1-β) D-n 2 n α n e n = 2[2e(1-β) α-1 α] n

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