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Improved bounds on equilibria solutions in the network design game
Authors:Akaki Mamageishvili  Matú? Mihalák  Simone Montemezzani
Institution:1.Department of Management, Technology and Economics,ETH Zurich,Zurich,Switzerland;2.Department of Data Science and Knowledge Engineering,Maastricht University,Maastricht,The Netherlands;3.Department of Mathematics,ETH Zurich,Zurich,Switzerland
Abstract:In the network design game with n players, every player chooses a path in an edge-weighted graph to connect her pair of terminals, sharing costs of the edges on her path with all other players fairly. It has been shown that the price of stability of any network design game is at most \(H_n\), the n-th harmonic number. This bound is tight for directed graphs.For undirected graphs, it has only recently been shown that the price of stability is at most \(H_n \left( 1-\frac{1}{\Theta (n^4)} \right) \), while the worst-case known example has price of stability around 2.25. We improve the upper bound considerably by showing that the price of stability is at most \(H_{n/2} + \varepsilon \) for any \(\varepsilon \) starting from some suitable \(n \ge n(\varepsilon )\).We also study quality measures of different solution concepts for the multicast network design game on a ring topology. We recall from the literature a lower bound of \(\frac{4}{3}\) and prove a matching upper bound for the price of stability. Therefore, we answer an open question posed by Fanelli et al. (Theor Comput Sci 562:90–100, 2015). We prove an upper bound of 2 for the ratio of the costs of a potential optimizer and of an optimum, provide a construction of a lower bound, and give a computer-assisted argument that it reaches 2 for any precision. We then turn our attention to players arriving one by one and playing myopically their best response. We provide matching lower and upper bounds of 2 for the myopic sequential price of anarchy (achieved for a worst-case order of the arrival of the players). We then initiate the study of myopic sequential price of stability and for the multicast game on the ring we construct a lower bound of \(\frac{4}{3}\), and provide an upper bound of \(\frac{26}{19}\). To the end, we conjecture and argue that the right answer is \(\frac{4}{3}\).
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