# Efficient Contention Resolution Protocols for Selfish Agents Amos Fiat, Joint work with Yishay Mansour and Uri Nadav Tel-Aviv University, Israel Workshop.

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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

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

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

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

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 19 0.05 Blocking no one gets served

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

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

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

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

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

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

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

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

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]

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

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?

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

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

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

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

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

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

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

(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 =

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

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

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

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

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

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

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

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

> 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 >

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

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

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

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

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

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

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

Two users : Equilibrium Always transmit: Best response is to Quiescence 12 12 2-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

Two users : Equilibrium Always transmit: Best response is to be quiescent 12 12 2-agents: Eq. q = ½, minimizes time to first success

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

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

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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