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On Approximately Fair Allocations of Indivisible Goods. Richard Lipton Vangelis Markakis. Elchanan Mossel Amin Saberi. Georgia Tech AUEB.
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On Approximately Fair Allocations of Indivisible Goods Richard Lipton Vangelis Markakis Elchanan Mossel Amin Saberi Georgia Tech AUEB U. C. Berkeley Stanford
Cake-cutting problems Divide the cake among a set of people in a fair manner Empirically: since Pharaoh times (land division) Mathematical approaches: [Steinhaus, Banach, Knaster ’48] Fairness measure: Envy[Foley ’67] Infinitely divisible cakes: Envy-free partitions exist Cake-cutting procedures: minimize # cuts, achieve additional fairness criteria [Brams, Taylor ’96, Robertson, Webb ‘98]
Discrete version Set of indivisible goods M = {1, 2, …, m} Set of agents N = {1, 2, …, n}
Model For agent p: utility function : (monotone) • Special cases: • Additive utilities (e.g. probability measures) • Same utility for every agent.
What is fair? • Proportionality [Steinhaus - Banach - Knaster ’48] • Envy-freeness [Foley ’67, Varian ‘74] • Max-min fairness [Dubins - Spanier ’61] • Equitability • …..
Fairness Concept Given an allocation A = (A1,…,An): Envy of p for q: Envy of A: Envy-free allocations may not exist Goal:Algorithms with upper bounds on the envy
Outline • Existence of allocations with bounded envy • Optimization problems: positive and negative results • Incentive Compatibility
Outline • Existence of allocations with bounded envy • Optimization problems: positive and negative results • Incentive Compatibility
Additive Utilities Theorem[Dall’Aglio - Hill ’03]: There exists an allocation A with e(A) ≤(2n)3/2. Proof: probability measure on [0,1], Tools: convexity arguments, envy seen as the distance between a certain space and its convex hull.
A Tight Bound [Dall’Aglio - Hill ’03]:e(A) ≤(2n)3/2 1 good, 2 players e(A) Theorem: We can compute in time O(mn3) an allocation A, such that e(A) ≤.
Proof A: allocation of a subset of the goods S M. G(A) = (V, E) : envy graph of A • V = {agents} • pq E iff p envies q in A. ● ● A5 ● A1 ● A4 A = (A1, A2,…,A5,…) A2 A3 ● ● ● ● ●
● ● A1 ● ● A2 A5 A3 A4 ● ● ● ● ● • Claim: For any allocation A, there exists an allocation B s.t.: • e(B) ≤ e(A). • envy-graph of B is acyclic ( i with in-degree = 0). ● A5 ● ● A1 ● A4 A2 A3 ● ● ● ● ● # of edges decreases Envy does not increase
Algorithm At step i: • Eliminate all the cycles from the envy graph. • Give good i to an agent that no-one envies (any node with in-degree = 0). □
Remarks • Bound is tight • Nonadditive utilities maximum marginal utility • Cyclic swaps: used in finding theater sponsors in ancient Greece, (2-cycles)!
Outline • Existence of allocations with bounded envy • Optimization problems: positive and negative results • Incentive Compatibility
Optimization Problem 1 [envy]: Find an allocation A that minimizes the envy: Problem 2[envy-ratio]: Find an allocation A that minimizes the ratio: Polynomial time algorithms?
Hardness Results Both problems are NP-hard. Proof: Partition; even if n = 2 and both players have the same utility function. Envy: Also hard to approximate; even for the above case.
Additive Utilities Assume agents have the same utility function Value of good Envy-ratio(A) =
Relations with Job Scheduling People Processors Goods Jobs • [Graham ’69]: • Order the goods in decreasing value. • Give next good to the person with the minimum current bundle. [Coffman-Langston ’84]: Graham’s algorithm achieves an approximation factor of 1.4 for the envy-ratio problem.
Polynomial Time Approximation Schemes PTAS: > 0, algorithm A with cost (1 + )OPT in time poly(| I |), instance I PTAS’s in job scheduling: [Hochbaum, Shmoys ’87]: Makespan [Woeginger ’97]: Maximize min. completion time [Alon, Azar, Woeginger, Yadid ’98]: Generalizations
A PTAS for the envy-ratio problem Theorem: The envy-ratio problem admits a Polynomial Time Approximation Scheme. Proof outline: • Rounding step ( I IR ). • Solve IR optimally: IP with constant # of variables • Transform allocation of rounded instance to an allocation in I.
Step 1: Rounding (I IR) Let L be the average utility: Rounding parameter: integer constant • 3 types of goods: • Large: • Medium: • Small:
Step 1: Rounding (I IR) Large: give to some agent, remove agent We may assume there are no large goods in I Claim: There exists an optimal solution in which every large good is assigned to a person with no other goods in her bundle.
Step 1: Rounding (I IR) • Large: WLOG no large goods in I • Medium: round to next integer multiple of • (ignore some of the least significant digits) • Small: merge together and round:
Step 1: Rounding (I IR) • Large: WLOG no large goods in I • Medium: round to next integer multiple of • (ignore some of the least significant digits) • Small: merge together and round:
Step 2: Solve IR optimally Constant number of distinct values for the goods in IR : Claim: optimal allocation A in IR s.t. # goods in #distinct bundles with 2λ goods is constant (exp(λ) but still constant)
Step 2: Solve IR optimally Integer variable XS: # agents with bundle S, for each S with 2λ goods • For , solve the decision problem: • Is there an allocation A = (A1,…,An) with ? • Integer program, constant number of variables Lenstra’s algorithm • Repeat only for a constant number of pairs (t1, t2). • Pick solution with best envy-ratio.
Step 3 (IR I) OPTR: Optimal solution of the rounded instance. Lemma 1: Given an optimal solution of IR, we can find an allocation in I, B = (B1,…,Bn), such that: Lemma 2: OPTR OPT
Finally… Which turns out to be:
Non-additive utilities Input: exponential in size Use only polynomial amount of input? (query model) Theorem 3: Any deterministic algorithm that computes a finite approximation to minimum envy or minimum envy-ratio needs an exponential number of queries for the players’ utilities. Proof: Counting argument, similar to [Nisan-Segal ’03] Note: Not dependent on any complexity theory assumption.
Related Work and Extensions • Envy-ratio • Additive non-identical utilities: O(m)-approximation • Nonadditive (e.g. submodular) ? • Max-min fairness: • [Bezakova, Dani ’05, Saberi, Asadpour ’07]: new approximations + hardness results
Incentive Compatibility So far we have assumed that players report their true utilities. Definition: An algorithm is truthful if being honest is always a dominant strategy for every player. Theorem 4: An algorithm that outputs a minimum envy allocation is not truthful.
Conclusions • There exist allocations, in which the envy is bounded by the maximum marginal utility. • Minimizing the envy is hard in general. • If all players have the same (additive) utility function the envy ratio can be well approximated. • Any algorithm that computes a minimum envy allocation is not truthful.
Economic Theory: models and solution concepts Rationality, fairness, incentive compatibility,… Mathematically rich; however mostly non-constructive Discrete math and theory of algorithms: Dealing with indivisibilities Computational complexity Post-mortem
Post-mortem Finding efficient algorithms for computing / approximating economic solution concepts: • Fair division (partially here, [DH ’88, BD ’05, AS ’07]) • Nash Equilibria [P ’94, LMM ’03, LM ’04, PT ‘04, DMP ’06, BBM ’07] • Market Equilibria [DPS ’02, DPSV ’02, JMS ’03, DV ’03] • Cost Sharing [MS ’97, FPS ’00, JV ‘01] • ……
Proof of Theorem 4 Proof: Construction of an example in which any such algorithm will fail.
Proof of Theorem 4 Proof: Construction of an example in which any such algorithm will fail. By misreporting Homer will receive the biscuit and more eggs than before.