Some Research Problems in Algorithmic Game Theory: Incentive compatible communications Envy Free makespan Grad Student Research Seminar Amos Fiat Tel Aviv University November 11, 2010

About Doing Research, work?? Flee from a Burning Theatre Envy and Makespan Network design games Combinatorial Auctions

First Subject: How to Escape from a burning Theatre

Contention: Broadcast Channel 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, I am better off transmitting all the time, until success 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 80’s To learn more

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 Equilibrium: Following the protocol is best response Strategy: Transmission probability is a function of the number of pending agents k and current waiting time t Protocol: Symmetric equilibrium

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 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

Two users : Equilibrium Always transmit: Best response is to be quiescent 12 12

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 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 Main Intuition 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)

“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 Deadlines:

2 agents 1 Slot before deadline Suppose a non-blocking equilibrium exist: – Transmission probability: q < 1 Deadline Let Lisa play according to protocol If Bart plays: Quiescent: cost is 1 Transmit: expected cost is q Non-blocking equilibrium does not exists Transmit is dominant strategy Slot #17

Deadline Cost – Few slots Theorem: In a symmetric equilibrium, whenever there aremore 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 Bart plays quiescent With probability q Lisa will transmit and leave q = 1-q ) q = ½ Bart plays transmit With probability 1-q Lisa will play quiescent Slot #16Slot #17

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

Solving with MATHEMATICA q 20 ( t ) : Transmission probability when 20 agents are pending as a function of the time t, in equilibrium Time Transmission Probability deadline Blocking

Solving with MATHEMATICA q k (4 k ) : Transmission probability when k agents are pending at time 4 k, before deadline, in equilibrium # agents

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

  (t+1) +(1-  ) C k,t+1  C k-1,t+1 + (1 -  ) C k,t+1 Equilibrium Equations * 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,t+1 -C k-1,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,t+1 -C k-1,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,t+1 -  (t+1)+  (t+1)-C k-1,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,t+1 -C k-1,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,t+1 -C k-1,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,t+1 -  (t+1)+  (t+1)-C k-1,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,t+1 -C k-1,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,t+1 -C k-1,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,t+1 -  (t+1)+  (t+1)-C k-1,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,t+1 -C k-1,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,t+1 -C k-1,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,t+1 -  (t+1)+  (t+1)-C k-1,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,t+1 -C k-1,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,t+1 -C k-1,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,t+1 -  (t+1)+  (t+1)-C k-1,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,t+1 -C k-1,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,t+1 -C k-1,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,t+1 -  (t+1)+  (t+1)-C k-1,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,t+1 -C k-1,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,t+1 -C k-1,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,t+1 -  (t+1)+  (t+1)-C k-1,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 losing one agent <1/2 2/ k >

Return to Deadline F k,t =  F k-1,t+1 + (1-  ) F k,t+1 We seek an upper bound on C n,0 = F n,0 Recall: Observation: –Either transmission probability in [ 1/k,2/k ] –Or, limited benefit from loosing 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 descendants One 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 Transmission probability 2 [1/n, 2/n ] < 0.8 < 0.3

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

Open Problems I: Contention Prove the magical 4k threshold (!!!) Extend to more general settings, multiple packets Justify TCP/IP (Congestion vs. Contention)

New Subject: Makespan and Envy

Mechanism Design: Allocation problems Set U of 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= ): receives agent valuations for object bundles as input Returns: allocation a and payments p for the agents

Mechanisms for allocation problems n agents, m items v i (S) – valuation of set of items S to agent i Mechanism produces – allocation a = (a 1,a 2,…,a m ) and – prices (p 1,p 2,…,p m ). Utility of player i: v i (a i ) - p i

Truthful mechanism Intuition: agent i whose valuation is v i would prefer “telling the truth” v i to the mechanism rather than any possible “lie” v’ i Mechanism is truthful (=incentive compatible): – If a = f(v i, v −i ) and a’= f (v’ i, v -i ), – then v i (a) − p i (v i, v −i ) ≥ v i (a’) − p i (v’ i, v −i ).

Envy freeness: no one wants to switch places with another. Envy freeness and Justice: – Rawls (A Theory of Justice ), – 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 v i (S) – valuation of agent i for set S Mechanism gives an allocation (a 1,a 2,…,a m ) and prices (p 1,p 2,…,p m ). Mechanism is envy-free if: v i (a i ) – p i ≥ v i (a k ) – p k

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 v i (a i ) Allocation is locally efficient if the sum of valuations is maximized over all permutations of the assignments (forget payments) ∑ i v i (a i ) ≥ ∑ i v i (a π(i) )

Characterizations Hartline et. al : a) If allocation is locally efficient, then there exist payments for this allocation function that make it (and payments) envy-free b) Allocation of every envy-free mechanism is locally efficient

Characterizations allocation is locally efficient exist envy-free mechanism

Proof (one way) Allocation of every envy-free mechanism is locally efficient envy-free => v i (a i ) – p i ≥ v i (a π(i) ) – p π(i) ∑v i (a i ) – ∑p i ≥ ∑v i (a π(i) ) – ∑p π(i) ∑v i (a i ) – ∑v i (a π(i) ) ≥ ∑p i - ∑p π(i) = 0

VCG = Locally Efficient VCG Makespan 4-4ε Envy Free (and Incentive Compatible) T 1T 2T 3T 4 M 11-ε M M M 41111

Another Locally Efficient Assignment T 1T 2T 3T 4 M 11-ε M M M There is no permutation that can decrease sum of costs

Envy Free Mechanism: Packing Bundles VCG allocation is locally efficient, but we could do better by restricting the bundles. Ergo, “what jobs do we put together in a bundle?” Start with approximation to optimal

Phase 1, Subphase 1: Permutation Compute the permutation that minimizes the sum of the loads for these specific bundles (cannot break a bundle apart) Can be done in polynomial time - weighted matching problem.

A-Opt Every machine may have multiple jobs Locally efficient on A-Opt bundles 2 A-Opt

A-Opt Locally efficient on A-Opt bundles On each machine can be more then onejob 2 A-Opt

Remaining bundles of A-Opt 2 A-Opt

Remaining bundles of A-Opt 2 A-Opt

Algorithm Start from A-Opt Calculate permutation to minimize the sum of the loads – locally efficient Put aside the bundles assigned to machines with load > 2 A-Opt.

Phase 1, Multiple Subphases: Bundles on short machines Take the bundles left over and (re) compute the assignment minimizing the sum of loads for them. Again some of the bundles may be on machines with load 2 A-Opt or more. Put these aside too, and repeat. In total, we will put aside no more than m/2 bundles during all subphases.

The 1 st phase ended when with makespan of remaining bundles ≤ 2 A-Opt We have a 1 st assignment of bundles to machines (those not put aside) Repeat the process with the ≤ m/2 bundles put aside, now – no more than m/4 bundles will be put aside. Phases

Combine the bundles assigned to each machine, This is also locally efficient Phase 1 Phase 2 Final assignment

Log m phases First phase assigns at least m/2 bundles (at most m/2 left unassigned) After second phase - at most m/4 bundles unassigned So we have no more than log m many phases

Log m makespan approximation The bundles assigned in the end of a phase are assigned to machines of load no more than 2A-Opt The load of the union of all such bundles assigned to any specific machine is therefore no more than O(log m) times A-Opt.

Lower Bound log m /log log m T 1T 2T 3T n M11∞∞∞∞∞∞∞ M213/141∞∞∞∞∞∞ M3M312/1411/121∞∞∞∞∞ M411/1410/129/101∞∞∞∞ M510/149/128/107/81∞∞∞ …9/148/127/106/85/61∞∞ 8/147/126/105/84/63/41∞ M nM n1/2 1 M n M n … M n + log( log m / log log m ) 4 = log n/ c loglog n

Lower bound log m / log log m If every task assigned to the first n-1 machines goes to the next machine, the load drops by log n ( ≈ log m) To make room, jobs in the bottom part need to drop too They will drop if the increase in cost is bounded by log m

Summary – EF makespan minimization We showed almost tight bounds for envy-free makespan minimization – open problem: close gap Homework: – Prove that locally efficient implies envy freeness (that there exists prices that make the allocation envy free) – Prove that the union of of locally efficient assignements is locally efficient