CS 684 Algorithmic Game Theory Scribe: Eva Tardos Instructor: Eva Tardos Monday, Jan 26- Friday, Febr 6, 2004 1 Simple Games and some definitions We start with discussing a few simple games. For small and simple games, the easiest way to describe a game is by a payoff matrix. Consider for example, the game Battle of Sexes.Thetwo players (a boy and a girl) want to go to either a softball or a baseball game. One player prefers softball the other prefers baseball, but they both prefer to be with each other, more than they care about which game to attend. This can be expressed by a simple game matrix: both players have two possible options (referred to as strategies), namely they can go to the softball or to the baseball game. In the following matrix we give the payoff for each player depending on what they both choose to do, where the first and second numbers represent the payoff for the two players respectively. B S B 2, 1 0, 0 S 0, 0 1, 2 A pure strategy for a player is a choice of one of the options he/she can do. We say that strategies si for each player i form a Nash equilibrium if no player has the unilateral incentive to change his/her strategy. More formally, if there is no player i and strategys ¯i, such that player i switching to strategys ¯i while all other players j staying with their strategies sj would improve i’s payoff. The idea is that a Nash equilibrium is a stable state of the game. If the players agree to play this equilibrium, then neither of them has an incentive to deviate from this agreement. The above game of the battle of sexes has two equilibria, either both players should attend the baseball game, or they should both attend the softball game. Not all games have such Nash equilibria using pure strategies, to this end consider the matching pennies, where the two players put out a penny “head” or “tail”. One wants the pennies to match, the others for them not to match. The payoff matrix is H T H −1, 1 1, −1 T 1, −1 −1, 1 Later in the course, we will talk about randomized equilibria, but for now, we just want to point out that this game has no deterministic (pure strategy) Nash equilibrium. Another interesting game is the coordination game, with the following payoffs. C D C 2, 2 0, 0 D 0, 0 1, 1 This game again has two Nash equilibria: either both players agree to coordinate (play strategy C), or they both defect (play strategy D). Both playing strategy D seems like a bad idea, as they both get lower payoff, but, it is also a Nash equilibrium, as no player has the incentive unilaterally change his strategy. Finally, maybe the most famous of these small two person games is the prisoner’s dilemma: C D C 3, 3 0, 4 D 4, 0 1, 1 In this game the usual story is about two criminals cooperating with each other, or defecting, and telling the truth to the police. The only Nash equilibrium in this game is for both players to defect, as no matter what the other player does, each player is a little better off defecting. However, in a way they would be both better off cooperating. To make the issues that came up in these games precise, we will define the social value of the solution to be the sum of all players payoffs. So in the prisoner’s dilemma, the solution with best social value is to have both players cooperate (for a value of 6), while the only Nash equilibrium has social value of 2. In the Battle of Sexes game both Nash equilibria has social value of 3 (which is also best possible outcome), while in the coordination game one Nash had a social value of 4, while the other one had a value of 2. Note that in place of social value we can (and will) also consider other objective functions. One natural alternate function that is very common in CS, is fairness, usually formalized with the max-min objective. So the goal is to find a solution where all players have a high payoff. So the min-max value of both Nash equilibria in the battle of sexes game is 1, as one of the players is only getting a payoff of 1. In this course we will consider games motivated by some computer science context. Examples will include routing on the Internet, load balancing of servers, cost-sharing for network design among others. For the first part of the course, we will focus on Nash equilibria, our main questions will be the following. • Does the game have a pure Nash equilibrium? is it unique? • How does the (social) value of a Nash equilibrium compare to the best possible outcome? • Can we show that the value of the best Nash equilibrium is good compared to the best possible outcome. 2 Load Balancing Game The first game we will consider is the following load balancing game. There are m servers, and n jobs. Think of the servers as serving Web pages (i.e., this group of servers is what may be used to support cnn.com), and the jobs as requests for some pages. A job j will have a load or job size pj, and a subset Sj of servers that it is allowed to go on (say where the appropriate information is available, or servers that is physically near enough to the request). We will think of each job as a selfish player. A solution is an assignment to jobs to servers, and the load of a server i is the sum of the loads of the set of jobs assigned to i. More formally, let M be an assignment of jobs to servers, so (j, i) ∈ M if job j is assigned to server i. Then we define the load of server i as Li = pj. j:(j,i)∈M The higher the load is the worse the response rate of the server will be. If all servers are identical, then we has that the assignment is at Nash equilibrium, if for all jobs j,ifj is assigned to server i and k ∈ Sj is an other server that could serve this job, than Li ≤ Lk + pj, that is the load on server i is no worse, than the load on the other server k would be if jobs j would switch to it. Later we will consider cases when the servers are not identical. Assume each server i has a load dependent response time: let ri(x) be the response time on server i as a function of the load x on the server. We assume that the function is nonnegative, and monotone increasing in the load x. With such response time as assignment is a Nash equilibrium if for all jobs j,ifj is assigned to server i and k ∈ Sj is an other server that could serve this job, than ri(Li) ≤ rk(Lk + pj ). First we want to show that in this load balancing game a pure strategy Nash equilibrium always exists. Theorem 1 If the response time of each server i is a nonnegative, and monotone function of the load, then the game has a pure strategy Nash equilibrium. Proof. The idea is to start with any assignment, and let the jobs migrate as they desire. We will show that this process terminates. When no job wants to move that is a Nash by definition. Consider an assignment M, and its set of response times, {ri(Li)}.Ifjobj migrates from server i to server k,thenLi decreases while Lk increases. To show that this migration process terminates we need to find some quantity that improves. We will not need to worry about how much the improvement is: As long as there is some improvement the process cannot cycle, the same assignment cannot be reached again. This implies that after going through possibly all assignments, the process will terminate. First note that the worst response time maxi ri(Li) is monotone improving as jobs migrate. Unfortunately, this quantity can be unchanged, if either the job moving was not on the server with worst response time, or if there are many servers with equal response time. The idea is as follows. When the worst response time is unchanged, than consider the second highest response time. Now this one either decreases or is unchanged. More generally, assume that a job j moves from server i to some other server k. Assume that we order the servers so that r1(L1) ≥ r2(L2) ≥ ...≥ rm(Lm), and that server i is the last among equal response time servers. We need to notice two facts. • In the above order k>i • ri(Li) >rk(Lk + pj) by the definition of a selfish move for job j. Theresponsetimeonservers with = i, k are not effected by the change. If we reorder the servers so that they will again be in decreasing order of response time, then the first i − 1servers will remain the same, and the new ith server, will either be the current server i,orserveri +1, or server k. In either case the ordered sequence of response times decreased, as the first i − 1 coordinates are unchanged, and the ith coordinate decreased. Note that this proof provides a finite algorithm to find a Nash equilibrium, but the algorithm can be rather slow.