Routing in Max-Min Fair Networks: A Game Theoretic Approach

Transcription

1 Routing in Max-Min Fair Networks: A Game Theoretic Approach Dejun Yang, Guoliang Xue, Xi Fang, Satyajayant Misra and Jin Zhang Arizona State University New Mexico State University

2 Outline/Progress of the Talk Background MAXBAR Problem Existence of Nash Equilibrium Best Response Routing Convergence to a Nash Equilibrium Numerical Results Conclusions 2/30

3 Background How can a user find a route with maximum bandwidth? Will the network converge to a stable state or oscillate forever? 3/30

4 Max-Min Fair Bandwidth Allocation: Definition An allocation among multiple routes is Max-Min Fair if it is not possible to increase the rate of any route without decreasing a rate of lower value allocated to another route. 4/30

5 Max-Min Fair Bandwidth Allocation: Example s 2 t 2 s 3 t 3 X Y Z s 1 t 1 s 1 -X-Y-Z-t 1, s 2 -X-Y-t 2, s 3 -Y-Z-t 3 b(x, Y) = 2, b(y, Z) = 4 5/30

6 Max-Min Fair Bandwidth Allocation: Illustration s 1 t 3 Global bottleneck is the link with least equal share, i.e., min e b(e)/u(e). s 2 s a 11 8 b c 5 d 5 30 t 1 t 2 (d, t 2 ) is the global bottleneck and b(d, t 2 ) = 3 b(p 2 ) = 3 (a, b) is the global bottleneck and b(a, b)/2 = 4 b(p 1 ) = b(p 3 ) = 4 6/30

7 Outline/Progress of the Talk MAXBAR Problem Existence of Nash Equilibrium Best Response Routing Convergence to a Nash Equilibrium Numerical Results Conclusions 7/30

8 Problem Formulation MAXimal-BAndwidth Routing (MAXBAR) Problem: a problem of routing in the max-min fair network with multiple (K) users, where each user aims to maximize the bandwidth of its own path. 8/30

9 Non-cooperative Game Formulation Player: User Strategy: Path Payoff: Bandwidth of the path 9/30

10 Concepts Best Response Given other users strategies P 1,, P i-1, P i+1,, P K, select a path P i to maximize b(p i ) Nash Equilibrium No user can find a better strategy than its current one Price of Anarchy Total payoff of worst NE/Social Optimum 10/30

11 Outline/Progress of the Talk Existence of Nash Equilibrium Best Response Routing Convergence to a Nash Equilibrium Numerical Results Conclusions 11/30

12 Existence of a Nash Equilibrium (1) s 1 t t 6 1 s a b c a b c s t 3 s t 3 t 2 t 1 s 3 s 3 Before player 2 changes: b(p 1 ) = 4, b(p 2 ) = 2, b(p 3 ) = 4. After player 2 changes: b(p 1 ) = 4, b(p 2 ) = 4, b(p 3 ) = 6. (2, 4, 4) < (4, 4, 6) 12/30

13 Existence of a Nash Equilibrium (2) Sort the bandwidth values in lexicographic order b = (b 1, b 2,, b K ), where b i b i+1 for 1 i K-1. b < b if b 1 < b 1 or b k = b k for 1 k < μ and b μ < b μ for some 1 < μ K. 13/30

14 Existence of a Nash Equilibrium (3) Let b be the bandwidth vector. Let b be the bandwidth vector after a player changes its path. Then b > b. Each player has a finite number of paths. There are a finite number of bandwidth vectors corresponding to different path configuration. There exists at least one Nash Equilibrium whose vector is the largest lexicographically. 14/30

15 Price of Anarchy Bad news: POA 1/K s 1 s 2 t 1 t 2 t K sk t 3 s 3 Good news: POA 1/K Conclusion: POA = 1/K 15/30

16 Outline/Progress of the Talk Best Response Routing Convergence to a Nash Equilibrium Numerical Results Conclusions 16/30

17 Best Response Routing (1) If each player can find its best response efficiently, then the process will converge to an NE! How does a player finds its best response strategy? Exhaust search Too expensive! If we can predict the achievable bandwidth on each link, it would be a different story! 17/30

18 Best Response Routing (2) s a 3 b 3 4 t 1 s a 3 b 3 4 t 1 s 2 s c 9 d 10 1 t 2 t 3 s 2 s c 9 d 10 1 t 2 t 3 Should Player 1 change its path? Equal allocation: each player gets 3 on (c, d) Actually, b(p 1 ) = 4 if Player 1 changes its path. What is wrong? 18/30

19 Best Response Routing (3) Player 3 Player 2 Player What about adding another Player 4? Player 1 should not lose bandwidth 1 < 11/4 Player 2 should not lose bandwidth either 3 < 10/3 Player 4 and Player 3 compete for the rest bandwidth each will get 7/2 19/30

20 Best Response Routing (4) Observed available bandwidth: where and 20/30

21 Best Response Routing (4) Player 3 Player 2 Player = 21/30

22 Best Response Routing (5) When a player plans to change its path, it does the following four steps in the giving order: Extract its current path from the network. Calculate the observed available bandwidth for each link in the resulting network. Find a widest path according to the observed bandwidth. Make a decision: if the observed bandwidth of the new path is greater than its current bandwidth, it switches to the new path (change); otherwise, it keeps the same path (no change). 22/30

23 Outline/Progress of the Talk Convergence to a Nash Equilibrium Numerical Results Conclusions 23/30

24 Converging to a Nash Equilibrium How fast can the process converge to an NE? Recall b=(b 1, b 2,, b K ), in lexicographical order. Good news: b 1 has at most Km possible values: m edges, shared by no more than K users. Bad news: b 2 may have more than Km possible values. For each fixed b 1, b 2 has at most Km possible values. b can increase at most (Km) K times during the process. Converges to a Nash Equilibrium in O((K log K + Km) (Km) K ) 24/30

25 Outline/Progress of the Talk Numerical Results Conclusions 25/30

26 Numerical Results Comparison Game Based Algorithm (GBA) Independent Maximization Algorithm (IMA) Sequential Reservation Algorithm (SRA) Set up BRITE: α=0.15, β= X X160, 80X320, 120X480, and 160X640 26/30

27 Numerical Results Total bandwidth 27/30

28 Numerical Results Bandwidth disparity ratio Number of iterations 28/30

29 Outline/Progress of the Talk Conclusions 29/30

30 Conclusions We have studied the dynamics of maximal bandwidth routing in a max-min fair network as a non-cooperative game. We have proved the existence of an NE and designed a best response based algorithm that converges to an NE A key concept is the observed available bandwidth, which enables a user to find its best response move efficiently. *****Thank You***** 30/30