1. THE PRICE OF STOCHASTIC ANARCHY Christine ChungUniversity of Pittsburgh Katrina LigettCarnegie Mellon University Kirk PruhsUniversity of Pittsburgh Aaron RothCarnegie Mellon University

2. Time Needed Machine 1 Machine 2 Job 1 Job 2 Job 3 Load Balancing on Unrelated Machines 2 Machine 1 Machine 2 n players, each with a job to run, chooses one of m machines to run it on Each players goal is to minimize her jobs finish time. NOTE: finish time of a job is equal to load on the machine where the job is run.

3. Time Needed Machine 1 Machine 2 Job 1 Job 2 Job 3 Load Balancing on Unrelated Machines 3 Machine 1 Machine 2 n players, each with a job to run, chooses one of m machines to run it on Each players goal is to minimize her jobs finish time. NOTE: finish time of a job is equal to load on the machine where the job is run.

4. n players, each with a job to run, chooses one of m machines to run it on Each players goal is to minimize her jobs finish time. NOTE: finish time of a job is equal to load on the machine where the job is run. Time Needed Machine 1 Machine 2 Job 1 Job 2 Job 3 Load Balancing on Unrelated Machines 4 Machine 1 Machine 2

5. Time Needed Machine 1 Machine 2 Job 1 Job 2 Job 3 Load Balancing on Unrelated Machines 5 Machine 1 Machine 2 n players, each with a job to run, chooses one of m machines to run it on Each players goal is to minimize her jobs finish time. NOTE: finish time of a job is equal to load on the machine where the job is run.

6. Unbounded Price of Anarchy in the Load Balancing Game on Unrelated Machines 6 Price of Anarchy (POA) measures the cost of having no central authority. Let an optimal assignment under centralized authority be one in which makespan is minimized. POA = (makespan at worst Nash)/(makespan at OPT) Bad POA instance: 2 players and 2 machines (L and R). OPT here costs δ. Worst Nash costs 1. Price of Anarchy: L job 2 job 1 R δ 1 1 δ δ 1

7. Drawbacks of Price of Anarchy A solution characterization with no road map. If there is more than one Nash, dont know which one will be reached. Strong assumptions must be made about the players: e.g., fully informed and fully convinced of one anothers rationality. Nash are sometimes very brittle, making POA results feel overly pessimistic. 7

8. Evolutionary Game Theory 8 Young (1993) specified a model of adaptive play.

9. Evolutionary Game Theory 9 Young (1993) specified a model of adaptive play that allows us to predict which solutions will be chosen in the long run by self-interested decision-making agents with limited info and resources. I dispense with the notion that people fully understand the structure of the games they play, that they have a coherent model of others behavior, that they can make rational calculations of infinite complexity, and that all of this is common knowledge. Instead I postulate a world in which people base their decisions on limited data, use simple predictive models, and sometimes do unexplained or even foolish things. – P. Young, Individual Strategy and Social Structure, 1998

10. Evolutionary Game Theory 10 Young (1993) specified a model of adaptive play. Adaptive play allows us to predict which solutions will be chosen in the long run by self-interested decision-making agents with limited info and resources.

11. In each round of play, each player uses some simple, reasonable dynamics to decide which strategy to play. E.g., imitation dynamics Sample s of the last mem strategies I played Play the strategy whose average payoff was highest (breaking ties uniformly at random) best response dynamics Sample the other players realized strategy in s of the last mem rounds. Assume this sample represents the probability distribution of what the other player will play the next round, and play a strategy that is a best response (minimizes my expected cost). Adaptive Play Example 11 L job 2 job 1 R δ 1 1 δ

12. In each round of play, each player uses some simple, reasonable dynamics to decide which strategy to play. E.g., imitation dynamics Sample s of the last mem strategies I played Play the strategy whose average payoff was highest (breaking ties uniformly at random) best response dynamics Sample the other players realized strategy in s of the last mem rounds. Assume this sample represents the probability distribution of what the other player will play the next round, and play a strategy that is a best response (minimizes my expected cost). Adaptive Play Example 12 L job 2 job 1 R δ 1 1 δ

13. Let mem = 4. If s = 3, each player randomly samples three past plays from the memory, and picks the strategy among them that worked best (yielded the highest payoff). LLRR LLLL Adaptive Play Example: a Markov process 13 LLLL player 1 player 2 3/4 1/4 1 LLLR LLLL LLLR RRRR LRRR RRRR... LLRL LLLL (Then there are 2^8 = 256 total states in the state space.) 1 L job 2 job 1 R δ 1 1 δ

14. Absorbing Sets of the Markov Process 14 An absorbing set is a set of states that are all reachable from one another, but cannot reach any states outside of the set. In our example, we have 4 absorbing sets: But which state we end up in depends on our initial state. Hence we perturb our Markov process as follows: During each round, each player, with probability ε, does not use imitation dynamics, but instead chooses a machine at random. 1 RRRR 1 LLLL 1 RRRR 1 LLLL NASH OPT L job 2 job 1 R δ 1 1 δ

15. Stochastic Stability 15 The perturbed process has only one big absorbing set (any state is reachable from any other state). Hence we have a unique stationary distribution μ ε (where μ ε P = μ ε ). The probability distribution μ ε is the time-average asymptotic frequency distribution of P ε. A state z is stochastically stable if L job 2 job 1 R δ 1 1 δ

16. Finding Stochastically Stable States 16 Theorem (Young, 1993): The stochastically stable states are those states contained in the absorbing sets of the unperturbed process that have minimum stochastic potential. 1 RRRR 1 LLLL 1 RRRR 1 LLLL L job 2 job 1 R δ 1 1 δ

17. Finding Stochastically Stable States 17 Theorem (Young, 1993): The stochastically stable states are those states contained in the absorbing sets of the unperturbed process that have minimum stochastic potential. RRRR LLLL RRRR LLLL 1 RRRR LRLR LRLR LRLR LRLR LLLL RRRRL LLL 3 L job 2 job 1 R δ 1 1 δ

18. Theorem (Young, 1993): The stochastically stable states are those states contained in the absorbing sets of the unperturbed process that have minimum stochastic potential. = cost of min spanning tree rooted there Finding Stochastically Stable States 18 RRRR LLLL RRRR LLLL 1 3 2 6 L job 2 job 1 R δ 1 1 δ

19. Finding Stochastically Stable States 19 Theorem (Young, 1993): The stochastically stable states are those states contained in the absorbing sets of the unperturbed process that have minimum stochastic potential. RRRR LLLL RRRR LLLL 3 2 1 6 L job 2 job 1 R δ 1 1 δ 6 = cost of min spanning tree rooted there

20. Finding Stochastically Stable States 20 Theorem (Young, 1993): The stochastically stable states are those states contained in the absorbing sets of the unperturbed process that have minimum stochastic potential. RRRR LLLL RRRR LLLL 3 1 1 6 5 L job 2 job 1 R δ 1 1 δ 6 = cost of min spanning tree rooted there

21. Finding Stochastically Stable States 21 Theorem (Young, 1993): The stochastically stable states are those states contained in the absorbing sets of the unperturbed process that have minimum stochastic potential. RRRR LLLL RRRR LLLL 2 1 1 6 5 4 Stochastically Stable! L job 2 job 1 R δ 1 1 δ 6 = cost of min spanning tree rooted there

22. Recap: Adaptive Play Model Assume the game is played repeatedly by players with limited information and resources. Use a decision rule (aka learning behavior or selection dynamics) to model how each player picks her strategy for each round. This yields a Markov Process where the states represent fixed-sized histories of game play. Add noise (players make mistakes with some small positive probability and dont always behave according to the prescribed dynamics) 22

23. 23 Stochastic Stability The states in the perturbed Markov process with positive probability in the long-run are the stochastically stable states (SSS). In our paper, we define the Price of Stochastic Anarchy (PSA) to be 23

24. Recall bad instance: POA = 1/ δ (unbounded) But the bad Nash in this case is not a SSS. In fact, OPT is the only SSS here. So PSA = 1 in this instance. Our main result: For the game of load balancing on unrelated machines, while POA is unbounded, PSA is bounded. Specifically, we show PSA m(Fib (n) (mn+1)), which is m times the (mn+1)th n-step Fibonacci number. We also exhibit instances of the game where PSA > m. PSA for Load Balancing (m is the number of machines, n is the number of jobs/players) Ω(m) PSA mFib (n) (mn+1) 24 L job 2 job 1 R δ 1 1 δ

25. In the game of load balancing on unrelated machines, we found that while POA is unbounded, PSA is bounded. Indeed, in the bad POA instances for many games, the worst Nash are not stochastically stable. Finding PSA in these games are interesting open questions that may yield very illuminating results. PSA allows us to determine relative stability of equilibria, distinguishing those that are brittle from those that are more robust, giving us a more informative measure of the cost of having no central authority. Closing Thoughts 25

26. You might notice in this game that if players could coordinate or form a team, they would play OPT. Instead of being unbounded, [AFM2007] have shown the strong price of anarchy is O(m). We conjecture that PSA is also O(m), i.e., that a linear price of anarchy can be achieved without player coordination. Conjecture 26 L job 2 job 1 R δ 1 1 δ