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Bottleneck Routing Games in Communication Networks Ron Banner and Ariel Orda Department of Electrical Engineering Technion- Israel Institute of Technology
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Selfish Routing u Often (e.g., large-scale networks, ad hoc networks) users pick their own routes. No central authority. u Network users are selfish. Do not care about social welfare. Want to optimize their own performance. u Major Question: how much does the network performance suffer from the lack of global regulation?
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Selfish Routing: Quantifying the Inefficiency u A flow is at Nash Equilibrium if no user can improve its performance. May not exist. May not be unique. u The price of anarchy: The worst-case ratio between the performance of a Nash equilibrium and the optimal performance. u The price of stability: The worst-case ratio between the performance of a best Nash equilibrium and the optimal performance.
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Cost structures of flows Additive Metrics (path performance= sum of link performances) E.g., Delay, Jitter, Loss Probability. Considerable amount of work on related routing games: [Orda, Rom & Shimkin, 1992]; [Korilis, Lazar & Orda, 1995]; [Roughgarden & Tardos, 2001]; [Altman, Basar, Jimenez & Shimkin, 2002]; [Kameda, 2002]; [La & Anantharam, 2002]; [Roughgarden, 2005]; [Awerbuch, Azar & Epstein, 2005]; [Even-Dar & Mansour, 2005]; … u Bottleneck Metrics (path performance = worst performance of a link on a path). No previous studies in the context of networking games!
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Bottleneck Routing Games (examples) u Wireless Networks: Each user maximizes the smallest battery lifetime along its routing topology. u Traffic bursts: Each user maximizes the smallest residual capacity of the links they employ. u Traffic Engineering: Each user minimizes the utilization of the most utilized buffer Avoids deadlocks and packet loss. Each user minimizes the utilization of the most utilized link. Avoids hot spots. u Attacks: usually aimed against the links or nodes that carry the largest amount of traffic. Each user minimizes the maximum amount of traffic that a link transfers in its routing topology.
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Model u A set of users U={u 1, u 2,…, u N }. u For each user, a positive flow demand u and a source- destination pair (s u,t u ). u For each link e, a performance function q e (∙). q e (∙) is continuous and increasing for all links. u Routing model Splittable Unsplittable
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Model (cont.) u User behavior Users are selfish. Each minimizes a bottleneck objective: u Social objective Minimize the network bottleneck:
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Questions u Is there at least one Nash Equilibrium? u Is the Nash equilibrium always unique? u How many steps are required to reach equilibrium? u What is the price of anarchy? u When are Nash equilibria socially optimal?
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Existence of Nash Equilibrium u Theorem: An Unsplittable Bottleneck Game admits a Nash equilibrium Very simple proof. u Theorem: A Splittable Bottleneck Game admits a Nash Equilibrium. Complex proof. Splittable bottleneck games are discontinuous! why Hence, standard proof techniques cannot be employed!
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Questions Is there at least one Nash Equilibrium? Yes! u Is the Nash equilibrium unique? u How many steps are required to reach equilibrium? u What is the price of anarchy? u When are Nash equilibria socially optimal?
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Non-uniqueness of Nash Equilibria s t u (f p1 =1, f p2 =0) & (f p1 =0, f p2 =1) are Unsplittable Nash flows. u (f p1 =0.5, f p2 =0.5) & (f p1 =0.25, f p2 =0.75) are Splittable Nash flows. u I.e.: at least two different Nash flows for each routing game. e2e2 e1e1 e3e3 p1p1 p2p2 = 1 q e (f e )=f e for each e in E.
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Questions Is there at least one Nash Equilibrium? Yes! Is the Nash equilibrium always unique? No! u How many steps are required to reach equilibrium? u What is the price of anarchy? u When are Nash equilibria socially optimal?
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Convergence time (unsplittable case) u Theorem: the maximum number of steps required to reach Nash equilibrium is u For O(1) users, convergence time is polynomial.
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Unbounded convergence time (splittable case) T1T1 T2T2 S1S1 S2S2 = 2 q e (f e )=f e for each e in E
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Questions Is there at least one Nash Equilibrium? Yes! Is the Nash equilibrium always unique? No! How many steps are required to reach equilibrium? Unsplittable: Splittable: ∞ u What is the price of anarchy? u When are Nash equilibria socially optimal?
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Unbounded Price of Anarchy (unsplittable case) A = B = 2∙ ST Network Bottleneck Optimal flow Nash flow Price of anarchy
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Unbounded Price of Anarchy (splittable case) Network Bottleneck Nash flow Optimal flow Price of anarchy S2S2 S1S1 T2T2 T1T1 q e (f e )=2 fe for each e in E. B=B= A =
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Questions Is there at least one Nash Equilibrium? Yes! Is the Nash equilibrium always unique? No! How many steps are required to reach equilibrium? Unsplittable: Splittable: ∞ What is the price of anarchy? ∞ u When are Nash equilibria socially optimal?
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Optimal Nash Equilibria (unsplittable case) u Theorem: The price of stability is 1. u Good news Selfish users can agree upon an optimal solution. Such solutions can be proposed to all users by some centralized protocol. u Bad news We prove that finding such an optimal Nash equilibrium is NP-hard.
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Optimal Nash Equilibria (splittable case) u Theorem: A Nash flow is optimal if all users route their traffic along paths with a minimum number of bottlenecks. S2S2 S1S1 T2T2 T1T1 q e (f e )=f e for each e in E. B = 1 A = 1 User B is not routing along paths with minimum number of bottlenecks
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Questions Is there at least one Nash Equilibrium? Yes! Is the Nash equilibrium always unique? No! How many steps are required to reach equilibrium? Unsplittable: Splittable: ∞ What is the price of anarchy? ∞ When Nash equilibriums are socially optimal? Unsplittable: each best Nash equilibrium (though NP-hard to find). Splittable: each Nash equilibrium with users that exclusively route over paths with a minimum number of bottlenecks.
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Some more results… u Unsplittable: link performance functions of q e (x)=x p Price of anarchy is O(|E| p ). This result is tight! u Splittable: Nash equilibrium with users that exclusively route over paths with minimum number of bottlenecks. The average performance (across all links) is |E| times larger than the minimum value. This result is tight!
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Conclusions u Bottleneck games emerge in many practical scenarios. (yet, they haven't been considered before). u A Nash equilibrium in a bottleneck game: Always exists Can be reached in finite time with unsplittable flows Might be very inefficient.
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Conclusions (cont.) u BUT, by proper design, Nash equilibria can be optimal! Unsplittable: any best equilibrium. Splittable: any equilibrium with users that route over paths with minimum number of bottlenecks. u With these findings, it is possible to optimize overall network performance. Steer users to choose particular Nash equilibria. Unsplittable: propose a stable solutions to all users. Splittable: provide incentives (e.g., pricing) for minimizing the number of bottlenecks.
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Questions?
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Splittable bottleneck games are discontinuous! ST = 1 q e (fe)=f e +2 q e (fe)=f e e1e1 e2e2 Flow configuration Cost
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