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Inefficiency of pure Nash equilibria in network congestion games: the impact of symmetry and network structure

Published: 06 September 2024 Publication History

Abstract

We study the inefficiency of pure Nash equili bria in symmetric unweighted network congestion games. We first explore the impact of symmetry on the worst-case PoA of network congestion games. For polynomial delay functions with highest degree p, we construct a family of symmetric congestion games over arbitrary networks which achieves the same worst-case PoA of asymmetric network congestion games. Next, we explore the impact of network structure within the class of symmetric network congestion games. For delay functions in class 𝒟, we introduce a quantity y(𝒟) and we show that the PoA is at most y(𝒟) in games defined over series-parallel networks. Thus, for polynomial delays with highest degree p, the PoA cannot exceed 2p+1 - 1, which is significantly smaller than the worst-case PoA in games defined over arbitrary networks. Moreover, we prove that the worst-case PoA quickly degrades from sub-linear to exponential when relaxing the network topology form extension-parallel to series-parallel.
Finally, we consider measuring the social cost as the maximum players’ cost. For polynomial delays of maximum degree p we show that the worst-case PoA is significantly smaller than that of general symmetric congestion games, but dramatically larger than that of games defined over extension-parallel networks.

1 Introduction

In a non-cooperative game, rational players act selfishly to maximize their utility. The players influence each other’s behavior, since the quality of each player’s strategy depends on the other players’ actions. The notion of Nash equilibrium, where no player can improve their cost by unilaterally changing strategy, is the best-known solution concept for predicting a stable outcome of a game. However, since the players act selfishly and independently in a non-cooperative fashion, a Nash equilibrium might be far from minimizing the social cost. The inefficiency of a Nash equilibrium can be measured by comparing its social cost against the minimum social cost that could be achieved. Precisely, the Price of Anarchy (PoA), introduced by Koutsoupias and Papadimitriou [26], is the largest ratio between the cost of a Nash equilibrium and the minimum social cost.
In this article, we study network congestion games, where each player aims at selecting a cheapest path from an origin to a destination, but the cost of each arc is non-decreasing with respect to the total number of players using it. These games are commonly used to model problems in large-scale networks such as routing in communication networks and traffic planning in road networks [24, 31], and represent a simple, yet powerful paradigm for selfish resource sharing.
We focus on the inefficiency of pure Nash equilibria. Unlike (mixed) Nash equilibria, where each player selects a probability distribution on their strategy set, in a pure Nash equilibrium (PNE) each player selects exactly one strategy from their strategy set. Pure Nash equilibria are not guaranteed to exist in general, but congestion games always admit one [32]. We consider two measures of social cost: the total cost, which is the sum of all players’ costs, and the maximum cost, which is the maximum cost of a player in a strategy profile.
Several variants of network congestion games have been studied in the literature, which depends on the combination of a number of parameters.
While some parameters seem to only marginally affect the PoA, the impact that symmetry and network structure have on the PoA is still not completely understood. For example, Bhawalkar et al. [5, 6] proved that symmetry does not impact the PoA in general congestion games, but leave as an open question to establish whether this is still true for network congestion games. Correa et al. [10] partially answered this question by showing that, for affine delay functions, the worst-case PoA of symmetric network congestion games is as large as that of asymmetric network congestion games, when the number of players goes to infinity. In this article, we establish that, for polynomial delay functions of highest degree p, the PoA of symmetric network congestion games is very close to the PoA in asymmetric network congestion games.
Moreover, Aland et al. [1, 2] left as an open direction the problem of characterizing “what structures provide immunity against a high PoA and what structures cause it”. The approaches that have been proposed for general network congestion games [1, 2, 3, 4, 9], later unified in the smoothness framework of Roughgarden [35, 36], cannot be used to derive stronger bounds that hold in the presence of special network structures. The two main network structures for which stronger bounds on the PoA have been provided are parallel-links networks [7, 8, 19, 22, 28, 38] and extension parallel networks [15, 16]. On the other hand, as remarked in both [1, 2] and [15, 16], the PoA in the larger class of series-parallel networks is not well understood. In this article, we focus on two-terminal series-parallel networks, and we provide upper and lower bounds on the worst-case PoA for (atomic, unweighted) symmetric network congestion games. Two-terminal series-parallel networks can be recognized in linear-time [39] and are relevant in many applications, such as for problems on electric networks, scheduling, and compiler optimization. Previous works have highlighted some strong properties of network congestion games defined over series-parallel networks, such as the existence of strong equilibria [25] and optimal tolls [18, 30].

1.1 Our Contributions

First, we consider the total players’ cost. For asymmetric network congestion games with polynomial delay functions of highest degree p, Aland et al. provided a tight bound for the worst-case PoA which is in \([ \left\lfloor \Phi _p \right\rfloor ^{p+1}, \Phi _p^{p+1}]\), where \(\Phi _p \in \Theta (p/\ln p)\) [1, 2]. Our first goal is to determine whether in symmetric network congestion games the same worst-case PoA can be achieved. In Section 3.1, we answer this question in the affirmative.
Theorem 1.1.
The PoA of symmetric network congestion games with polynomial delay functions of the highest degree p is at least \(\left\lfloor \Phi _p \right\rfloor ^{p+1}\).
Theorem 1.1 indicates that symmetry has little impact on the worst-case PoA of network congestion games.
In Section 3.2, we focus on symmetric congestion games defined over series-parallel networks. Let \(\mathcal {D}\) be a class of nonnegative and non-decreasing functions. We introduce a new parameter \(y(\mathcal {D})\) defined as
\begin{equation} y(\mathcal {D}) = \sup _{d \in \mathcal {D},~x\in \mathbb {N}^+}\frac{(x+1)d(x+1) - xd(x)}{d(x)}, \end{equation}
(1)
which intuitively can be used to upper bound by what percentage the cost of an arc increases when one more player uses the arc. Note that \(y(\mathcal {D}) \ge 1\) because \(d(x) = (x+1)d(x) - xd(x) \le (x+1)d(x+1) - xd(x).\) Our main result shows that the worst-case PoA in series-parallel networks is at most \(y(\mathcal {D})\).
Theorem 1.2.
In a symmetric (unweighted) network congestion game on a series-parallel \((s,t)\)-network with delays functions in class \(\mathcal {D}\), the PoA w.r.t. the total players’ cost is at most \(y(\mathcal {D})\).
The above result has interesting implications when \(\mathcal {D}\) is the class of polynomial functions with nonnegative coefficients and highest degree p. We show that in this case \(y(\mathcal {D})\) is at most \(2^{p+1} -1\). Our result indicates a significant drop of the worst-case PoA, which decreases from \(\Theta (p/\ln p)^{p+1}\) in symmetric games over arbitrary networks (by Theorem 1.1) to \(O(2^{p+1})\) in symmetric games over series-parallel networks.
In Section 3.3, we also provide a lower bound on the worst-case PoA in symmetric congestion games defined over series-parallel networks. This bound is independent of \(y(\mathcal {D})\).
Theorem 1.3.
The worst-case PoA w.r.t. the total players’ cost of a symmetric (unweighted) network congestion game on a series-parallel \((s,t)\)-network, where the delay functions are polynomials with non-negative coefficients and highest degree p, is at least
\begin{equation} \frac{1}{1+ l^2\sqrt [2^{{p}}]{r} - rl - \sqrt [2^{{p}}]{r} +r}, \end{equation}
(2)
where \(r = (\frac{2}{2^{p+1}-1}) ^{\frac{2^{{p}}}{2^{{p}}-1}}\) and \(l = \frac{1}{2}r^{1 - \frac{1}{2^{{p}}}}\).
We finally prove that our lower bound is in \(\Omega ({\frac{2^p}{p}})\), thus also in \(\Omega (2^{c p})\) for each \(c\in (0,1)\), which almost asymptotically matches the upper bound of \(2^{p+1}-1\). Since the worst-case PoA in extension-parallel networks (a subclass of series-parallel networks) is in \(\Theta (p/ \ln p)\) [15, 16, 17], our result shows that the PoA dramatically increases when relaxing the network topology from extension-parallel to series-parallel.
In Table 1, we provide a numerical comparison of the bounds established in Theorems 1.11.3, and the worst-case PoA in extension-parallel networks.
Table 1.
 LB arbitraryUB series-parallelLB series-parallelPoA ext-parallel
\(p\)(Theorem 1.1)(Theorem 1.2)(Theorem 1.3)[15, 16]
112 [23]1.4211.333
2871.9381.626
316152.8841.896
4243314.5482.151
5729637.4912.394
62,18712712.7472.630
765,53625522.2282.858
8262,14451139.483.081
\(\vdots\)\(\vdots\)\(\vdots\)\(\vdots\)\(\vdots\)
 \(\left\lfloor \Phi _p \right\rfloor ^{p+1} \in \Theta (p/ \ln p)^{p+1}\)\(2^{p+1} - 1\)\(\Omega (2^{p}/p)\)\(\rho (\mathcal {D}) \in \Theta (p/ \ln p)\)
Table 1. Numerical Comparison of Our Bounds on the (Pure) PoA of Symmetric Series-Parallel Network Congestion Games with Polynomial Delays of Highest Degree \(p\) with: (i) the Lower Bound on the Worst-Case PoA of Symmetric Congestion Games on Arbitrary Networks Established in Theorem 1.1 (First Column); and (ii) the Worst-Case PoA of Symmetric Extension-Parallel Congestion Games [15, 16, 17] (last column). The Upper Bound of 2 for Series-Parallel Networks and Affine Delays is from [23]
In Section 4, we consider measuring the social cost of a strategy profile as the maximum players’ cost. This variant of the social cost expresses the goal that a central authority might have to maximize fairness by minimizing the cost of the most disadvantaged player. We first consider arbitrary delay functions. To bound the PoA in this setting, introduce a new parameter \(z(\mathcal {D})\) defined as
\begin{equation} z(\mathcal {D}) = \sup _{d \in \mathcal {D},~x\in \mathbb {N}^+}\frac{d(x+1)}{d(x)}. \end{equation}
(3)
We then prove that the worst-case PoA in series-parallel networks is at most \(y(\mathcal {D})z(\mathcal {D})\).
Theorem 1.4.
In a symmetric (unweighted) network congestion game on a series-parallel \((s,t)\)-network with delays functions in class \(\mathcal {D}\), the PoA w.r.t. the maximum players’ cost is at most \(z(\mathcal {D})y(\mathcal {D})\).
When \(\mathcal {D}\) is the class of polynomial functions with nonnegative coefficients and maximum degree p we obtain that \(z(\mathcal {D})\) is upper bounded by \(2^p\), thus, the PoA is at most \(2^{2p+1} - 2^p\). Since the worst-case PoA for general symmetric congestion games and polynomial delays is in \(p^{\Theta (p)}\) [9], our result shows a significant drop in the PoA in series-parallel networks.
Finally, we show that the lower bound on the PoA w.r.t. the total players’ cost also yields a valid lower bound when considering the maximum players’ cost. We say that a class of networks \(\mathcal {N}\) is closed under series compositions if the series composition of two networks \(G^1\) and \(G^2\) in \(\mathcal {N}\) still belongs to \(\mathcal {N}\).
Theorem 1.5.
Let \(\mathcal {N}\) be a class of networks closed under series compositions and let G be a network in \(\mathcal {N}\). Then the worst-case PoA with respect to the maximum social cost of a symmetric (unweighted) network congestion game defined over G is at least the worst-case PoA with respect to the total social cost.
For series-parallel networks and polynomial delays with nonnegative coefficients and maximum degree p, Theorem 1.5 implies that the worst-case PoA is in \(\Omega (2^{p}/p)\). This is in stark contrast with the result of [13], establishing that the PoA in extension-parallel networks is 1, i.e., any PNE is also a social optimum w.r.t. the maximum players’ cost. Thus, relaxing the network topology from extension-parallel to series-parallel dramatically increases the inefficiency of pure Nash equilibria. The reason for this is that the key network operations that we need to allow are the series compositions, which are forbidden for extension-parallel networks.

1.2 Further Related Work

Total cost. There is a rich literature concerning the PoA in network congestion games where the social cost is the players’ total cost. Many variants of network congestion games arise from considering different parameters and their combinations. As we shall see, the impact that the network structure has on the inefficiency of pure Nash equilibria varies significantly based on the combination of these parameters.
The first distinction is between atomic and non-atomic congestion games. In non-atomic congestion games, the number of players is infinite and each player controls an infinitesimal amount of flow. For these games, Roughgarden [33] proved that the PoA is independent of the network structure and equal to \(\rho (\mathcal {D})\), where \(\rho\) depends on the class of delay functions \(\mathcal {D}\) [37].
For atomic games, where each player controls a non-negligible amount of flow, network structure affects the PoA differently, depending on whether all the players have the same effect on congestion. In weighted congestion games, where the effect of each player on congestion is proportional to the player’s weight, the worst-case PoA is already achieved by very simple networks consisting of only parallel links [6] when \(\mathcal {D}\) is the class of polynomial functions with nonnegative coefficients and highest degree p. In contrast, in unweighted congestion games the effect of network structure seems significant. For asymmetric congestion games defined over general networks and in the case where \(\mathcal {D}\) is the class of polynomial functions with nonnegative coefficients, Christodoulou and Koutsoupias [9] showed that the PoA is in \(p^{\Theta (p)}\) (see also [3, 4]). Aland et al. [1, 2] later obtained exact values for the worst-case PoA. These exact values admit a lower bound of \(\left\lfloor \Phi _p \right\rfloor _p^{p+1}\) and an upper bound of \(\Phi _p^{p+1}\), where \(\Phi _p\in \Theta (p/ \ln p)\) is the unique nonnegative real solution to \((x + 1)^p= x^{p+1}\). For symmetric congestion games the PoA is again \(p^{\Theta (p)}\) [3, 4, 9]. The worst case PoA drops significantly in the presence of special structure. Lücking et al. [27, 28] studied symmetric congestion games on parallel links and proved that the PoA is \(4/3\) for linear functions. Later Fotakis [15, 16] extended this result by proving an upper bound of \(\rho (\mathcal {D})\) for the larger class of extension parallel networks with delays in class \(\mathcal {D}\). Moreover, this upper bound is tight [14, 17]. It is known that, for the class of polynomial delays with nonnegative coefficients and highest degree p, \(\rho (\mathcal {D}) \in \Theta ({p}/{\ln p})\). This indicates that there is a huge gap between the worst-case PoA in general networks and in extension-parallel networks.
The PoA in symmetric series-parallel network congestion games has been recently investigated only for the specific case of affine delay functions [23], and it has been shown that the worst-case PoA is between \(27/19\) and 2 [23], which is strictly worse than the PoA of \(4/3\) in extension-parallel networks [15, 16], and strictly better than the PoA of \(5/2\) in general networks [10]. One key step to prove the upper bound in [23] consists in using the following inequality introduced in [15, 16]
\begin{equation} \frac{\text{cost}(f)}{\rho (\mathcal {D})} \le \text{cost}(o) + \Delta (f,o), \end{equation}
(4)
where \(\text{cost}(f)\) and \(\text{cost}(o)\) denote the total cost of a PNE flow f and of a social optimum flow o, respectively, and \(\Delta (f,o)\), is a quantity that depends on the difference \(o-f\). For series-parallel networks with affine delays, Hao and Michini [23] prove that \(\Delta (f,o) \le 1/4 \text{cost}(f)\). This approach cannot be further extended to polynomial delays of maximum degree p, because we would obtain \(\Delta (f,o) \le \alpha (p) \text{cost}(f)\), where \(\alpha (p)\) is a function of p that exceeds \(1/\rho (\mathcal {D})\) for large p. Thus, an extension of the approach in [23] would provide an inconsequential bound. A summary of the currently best-known results on the PoA of different unweighted network congestion games classes is provided in Table 2.
Table 2.
Table 2
AtomicSymmetricDelaysnetwork structurePoA
NN\(\mathcal {D}\)\(=\rho (\mathcal {D})\in \Theta (p/ \ln p)\)[33]
YNAffine\(= 5/2\)[9]
YYAffine\(= \frac{5N - 2}{ 2N + 1}\)[9, 10]
YYAffineext-para\(= 4/3\)[15, 16]
YYAffineser-para\(\in [27/19, 2]\)[23]
YNPoly-\(p\)\(\in [\left\lfloor \Phi _p \right\rfloor ^{p+1}, \Phi _{p}^{p+1}]\)[1, 2]
YYPoly-\(p\)\(\ge \left\lfloor \Phi _p \right\rfloor ^{p+1} \in \Theta (p/ \ln p)^{p+1}\)[this work]
YYPoly-\(p\)ext-para\(=\rho (\mathcal {D}) \in \Theta (p/ \ln p)\)[15, 16]
YYPoly-\(p\)ser-para\(\le y(\mathcal {D}) = 2^{p+1}-1\)[this work]
YYPoly-\(p\)ser-para\(\in \Omega (\frac{2^{p}}{p})\)[this work]
Table 2. Summary of Related Work and of Our Main Results for the Worst-Case PoA with Respect to the Total Players’ Costs. In the Column “network structure”, We write “-” When no Assumption on the network structure is Made
Although our focus here is only on pure Nash equilibria, previous works have also investigated the PoA of mixed Nash equilibria. For both atomic and non-atomic congestion games in arbitrary networks, the upper bounds on the PoA of pure Nash equilibria also hold for mixed Nash equilibria [1, 2, 33, 36]. In the symmetric case, previous results have shown that the network structure can limit the inefficiency of mixed Nash equilibria. Specifically, Gairing et al. [21] showed that the (mixed) PoA for symmetric unweighted network congestion games on parallel links with polynomial delay functions of highest degree p is at most \(B_{p+1}\), where \(B_i\) denotes the Bell number of order i. They further established that this bound is asymptotically tight as the number of players and resources goes to infinity. While the (mixed) PoA bound for congestion games on parallel links is lower than that for arbitrary networks, it exceeds the upper bound presented in Theorem 1.3 for pure Nash equilibria. This implies that our upper bound on the (pure) PoA of symmetric congestion games over series-parallel networks does not hold for mixed Nash equilibria.
Maximum cost. The PoA with respect to the maximum players’ cost has received less attention. In the non-atomic setting, Roughgarden [34] showed that the worst-case PoA is \(n-1\), where n is the number of nodes in the network.
In the atomic setting, Koutsoupias and Papadimitriou [26] first studied weighted congestion games with linear delay functions on m-parallel links. For the case where the delay functions on the parallel links are all identical, they provided a lower bound in \(\Omega (\frac{\log m}{\log \log m})\) and an upper bound in \(O(\sqrt {m\log m})\) on the worst-case PoA for both pure and mixed Nash equilibria. Later Czumaj and Vöcking [12] extended this framework to parallel links with different linear delay functions, and they established an asymptotically tight bound on the worst-case PoA in \(\Theta (\frac{\log m}{\log \log \log m})\) for mixed Nash equilibria and in \(\Theta (\frac{\log m}{\log \log m})\) for pure Nash equilibria. The m-parallel links model has also been studied in [11, 20, 29]. Christodoulou and Koutsoupias [9] investigated the PoA of pure Nash equilibria in general unweighted congestion games. In the symmetric case, they showed that the worst-case PoA is \(5/2\) for affine delays and \(p^{\Theta (p)}\) for polynomial delays of maximum degree p. In the asymmetric case, for games with N players, they proved that the worst-case PoA is in \(\Theta (\sqrt {N})\) for affine delays and in \(\Omega (N^\frac{p}{p+1})\) and \(O(N)\) for polynomial delays of maximum degree p.
Epstein et al. [13] characterized efficient network topologies, i.e., network topologies such that, for any class of non-decreasing delay functions, every PNE is also a social optimum. For unweighted symmetric network congestion games they established that extension-parallel networks are efficient, implying that on these networks the PoA is 1. They also proved that this result is tight, i.e., it does not hold when further relaxing the network topology.

2 Preliminaries

Notation. Let \(G=(V,E)\) be an \((s,t)\)-network, i.e., a network with source s and sink t. Directed paths will be simply referred to as paths. A path from node u to node v is called a \((u,v)\)-path. We will only consider simple paths, i.e., paths that do not traverse any node multiple times. Paths and cycles of G are regarded as sequences of arcs, thus we may, for example write \(e \in p\) for a path p. An \((s,t)\)-flow is an assignment of values to the arcs of G such that, at each node u other than s and t, the sum of the values of the arcs entering u equals the sum of the values of the arcs leaving u. The value of the \((s,t)\)-flow is the sum of the values of the arcs entering t. We say a path p is contained in an \((s,t)\)-flow f if for all \(e \in p\), we have \(f_e \gt 0\). For \(n \in {\mathbb {N}}\), we denote by \([n]\) the set \(\lbrace 1, \ldots , n\rbrace\).
Network congestion games. Let \(G=(V,E)\) be an \((s,t)\)-network. We consider an unweighted network congestion game on G with N players. The strategy set \(X^i\) of player i is the set \(\mathcal {P}\) of \((s,t)\)-paths in G. Since all the players have the same origin and destination, their strategy sets all coincide with \(\mathcal {P}\) and the game is called symmetric. A state of the game is a strategy profile \(P=(p^1, \ldots , p^N)\) where \(p^i \in \mathcal {P}\) is the \((s,t)\)-path chosen by player i, for \(i \in [N]\). The set of states of the game is denoted by \(X = X^1 \times \cdots \times X^N\). Each state \(P=(p^1, \ldots , p^N) \in X\) induces an \((s,t)\)-flow \(f=f(P) = \chi ^1+ \cdots + \chi ^N\) of value N, where \(\chi ^i\) is the incidence vector of \(p^i\) for all \(i \in [N]\). We say that the \((s,t)\)-paths \(p^1, \ldots , p^N\) are a decomposition of the \((s,t)\)-flow f if they induce flow f. Note that an \((s,t)\)-flow f of value N can correspond to several states since there might be multiple decompositions of f into N \((s,t)\)-paths.
For each \(e\in E\), we have a nondecreasing delay function \(d_e : [N] \rightarrow {\mathbb {R}}_{\ge 0}\). Each player using e incurs a cost equal to \(d_e(f_e)\), i.e., the cost of e depends on the total number of players that use e in f. Since \(d_e\) is a nondecreasing function, \(d_e(j+1)\ge d_e(j)\) for \(j \in [N-1]\), which models the effect of congestion. We denote the cost of a path p in G with respect to a flow f by \(\text{cost}_f(p) = \sum _{e \in p} d_e(f_e)\). Thus, the cost incurred by player i in state P is \(\text{cost}_f(p^i)\). We also define \(\text{cost}_f^+(p) = \sum _{e \in p} d_e(f_e+1)\). Finally, the cost of flow f in G is denoted by \(\text{cost}(f) = \sum _{e \in E} f_e d_e(f_e)\). The total cost of a state P, denoted by \(\text{tot}(P)\), is the sum of all players’ costs. Clearly, \(\text{tot}(P)\) coincides with the cost of the flow \(f(P)\):
\begin{equation*} \text{tot}(P) = \sum _{i \in [N]} \text{cost}_{f(P)} (p^i) = \text{cost}(f(P)). \end{equation*}
We also define the maximum cost of P, denoted by \(\max (P)\) as the maximum cost of a player in P:
\begin{equation*} \max (P) = \max _{i \in [N]} \text{cost}_{f(P)} (p^i). \end{equation*}
Pure Nash equilibria and social optima. A PNE is a state \((p^1, \ldots , p^i, \ldots , p^N)\) inducing an \((s,t)\)-flow f such that, for each \(i \in [N]\) we have
\begin{equation*} \text{cost}_{f}(p^i) \le \text{cost}_{\tilde{f}}(\tilde{p}^i) \qquad \forall (p^1, \ldots , \tilde{p}^i, \ldots , p^N) \in X \text{ inducing $(s,t)$-flow $\tilde{f}$}. \end{equation*}
A PNE represents a stable outcome of the game since no player \(i \in [N]\) can improve their cost if she unilaterally changes strategy by selecting a different \((s,t)\)-path \(\tilde{p}^i\). With a slight abuse of terminology, we say that an \((s,t)\)-flow f is a PNE if there exists a PNE \(P=(p^1, \ldots , p^N)\in X\) such that \(f = f(P)\), i.e., f is the flow induced by P. On the other hand, we are also interested in a social optimum. We consider two definitions of social optimum, which depend on whether we measure the cost of a state P according to \(\text{tot}(P)\) or \(\max (P)\). In the first case, a social optimum is a state that minimizes \(\text{tot}(P) = \text{cost}(f(P))\) over all the states \(P\in X\). With a slight abuse of terminology, we say that an \((s,t)\)-flow o is a social optimum if o minimizes \(\text{cost}(g)\) over all integral \((s,t)\)-flows g of value N. In the second case, a social optimum is a state that minimizes \(\max (P)\) over all the states \(P\in X\). In other words, the social optimum is a state where the maximum player’s cost is minimized.
Price of Anarchy. To measure the inefficiency of pure Nash equilibria, we use the definition of (pure) Price of Anarchy. The (pure) PoA is the maximum ratio between the cost of a PNE and the cost of a social optimum. In other words, to compute the PoA we consider the “worst” PNE, i.e., a PNE whose cost is as large as possible. For simplicity, from now on we will refer to the pure PoA as PoA.
We consider two definitions of PoA, which depend on whether we measure the cost of a state P according to \(\text{tot}(P)\) or \(\max (P)\). In the first case, the PoA is the maximum ratio \(\frac{\text{cost}(f)}{\text{cost}(o)}\) such that o is a social optimum flow and f is a PNE flow. In the second case, the PoA is the maximum ratio \(\frac{\max (P_f)}{\max (P_o)}\) such that \(P_o\) is a social optimum state and \(P_f\) is a PNE.
Series-parallel networks. An \((s,t)\)-network is series-parallel if it consists of either a single arc \((s,t)\) or of two series-parallel networks composed either in series or in parallel. The parallel composition of two networks \(G_1\) and \(G_2\) is an \((s,t)\)-network obtained from the union of \(G_1\) and \(G_2\) by identifying the source of \(G_1\) and the source of \(G_2\) into s, and by identifying the sink of \(G_1\) and the sink of \(G_2\) into t. The series composition of \(G_1\) and \(G_2\) is an \((s,t)\)-network obtained from the union of \(G_1\) and \(G_2\) by letting s be the source of \(G_1\), t be the sink of \(G_2\), and by identifying the sink of \(G_1\) with the source of \(G_2\). We remark that series-parallel networks are a superclass of parallel-link networks and extension-parallel networks, for which the PoA has been previously studied. An \((s,t)\)-network is extension-parallel if it consists of a single arc \((s,t)\) or of an extension-parallel network and a single arc composed either in series or in parallel.

3 Total Cost

3.1 Lower Bound on the PoA in Arbitrary Networks

In this section, we will prove the lower bound on the PoA stated in Theorem 1.1. Specifically, we construct a family of symmetric network congestion game instances that asymptotically achieves this lower bound.
Let \(g_p(x) = (x+1)^p- x^{p+1}\). We recall that \(\Phi _p\) is the unique nonnegative real solution to \(g_p(x) = 0\) [1, 2]. Clearly, \(g_p(0) = 1 \gt 0\). Moreover, by Lemma 5.2 in [1, 2], we know that \(g_p(x)\) has exactly one local maximum \(\xi\) in \({\mathbb {R}}_+\), is strictly increasing in \([0,\xi)\) and strictly decreasing in \((\xi ,\infty)\). As a result, for \(0 \le x \le \phi _p\) we have \(g_p(x) \ge 0\) and for \(x \ge \phi _p\) we have \(g_p(x) \le 0\). Next, we prove two additional properties of \(\phi _p\).
Lemma 3.1.
For every positive integer p and for all \(k = 0, 1, \ldots , p-1\), we have \((\Phi _{p-1}+1)^k \le \Phi _{p-1}^{k+1}\).
Proof.
We proceed by backward induction. The base case is \(k=p-1\). In this case, by the definition of \(\Phi _p\) we know that \((\Phi _{p-1}+1)^{p-1} = \Phi _{p-1}^{p}\). We now assume that the claim holds for all \(k \in \lbrace 2, \ldots , p-1\rbrace\), and we prove that the claim also holds for \(k-1\). We have
\begin{equation*} (\Phi _{p-1}+1)^{k} \le \Phi _{p-1}^{k+1} \end{equation*}
By dividing both terms by \(\Phi _{p-1}\gt 0\) we obtain
\begin{equation*} \qquad \qquad (\Phi _{p-1}+1)^{k-1} = \frac{(\Phi _{p-1}+1)^{k}}{\Phi _{p-1}+1} \le \frac{(\Phi _{p-1}+1)^{k}}{\Phi _{p-1}} \le \frac{\Phi _{p-1}^{k+1}}{\Phi _{p-1}}=\Phi _{p-1}^{k}. \end{equation*}
 □
Lemma 3.2.
For every positive integer p we have \(\Phi _{p-1} \le \Phi _p \le \Phi _{p-1} + 1\).
Proof.
We first prove that \(\Phi _{p-1} \le \Phi _p\) by showing that \(g_p (\Phi _{p-1}) = (\Phi _{p-1} + 1)^p- \Phi _{p-1}^{p+1} \gt 0\). Since \(\Phi _{p-1} \gt 0\) for every positive integer p, we equivalently show
\begin{equation*} \frac{(\Phi _{p-1} + 1)^p}{\Phi _{p-1}} - \frac{\Phi _{p-1}^{p+1}}{\Phi _{p-1}} \gt \frac{(\Phi _{p-1} + 1)^p}{\Phi _{p-1} + 1} - \frac{\Phi _{p-1}^{p+1}}{\Phi _{p-1}} = (\Phi _{p-1} + 1)^{p-1} - \Phi _{p-1}^{p} =0. \end{equation*}
Next, we prove that \(\Phi _p \le \Phi _{p-1} + 1\). To this purpose, we show that \(g_p(\Phi _{p-1} + 1) \lt 0\) which implies \(\Phi _{p-1} + 1 \gt \Phi _p\). Precisely, our goal is to prove
\begin{align} (\Phi _{p-1} + 2)^p\lt (\Phi _{p-1} + 1)^{p+1}. \end{align}
(5)
We rewrite the left-hand-side of (5) as follows:
\begin{align*} (\Phi _{p-1} + 1 + 1)^p&= \sum _{k=0}^p\binom{p}{k} (\Phi _{p-1} + 1)^k \\ &= (\Phi _{p-1} +1)^p+ \sum _{k=0}^{p-1} \binom{p}{k} (\Phi _{p-1} + 1)^k \\ &\le (\Phi _{p-1} +1)^p+ \sum _{k=0}^{p-1} \binom{p}{k} \Phi _{p-1}^{k+1} \\ &= (\Phi _{p-1} +1)^p+ \sum _{k=0}^{p} \binom{p}{k} \Phi _{p-1}^{k+1} - \Phi _{p-1}^{p+1} \\ &= (\Phi _{p-1} +1)^p+ \Phi _{p-1}(\Phi _{p-1} +1)^p- \Phi _{p-1}^{p+1}\\ &= (\Phi _{p-1}+1)^{p+1} - \Phi _{p-1}^{p+1} \\ &\lt (\Phi _{p-1}+1)^{p+1}, \end{align*}
where the first inequality holds by Lemma 3.1. □
Now we are ready to prove Theorem 1.1, which is restated next for convenience.
Theorem 1.1.
The PoA of symmetric network congestion games with polynomial delay functions of highest degree p is at least \(\left\lfloor \Phi _p \right\rfloor ^{p+1}\).
Proof.
We provide an example of an unweighted symmetric congestion game with delays in \(\operatorname{Poly-}p\) that is defined over a network that is not series-parallel, and whose PoA asymptotically goes to \(\left\lfloor \Phi _p \right\rfloor ^{p+1}\) when the number of players is large. In Figure 1, we provide an instance with three players and \(p= 2\). In this construction there are \(N \ge \max \lbrace \left\lfloor \Phi _p \right\rfloor +1, 2 \left\lfloor \Phi _p \right\rfloor -2\rbrace\) players, i.e., N is lower bounded by a constant that depends on p. Let p be a positive integer. The network G has \(N(\left\lfloor \Phi _p \right\rfloor + N) +2\) nodes: the source s, the sink t, and N rows of \(\left\lfloor \Phi _p \right\rfloor + N\) nodes. The nodes in row i are denoted by \(v_{i,0},v_{i,1}, \ldots , v_{i,\left\lfloor \Phi _p \right\rfloor + N - 1}\). In the following, for two integers h and k we denote by \(h + k\) their sum modulo N. The network G has N arcs \(a_i=(s, v_{i,0})\) and N arcs \(b_i=(v_{i,\left\lfloor \Phi _p \right\rfloor + N - 1},t)\) for all \(i \in [N]\), having delay 0.
Fig. 1.
Fig. 1. An instance from our construction with \(N=3\) and delay functions in \(\operatorname{Poly-}p\) with \(p = 2\) or 3.
For all \(i \in [N]\) and \(j \in [\left\lfloor \Phi _p \right\rfloor + N -1]\) there is an arc \(e_{ij}\) from \(v_{i, j-1}\) to \(v_{i, j}\). The delay function associated to the arc is \(x^p\) for \(\left\lfloor \Phi _p \right\rfloor \le j \le N\) and \(c_j x^p\) for \(1 \le j \le \left\lfloor \Phi _p \right\rfloor -1\) and \(N+1 \le j \le \left\lfloor \Phi _p \right\rfloor + N - 1\). For each j such that \(1 \le j \le \left\lfloor \Phi _p \right\rfloor -1\), the coefficient \(c_j\) is computed as the solution of
\begin{equation*} c_j (j+1)^p= \sum _{i=j}^{\left\lfloor \Phi _p \right\rfloor -1} c_i \cdot i^p+ j \cdot \left\lfloor \Phi _p \right\rfloor ^p, \end{equation*}
and for each \(N+1 \le j \le \left\lfloor \Phi _p \right\rfloor + N - 1\) we set \(c_j = c_{\left\lfloor \Phi _p \right\rfloor +N -j}\). For all \(i \in [N]\) and \(j \in [\left\lfloor \Phi _p \right\rfloor + N -1]\) there is an arc \(e_{ij}\) from \(v_{i, j-1}\) to \(v_{i, j}\). The delay functions are
\begin{equation} d_{e_{ij}}= {\left\lbrace \begin{array}{ll} c_j x^p, & \text{if}\ 1 \le j \le \left\lfloor \Phi _p \right\rfloor -1 \\ x^p, & \text{if}\ \left\lfloor \Phi _p \right\rfloor \le j \le N \\ c_{\left\lfloor \Phi _p \right\rfloor +N -j} x^p, & \text{if}\ N+1 \le j \le \left\lfloor \Phi _p \right\rfloor + N - 1, \end{array}\right.} \end{equation}
(6)
where the coefficient \(c_j\) is computed as the solution of
\begin{equation*} c_j (j+1)^p= \sum _{i=j}^{\left\lfloor \Phi _p \right\rfloor -1} c_i \cdot i^p+ j \cdot \left\lfloor \Phi _p \right\rfloor ^p. \end{equation*}
Finally, for all \(i \in [N]\), \(j \in [\left\lfloor \Phi _p \right\rfloor , \left\lfloor \Phi _p \right\rfloor + N - 2]\) there is an arc \(g_{ij}\) from \(v_{i, j}\) to \(v_{i+1, j-\left\lfloor \Phi _p \right\rfloor +1}\) of constant delay 0. Note that for all \(h \in [N]\) the only arc going to \(v_{h,0}\) is \(a_h\) and the only arc going out from \(v_{h,N + \left\lfloor \Phi _p \right\rfloor - 1}\) is \(b_h\).
We define the state \(P= \lbrace P^1, \ldots , P^N\rbrace\) where the \((s,t)\)-path \(P^i\) chosen by player i is
\begin{align*} & a_i, e_{i,1}, e_{i,2}, \ldots , e_{i,\left\lfloor \Phi _p \right\rfloor }, g_{i,\left\lfloor \Phi _p \right\rfloor },\\ & e_{i+1,2}, e_{i+1,3}, \ldots , e_{i+1, \left\lfloor \Phi _p \right\rfloor +1}, g_{i+1, \left\lfloor \Phi _p \right\rfloor +1},\\ & \ldots , \\ & e_{i+N-1, N}, e_{i+N-1,N+1}, \ldots , e_{i+N-1, N+\left\lfloor \Phi _p \right\rfloor -1}, b_{i+N-1} \end{align*}
which selects \(\left\lfloor \Phi _p \right\rfloor\) consecutive arcs in each row. Note that the players’ strategies have all the same structure: by construction, the path \(P^i\) chosen by player i traverses row i first, then row \(i+1\), and lastly row \(i-1\). Since each arc with delay function \(x^p\) is used by \(\left\lfloor \Phi _p \right\rfloor\) players, and each arc with delay function \(c_j x^p\) for \(j \in [\left\lfloor \Phi _p \right\rfloor ]-1\) is used by j players, we conclude that the total players’ cost in P is equal to \(N((N-\left\lfloor \Phi _p \right\rfloor +1) \cdot \left\lfloor \Phi _p \right\rfloor \cdot \left\lfloor \Phi _p \right\rfloor ^p+ 2 \cdot \sum _{j=1}^{\left\lfloor \Phi _p \right\rfloor -1} c_j \cdot j^p)\).
We also define a state \(P^*= \lbrace P^{*1}, \ldots , P^{*N}\rbrace\) where player i selects the path \(P^{*i}\) that only traverses row i:
\begin{equation*} a_i, e_{i,1},e_{i,2}, \ldots , e_{i,N+\left\lfloor \Phi _p \right\rfloor -1},b_i. \end{equation*}
Since each arc \(e_{ij}\) is used by only one player, the total players’ cost in \(P^*\) is equal to \(N((N - \left\lfloor \Phi _p \right\rfloor +\) \(1) + 2 \cdot \sum _{j=1}^{\left\lfloor \Phi _p \right\rfloor -1} c_j)\). We remark that the cost of \(P^*\) is an upper bound on the cost of a social optimal state.
We will next prove that P is a PNE. This will imply that for \(N \rightarrow \infty\) the PoA is at least \(\left\lfloor \Phi _p \right\rfloor ^{p+1}\). Specifically,
\begin{align*} \frac{\text{cost}(P)}{\text{cost}(SO)} &\ge \frac{\text{cost}(P)}{\text{cost}(P^*)} \\ &= \lim _{N \rightarrow +\infty }\frac{(N-\left\lfloor \Phi _p \right\rfloor +1) \cdot \left\lfloor \Phi _p \right\rfloor \cdot \left\lfloor \Phi _p \right\rfloor ^p+ 2 \cdot \sum _{j=1}^{\left\lfloor \Phi _p \right\rfloor -1} c_j \cdot j^p}{(N - \left\lfloor \Phi _p \right\rfloor + 1) + 2 \cdot \sum _{j=1}^{\left\lfloor \Phi _p \right\rfloor -1} c_j}\\ &=\lim _{N \rightarrow +\infty } \frac{(N-\left\lfloor \Phi _p \right\rfloor +1) \cdot \left\lfloor \Phi _p \right\rfloor \cdot \left\lfloor \Phi _p \right\rfloor ^p}{(N - \left\lfloor \Phi _p \right\rfloor + 1)+ 2 \cdot \sum _{j=1}^{\left\lfloor \Phi _p \right\rfloor -1} c_j}\\ &=\left\lfloor \Phi _p \right\rfloor ^{p+1}. \end{align*}
To show that P is a PNE, we prove that every player i is not able to decrease their cost by deviating to another \((s,t)\)-path. Upon renaming the rows’ and the players’ indices (every index \(j\in \lbrace i,\ldots , N\rbrace\) is replaced by \(j-i+1\) and every index \(j\in \lbrace 1,\ldots , i-1\rbrace\) is replaced by \(N+j-i+1\)), we have only to examine the strategy chosen by player 1:
\begin{equation*} P^1= a_1, P^1_1,g_{1,\left\lfloor \Phi _p \right\rfloor }, P^1_2,g_{2, \left\lfloor \Phi _p \right\rfloor +1},\ldots , P^1_i,g_{i, \left\lfloor \Phi _p \right\rfloor +i-1}, \ldots , P^1_N,b_{N}. \end{equation*}
where for \(i \in [N]\)
\begin{equation*} P^1_i = e_{i, i}, e_{i, i+1}, \ldots , e_{i, i+\left\lfloor \Phi _p \right\rfloor - 1} \end{equation*}
Let f denote the flow induced by strategy profile P. To show that P is a PNE, we need the following claim:
Claim 1. For \(i \in [N]\), \(j \in [N-1]\), we have \(d_{e_{i,j}}(f_{e_{i,j}}+1) \ge \text{cost}(P^1_{j})\) and \(d_{e_{i,\left\lfloor \Phi _p \right\rfloor + N - j}}(f_{e_{i,\left\lfloor \Phi _p \right\rfloor + N - j}}+1) \ge \text{cost}(P^1_{N - j + 1})\).
Proof of claim. Let \(i \in [N]\) and \(j \in [N-1]\). We prove \(d_{e_{i,j}}(f_{e_{i,j}}+1) \ge \text{cost}(P^1_{j})\). To show \(d_{e_{i,\left\lfloor \Phi _p \right\rfloor + N - j}}(f_{e_{i,\left\lfloor \Phi _p \right\rfloor + N - j}}+1) \ge \text{cost}(P^1_{N - j + 1})\) the proof is analogous.
Case(i): \(i \in [N]\) and \(j \in [\left\lfloor \Phi _p \right\rfloor -1]\)
We have \(d_{e_{i,j}}(x) = c_{j} \cdot x^p\) and \(f_{e_{i,j}}=j\). Thus, we obtain
\begin{align*} d_{e_{i,j}}(f_{e_{i,j}}+1) &= c_{j} \cdot (j + 1)^p\\ &= \sum _{i=j}^{\left\lfloor \Phi _p \right\rfloor -1} c_i \cdot i^p+ j \cdot \left\lfloor \Phi _p \right\rfloor ^p\\ &= \text{cost}(P^1_{j}), \end{align*}
where the second equality follows the definition of \(c_j\).
Case(ii): \(i \in [N]\) and \(j \in [\left\lfloor \Phi _p \right\rfloor , N - \left\lfloor \Phi _p \right\rfloor +1]\)
In this case, all the \(\left\lfloor \Phi _p \right\rfloor\) arcs in \(P^1_j\) have cost \(x^p\) and are used by \(\left\lfloor \Phi _p \right\rfloor\) players. As a consequence, \(\text{cost}(P^1_j) = \left\lfloor \Phi _p \right\rfloor ^{p+1}\) and \(d_{e_{i,j}}(f_{e_{i,j}}+1) = (\left\lfloor \Phi _p \right\rfloor +1)^p\). Since \(0 \le \left\lfloor \Phi _p \right\rfloor \le \Phi _{p}\) we have \(g_{p}(\left\lfloor \Phi _p \right\rfloor) \ge 0\) and we conclude that the claim holds.
Case(iii): \(i \in [N]\) and \(j \in [N - \left\lfloor \Phi _p \right\rfloor + 2, N -1]\)
Since \(N \ge \max \lbrace \left\lfloor \Phi _p \right\rfloor +1, 2 \left\lfloor \Phi _p \right\rfloor -2\rbrace\), we have \(\left\lfloor \Phi _p \right\rfloor \le j \le N\), thus, \(e_{i,j}\) has delay \(x^p\) and is used by \(\left\lfloor \Phi _p \right\rfloor\) players. We obtain
\begin{equation*} d_{e_{i,j}}(f_{e_{i,j}}+1) = (\left\lfloor \Phi _p \right\rfloor +1)^p\ge \left\lfloor \Phi _p \right\rfloor ^{p+1}, \end{equation*}
where the inequality follows from the fact that \(0 \le \left\lfloor \Phi _p \right\rfloor \le \Phi _p\) and \(g_{p}(\left\lfloor \Phi _p \right\rfloor) \ge 0\). Next, we prove \(\left\lfloor \Phi _p \right\rfloor ^{p+1} \ge \text{cost}(P^1_{j})\). We recall that \(P^1_j\) consists of \(\left\lfloor \Phi _p \right\rfloor\) arcs. In this case, the first \(k = N-j+1\) have delay \(x^p\) and are used by \(\left\lfloor \Phi _p \right\rfloor\) players, while the last \(\left\lfloor \Phi _p \right\rfloor - k\) arcs are such that, for \(\ell \in [\left\lfloor \Phi _p \right\rfloor -k]\) and \(j = N + \ell\), arc \((i,j)\) has delay \(c_{N + \ell } x^p = c_{\left\lfloor \Phi _p \right\rfloor - \ell } x^p\) and is used by \(\left\lfloor \Phi _p \right\rfloor - \ell\) players. Thus,
\begin{align*} \text{cost}(P^1_{j}) &= \sum _{\ell =1}^{\left\lfloor \Phi _p \right\rfloor -k} c_{\left\lfloor \Phi _p \right\rfloor - \ell } \cdot (\left\lfloor \Phi _p \right\rfloor - \ell)^p + k \cdot \left\lfloor \Phi _p \right\rfloor ^p\\ &= \sum _{\ell =k}^{\left\lfloor \Phi _p \right\rfloor -1} c_{\ell } \cdot \ell ^p + k \cdot \left\lfloor \Phi _p \right\rfloor ^p. \end{align*}
We next show that for \(\ell \in [k, \left\lfloor \Phi _p \right\rfloor -1]\) we have \(c_{\ell } \cdot \ell ^p \le \left\lfloor \Phi _p \right\rfloor ^p\), which directly implies our claim. We proceed by backward induction. First, we show the base case, where \(\ell =\left\lfloor \Phi _p \right\rfloor -1\). By the definition of \(c_\ell\), we have
\begin{equation*} c_{\left\lfloor \Phi _p \right\rfloor -1} \cdot \left\lfloor \Phi _p \right\rfloor ^p = c_{\left\lfloor \Phi _p \right\rfloor -1}\cdot (\left\lfloor \Phi _p \right\rfloor -1)^p + (\left\lfloor \Phi _p \right\rfloor -1)\cdot \left\lfloor \Phi _p \right\rfloor ^p. \end{equation*}
Thus, we obtain
\begin{equation*} c_{\left\lfloor \Phi _p \right\rfloor -1} = \frac{(\left\lfloor \Phi _p \right\rfloor -1)\cdot \left\lfloor \Phi _p \right\rfloor ^p}{\left\lfloor \Phi _p \right\rfloor ^p - (\left\lfloor \Phi _p \right\rfloor -1)^p}. \end{equation*}
To prove our claim our goal is to show that
\begin{equation*} \frac{(\left\lfloor \Phi _p \right\rfloor -1)^{p+1}\cdot \left\lfloor \Phi _p \right\rfloor ^p}{\left\lfloor \Phi _p \right\rfloor ^p - (\left\lfloor \Phi _p \right\rfloor -1)^p} \le \left\lfloor \Phi _p \right\rfloor ^p, \end{equation*}
or equivalently
\begin{equation*} \left\lfloor \Phi _p \right\rfloor ^p \ge (\left\lfloor \Phi _p \right\rfloor -1)^{p+1} + (\left\lfloor \Phi _p \right\rfloor -1)^p= \left\lfloor \Phi _p \right\rfloor \cdot (\left\lfloor \Phi _p \right\rfloor -1)^p. \end{equation*}
We rewrite the above condition as
\begin{equation} (\left\lfloor \Phi _p \right\rfloor -1)^p \le \left\lfloor \Phi _p \right\rfloor ^{p-1}. \end{equation}
(7)
To prove that (7) is satisfied, we first observe that \(\left\lfloor \Phi _{p-1} \right\rfloor ^{p} \le (\left\lfloor \Phi _{p-1} \right\rfloor +1)^{p-1}\) because \(0 \le \left\lfloor \Phi _{p-1} \right\rfloor \le \Phi _{p-1}\) implies \(g_{p-1} (\left\lfloor \Phi _{p-1} \right\rfloor)\ge 0\). Moreover, \(\Phi _p - \Phi _{p-1} \le 1\) by Lemma 3.2, which implies \(\left\lfloor \Phi _p \right\rfloor - \left\lfloor \Phi _{p-1} \right\rfloor \le 1\) for all \(p \in \mathbb {Z}^+\). Note that \(\Phi _1 = \frac{1 + \sqrt {5}}{2} \gt 1\) thus, by Lemma 3.2, \(\Phi _p \gt 1\) for every positive integer p. Thus \(0 \le \left\lfloor \Phi _p \right\rfloor - 1 \le \left\lfloor \Phi _{p-1} \right\rfloor \le \Phi _{p-1}\) and \(g_{p-1}(\left\lfloor \Phi _p \right\rfloor - 1) \ge 0\) implies (7).
Next, we assume our claim holds for \(\ell \in [k+1, \left\lfloor \Phi _p \right\rfloor -1]\) and we prove the claim still holds for \(\ell -1\). By definition we have
\begin{equation*} c_j (j+1)^p = \sum _{i=j}^{\left\lfloor \Phi _p \right\rfloor -1} c_i \cdot i^p + j \cdot \left\lfloor \Phi _p \right\rfloor ^p, \end{equation*}
\begin{align*} c_{\ell -1} \cdot \ell ^p &= \sum _{h=\ell -1}^{\left\lfloor \Phi _p \right\rfloor -1} c_h \cdot h^p + (\ell -1) \cdot \left\lfloor \Phi _p \right\rfloor ^p\\ & = c_{\ell -1} \cdot (\ell -1)^p + \sum _{h=\ell }^{\left\lfloor \Phi _p \right\rfloor -1} c_h \cdot h^p + (\ell -1) \cdot \left\lfloor \Phi _p \right\rfloor ^p\\ & \le c_{\ell -1} \cdot (\ell -1)^p + (\left\lfloor \Phi _p \right\rfloor -\ell) \left\lfloor \Phi _p \right\rfloor ^p + (\ell -1) \cdot \left\lfloor \Phi _p \right\rfloor ^p\\ & = c_{\ell -1} \cdot (\ell -1)^p + (\left\lfloor \Phi _p \right\rfloor -1) \left\lfloor \Phi _p \right\rfloor ^p. \end{align*}
Thus,
\begin{equation*} c_{\ell -1} = \frac{(\left\lfloor \Phi _p \right\rfloor -1) \cdot \left\lfloor \Phi _p \right\rfloor ^p}{\ell ^p - (\ell -1)^p}. \end{equation*}
To prove our claim our goal is to show that
\begin{equation*} \frac{(\left\lfloor \Phi _p \right\rfloor -1) \cdot \left\lfloor \Phi _p \right\rfloor ^p \cdot (\ell -1)^p}{\ell ^p - (\ell -1)^p} \le \left\lfloor \Phi _p \right\rfloor ^p, \end{equation*}
that is equivalent to
\begin{equation*} \left\lfloor \Phi _p \right\rfloor \cdot (\ell -1)^p \le \ell ^p, \end{equation*}
and in turn also to
\begin{equation} \left(\frac{\ell -1}{\left\lfloor \Phi _p \right\rfloor }\right)^p \cdot \left\lfloor \Phi _p \right\rfloor ^{p+1} \le \left(\frac{\ell }{\left\lfloor \Phi _p \right\rfloor +1}\right)^p(\left\lfloor \Phi _p \right\rfloor +1)^p. \end{equation}
(8)
We conclude that (8) is satisfied since (i) \(0 \le \left\lfloor \Phi _p \right\rfloor \le \Phi _p\), thus \(g_p(\left\lfloor \Phi _p \right\rfloor) \ge 0\) implies \(\left\lfloor \Phi _p \right\rfloor ^{p+1} \le (\left\lfloor \Phi _p \right\rfloor +1)^p\) and (ii) \(\frac{\ell -1}{\left\lfloor \Phi _p \right\rfloor }\le \frac{\ell }{\left\lfloor \Phi _p \right\rfloor +1}\) and since \(\ell \le \left\lfloor \Phi _p \right\rfloor +1\). Thus, Claim 1 is proved. \(\diamond\)
Now we show that player 1 is not able to strictly decrease their cost by deviating to another \((s,t)\)-path. We provide an algorithm whose input is an arbitrary \((s,t)\)-path \(\bar{P}^1\) and whose output is an alternative \((s,t)\)-path that player 1 could choose to achieve a lower cost. We will show that by applying this algorithm repeatedly, the output will finally become \(P^1\). First, we introduce some notations. We define the arc costs \(w: E \rightarrow \mathbb {{\mathbb {R}}}_{\ge 0}\) as \(w_e = d_e(f_e)\) if \(e\in P^1\) and \(w_e= d_e(f_e+1)\) if \(e \notin P^1\). For any path q, let \(w(q) = \sum _{e \in q}w_e\). Given a path q, let \(q_{u,v}\), where \(u,v \in q\), denotes the subpath of q between nodes \(u,v\).
The algorithm receives in input a simple \((s,t)\)-path \(\bar{P}^1\) and returns in output another simple \((s,t)\)-path \(\hat{P}^1\) such that \(w(\hat{P}^1) \le w(\bar{P}^1)\). Let \(v_1\) denote the last node in \(\bar{P}^1\) such that \(\bar{P}^1_{s,v_1}\) coincides with \(P^1_{s, v_1}\). Note that we could have \(v_1=s\). If \(v_1 = t\), then \(P^1\) and \(\bar{P}^1\) coincide. In this case, the algorithm stops and returns \(P^1\). If \(v_1 \ne t\), then the algorithm determines the first node \(v_2\) occurring after \(v_1\) in \(\bar{P}^1_{v_1,t}\), such that \(v_2\) also belongs to \(P^1\). Finally, the algorithm identifies the last node \(v_3\) in \(\bar{P}^1_{v_2,t}\) such that \(\bar{P}^1_{v_2, v_3}\) coincides with \(P^1_{v_2,v_3}\) and outputs \(\hat{P}^1 = P^1_{s, v_3}, \bar{P}^1_{v_3, t}\). If \(\hat{P}^1\) is not simple, then we make it simple by eliminating the cycles. If the algorithm returns a path \(\hat{P}^1\) different from \(P^1\), then the algorithm is applied again by setting as input \(\bar{P}^1= \hat{P}^1\).
We first argue that by repeatedly applying this algorithm we will finally obtain \(P^1\) in output. In fact, either \(v_1 = t\), or \(\bar{P}^1_{v_3, t}\) is strictly contained in \(\bar{P}^1_{v_1, t}\), since \(v_3\) occurs after \(v_1\) in \(\bar{P}^1\). Thus, at the next iteration, when we set as input \(\bar{P}^1= \hat{P}^1\), the number of arcs in the new \(\bar{P}^1_{v_1,t}\) strictly decreases, since node \(v_1\) of the current iteration coincides with node \(v_3\) of the previous iteration. Next we show that \(w(\hat{P}^1) \le w(\bar{P}^1)\).
Case(i): \(v_1 = s\). In this case,
\begin{equation*} \bar{P}^1 = \bar{P}^1_{s,v_2}, P^1_{v_2, v_3}, \bar{P}^1_{v_3, t} \quad \text{and} \quad \hat{P}^1 = P^1_{s,v_2}, P^1_{v_2, v_3}, \bar{P}^1_{v_3, t}. \end{equation*}
Thus, we need to show that \(w(\bar{P}^1_{s,v_2}) \ge w(P^1_{s, v_2})\). Since \(v_2\) is the first node occurring after s in \(\bar{P}^1\) that also belongs to \(P^1\), by the definition of the weights \(w_e\) we have \(w_e = d_e(f_e + 1)\) for each \(e \in \bar{P}^1_{s, v_2}\) and \(w_e = d_e(f_e)\) for each \(e \in P^1_{s, v_2}\). Suppose that \(v_2 = v_{i,j}\). Then every path from s to \(v_{i,j}\) must traverse an arc \(e_{k,\ell }\) with \(k\in [N]\) for each \(\ell \in [j]\). Thus we have \(w(\bar{P}^1_{s,v_{i,j}}) \ge \sum _{\ell =1}^{j} d_{e_{1,\ell }}(f_{e_{1,\ell }}+1)\), where we arbitrarily picked \(k =1\) for each \(\ell \in [j]\). By Claim 3.1
\begin{equation} \sum _{\ell =1}^{\min \lbrace N,j\rbrace } \text{cost}\left(P^1_\ell \right) \le \sum _{\ell =1}^{j} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right) \le w\left(\bar{P}^1_{s,v_{i,j}}\right). \end{equation}
(9)
Note that since \(v_{i,j}\) belongs to \(P^1_i\), it must be \(i -1 \le j \le i + \left\lfloor \Phi _p \right\rfloor -1\). If \(j=i-1 \le N\), then \(v_2\) is the first node in \(P^1_i\). Note that i cannot be 1 because the only arc goes into node \(v_{1,0}\) is \(a_1\), which belongs to \(P^1\). If \(v_2=v_{1,0}\) then \(a_1 \in P^1\) contradicts to the definition of \(v_2\). Then we have
\begin{equation} w\left(P^1_{s, v_{i,j}}\right) = \sum _{\ell =1}^{i-1} \text{cost}\left(P^1_\ell \right) = \sum _{\ell =1}^{j} \text{cost}\left(P^1_\ell \right). \end{equation}
(10)
If \(i \le j \le i + \left\lfloor \Phi _p \right\rfloor -1\), we have
\begin{equation} w\left(P^1_{s, v_{i,j}}\right) \le \sum _{\ell =1}^{i} \text{cost}\left(P^1_\ell \right) \le \sum _{\ell =1}^{\min \lbrace N,j\rbrace } \text{cost}\left(P^1_\ell \right). \end{equation}
(11)
By combining Equations (9), (10), and (11) we obtain that
\begin{equation*} w \left(P^1_{s, v_{i,j}}\right) \le \sum _{\ell =1}^{\min \lbrace N, j\rbrace } \text{cost}\left(P^1_\ell \right) \le w \left(\bar{P}^1_{s,v_{i,j}}\right). \end{equation*}
Case(ii): \(v_2 = t\). In this case,
\begin{equation*} \bar{P}^1 = P^1_{s,v_1}, \bar{P}^1_{v_1, t} \quad \text{and} \quad \hat{P}^1 = P^1_{s,v_1}, P^1_{v_1, t} = P^1. \end{equation*}
Thus, we need to show that \(w(\bar{P}^1_{v_1,t}) \ge w(P^1_{v_1, t})\). Since \(v_2=t\) is the first node occurring after \(v_1\) in \(\bar{P}^1\) that also belongs to \(P^1\), by the definition of the weights \(w_e\) we have \(w_e = d_e(f_e + 1)\) for each \(e \in \bar{P}^1_{v_1, t}\) and \(w_e = d_e(f_e)\) for each \(e \in P^1_{v_1, t}\). Suppose that \(v_1 = v_{i,j}\). Then every path from s to \(v_{i,j}\) must traverse an arc \(e_{k,\ell }\) with \(k\in [N]\) for each \(\ell \in [j+1, N+\left\lfloor \Phi _p \right\rfloor -1]\). Thus, we have \(w(\bar{P}^1_{v_{i,j},t}) \ge \sum _{\ell =j+1}^{N+\left\lfloor \Phi _p \right\rfloor -1} d_{e_{1,\ell }}(f_{e_{1,\ell }}+1)\), where we arbitrarily picked \(k =1\) for each \(\ell \in [j]\). By the second inequality of Claim 3.1
\begin{equation} \sum _{\ell =\max \lbrace j-\left\lfloor \Phi _p \right\rfloor +2, 1\rbrace }^{N} \text{cost}\left(P^1_\ell \right) \le \sum _{\ell =j+1}^{N+\left\lfloor \Phi _p \right\rfloor -1} d_{e_{1,\ell }} \left(f_{e_{1,\ell }}+1\right) \le w\left(\bar{P}^1_{s,v_{i,j}}\right). \end{equation}
(12)
Note that since \(v_{i,j}\) belongs to \(P^1_i\), it must be \(i -1 \le j \le i + \left\lfloor \Phi _p \right\rfloor -1\). If \(j= i+\left\lfloor \Phi _p \right\rfloor -1\), then \(v_1\) is the last node of \(P^1_i\). Note that i cannot be N because the only arc goes into node \(v_{N,N+\left\lfloor \Phi _p \right\rfloor -1}\) is \(b_N\), which belongs to \(P^1\). If \(v_1=v_{N,N+\left\lfloor \Phi _p \right\rfloor -1}\) then \(b_N \in P^1\) contradicts to the definition of \(v_1\). Then we have
\begin{equation} w\left(P^1_{v_{i,j}, t}\right) = \sum _{\ell =i+1}^{N} \text{cost}\left(P^1_\ell \right) \le \sum _{\ell =\max \lbrace j-\left\lfloor \Phi _p \right\rfloor +2, 1\rbrace }^{N} \text{cost}\left(P^1_\ell \right). \end{equation}
(13)
If \(i-1 \le j \le i + \left\lfloor \Phi _p \right\rfloor -2\), we have
\begin{equation} w\left(P^1_{v_{i,j}, t}\right) \le \sum _{\ell =i}^{N} \text{cost}\left(P^1_\ell \right) \le \sum _{\ell =\max \lbrace j-\left\lfloor \Phi _p \right\rfloor +2, 1\rbrace }^{N} \text{cost}\left(P^1_\ell \right). \end{equation}
(14)
By combining Equations (12), (13), and (14) we obtain that
\begin{equation*} w\left(P^1_{v_{i,j}, t}\right) \le \sum _{\ell =\max \lbrace j-\left\lfloor \Phi _p \right\rfloor +2, 1\rbrace }^{N} \text{cost}\left(P^1_\ell \right) \le w\left(\bar{P}^1_{v_{i,j},t}\right). \end{equation*}
Case(iii): \(v_1 \ne s\) and \(v_2 \ne t\). Without loss of generality, let \(v_1= v_{i,j}\) and \(v_2= v_{h,k}\). Note that \(v_1\) belongs to \(P^1_i\), thus \(i -1 \le j \le i + \left\lfloor \Phi _p \right\rfloor - 1\), and \(v_2\) belongs to \(P^1_h\), thus \(h -1 \le k \le h + \left\lfloor \Phi _p \right\rfloor - 1\). Finally, note that \(k \gt j\) if \(i = h\). Thus, \(\bar{P}^1 = (P^1_{s,v_1}, \bar{P}^1_{v_1,v_2}, P^1_{v_2, v_3}, \bar{P}^1_{v_3, t})\) and \(\hat{P}^1 = (P^1_{s,v_1}, P^1_{v_1, v_2}, P^1_{v_2, v_3}, \bar{P}^1_{v_3, t})\). So we only need to show that
\begin{equation*} w\left(\bar{P}^1_{v_1,v_2}\right) \ge w\left(P^1_{v_1, v_2}\right). \end{equation*}
Since \(v_2\) is the first node occurring after s is \(\bar{P}^1\) that also belongs to \(P^1\), by the definition of the weights \(w_e\) we have \(w_e = d_e(f_e + 1)\) for each \(e \in \bar{P}^1_{v_1, v_2}\) and \(w_e = d_e(f_e)\) for each \(e \in P^1_{v_1, v_2}\).
Subcase(iii).1: \(j = i + \left\lfloor \Phi _p \right\rfloor - 1\), i.e., \(v_{i,j}\) is the last node in \(P^1_i\), and \(h \le k \le h + \left\lfloor \Phi _p \right\rfloor -1\). Thus, \(h \gt i\) and \(h \ne k-1\), otherwise \(\bar{P}^1_{v_1, v_2}\) will intersect with \(P^1_{v_1,v_2}\) before \(v_2\). Since \(g_{i,j} \in P^1_{v_1,v_2}\), we conclude that the first arc in \(\bar{P}^1_{v_1, v_2}\) is \(e_{i, i+\left\lfloor \Phi _p \right\rfloor }\). And since \(e_{h,k} \in P^1_{v_1,v_2}\), we conclude that the last arc in \(\bar{P}^1_{v_1, v_2}\) is \(g_{h-1, k + \left\lfloor \Phi _p \right\rfloor - 1}\). Note that because \(k + \left\lfloor \Phi _p \right\rfloor - 1 \ge h + \left\lfloor \Phi _p \right\rfloor - 1 \ge i + \left\lfloor \Phi _p \right\rfloor\), then every path begin with \(e_{i, i+\left\lfloor \Phi _p \right\rfloor }\) and end with \(g_{h-1, k + \left\lfloor \Phi _p \right\rfloor - 1}\) must traverse an arc \(e_{m,\ell }\) with \(m\in [N]\) for each \(\ell \in [i+\left\lfloor \Phi _p \right\rfloor , k + \left\lfloor \Phi _p \right\rfloor - 1]\). We arbitrarily picked \(m =1\) for each \(\ell\), then we can conclude that
\begin{equation} w\left(\bar{P}^1_{v_1, v_2}\right) \ge \sum _{\ell = i+\left\lfloor \Phi _p \right\rfloor }^{k + \left\lfloor \Phi _p \right\rfloor - 1} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right) \ge \sum _{\ell = i+\left\lfloor \Phi _p \right\rfloor }^{h + \left\lfloor \Phi _p \right\rfloor - 1} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right). \end{equation}
(15)
Then according to the second inequality of Claim 3.1, we have
\begin{equation} \sum _{\ell = i+\left\lfloor \Phi _p \right\rfloor }^{h + \left\lfloor \Phi _p \right\rfloor - 1} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right) \ge \sum _{\ell =i+1}^h w\left(P^1_\ell \right). \end{equation}
(16)
Recall that \(v_1=v_{i,j}\) is the last node in \(P^1_i\), which implies that \(P^1_{v_1, v_2}\) contains subpath \(P^1_{i+1}, \ldots , P^1_{h-1}\) and part of subpath \(P^1_h\). So we can conclude that
\begin{equation} \sum _{\ell =i+1}^h \text{cost}\left(P^1_\ell \right) \ge w\left(P^1_{v_1, v_2}\right) \end{equation}
(17)
By combining inequalities (15), (16), and (17) we have \(w(\bar{P}^1_{v_1,v_2}) \ge w(P^1_{v_1, v_2}).\)
Subcase(iii).2: \(i-1 \le j \le i + \left\lfloor \Phi _p \right\rfloor - 2\) and \(k = h-1\), i.e., \(v_{h,k}\) is the first node in \(P^1_h\). Thus, we have \(h \gt i\). Since \(e_{i,j+1} \in P^1_{v_1,v_2}\), we conclude that the first arc in \(\bar{P}^1_{v_1, v_2}\) is \(g_{i, j}\). And since \(g_{h-1,k+\left\lfloor \Phi _p \right\rfloor -1}\), who ends with \(v_{h,k}\), belongs to \(P^1_{v_1,v_2}\), we conclude that the last arc in \(\bar{P}^1_{v_1, v_2}\) is \(e_{h, h-1}\). Note that because \(g_{i,j}\) ends with node \(v_{i+1, j-\left\lfloor \Phi _p \right\rfloor +1}\) and \(h-1 \ge j - \left\lfloor \Phi _p \right\rfloor + 2\), then every path begin with \(g_{i, j}\) and end with \(e_{h, h-1}\) must traverse an arc \(e_{m,\ell }\) with \(m\in [N]\) for each \(\ell \in [j - \left\lfloor \Phi _p \right\rfloor + 2, h - 1]\). We arbitrarily picked \(m =1\) for each \(\ell\), then we can conclude that
\begin{equation} w\left(\bar{P}^1_{v_1, v_2}\right) \ge \sum _{\ell = j - \left\lfloor \Phi _p \right\rfloor + 2}^{h - 1} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right) \ge \sum _{\ell = i}^{h - 1} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right). \end{equation}
(18)
Then according to the first inequality of Claim 3.1, we have
\begin{equation} \sum _{\ell = i}^{h - 1} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right) \ge \sum _{\ell =i}^{h-1} w\left(P^1_\ell \right). \end{equation}
(19)
Recall that \(v_2=v_{h,k}\) is the first node in \(P^1_i\), which implies that \(P^1_{v_1, v_2}\) contains subpath \(P^1_{i+1}, \ldots , P^1_{h-1}\) and part of subpath \(P^1_i\). So we can conclude that
\begin{equation} \sum _{\ell =i}^{h-1} \text{cost}\left(P^1_\ell \right) \ge w\left(P^1_{v_1, v_2}\right) . \end{equation}
(20)
By combining inequalities (18), (19), and (20) we have \(w(\bar{P}^1_{v_1,v_2}) \ge w(P^1_{v_1, v_2}).\)
Subcase(iii).3: \(i-1 \le j \le i + \left\lfloor \Phi _p \right\rfloor - 2\) and \(h \le k \le h + \left\lfloor \Phi _p \right\rfloor -1\). Thus, we have \(h \ge i\) and \(k \gt j\) if \(h=i\). Since \(e_{i,j+1} \in P^1_{v_1,v_2}\), we conclude that the first arc in \(\bar{P}^1_{v_1, v_2}\) is \(g_{i, j}\). And since \(e_{h,k} \in P^1_{v_1,v_2}\), we conclude that the last arc in \(\bar{P}^1_{v_1, v_2}\) is \(g_{h-1, k + \left\lfloor \Phi _p \right\rfloor - 1}\). Note that because \(g_{i,j}\) ends with node \(v_{i+1, j-\left\lfloor \Phi _p \right\rfloor +1}\) and \(k + \left\lfloor \Phi _p \right\rfloor - 1 \ge j - \left\lfloor \Phi _p \right\rfloor + 2\), then every path begin with \(g_{i, j}\) and end with \(g_{h-1, k + \left\lfloor \Phi _p \right\rfloor - 1}\) must traverse an arc \(e_{m,\ell }\) with \(m\in [N]\) for each \(\ell \in [j - \left\lfloor \Phi _p \right\rfloor + 2, k + \left\lfloor \Phi _p \right\rfloor - 1]\). We arbitrarily picked \(m =1\) for each \(\ell\), then we can conclude that
\begin{equation} w\left(\bar{P}^1_{v_1, v_2}\right) \ge \sum _{\ell = j - \left\lfloor \Phi _p \right\rfloor + 2}^{k + \left\lfloor \Phi _p \right\rfloor - 1} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right) \ge \sum _{\ell = i}^{h} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right). \end{equation}
(21)
Then according to the first inequality of Claim 3.1, we have
\begin{equation} \sum _{\ell = i}^{h} d_{e_{1,\ell }}\left(f_{e_{1,\ell }}+1\right) \ge \sum _{\ell =i}^{h} w\left(P^1_\ell \right). \end{equation}
(22)
Because \(v_1 \in P^1_i\) and \(v_2 \in P^1_h\), so \(P^1_{v_1, v_2}\) contains subpath \(P^1_{i+1}, \ldots , P^1_{h-1}\) and part of subpaths \(P^1_i\) and \(P^1_h\). So we can conclude that
\begin{equation} \sum _{\ell =i}^{h} \text{cost}\left(P^1_\ell \right) \ge w\left(P^1_{v_1, v_2}\right). \end{equation}
(23)
By combining inequalities (21), (22), and (23) we have \(w(\bar{P}^1_{v_1,v_2}) \ge w(P^1_{v_1, v_2}).\) □

3.2 Upper Bound on the PoA in Series-Parallel Networks

In this section, we prove Theorem 1.2, which is restated next for convenience.
Theorem 1.2.
In a symmetric (unweighted) network congestion game on a series-parallel \((s,t)\)-network with delays functions in class \(\mathcal {D}\), the PoA w.r.t. the total players’ cost is at most \(y(\mathcal {D})\).
First, we need to introduce some necessary notation and properties of series-parallel networks. In the following, we denote by f and o a PNE and a social optimum, respectively, of the series-parallel network congestion game. We consider the network \(G(o-f)\) introduced in [15, 16]. Precisely, the node set of \(G(o-f)\) is V, and the arc set is \(E(o-f) = \lbrace (u,v): (e=(u,v) \in E \text{ and}\ o_e - f_e \gt 0) \text{ or}\ (e=(v,u) \in E\ \text{and}\ o_e -f_e \lt 0)\rbrace\). \(G(o-f)\) is a collection of simple cycles \(\lbrace C_1, \ldots , C_h\rbrace\) such that each \(C_i\) carries \(s_i\) units of flow. As previously done in [15, 16], for each \(i \in [h]\), we define \(C_i^+ = \lbrace e= (u,v) \in E: (u,v) \in C_i, o_e \gt f_e\rbrace\) and \(C_i^- = \lbrace e= (u,v) \in E: (v,u) \in C_i, o_e \lt f_e\rbrace\).
Recall the definition of \(y(\mathcal {D})\) from Section 1.
\begin{equation*} y(\mathcal {D}) = \sup _{d \in \mathcal {D},~x\in \mathbb {N}^+}\frac{(x+1)d(x+1) - xd(x)}{d(x)}, \end{equation*}
In the next four lemmas, we will assume that there exists an index \(i \in [h]\) such that \(C^+_i\) is an \((s,t)\)-path, and we will prove that the PoA is at most \(y(\mathcal {D})\). Later, we will relax this assumption. Observe that, by definition, \(C^+_i\) is contained in o. In the next lemma, we prove that the cost of \(C^+_i\) with respect to o is at least the average players’ cost in the PNE f, that is, \(\text{cost}(f)/N\).
Lemma 3.3.
If \(C^+_i\), \(i \in [h]\), is an \((s,t)\)-path, then \(\text{cost}_o(C^+_i) \ge {\text{cost}(f)}/{N}\).
Proof.
The cost of \(C^+_i\) with respect to flow o satisfies:
\begin{align*} \text{cost}_o(C^+_i) &= \sum _{e \in C^+_i} d_e(o_e) \ge \sum _{e \in C^+_i} d_e(f_e + 1) \ge \frac{\text{cost}(f)}{N}. \end{align*}
The first inequality holds since for every \(e \in C^+_i\), we have \(o_e \ge f_e +1\). Next, we show that the second inequality holds. Denote by \(P^*\) the set of N \((s,t)\)-paths in the PNE inducing f. Clearly \(\max \left\lbrace \text{cost}_{f}(\pi): \pi \in P^*\right\rbrace \ge \frac{\text{cost}(f)}{N}\). By contradiction, suppose that \(\sum _{e \in C^+_i} d_e(f_e + 1) \lt \frac{\text{cost}(f)}{N}\). We would obtain that \(\max \left\lbrace \text{cost}_{f}(\pi): \pi \in P^*\right\rbrace \gt \text{cost}_{f}^+(C_i^+)\), thus, one player would prefer to change their strategy into \(C_i^+\). This contradicts the fact that f is a PNE. □
In the next lemma, we contemplate adding one unit of flow on an arbitrary \((s,t)\)-path p contained in o, and we lower bound the corresponding increase of the total cost. This will be crucial to derive a lower bound on \(\text{cost}_o(p)\) that will be used to relate \(\text{cost}(f)\) and \(\text{cost}(o)\).
Lemma 3.4.
Suppose that there exists an index \(i \in [h]\), such that \(C^+_i\) is an \((s,t)\)-path. Then every \((s,t)\)-path p contained in o satisfies
\begin{equation*} \sum _{e \in p} \left((o_e + 1)d_e(o_e + 1) - o_ed_e(o_e)\right) \ge \frac{\text{cost}(f)}{N}. \end{equation*}
Proof.
We will prove this by contradiction. Assume that there is an \((s,t)\)-path p contained in o such that
\begin{equation} \sum _{e \in p} (o_e + 1)d_e(o_e + 1) - \sum _{e \in p} o_ed_e(o_e) \lt \frac{\text{cost}(f)}{N}. \end{equation}
(24)
We define a new state \(o^{\prime }\) obtained from o by deviating one unit of flow from \(C^+_i\) to p. Let \(S=C^+_i \cap p\). First, the cost difference between \(o^{\prime }\) and o is
\begin{align*} \text{cost}(o^{\prime }) - \text{cost}(o) = & \; \; \sum _{e \in C^+_i \setminus S}((o_e - 1)d_e(o_e - 1)- o_ed_e(o_e)) \\ &+ \sum _{e \in p \setminus S}((o_e+1)d_e(o_e+1) - o_ed_e(o_e)). \end{align*}
Observe that, since the delay functions are non-decreasing, we have \(d_e(o_e - 1) \le d_e(o_e)\) for all \(e \in C^+_i\), thus,
\begin{align*} \text{cost}(o^{\prime }) - \text{cost}(o) &\le \sum _{e \in C^+_i \setminus S}((o_e - 1)d_e(o_e)- o_ed_e(o_e)) \\ &+ \sum _{e \in p \setminus S}((o_e+1)d_e(o_e+1) - o_ed_e(o_e))\\ &= - \sum _{e \in C^+_i \setminus S}d_e(o_e) + \sum _{e \in p \setminus S}((o_e+1)d_e(o_e+1) - o_ed_e(o_e)). \end{align*}
Moreover, we have \(d_e(o_e+1) \ge d_e(o_e)\) for all \(e \in S\), thus,
\begin{align*} 0 &\le \quad \sum _{e \in S}(o_e + 1)(d_e(o_e+1) - d_e(o_e))\\ &=- \sum _{e \in S}d_e(o_e) + \sum _{e \in S}((o_e+1)d_e(o_e+1) - o_ed_e(o_e)). \end{align*}
By summing up these two inequalities we get
\begin{align*} \text{cost}(o^{\prime }) - \text{cost}(o) &\le - \sum _{e \in C^+_i}d_e(o_e) + \sum _{e \in p}((o_e+1)d_e(o_e+1) - o_ed_e(o_e)). \end{align*}
By Lemma 3.3, since \(C^+_i\) is an \((s,t)\)-path, we have \(\text{cost}_o(C^+_i) =\sum _{e \in C^+_i}d_e(o_e) \ge \frac{\text{cost}(f)}{N}\). Thus, by (24) we obtain \(\text{cost}(o^{\prime }) - \text{cost}(o) \lt 0\), which contradicts the fact that o is a social optimum. □
By using Lemma 3.4, we can derive a lower bound on \(\text{cost}_o(p)\) similar to the lower bound on \(\text{cost}_o(C_i^+)\) stated in Lemma 3.3, but with an extra factor of \(y(\mathcal {D})\).
Lemma 3.5.
Suppose there exists an index \(i \in [h]\) such that \(C^+_i\) is an \((s,t)\)-path, and let P be any decomposition of o. Then for every \(p \in P\),
\begin{equation*} y(\mathcal {D})\text{cost}_o(p) \ge \frac{\text{cost}(f)}{N}. \end{equation*}
Proof.
Since P is a decomposition of o, for each \(p \in P\) we have \(o_e \gt 0\) for all \(e \in p\). Then we have
\begin{align*} y(\mathcal {D})\text{cost}_o(p) &= \sum _{e \in p} y(\mathcal {D}) d_e(o_e) \ge \sum _{e \in p}((o_e+1)d_e(o_e+1) - o_ed_e(o_e)) \ge \frac{\text{cost}(f)}{N}, \end{align*}
where the first inequality follows the definition of \(y(\mathcal {D})\) stated in Equation (1) and the second inequality follows from Lemma 3.4. □
Finally, under the assumption that there exists a path \(C^+_i\) from s to t, we are ready to prove that the PoA is at most \(y(\mathcal {D})\).
Lemma 3.6.
If there exists an index \(i \in [h]\) such that \(C^+_i\) is an \((s,t)\)-path, then \(\text{cost}(f) \le y(\mathcal {D}) \text{cost}(o)\).
Proof.
By Lemma 3.5, we know that given an arbitrary decomposition P of the social optimal flow o, for all \(p \in P\), we have \(y(\mathcal {D}) \text{cost}_o(p) \ge \frac{\text{cost}(f)}{N}\). Then we can conclude that
\begin{align*} y(\mathcal {D}) \text{cost}(o) &= \sum _{p \in P} y(\mathcal {D}) \text{cost}_o(p) \ge |P|\frac{\text{cost}(f)}{N} = \text{cost}(f), \end{align*}
where the last equality follows from the fact that \(|P| = N\). This implies that \(\text{cost}(f) \le y(\mathcal {D})\) \(\text{cost}(o)\). □
We now relax the assumption that there exists a path \(C^+_i\) from s to t. In order to do this, we will exploit the structure of series-parallel networks. If G is series-parallel, it is known that for each \(i \in [h]\) \(C_i^+\) and \(C_i^-\) are two internally disjoint paths in G from a node \(u_i\) to a node \(v_i\) [15, 16]. For each \(i \in [h]\), we identify the pair of nodes \(u_i,v_i\) and we define
\begin{align*} V_i &= \lbrace w \in V: \text{there is a $(u_i,v_i)$-path containing $w$} \rbrace ,\\ E_i &= \lbrace e \in E: \text{there is a $(u_i,v_i)$-path containing $e$} \rbrace , \end{align*}
and we let \(\mathcal {L} = \lbrace E_{1}, \ldots , E_{h} \rbrace\).
Before stating the next lemma, we recall that a collection of sets \(\lbrace S_1, S_2, \ldots , S_k\rbrace\) is a laminar family if for every \(i,j \in [k]\) we have either \(S_i \cap S_j = \emptyset\) or \(S_i \subseteq S_j\) or \(S_j \subseteq S_i\).
Lemma 3.7.
If G is series-parallel, then \(\mathcal {L} = \lbrace E_{1}, \ldots , E_{h} \rbrace\) is a laminar family.
Proof.
We prove this lemma by showing that if \(E_i \cap E_j \ne \emptyset\) for some i and j in \([h]\), then \(E_i \subseteq E_j\) or \(E_j \subseteq E_i\). We proceed by induction on \(|E|\).
The base case as \(|E| = 2\). If the two arcs of G are composed in series, then there are no cycles. If they are composed in parallel, then there is only one cycle, i.e., \(i=j\), and \(E_{i} = E_{j} = E\). This implies that the lemma holds for the base case. Now we assume that when \(|E| \le t\), the lemma holds. When \(|E| = t+1\), since G is series-parallel, it can be decomposed either in series or in parallel.
Suppose that G can be decomposed in series into \(G_1\) and \(G_2\). We first show that \(E_i\) and \(E_j\) are both contained either in the arc set of \(G_1\) or in the arc set \(G_2\). In fact, \(E_i\) cannot have arcs both in \(G_1\) and in \(G_2\), otherwise \(C_i^+\) and \(C_i^-\) would not be internally disjoint paths. Thus, \(E_i\) is contained either in the arc set of \(G_1\) or in the arc set \(G_2\). Similarly, \(E_j\) is contained either in the arc set of \(G_1\) or in the arc set \(G_2\). Moreover, \(E_i\) and \(E_j\) cannot belong to different components, otherwise, we would have \(E_i \cap E_j = \emptyset\). Thus, \(E_i\) and \(E_j\) both belong to the same component. Assume without loss of generality that this is \(G_1\). Since the number of arcs of \(G_1\) is at most t, by the inductive hypothesis we obtain that \(E_i \subseteq E_j\) or \(E_j \subseteq E_i\), thus, the claim is proven in this case.
Now suppose that G can be decomposed in parallel into \(G_1\) and \(G_2\). If \(E_i\) and \(E_j\) are both contained either in the arc set of \(G_1\) or in the arc set of \(G_2\), then by induction the claim holds. If \(E_i\) is contained in the arc set of one component, say \(G_1\), and \(E_j\) is contained in the arc set of the other component \(G_2\), then \(E_i \cap E_j = \emptyset\), a contradiction. Thus, at least one among \(E_i\) and \(E_j\) has arcs both in \(G_1\) and in \(G_2\). Without loss of generality, suppose \(E_i\) does. We prove that \(C_i^+\) and \(C_i^-\) are (internally disjoint) \((s,t)\)-paths. By contradiction, suppose that \(C_i^+\) and \(C_i^-\) are \((s_i,t_i)\)-paths such that \(s_i \ne s\) or \(t_i \ne t\). Note that \(s_i\) and \(t_i\) are either both in \(G_1\) or both in \(G_2\). Suppose w.l.o.g. they are both in \(G_1\). Then each \((s_i, t_i)\)-path cannot contain any arc in \(G_2\). Because \(C_i^+\) and \(C_i^-\) are \((s,t)\)-paths, by the definition of \(E_i\), we have \(E_i = E\). Thus we conclude that \(E_j \subseteq E_i\), which proves the claim in this case. □
By Proposition 1 in [15, 16], if w and \(w^{\prime }\) are two nodes in \(V_i\) such that there exist two internally disjoint \((w,w^{\prime })\)-paths \(p_1\) and \(p_2\), then every \((s,t)\)-path having an arc in common with \(p_1\) contains both w and \(w^{\prime }\) and intersects \(p_2\) only at w and \(w^{\prime }\). This implies that each \((s,t)\)-path going through \(u_i\) also goes through \(v_i\). As a consequence, for each \(i \in [h]\) the sub-vectors of f and o that are indexed by the arcs of \(E_i\), denoted by \(f(E_i)\) and \(o(E_i)\), respectively, both define \((u_i,v_i)\)-flows in the subgraph \(G_i=(V_i,E_i)\). Define a network congestion game on \(G_i\), where each arc \(e \in E_i\) has the same delay \(d_e\) as in G, and the number of players \(N_i\) is equal to the value of flow \(f(E_i)\).
Lemma 3.8.
If G is series-parallel and \(E_i\) is a maximal set in \(\mathcal {L}\), then in the network congestion game defined on \(G_i\), \(f(E_i)\) and \(o(E_i)\) are a PNE flow and a social optimum flow, respectively.
Proof.
Let \(N_i\) be the flow value of \(f(E_i)\). First, we show that \(o(E_i)\) also has value \(N_i\). Recall that \(G(o-f)\) is a collection of cycles \(\lbrace C_1, \ldots , C_h \rbrace\) and each \(C_i\) carries \(s_i\) units of flow. By the definition of \(G(o-f)\) we can change f into o as follows: for \(j \in [h]\), decrease the flow on \(C^-_j\) by \(s_j\) and increase the flow on \(C^+_j\) by \(s_j\). By Lemma 3.7, \(\mathcal {L}\) is a laminar family, thus, for each \(j \in [h]\), the paths \(C^-_j\) and \(C^+_j\) are either both in \(G_i\) or neither of them in \(G_i\), i.e., either \(E_{j} \subseteq E_i\), or \(E_{j} \cap E_i = \emptyset\). Thus, each step does not change the flow value on \(G_i\). We can conclude that when the procedure ends, the flow value \(o(E_i)\) equals the flow value of \(f(E_i) = N_i\).
Next, we show that \(f(E_i)\) is a PNE flow on \(G_i\). By contradiction, suppose that \(f(E_i)\) is not a PNE flow on \(G_i\). This implies that in each decomposition of \(f(E_i)\) into \(N_i\) \((u_i,v_i)\)-paths there is always one player who can decrease their cost by deviating their strategy to another \((u_i,v_i)-\)path in \(G_i\). This implies that in each decomposition of f into N \((s,t)\)-paths there is always one player that can unilaterally deviate and decrease their cost. This contradicts to that f is a PNE flow.
Finally, we show that \(o(E_i)\) is a social optimum on \(G_i\). By contradiction, suppose that there is another flow \(o^{\prime }(E_i)\) in \(G_i\) of value \(N_i\) such that \(\text{cost}(o^{\prime }(E_i)) \lt \text{cost}(o(E_i))\). Then we can construct a flow \(o^{\prime \prime }\) such that \(o^{\prime \prime }_e = o_e\) for all \(e \in E \setminus E_i\) and \(o^{\prime \prime }_e = o^{\prime }_e\) for all \(e \in E_i\). Then \(\text{cost}(o^{\prime \prime }) \lt \text{cost}(o)\), contradicting the fact that o is the social optimum. □
We now consider the networks \(G_i\), \(i \in [h]\), having node set \(V_i\) and arc set \(E_i\).
Lemma 3.9.
If G is series-parallel and \(E_i\) is a maximal set in \(\mathcal {L}\), then
\begin{equation*} \text{cost}(f(E_i)) \le y(\mathcal {D}) \text{cost}(o(E_i)). \end{equation*}
Proof.
According to Lemma 3.8, the congestion game with \(N_i\) players on the two terminal-series parallel network \(G_i\) is such that \(f(E_i)\) is a PNE and \(o(E_i)\) is a social optimum. Note that \(u_i\) and \(v_i\) are, respectively, the source and the sink of \(G_i\). Since \(C^+_i\) is a \((u_i,v_i)\)-path, by Lemma 3.6, we conclude that the lemma holds. □
We are finally ready to prove Theorem 1.2, i.e., in a symmetric network congestion game defined over a series-parallel network with delay functions in class \(\mathcal {D}\), the PoA is at most \(y(\mathcal {D})\).
Proof of Theorem 1.2. Consider the PNE flow f, the social optimum flow o and the laminar family \(\mathcal {L}\) defined previously in this section. We will prove that, since G is series-parallel, then \(\text{cost}(f) \le y(\mathcal {D}) \text{cost}(o)\). Let \(E_{C_1}, \ldots , E_{C_l}\) be the maximal sets in \(\mathcal {L}\) and denote by \(E(\mathcal {L})\) their union. We rewrite \(\text{cost}(f)\) as follows:
\begin{align*} \text{cost}(f) = \sum _{e \notin E(\mathcal {L})} f_e d_e(f_e) + \sum _{e \in E(\mathcal {L})} f_ed_e(f_e). \end{align*}
Note that for each arc \(e \notin E(\mathcal {L})\) we have \(f_e = o_e\). Moreover, \(E_{C_1}, \ldots , E_{C_l}\) are a partition of \(E(\mathcal {L})\), since they are maximal members of \(\mathcal {L}\) that are pairwise disjoint. Thus, we can rewrite the above expression as
\begin{align*} \text{cost}(f) &= \sum _{e \notin E(\mathcal {L})} o_ed_e(o_e) + \sum _{i=1}^l \sum _{e \in E_{C_i}} f_ed_e(f_e)\\ &\le y(\mathcal {D}) \sum _{e \not\in E(\mathcal {L})} o_ed_e(o_e) + y(\mathcal {D})\sum _{i=1}^l \sum _{e \in E_{C_i}} o_ed_e(o_e) = y(\mathcal {D}) \text{cost}(o), \end{align*}
where the inequality follows from the fact that \(y(\mathcal {D}) \ge 1\) and from Lemma 3.9. \(\square\)
Let \(\operatorname{Poly-}p\) be the class of polynomial delay functions with maximum degree p, which are of the form \(\sum _{j=0}^pa_j x^j\), with \(a_j \ge 0\) for \(j=0,\ldots , p\).
Lemma 3.10.
For the class of polynomial delay functions \(\operatorname{Poly-}p\) it holds that \(y(\operatorname{Poly-}p) \le 2^{p+ 1} - 1\).
Proof.
By using the definition of \(y(\operatorname{Poly-}p)\) in (1) we have that for any \(x \in {\mathbb {N}}^+\)
\begin{align} y(\operatorname{Poly-}p) &= \sup _{a_0, \ldots ,a_p \in \mathbb {R}_{\ge 0},~x\in \mathbb {N}^+} \frac{(x+1)\sum _{j=0}^pa_j (x+1)^j - x \sum _{j=0}^pa_j x^j}{\sum _{j=0}^pa_j x^j} \nonumber \nonumber\\ & = \sup _{a_0, \ldots ,a_p \in \mathbb {R}_{\ge 0},~x\in \mathbb {N}^+} \frac{\sum _{j=0}^p\left(a_j \left((x+1)^{j+1} -x^{j+1} \right)\right)}{\sum _{j=0}^pa_j x^j}. \end{align}
(25)
We now exploit the fact that given two collections of nonnegative real numbers \(b_0,\ldots , b_p\) and \(c_0,\ldots , c_p\), we have
\begin{equation*} \frac{\sum _{j=0}^pb_j}{\sum _{j=0}^pc_j} \le \max _{j=0,\ldots ,p} \frac{b_j}{c_j}. \end{equation*}
As a consequence, we can upper bound (25) by
\begin{equation} \max _{j \in \lbrace 0,\ldots , p\rbrace , x \in \mathbb {N}^+} \frac{(x+1)^{j+1} - x^{j+1}}{x^j}. \end{equation}
(26)
We now upper bound the numerator of the above expression as follows:
\begin{align*} (x+1)^{j+1} - x^{j+1} &= \sum _{k=0}^{j+1}\binom{j+1}{k}x^{j+1-k} - x^{j+1} \le \sum _{k=1}^{j+1}\binom{j+1}{k}x^{j}, \end{align*}
where the inequality follows from the fact that \(j+1 \ge 1\) and \(x \in {\mathbb {N}}^+\). From (26) we then obtain
\begin{align*} y(\operatorname{Poly-}p) \le \max _{j \in \lbrace 0,\ldots , p\rbrace } \sum _{k=1}^{j+1}\binom{j+1}{k} = \max _{j \in \lbrace 0,\ldots , p\rbrace } \sum _{k=0}^{j+1}\binom{j+1}{k} -1 = 2^{p+1} -1. \end{align*}
 □
By Theorem 1.2 and Lemma 3.10, we obtain that the PoA of series-parallel network congestion games with polynomial delay functions with highest degree is p is at most \(2^{p+ 1} - 1\).

3.3 Lower Bound on the PoA in Series-Parallel Networks

In this section, we illustrate how to construct a family of instances that asymptotically achieve the lower bound on the PoA stated in Theorem 1.3. This construction is an extension to polynomial delays of the construction proposed in [23] for affine delays. Let \(\lbrace q_1, \ldots , q_N \rbrace\) be an ordered sequence of positive numbers such that \(\sum _{i=1}^N q_i=1\) and \(q_{i+1} = \frac{1}{2^p}\sum _{j=1}^i \frac{q_j}{i}\) for \(i \in [N-1]\). Let \(m \in [N-1]\). We define a new sequence \(\lbrace s_1, \ldots , s_N \rbrace\) by averaging \(\lbrace q_1, \ldots , q_m\rbrace\). Precisely, \(s_1 = \cdots = s_m = \frac{\sum _1^m q_i}{m}\) and \(s_j = q_j\) for \(j \ge m+1\). We construct a series-parallel \((s,t)\)-network G with delays in \(\operatorname{Poly-}p\), and an \((s,t)\)-flow f of value N recursively. Let \(G_m\) be a single \((s,t)\)-arc with flow \(f_m\) of value m and delay equal to \(\frac{s_1 x}{m}\). For every \(i \in [m, N-1]\), we construct \(G_{i+1}\) and \(f_{i+1}\) using \(G_i\) and \(f_i\) as follows: we compose in parallel \(G_i\) and a new \((s,t)\)-arc with flow value 1 and delay function \(s_{i+1}x^p\) and call the new network \(\tilde{G}_i\) and the new \((s,t)\)-flow \(\tilde{f}_i\). Next, we compose in series \(i+1\) copies of \(\tilde{G}_i\) with flow \(\tilde{f}_i\) to get \(G_{i+1}\) and \(f_{i+1}\). We also divide the delay functions by \({i+1}\). Then we set \(f = f_N\). Finally, we compose \(G_N\) in parallel with m new \((s,t)\)-arcs \(e_1,\ldots , e_m\) with delay function \(\frac{1}{N} x^p\) to get G. By construction, G is a series-parallel network with polynomial delay functions having non-negative coefficients and maximum degree p.
To prove Theorem 1.3, we first show that f is a PNE. Then we define a new \((s,t)\)-flow h that is obtained from f by deviating \(k \in [m]\) units of flows from the most expensive \((s,t)\)-paths in f to the k parallel \((s,t)\)-arcs in G with delay function \(\frac{1}{N} x^p\). The parameters r and l in (2) are defined as \(r = \frac{m}{N}\), \(l = \frac{k}{m}\). The complete proof of Theorem 1.3 is given in the appendix.
We now argue that the worst case PoA is in \(\Omega ({2^p}/{p})\). By substituting the expression of l in the denominator of (2), we obtain
\begin{equation} 1+ l^2\sqrt [2^{{p}}]{r} - rl - \sqrt [2^{{p}}]{r} +r = 1 - \frac{1}{4}r^{2 - \frac{1}{2^{{p}}}} + r - r^{\frac{1}{2^{{p}}}}. \end{equation}
(27)
Since \(r,l \in [0,1]\), we can upper bound the above expression with
\begin{align*} 1 + r - r^{\frac{1}{2^{{p}}}} &= 1 + \left(\frac{2}{2^{p+1} - 1}\right)^{\frac{2^{{p}}}{2^{{p}}-1}} - \left(\frac{2}{2^{p+1} - 1}\right)^{\frac{1}{2^{{p}}-1}}\\ &\le 1 + \left(\frac{2}{2^{p+1} - 1}\right)^{\frac{2^{{p}}}{2^{{p}}-1}} - \left(\frac{1}{2^{p}}\right)^{\frac{1}{2^{{p}}-1}} \le 1 - \left(1 - \frac{1}{2^{p}}\right)\left(\frac{1}{2^{p} - 1}\right)^{\frac{1}{2^{{p}}-1}}. \end{align*}
Finally, we have that \(\lim _{p\rightarrow \infty } \frac{1 - \left(1 - \frac{1}{2^{{p}}}\right)\left(\frac{2}{2^{{{p}}+1} - 1}\right)^{\frac{1}{2^{{p}}-1}}}{\frac{p}{2^{{p}}}} = 0\), proving that (27) is in \(O\left({p}/{2^p}\right)\), which implies that when N goes to infinity the PoA is at least in \(\Omega \left({2^p}/{p}\right)\).

4 Maximum Cost

In this section, we measure the social cost of a state P as the maximum players’ cost in P, and we derive an upper bound and a lower bound on the PoA with respect to this notion of cost. Recall that given any state P, \(\text{tot}(P)\) is the total cost of P and \(\max (P)\) is the maximum cost of a player in P.
We first prove the upper bound on the PoA stated in Theorem 1.4, which is restated next for convenience.
Theorem 1.3.
In a symmetric (unweighted) network congestion game on a series-parallel \((s,t)\)-network with delays functions in class \(\mathcal {D}\), the PoA w.r.t. the maximum players’ cost is at most \(z(\mathcal {D})y(\mathcal {D})\).
Proof.
Let \(P_{o}\) be the social optimum with respect to the total cost, and let \(P_{\hat{o}}\) be the social optimum with respect to the maximum cost. Let \(P_f= \lbrace p^1_f, \ldots , p^N_f \rbrace\) be an arbitrary PNE. We will show that \(\max (P_f) \le z(\mathcal {D})y(\mathcal {D})\max (P_{\hat{o}})\).
Because \(P_f\) is a PNE and \(\max (P_f)\) is the cost of a player, we have \(\max (P_f) \le \text{cost}^+_f(p^i_f)\) for any \(i \in [N]\). Moreover, by (3), we have \(\text{cost}^+_f(p^i_f) \le z(\mathcal {D}) \text{cost}_f(p^i_f)\). In other words, the most expensive path in \(P_f\) has cost no greater than \(z(\mathcal {D})\) times the cost of any other path in \(P_f\). Thus, we can conclude that
\begin{equation*} N\cdot \max (P_f) \le \sum _{i=1}^N \text{cost}^+_f(p^i_f) \le z(\mathcal {D}) \sum _{i=1}^N \text{cost}_f(p^i_f) = z(\mathcal {D})\text{tot}(f), \end{equation*}
i.e., the most expensive path in \(P_f\) has cost no greater than \(z(\mathcal {D})\) times the average players’ cost in \(P_f\). Moreover,
\begin{align} z(\mathcal {D})\text{tot}(P_f) &\le z(\mathcal {D})y(\mathcal {D})\text{tot}(P_o) \end{align}
(28)
\begin{align} &\le z(\mathcal {D})y(\mathcal {D})\text{tot}(P_{\hat{o}}) \end{align}
(29)
\begin{align} &\le z(\mathcal {D})y(\mathcal {D})(N\cdot \max (P_{\hat{o}})) . \end{align}
(30)
Inequality (28) directly follows Theorem 1.2. Inequality (29) holds since \(P_o\) is the social optimum state with respect to the total cost, which implies that \(\text{tot}(P_o) \le \text{tot}(P_{\hat{o}})\). Inequality (30) holds because \(\max (P_{\hat{o}})\) is the maximum player’s cost in \(P_{\hat{o}}\). □
We now consider the class \(\operatorname{Poly-}p\) of polynomial delays with nonnegative coefficients and maximum degree p, and we prove that \(z(\operatorname{Poly-}p)\) is at most \(2^p\).
Lemma 4.1.
For the class of polynomial delay functions \(\operatorname{Poly-}p\) it holds that \(z(\operatorname{Poly-}p) \le 2^{p}\).
Proof.
By the definition of \(z(\operatorname{Poly-}p)\) in (3) we have that for any \(x \in {\mathbb {N}}^+\)
\[\begin{eqnarray*} z(\operatorname{Poly-}p) &= \max _{x \in {\mathbb {N}}^+} \frac{\sum _{j=0}^pa_j (x+1)^j}{\sum _{j=0}^pa_j x^j} \nonumber \nonumber \end{eqnarray*}\]
Note that given two collections of nonnegative real numbers \(b_0,\ldots , b_p\) and \(c_0,\ldots , c_p\), we have
\begin{equation*} \frac{\sum _{j=0}^pb_j}{\sum _{j=0}^pc_j} \le \max _{j=0,\ldots ,p} \frac{b_j}{c_j}. \end{equation*}
Thus,
\begin{align*} z(\operatorname{Poly-}p) &= \max _{x \in {\mathbb {N}}^+} \frac{\sum _{j=0}^pa_j (x+1)^j}{\sum _{j=0}^pa_j x^j} \le \max _{x \in {\mathbb {N}}^+} \max _{j=0,\ldots ,p} \frac{a_j (x+1)^j}{a_j x^j} \le 2^p. \end{align*}
 □
Finally, we prove that, for any class of delay functions, and as long as the network’s structure is preserved under series compositions, any lower bound on the PoA with respect to the total social cost is also valid when measuring the social cost in terms of the maximum players’ cost.
Theorem 1.4.
Let \(\mathcal {N}\) be a class of networks closed under series compositions and let G be a network in \(\mathcal {N}\). Then the worst-case PoA with respect to the maximum social cost of a symmetric (unweighted) network congestion game defined over G is at least the worst-case PoA with respect to the total social cost.
Proof.
We start with an instance of an atomic, unweighted, symmetric network congestion game on a \((s,t)\)-network G, where \(P_f\) is a PNE, \(P_o\) is a social optimum with respect to the total players’ cost, and the PoA is \({\text{cost}(P_f)}/{\text{cost}(P_o)}\). Our goal is to construct a new instance on a network \(G^{\prime }\), and to define a PNE \(P_{f^{\prime }}\) and a social optimum \(P_{o^{\prime }}\) with respect to the maximum players’ cost, such that
\begin{equation*} \frac{\max (P_{f^{\prime }})}{\max (P_{o^{\prime }})} = \frac{\text{cost}(P_f)}{\text{cost}(P_o)}. \end{equation*}
We construct \(G^{\prime }\) as follows. First, let \(G_1, \ldots , G_N\) be N duplicates of G and let \(G^{\prime }\) be the \((s,t)\)-network obtained by composing in series \(G_1,\ldots , G_N\). We remark that any network structure possessed by G is still valid for \(G^{\prime }\), by our assumption. Let \(P_f = \lbrace p^1_{f}, \ldots , p^N_{f} \rbrace\) and \(P_{o} = \lbrace p^1_{o}, \ldots , p^N_{o} \rbrace\). For each \(i\in [N]\), let \(P_{f_i} = \lbrace p^1_{f_i}, \ldots , p^N_{f_i} \rbrace\) and \(P_{o_i} = \lbrace p^1_{o_i}, \ldots , p^N_{o_i} \rbrace\) be the corresponding duplicates of \(P_f\) and \(P_o\) in \(G_i\), respectively. For each player \(i \in [N]\), we define the strategy \(p^i_{f^{\prime }}\) of player i in \(P_{f^{\prime }}\) by having the player choose the path \(p^{j(i)}_{f_j}\) in \(G_j\), where \(j(i)=(i+N-1)\mod {N}\). For example, the strategy of player 2 in \(P_{f^{\prime }}\) is obtained by composing in series the paths \(p_{f_1}^2, p_{f_2}^3, \ldots , p_{f_{N-1}}^N, p_{f_N}^1\). Analogously, we define the strategy \(p^i_{o^{\prime }}\) of player i in \(P_{o^{\prime }}\) by having the player choose the path \(p^{j(i)}_{o_j}\) in \(G_j\). It can be checked that \(P_{f^{\prime }} = \lbrace p^1_{f^{\prime }}, \ldots , p^N_{f^{\prime }} \rbrace\) is a PNE for the new instance defined on \(G^{\prime }\) (otherwise we would contradict that f is a PNE in the original instance). Similarly, it can be checked that \(P_{o^{\prime }} = \lbrace p^1_{o^{\prime }}, \ldots , p^N_{o^{\prime }} \rbrace\) is the social optimum in \(G^{\prime }\) with respect to the total cost (otherwise, we would contradict that o is a social optimum in the original instance).
Observe that, since in our construction we are permuting the players’ strategies, all the players have the same cost, both in \(P_{f^{\prime }}\) and in \(P_{o^{\prime }}\). Moreover, this cost is equal to \(\text{tot}(P_f)\) in \(P_{f^{\prime }}\) and to \(\text{tot}(P_o)\) in \(P_{o^{\prime }}\). Thus, \(\max (P_{f^{\prime }}) = \text{tot}(P_f)\) and \(\max (P_{o^{\prime }}) = \text{tot}(P_o)\). Now let \(\hat{f}\) and \(\hat{o}\) be the worst PNE and the social optimum in the new instance. We conclude that
\begin{equation*} \frac{\text{tot}(P_f)}{\text{tot}(P_o)} = \frac{\max (P_{f^{\prime }})}{\max (P_{o^{\prime }})} \le \frac{\max (P_{\hat{f}})}{\max (P_{\hat{o}})}, \end{equation*}
which implies the statement of this theorem. □

5 Conclusion

Our contributions fill a gap in the literature on the PoA of atomic, unweighted, symmetric network congestion games. We have investigated the impact of both symmetry and network structure on the worst-case PoA in network congestion games. Previous works had either addressed asymmetric games over general networks, or symmetric games over very simple network structures, such as parallel-link networks and extension-parallel networks. First, we considered symmetric network congestion games over arbitrary networks and showed that when the delay functions are polynomial, the worst-case PoA is very close to that of asymmetric network congestion games. This implies that symmetry does not influence the worst-case PoA significantly.
Then, we considered the class of series-parallel networks, corresponding to graphs with treewidth 2. These networks arise in many applications and understanding how their structure impacts the PoA in network congestion games could be the first step towards relating the worst-case PoA to the treewidth parameter. Our results indicate that, when restricting from symmetric games over arbitrary networks to symmetric games over series-parallel networks, the worst-case PoA significantly drops. On the other hand, the worst-case PoA quickly degrades when going from extension-parallel to series-parallel networks.
In this article, we have focused on symmetric games, but it is not clear if network structure could affect the PoA in asymmetric games. An open question is: in asymmetric (unweighted) congestion games over other special network structures, for example series-parallel networks, is the PoA still significantly smaller than that over arbitrary networks?

A Appendix

We here provide the complete proof of Theorem 1.3, which is restated next for convenience.
Theorem 1.3.
The worst-case PoA w.r.t. the total players’ cost of a symmetric (unweighted) network congestion game on a series-parallel \((s,t)\)-network, where the delay functions are polynomials with non-negative coefficients and highest degree p, is at least
\begin{equation*} \frac{1}{1+ l^2\sqrt [2^{{p}}]{r} - rl - \sqrt [2^{{p}}]{r} +r}, \end{equation*}
where \(r = (\frac{2}{2^{p+1}-1}) ^{\frac{2^{{p}}}{2^{{p}}-1}}\) and \(l = \frac{1}{2}r^{1 - \frac{1}{2^{{p}}}}\).
Consider the construction described in Section 3.3, which is represented in Figure 2. We will first show that this construction satisfies the properties stated in the next two lemmas.
Fig. 2.
Fig. 2. Given an input sequence \(\lbrace q_1, q_2, q_3\rbrace\), \(p\in {\mathbb {N}}^+\) and \(m=2\), we first average the first m numbers and get \(\lbrace s_1,s_2,s_3\rbrace\), where \(s_1 = s_2 = \frac{q_1+q_2}{2}\), \(s_3 = q_3\) and \(\bar{s} = \frac{s_1 + s_2 + s_3}{3}\). (2(f)) is the output network G and its corresponding PNE flow f. (2(c)), (2(d)), (2(e)) are the intermediate networks and flows according to our construction. (2(g)) is the flow h defined in the proof of Theorem 1.3 where \(k=1\).
Lemma A.1.
The \((s,t)\)-flow f has an \((s,t)\)-path \(\bar{p}\) with flow value m and \(\text{cost}_f(\bar{p}) = s_1\).
Proof.
We prove the lemma by induction on \(i \in \lbrace m,\ldots , N\rbrace\). The base case is \(i =m\). In this case, \(f_i = f_m\) is a flow of value m on a single \((s,t)\)-arc with delay function \(\frac{s_1 x}{m}\). The path \(\bar{p}^m\) defined by this arc has cost \(\text{cost}_{f_m}(\bar{p}^m) = s_1\).
Suppose that for each \(m \le i \lt N\) it holds that \(f_i\) has an \((s,t)\)-path \(\bar{p}^i\) with flow value m and \(\text{cost}_{f_i}(\bar{p}^i) = s_1\). We first construct \(\tilde{f}_i\) by composing in parallel \(f_i\) and a new \((s,t)\)-arc. Clearly, \(\bar{p}^i\) has still flow value m and \(\text{cost}_{\tilde{f}_i}(\bar{p}^i) = s_1\). Then we compose in series \(i+1\) copies of flow \(\tilde{f}_i\) to get \(f_{i+1}\) and we divide the delay functions by \(i+1\). The new \((s,t)\)-path \(\bar{p}^{i+1}\) is obtained by composing in series \(i+1\) copies of \(\bar{p}^i\). By construction, this path has flow value m and \(\text{cost}_{f_{i+1}}(\bar{p}^{i+1}) = s_1\). □
Lemma A.2.
The \((s,t)\)-flow f has cost 1, and it can be decomposed into N \((s,t)\)-paths \(\lbrace p^1,\ldots , p^N\rbrace\) that define a PNE in G. Moreover \(\text{cost}_f(p^i) = 1/N\) for all \(i\in [N]\), i.e., each player incurs the same cost.
Proof.
First, we show that \(f_N\) has cost \(\sum _{i=1}^N s_i=1\) and it can be decomposed into a PNE in \(G_N\) where each player incurs the same cost. We show this by induction on i. When \(i=m\), \(G_m\) is a single \((s,t)\)-arc, and \(f_m\) is an \((s,t)\)-flow of value m routed through this arc. Moreover, \(\text{cost}(f_m) = \frac{s_1 m}{m}m = \sum _{i=1}^m s_i\). Note that we cannot define any alternative flow in \(G_m\). Moreover, \(f_m\) admits a unique decomposition into N \((s,t)\)-paths, thus, \(f_m\) is a PNE flow where each player uses the same arc and incurs the same cost.
Now we assume that when \(i=k\), \(f_k\) has cost \(\sum _{i=1}^k s_i\), and it can be decomposed into a PNE in \(G_k\) where each player incurs the same cost. Our goal is to prove that the same holds for \(i = k+1\). Note that in our construction first we define \(\tilde{G}_k\) and \(\tilde{f}_k\) by composing in parallel \(f_k\) and a new \((s,t)\)-arc with delay \(s_{k+1} x\) and flow value 1. Thus, we first show that \(\tilde{f}_k\) is a PNE flow in \(\tilde{G}_k\). By the inductive hypothesis, flow \(f_k\) can be decomposed into a PNE in \(G_k\) where each player’s cost is \(\frac{1}{k}{\sum _{i=1}^k s_i}\). To define a decomposition of \(\tilde{f}_k\), we augment the decomposition of \(f_k\) by appending the extra \((s,t)\)-arc used to construct \(\tilde{G}_k\). Clearly, \(\text{cost}(\tilde{f}_{k}) = \text{cost}(f_{k}) + s_{k+1} = \sum _{i=1}^{k+1}s_i\). Moreover, (i) no player paying \(\frac{1}{k}{\sum _{i=1}^k s_i}\) has an incentive to deviate, since \(2^ps_{k+1} \ge \frac{1}{k} {\sum _{i=1}^k s_i}\), and (ii) the player paying \(s_{k+1}\) does not deviate since \(s_{k+1}\) is the minimum cost \((s,t)\)-path in \(\tilde{f}_k\). This shows that \(\tilde{f}_k\) is a PNE flow in \(\tilde{G}_k\). Recall that in our construction we define \(G_{k+1}\) and \(f_{k+1}\) by composing in series \(k+1\) copies of \(\tilde{G}_k\) with flow \(\tilde{f}_k\), and we divide all the delay functions by \(k+1\). Clearly, \(\text{cost}(f_{k+1})=\text{cost}(\tilde{f}_k)= \sum _{i=1}^{k+1} s_i\). We define a decomposition of \(f_{k+1}\) into \(k+1\) \((s,t)\)-paths as follows. Since there are \(k+1\) players and \(k+1\) identical copies of \(\tilde{G}_k\) composed in series, we let each player choose their original strategy in \(f_k\) in k components, and choose the extra arc used to define \(\tilde{G}_k\) in one component. Thus, in this decomposition of \(f_{k+1}\) each player incurs the same cost and no player has an incentive to deviate from their strategy.
Finally, we show that \(f=f_N\) is a PNE flow on G and \(\text{cost}(f) = \text{cost}(f_N) = \sum _1^N s_i = 1\). Recall that we construct G by composing in parallel \(G_N\) and m new \((s,t)\)-arcs \(e_1,\ldots , e_m\) with delay function \(\frac{1}{N} x^p\). Since in f every player incurs a cost equal to \(\frac{1}{N}\), no player has an incentive to deviate to an arc \(e_i\), \(i \in [m]\). Thus, f is a PNE flow on G. □
Define \(\mu (m,N) = \prod _{j=m}^{N-1} \frac{2^{p} j}{2^{p} j+1}\). We will need the results stated in the next two lemmas.
Lemma A.3.
Let \(\lbrace q_1, \ldots , q_N \rbrace\) be an ordered sequence of positive numbers such that \(\sum _{i=1}^N q_i=1\) and \(q_{i+1} = \frac{1}{2^p}\sum _{j=1}^i \frac{q_j}{i}\) for \(i \in [N-1]\). Then for every \(m \in [N]\) we have \(\sum _{i=1}^m q_i = \mu (m,N)\).
Proof.
We proceed by induction on m. The base case is \(m = N-1\). Since \(q_N = \frac{1}{2^p(N-1)}\sum _{j=1}^{N-1} q_j\), we have:
\begin{equation} \sum _{j=1}^{N-1} q_j = 1 - q_{N} = 1 - \frac{1}{2^p(N-1)}\sum _{j=1}^{N-1} q_j. \end{equation}
(31)
By Equation (31), we have \(\frac{2^p(N-1) + 1}{2^p(N-1)}\sum _{j=1}^{N-1} q_j \le 1\). This implies that \(\sum _{j=1}^{N-1} q_j \le \frac{2^p(N-1)}{2^p(N-1)+1} = \mu (N-1, N)\). Thus, the statement holds for the base case.
Next we assume that the statement holds for \(m \in \lbrace k, \ldots , N-1\rbrace\), and we prove that it also holds for \(m = k-1\). Based on our inductive hypothesis, \(\sum _{j=1}^{k} q_j \le \mu (k, N)\). Moreover, since \(q_k =\frac{1}{2^p(k-1)}\sum _{j=1}^{k-1} q_j\), we have
\begin{equation} \sum _{j=1}^{k-1} q_j = \sum _{j=1}^{k} q_j - q_{k} = \mu (k, N) - \frac{1}{2^p(k-1)}\sum _{j=1}^{k-1} q_j. \end{equation}
(32)
According to (32), we have \(\frac{2^p(k-1) + 1}{2^p(k-1)}\sum _{j=1}^{k-1} q_j = \mu (k, N)\). This implies that \(\sum _{j=1}^{k-1} q_j = \frac{2^p(k-1)}{2^p(k-1)+1}\mu (k,N) = \mu (k-1, N)\). Thus, the statement holds. □
Lemma A.4.
For \(m \in [N-1]\) we have \(\sqrt [2^p]{\frac{2^pm- (2^p- 1)}{2^pN- (2^p- 1)}} \le \mu (m,N)\).
Proof.
First we can equivalently write
\begin{equation*} \mu (m,N)= \prod _{j=m}^{N-1} \frac{2^{p}j}{2^{p}+1} = \sqrt [2^p]{\left(\prod _{j=m}^{N-1} \frac{2^{p}j}{2^{p}j+1}\right)^{2^p}}. \end{equation*}
We lower bound the argument of the square root as follows:
\begin{align*} \prod _{j=m}^{N-1} \left(\frac{2^{p}j}{2^{p}j+1}\right)^{2^p} &\ge \prod _{j=m}^{N-1} \left(\prod _{k=0}^{2^p- 1}\frac{2^{p}j - k}{2^{p}j+1 - k}\right) \\ &= \frac{2^{p}m-(2^p- 1)}{2^{p}N-(2^p- 1)}. \end{align*}
 □
We will now use the results stated in the above lemmas to prove Theorem 1.3.
Proof of Theorem 1.3. Consider the network congestion game on the network G defined above. By Lemma A.1, f has an \((s,t)\)-path \(\bar{p}\) with flow value m and \(\text{cost}_f(\bar{p}) = s_1\). For each arc e in \(\bar{p}\), let \(a_e x^p\) be the delay function of e. Note that \(\text{cost}_f(\bar{p}) = \sum _{e \in \bar{p}} a_{e}m = s_1\) implies that \(\sum _{e\in \bar{p}}a_e = \frac{s_1}{m}\). Recall that \(r=\frac{m}{n}\) and \(l = \frac{k}{m}\). Define h as the flow obtained from f by moving a subflow of value \((m-k)\) from p to the \((s,t)\)-arcs \(e_1,\ldots , e_{m-k}\), which have all delay function \(\frac{1}{N} x^p\). Then by construction we have
\begin{align} \text{cost}(f) - \text{cost}(h) &= m\text{cost}_f(\bar{p}) - \left(k \text{cost}_h(\bar{p}) + (m-k)\frac{1}{N}\right)\nonumber \nonumber\\ &= {s_1}{m} - \left(\frac{s_1}{m}k^2 + (m-k)\frac{1}{N}\right) \end{align}
(33)
\begin{align} &= \left(\frac{s_1}{m}m^2 - \frac{s_1}{m}k^2 - \frac{m-k}{m}m s_1\right) + \frac{m-k}{m}\left(m s_1 - \frac{m}{N}\right) \nonumber \nonumber\\ &= \left(\frac{s_1}{m}mk - \frac{s_1}{m}k^2\right) + \frac{m-k}{m}\left(\sum _{i=1}^m s_i - \frac{m}{N}\right) , \end{align}
(34)
where equality (33) holds since \(\sum _{e\in \bar{p}}a_e = \frac{s_1}{m}\). Equality (34) holds since the first m of \(s_i\) are equal.
By Lemma A.3 and Lemma A.4, we have
\begin{equation} \sum _{i=1}^m q_i - \frac{m}{N} = \mu (m,N) - \frac{m}{N} \ge \left[(\sqrt [2^{{p}}]{r} - r) - \epsilon \right]. \end{equation}
(35)
Now observe that
\begin{align} \frac{s_1}{m}m^2 =ms_1 = \frac{m}{N} + \left(\sum _1^m s_i - \frac{m}{N}\right) \ge r + \left[\left(\sqrt [2^{{p}}]{r} - r\right) - \epsilon \right] = \left(\sqrt [2^{{p}}]{r} - \epsilon \right), \end{align}
(36)
where the inequality follows from (35) and the fact that \(\sum _1^m s_i = \sum _1^m q_i\).
This implies
\begin{equation} {\frac{s_1}{m}mk - \frac{s_1}{m}k^2} = \left(l - l^2\right) {\frac{s_1}{m}m^2} \ge \left(l - l^2\right) \left(\sqrt [2^{{p}}]{r}- \epsilon \right), \end{equation}
(37)
where the inequality follows from (36).
From (34) and (37)we obtain
\begin{align} \text{cost}\left(f\right) - \text{cost}\left(h\right) &\ge \left(l - l^2\right)\left(\sqrt [2^{{p}}]{r}- \epsilon \right) + \left(1-l\right)\left(\sum _1^m s_i - \frac{m}{N}\right) \nonumber \nonumber\\ &\ge \left(l - l^2\right)\left(\sqrt [2^{{p}}]{r}- \epsilon \right) + \left(1-l\right)\left[\left(\sqrt [2^{{p}}]{r} - r\right) - \epsilon \right], \end{align}
(38)
where inequality (38) follows from (35). By Lemma A.2 we know that \(\text{cost}(f) = \sum _1^N s_i = 1\), thus, we obtain
\begin{align} \text{cost}\left(h\right) &\le 1- \left(l - l^2\right)\left(\sqrt [2^{{p}}]{r}- \epsilon \right) - \left(1-l\right)\left[\left(\sqrt [2^{{p}}]{r} - r\right) - \epsilon \right] \nonumber \nonumber\\ &= {1 + l^2 \sqrt [2^{{p}}]{r} - rl - \sqrt [2^{{p}}]{r} + r + \left(1-l^2\right)\epsilon } \end{align}
(39)
To obtain an upper bound on \(\text{cost}(h)\) we minimize the right-hand-side of (39) with respect to r and l. Observe that \(\epsilon \rightarrow 0\) when \(N \rightarrow \infty\), thus, we solve
\begin{align*} \min \;\; & l^2 \sqrt [2^{{p}}]{r} - rl - \sqrt [2^{{p}}]{r} + r \\ \text{s.t.} & r \in [0, 1),l \in [0,1], \end{align*}
which is achieved at \(r = (\frac{2}{2^{p+1}-1}) ^{\frac{2^{{p}}}{2^{{p}}-1}}\), \(l = \frac{1}{2}r^{1 - \frac{1}{2^{{p}}}}\). Since \(\frac{\text{cost}(f)}{\text{cost}(o)} \ge \frac{\text{cost}(f)}{\text{cost}(h)}\), we obtain a lower bound for the PoA. \(\square\)

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cover image ACM Transactions on Economics and Computation
ACM Transactions on Economics and Computation  Volume 12, Issue 3
September 2024
124 pages
EISSN:2167-8383
DOI:10.1145/3613631
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Publication History

Published: 06 September 2024
Online AM: 28 June 2024
Accepted: 03 May 2024
Revised: 23 February 2024
Received: 23 February 2023
Published in TEAC Volume 12, Issue 3

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  1. Congestion games
  2. series-parallel networks
  3. price of anarchy

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