The Price of Anarchy in Bilateral Network Formation in an Adversary Model

We study the bilateral version of the adversary network formation game introduced by the author in 2010. In bilateral network formation, a link is formed only if both endpoints agree on it and then both have to pay the link cost α>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha >0$$\end{document} for it. In the adversary model, the cost of each player comprises this building cost plus the expected number of other players to which she will lose connection if one link is destroyed randomly according to a known probability distribution. Two adversaries are considered: one chooses the link to destroy uniformly at random from the set of all built links, the other instead concentrates the probability measure on the built links which cause a maximum number of player pairs to be separated when destroyed. Pairwise stability (PS) and pairwise Nash equilibrium (PNE) are used as equilibrium concepts. For the first adversary, we prove convexity of cost, hence PS and PNE coincide. The main result is an upper bound of 10 on the price of anarchy for this adversary, for link cost α>12\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha > \frac{1}{2}$$\end{document}. New proof techniques are required to obtain this bound, compared to the version with unilateral link formation. For the second adversary, we also prove a bound tight up to constants, namely Θ(1+nα)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varTheta (1+\frac{n}{\alpha })$$\end{document}, where n is the number of players.