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Course: CS 684, Fall 2001
School: Cornell
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684 CS Algorithmic Game Theory Instructor: Eva Tardos Scribe: Gei-Tai Lin (gal24) September 23, 2005 1 Network Design with Discrete Potential Games Recall that users chose a path P si ti in a graph G. The delay incurred had to do with the number of users on every edge. le (x) = delay on edge if x users use it. Today we will explore network design with potential functions. In talking about networks we will...

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684 CS Algorithmic Game Theory Instructor: Eva Tardos Scribe: Gei-Tai Lin (gal24) September 23, 2005 1 Network Design with Discrete Potential Games Recall that users chose a path P si ti in a graph G. The delay incurred had to do with the number of users on every edge. le (x) = delay on edge if x users use it. Today we will explore network design with potential functions. In talking about networks we will think in terms of the cost instead of delay. The cost of traveling across a path is additive, akin to delay in the routing problem. We can think of the cost as some sort of maintenance cost to regularly send packets. In our network design, le (x) is the cost of sharing that edge e for each user. Therefore the cost of a total path is e le (xe ) where xe is the number of users sharing edge e. In our network all our users will pay the same amount to cross a given edge. Because users pay the same amount, we have fair cost sharing. We dene ce (x) as being the total cost for all users, and l e (x) = ce (x)/x all int x 1 We will assume le (x) is monotone decreasing (non-increasing), such that as we increase the number of people using an edge, the cost incurred by each user can only decrease. We will also assume that ce (x) is monotone increasing, such that the total cost cannot decrease if we add more users. See Figure 1. 2 Previous Lecture Recall from the previous lecture that a deterministic Nash exists within our routing game. Formally: Theorem 1 There exists a selection of path P 1 ..Pk that is a Nash equilibrium For this theorem, we considered the potential function: = e xe le (i) In the same way, even i=1 though we now have costs on the edges instead of delay, a deterministic Nash still exists within our network design. 2.1 Comparing the quality of Nash and Opt Now that we know that a deterministic Nash exists in our network, we want to compare the quality of the Nash to the quality of the optimal solution. We will dene quality to be the total cost for these network designs. total cost c(x) peruer cost l(x) x x Figure 1: Graph of total cost for x users (on left), and graph of per-user cost as a function of the number of users (on right). e ce (x) = e xe le (x) This is our natural measure of quality. Notice that this is the same expression as total delay in our routing problem. We consider the Nash to minimizing (x), and will compare it to some other possible c(x) (and implicitly, opt). We would like to argue that C(x) (x). For an individual edge, e (xe ) = le (1) + le (2) + le (3) ... + + le (xe ) ce (x) = xe le (xe ) From Figure 2 below , we can see clearly that as l e (x) monotone decreases, e (xe ) ce (xe ), and summing over all these edges would give C(x) (x). What we want to claim is that the functions are approximately equal. To do this, we will have to bound (x) C(x) (x) C(x) Where is some factor. We can turn to our xed cost example to gain some intuition into what is. For a xed cost ce the potential is = ce + ce /2 + ce /3 + ... + ce /xe = ce (1 + 1/2 + 1/3 + 1/xe ) ce ln(xe ) To be precise is somewhere between [ln(x e ), ln(xe + 1)] Theorem 2 C(x) (x) ln(k + 1)C(x) for all solutions x, where k is the number of users Proof Because l is monotone decreasing from our assumption in section 1, we saw that C . This satises the rst inequality. For the 2nd inequality, we can compare l e (1)+ ... +le (e) to xe le (xe ) xe le (xe ) = ce (x) xe x Figure 2: The total cost xe le (xe ) is the area of the small box. The potential function is the sum of the strips. For any i xe we have that le (i) = ce (i)/i ce (xe )/i, so we get e (xe ) = xe le (i) ce (xe )xe 1/i ce (x)ln(xe + 1) ce (xe )ln(k + 1) i=1 i=1 Recall that the left hand side of the inequality is the denition of , so if we take our conclusion for a single edge above and take the summation over all edges we nd that (x) ln(k + 1)C(x) Theorem 3 The Nash minimizing has total cost ln(k+1)*Opt Proof We have already shown that C C. This implies the solution minimizing is within factor of minimizing ...

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