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Applications of TU Games Vito Fragnelli University of Outline Eastern Piedmont Cost Allocation Problems Infrastructure Cost Games Short Survey Politecnico di Torino 24 April 2002 Cost Allocation Problems Alan and Bob Several agents want to rent a car have to carry out a common project they can rent the car for one day they can perform it separately each (Saturday for Alan at 70 €, Sunday for Bob at 40 €) or they can rent the car for the or they can do it jointly weekend (at 90 €) (if the total cost decreases) Allocate among the agents the total cost of the joint project The equal sharing (45 € each) will Allocation should be fair be rejected by Bob Cost Allocation Problems It is a pair N, c , where N = {1, ..., n} set of agents c cost function c (S ) is the cost of the project if it is jointly performed by the agents in S, satisfying their necessities N = {Alan, Bob} c (Alan) = 70, c (Bob) = 40, c (Alan, Bob) = 90 The agents not included in S cannot benefit from this project, so that they could have to carry out their own separate ones The solution is a vector x in Rn; x should be efficient : iN xi = c (N ) Cost Allocation Problems The Tennessee Valley Authority Problem (’30) Some interventions (dams and reservoirs) were planned in the Tennessee River valley in order to encourage the economy: • provide water for energy • control flooding • improve navigation The fair allocations should satisfy: • Stand-alone cost test no subset of agents S should be charged more than the cost of an alternative project specialized for S iS xi c (S ) • Incremental cost test no subset of agents S should be charged less than the additional cost including them in the project iS xi c (N ) - c (N \ S ) By efficiency the two properties are equivalent Cost Allocation Problems The Tennessee Valley Authority Problem (’30) • Does there exists a vector that satisfies the previous conditions? • How do select a single vector? Point solutions: Separable costs methods • ECA: Equal Charge Allocation • ACA: Alternate Costs Avoided • CGA: Cost Gap Allocation (-value) Cost Allocation Problems Separable costs methods • Separable cost of player i (marginal cost) mi = c (N ) - c (N \i ) • Non separable cost g(N ) = c (N ) - iN mi 1 ri c (i ) mi ECAi m i g(N ) n M i min { c (S ) - k S \ i m k | i S , S N } ri g(S ) c (S ) - i S m i ACAi m i g(N s.t. ) i N i c (N ) i N ri g i min g(S ) | i S gi CGAi m i g(N ) M (1 - )m i N g i Cost Allocation Problems Cost Allocations Problems and Game Theory A cost allocation problem N, c induces a cooperative cost game N, c • The cost game N, c has non-empty core? • How do select a single core allocation? Infrastructure Cost Games Some users need an infrastructure They are grouped according to different requests, corresponding to increasing levels of the infrastructure or of the facilities that build up the infrastructure, as g1, ..., gk An infrastructure has costs independent from the number of users (Building costs) and costs depending on the number of users (Maintenance costs) Infrastructure cost game = Building cost game + Maintenance cost game The Shapley Value is additive Infrastructure Cost Games Building Cost Game It is precisely an Airport Game (Littlechild – Thompson, 1973): assign to a subset of planes the cost of the larger strip required c (S ) = Cj(S) where j (S ) = max { j | S gj } N = g1 g2 g3 ; g1 = {1}; g2 = {2, 3}; g3 = {4}; c1 c2 c3 Sh(1) = c1 /4 Sh(2) = Sh(3) = c1 /4 + c2 /3 Sh(4) = c1 /4 + c2 /3 + c3 Infrastructure Cost Games Maintenance Cost Game If a player of kind i uses (and damages) a facility of level j he has to refund an amount Aij = k=i,j ij , where ii is the cost for a player of kind i to repair the facility to the level i and ik (k > i ) is the cost for a player of kind i to repair the facility from the level k - 1 to the level k c (S ) S gi Aij (S ) i 1, j (S ) Infrastructure Cost Games Maintenance Cost Game N = {1, 2, 3, 4} 11 12 13 222 223 33 Sh(1) = 11 + 3/4 12 + 1/2 13 Sh(2) = Sh(3) = 22 + 1/2 23 + 1/4 1/3 12 Sh(4) = 33 + 1/4 1/3 12 + 1/2 13 + 1/2 2 23 Other Applications Short survey • Voting games (power indices) • Bankruptcy games • Operations Research games Production & Linear Transformation games Network games - Flow games - Connection games (Spanning tree games, Forest games, Extension games, Fixed tree games) - Shortest Path games Sequencing, Assignment & Permutation games Inventory games Location games Transportation games Thank you for your attention! Bankruptcy Games A set of agents N = {1, ..., n} has claims d = (d1, ..., dn) over an estate E with: i=1,n di > E A feasible division x has to satisfy: • 0 xi di • i=1,n di = E It is possible to define a (pessimistic) game N, v whose core coincides with the set of feasible divisions: v (S ) = max {0, E - iS di } back Linear Production Games A set of agents N = {1, ..., n} has a set of resources b = (b1, ..., bn); they can produce a set of m goods according to a (fixed) technology represented by the matrix A; each unit of good j produce an income cj , j = 1, ..., m A group of agents S N can use their resources bS = iS bi to maximize their income It is possible to define a TU game as: v (S ) = max {c Tx s.t. Ax bS ; x 0 } back Flow Games A set of agents N = {1, ..., n} operates on a transportation network; each agent control a set of arcs A group of agents can use only the arcs they control in order to maximize the flow from the source s to the sink t N = {1, 2, 3} 2 3 s 1 t 2 2 v (1) = 0; v (2) = 1; v (3) = 0 v (1, 2) = 3; v (1, 3) = 0; v (2, 3) = 2; v (1, 2, 3) = 4 back Minimum Cost Spanning Tree Games A set of agents N = {1, ..., n} wants to be connected to a common source {0} building suitable connections at minimum cost N = {1, 2, 3} 3 4 2 3 1 2 4 5 1 Non monotonic game 0 v (1) = 5; v (2) = 1; v (3) = 4 v (1, 2) = 4; v (1, 3) = 8; v (2, 3) = 3; v (1, 2, 3) = 6 Monotonic game v (1) = 4; v (2) = 1; v (3) = 3 v (1, 2) = 4; v (1, 3) = 6; v (2, 3) = 3; v (1, 2, 3) = 6 back Sequencing Games A set of agents N = {1, ..., n} waits for a service; each agent i has a fixed service time ti at a fixed cost per unit of time ci Adjacent agents can reorder themselves for improving their global service time N = {I, II, III} t = (5, 3, 4) c = (5, 9, 8) u = (1, 3, 2) optimal order (Smith): II, III, I A are The two agentssingle agent v (I) = v (II) = v (III) = 0 cannot switch not adjacent v (I, II) = 30 = (5x9 – 3x5)+ v (I, III) = 0 : Agent II + (5x8 = (5x9 – 3x5)+prefers – 4x5)+ v (II, III) = 0 = (3x8 – 9x4)+ the switch v (I, II, III) = 50 back

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posted: | 10/18/2011 |

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