Transfer Implementation in Congestion Games

Abstract

We study an implementation problem faced by a planner who can influence selfish behavior in a roadway network. It is commonly known that Nash equilibrium does not necessarily minimize the total latency on a network and that levying a tax on road users that is equal to the marginal congestion effect each user causes implements the optimal latency flow. This holds, however, only under the assumption that taxes have no effect on the utility of the users. In this paper, we consider taxes that satisfy the budget balance condition and that are therefore obtained using a money transfer among the network users. Hence at every state the overall taxes imposed upon the users sum up to zero. We show that the optimal latency flow can be guaranteed as a Nash equilibrium using a simple, easily computable transfer scheme that is obtained from a fixed matrix. In addition, the resulting game remains a potential game, and the levied tax on every edge is a function of its congestion.

Appendix: Bounded Implementation

Let $Q$ be the matrix guaranteed by Theorem 1 and let $F^{Q,M}$ be the game obtained when the transfer to any edge $e$ is bounded by $M>0$. The new payoff from using edge $e$ is then,

$$\begin{aligned} F^{Q,M}_e(x)=l_e(x_e)+\sum _{f\ne e} q_{f}-\min \left\{ \left( q_e\frac{1-x_e}{x_e}\right) ,M\right\} . \end{aligned}$$

We note that the function $F^{Q,M}_e$ is Lipschitz continuous for every edge $e$. This in turn determines a Lipschitz continuous function for every route $i\in \mathcal {P}$.

For every edge $e$ such that $q_e>0,$ let $x^M_e=\frac{q_e}{M+q_e}$ and let $c_e=x^M_e(q_e+M)-q_e\ln (x^M_e).$ For every such $e$ define the function $\eta _e(x_e)$ as follows:

$$\begin{aligned} \eta _e(x_e)= {\left\{ \begin{array}{ll} x_e\Big (\sum _f q_f\Big )-q_e\ln (x_e) \quad \text { if }x_e\ge x^M_e\\ x_e\Big (\sum _{f\ne e} q_f\Big )-x_eM+c_e\ \ \text { if }x_e<x^M_e. \end{array}\right. } \end{aligned}$$

(7)

Note that by the choice $c_e$ the function $\eta _e(x_e)$ is continuous and differentiable. For any edge $e$ for which $q_e=0$ let $\eta _e(x_e)=x_e(\sum _{f\ne e} q_f)$. The resulting differentiable potential function $h^M$ of the game $F^{Q,M}$ may be defined by the functions $\{\eta _e(x_e)\}_{e\in E}$ as follows:

$$\begin{aligned} h^M(x)=\sum _{e\in E}\left[ \int _0^{x_e}l_e(x)\mathrm {d}x+\eta _e(x_e)\right] . \end{aligned}$$

Let $X^M=\{ x\in X : \forall e\in E,\ x_e\ge x_e^M \}.$ Note that $X^M$ approaches $X$ (in the Hausdorff distance) as $M$ grows. Since $\tau _e(x_e)$ are weakly convex, it follows that the potential $h^M$ is a weakly convex function and is strictly convex over $X^M$. Hence, as in Theorem 1, there exists $M_0>0$ such that the optimal latency $x^*$ is the unique equilibrium of $F^{Q,M}$ for every $M\ge M_0$.

The game $F^{Q,M}$ no longer satisfies the budget balance condition. That is, if we let $\tau ^M_e(x_e)=\sum _{f\ne e} q_{f}-\min \{(q_e\frac{1-x_e}{x_e}),M\}$ be the levied tax on edge $e$ in the game $F^{Q,M},$ then for $x\not \in X^M$ it holds that⁸

$$\begin{aligned} V^M(x)=\sum _{e\in E}x_e \tau ^M_e(x_e)>0. \end{aligned}$$

The budget balance condition does hold for any $x\in X^M$. Let $|\mathcal {P}|=m$. A revision protocol $\rho$ is a Lipschitz function,

$$\begin{aligned} \rho :X\times \mathbb {R}^m\rightarrow \mathbb {R}^{m\times m}_+. \end{aligned}$$

For each vector payoff $\pi \in \mathbb {R}^m$ over the routes, every state $x,$ and all pairs of distinct routes $i,j\in \mathcal {P},$ the function $\rho _{ij}(x,\pi )$ determines the switching rate of revision from route $i$ to route $j$.⁹ Any revision protocol and initial state $y\in \Delta (\mathcal {P})$ determines a differential equation $z:\mathbb {R}_+\rightarrow Y$ that describes the learning process as follows:

$$\begin{aligned} \forall i\in \mathcal {P},\ \dot{z}_i(t)=\sum _{j\in \mathcal {P}}z_j(t)\rho _{ji}(z(t),F^{Q,M}(z(t)))-z_i(t)\rho _{ij}(z(t),F^M(z(t))). \end{aligned}$$

(8)

$\{z(t)\}_{t\ge 0}$ naturally defines a flow on edges $\{x(t)\}_{t\ge 0}$.

Consider,

$$\begin{aligned} \int _{0}^\infty V^M(x(t))\mathrm {d}t. \end{aligned}$$

(9)

This expression represents the tax that is lost during the learning process. Under mild conditions on the revision protocol, the flow $\{x(t)\}_{t\ge 0}$ defined by Eq. (8) converges to the equilibrium $x^*$ of $F^{Q,M}$ for every $M\ge M_0$.¹⁰ Therefore, the flow $\{x(t)\}_{t\ge 0}$ must enter $X^M$ in a bounded time, independently of the initial conditions. Since $X^M$ approaches $X$ for large $M,$ this bounded time must approach zero as $M$ grows. Since $V^M(x)$ is bounded and zero on $X^M$, it must be the case that (9) goes to zero as $M$ increases.