A decomposition method for a class of convex generalized Nash equilibrium problems

Abstract

In this paper, we study a numerical approach to compute a solution of the generalized Nash equilibrium problem (GNEP). The GNEP is a potent modeling tool that has been increasingly developing in recent decades. Much of this development has centered around applying variational methods to the so-called GNSC, a useful but restricted subset of GNEP where each player has the same constraint set. One popular approach to solve the GNSC is to use the apparent separability of each player to build a decomposition method. This method has the benefit of being easily implementable and can be parallelized. Our aim in this paper is to show an extension of the decomposition method to a class of convex GNEP. We prove convergence of the proposed algorithm under a full convexity assumption. Then, we show numerical results on some examples to validate our approach and discuss the assumptions.

Appendices

Appendix 1: Proof of Theorem 1

Theorem 5

Let $\nu \in \{1,\dots ,N\}$. Assume that the GNEP satisfies Assumption 2.

Suppose that the tolerance and penalty parameters satisfy $\epsilon _k \rightarrow 0^+$, $\eta _k \rightarrow 0^+$ and $\rho _k \rightarrow 0^+$. Let $\{(x^k,\tilde{x}^k,\alpha ^k)\}$ be a sequence satisfying for all k: $x^k\in X,\tilde{x}^k \in X$, $\min (x^k-l_b,-\alpha ^k)=0$, $\min (u_b-x^k,\alpha ^k)=0$, $\Vert Q_\nu (x^k,\tilde{x}^k,\eta _k,\rho _k,\alpha ^k) \Vert \le \epsilon _k$ and $\Vert x^k - \tilde{x}^{k} \Vert \le C\epsilon _k$ for some constant $C>0$.

Then, if a limit point $x^*$ of the sequence $\{x^k\}$ is infeasible for (5) (i.e. $c^\nu (x^*)\ne 0$), it is a stationary point of the function

$$\Vert h^\nu (x),\max (g^\nu (x),0)\Vert ^2.$$

On the other hand, if a limit point $x^*$ is feasible for (5) (i.e. $c(x^*)=0$) and the constraint gradients

$$\begin{aligned}&\{\nabla g^\nu _{i}(x^*)\,(i \in \mathcal {I}_\nu (x^*)), \nabla x_i\,(i \in \mathcal {A}(x^*)) \} \\&\quad \cup \{\{\nabla h^\nu _{i}(x^*)\,(i \in \mathcal {E}_\nu (x^*)), \nabla x^{-\nu }_i\,(i=1,\dots ,n_{-\nu })\}\} \end{aligned}$$

are positively linearly independent, then $x^*$ is a KKT point for the problem (5). For such points, we have for any infinite subsequence $\mathcal {K}$ such that $\lim _{k \in \mathcal {K}} x^k =\lim _{k \in \mathcal {K}} \tilde{x}^k =x^*$ that $\lim _{k \in \mathcal {K}} \alpha ^k =\alpha ^*$ and

$$\begin{aligned} \lim \limits _{k\in \mathcal {K}} \,&\frac{\max (0,g^\nu _{i}(x^k))}{\eta _k} = \lambda _i^*, \ i=1,\dots ,m_\nu ,\nonumber \\ \lim \limits _{k\in \mathcal {K}} \,&\frac{h^\nu _{i}(x^k)}{\eta _k} = \mu _i^*, \ i=1,\dots ,p_\nu , \nonumber \\ \lim \limits _{k\in \mathcal {K}}\,&\frac{\tilde{x}_i^{k,-\nu } - x^{k,-\nu }_i}{\rho _k} = \xi _i^*, \ i=1,\dots ,n, \end{aligned}$$

(10)

where $(\lambda ^*,\mu ^*,\xi ^*,\alpha ^*)$ is a Lagrange multiplier for (5).

Proof

The condition $\Vert Q_\nu (x^k,\tilde{x}^k,\eta _k,\rho _k,\alpha ^k) \Vert \le \epsilon _k$ yields

$$\begin{aligned}&\left\| \nabla \theta _\nu (x^k)+ \sum _{i \notin \mathcal {I}_\nu (x^k)} \frac{\max (g^\nu _{i}(x^k),0)}{\eta _k}\nabla g^\nu _{i}(x^k) \right. +\sum _{i\notin \mathcal {E}_\nu (x^k)} \frac{h^\nu _{i}(x^k)}{\eta _k}\nabla h^\nu _{i}(x^k) \\&\qquad \qquad \left. + \sum _{i=1}^{n_{-\nu }} \frac{\tilde{x}^{k,-\nu }_i-x^{k,-\nu }_i}{\rho _k} \nabla x^{-\nu }_i + \alpha ^k\right\| \le \epsilon _k. \end{aligned}$$

By rearranging this expression (and in particular using $\Vert a\Vert -\Vert b\Vert \le \Vert a+b\Vert$) we obtain

$$\begin{aligned}&\left\| \sum _{i \notin \mathcal {I}_\nu (x^k)} \right. \left. \max (g^\nu _{i}(x^k),0) \nabla g^\nu _{i}(x^k) \right. +\sum _{i\notin \mathcal {E}_\nu (x^k)} h^\nu _{i}(x^k)\nabla h^\nu _{i}(x^k) \\&\qquad \qquad \left. + \sum _{i=1}^{n_{-\nu }} (\tilde{x}^{k,-\nu }_i-x^{k,-\nu }_i) \nabla x^{-\nu }_i \right\| \le \min (\eta _k,\rho _k)(\epsilon _k+\Vert \nabla \theta _\nu (x^k)+ \alpha ^k\Vert ). \end{aligned}$$

Let $x^*$ be a limit point of the sequence $\{ x^k \}$. Then there is a subsequence $\mathcal {K}$ such that $\lim _{k \in \mathcal {K}} x^k=\lim _{k \in \mathcal {K}} \tilde{x}^k=x^*$. When we take limits as $k\rightarrow \infty$ for $k \in \mathcal {K}$, the right-hand side approaches zero, because $(\eta _k,\rho _k) \rightarrow 0^+$ and $\epsilon _k \rightarrow 0^+$. From the corresponding limit on the left-hand side we obtain

$$\begin{aligned} \sum _{i \notin \mathcal {I}_\nu (x^*)} \max (g^\nu _{i}(x^*),0)\nabla g^\nu _{i}(x^*)+\sum _{i\notin \mathcal {E}_\nu (x^*)} h^\nu _{i}(x^*)\nabla h^\nu _{i}(x^*)=0. \end{aligned}$$

(11)

We can have $x^*$ infeasible, but in this case (11) implies that $x^*$ is a stationary point of the function $\Vert h_\nu (x),\max (g_\nu (x),0)\Vert ^2$.

If, on the other hand, the constraints gradient are positively linearly independent at a limit point $x^*$, we have from (11) that $x^*$ is feasible. Hence, the second and third condition in the KKT system (4) are satisfied. We need to check the first condition as well, and to show that the limit (10) holds. By definition of $Q_\nu$ we have that

$$\begin{aligned} \nabla \theta _\nu (x^k)-Q_\nu (x^k,\tilde{x}^k,\eta _k,\rho _k,\alpha ^k)=&\sum _{i \notin \mathcal {I}_\nu (x^k)} \lambda ^k_i\nabla g^\nu _{i}(x^k) \\&+\sum _{i\notin \mathcal {E}_\nu (x^k)} \mu ^k_i\nabla h^\nu _{i}(x^k)+ \sum _{i=1}^{n_{-\nu }} \xi _i^k \nabla x^{-\nu }_i + \alpha ^k, \end{aligned}$$

where $\lambda ^k_i=\frac{\max (g^\nu _{i}(x^k),0)}{\eta _k}$, $\mu ^k_i=\frac{h^\nu _{i}(x^k)}{\eta _k}$ and $\xi _i^k=\frac{\tilde{x}^{k,-\nu }_i-x^{k,-\nu }_i}{\rho _k}$. Let us prove by contradiction that the sequence $\{\lambda ^k,\mu ^k,\xi ^k,\alpha ^k \}$ is bounded. If it were not bounded, there would exist a subsequence such that

$$\begin{aligned} \frac{(\lambda ^k,\mu ^k,\xi ^k,\alpha ^k)}{\Vert \lambda ^k,\mu ^k,\xi ^k,\alpha ^k \Vert _\infty } \rightarrow (\lambda ,\mu ,\xi ,\alpha ) \ne 0 \end{aligned}$$

Dividing by $\Vert \lambda ^k,\mu ^k,\xi ^k,\alpha ^k \Vert _\infty$ and passing to the limit in the equation above,

$$\begin{aligned} 0= \sum _{i \notin \mathcal {I}_\nu (x^*)} \lambda _i\nabla g^\nu _{i}(x^*) +\sum _{i\notin \mathcal {E}_\nu (x^*)} \mu _i\nabla h^\nu _{i}(x^*)+ \sum _{i=1}^{n_{-\nu }} \xi _i \nabla x^{-\nu }_i + \alpha . \end{aligned}$$

By the positive linear independence assumption it follows that $(\lambda ,\mu ,\xi ,\alpha )=0$, which leads to a contradiction. Therefore, the sequence $\{\lambda ^k,\mu ^k,\xi ^k,\alpha ^k \}$ is bounded. Hence by taking the limit as $k \in \mathcal {K}$ goes to $\infty$, we find that $\lim _{k \in \mathcal {K}} \alpha ^k =\alpha ^*$ and

$$\begin{aligned} \lim \limits _{k\in \mathcal {K}}&\frac{\max (0,g^\nu _{i}(x^k))}{\eta _k} = \lambda _i^*, \ i=1,\dots ,m_\nu ,\\ \lim \limits _{k\in \mathcal {K}}&\frac{h^\nu _{i}(x^k)}{\eta _k} = \mu _i^*, \ i=1,\dots ,p_\nu , \\ \lim \limits _{k\in \mathcal {K}}&\frac{\tilde{x}_i^{k,-\nu } - x^{k,-\nu }_i}{\rho _k} = \xi _i^*, \ i=1,\dots ,n. \end{aligned}$$

By taking limits in $\Vert Q_\nu (x^k,\tilde{x}^k,\eta _k,\rho _k,\alpha ^k) \Vert \le \epsilon _k$, we conclude that

$$\begin{aligned} -\nabla \theta _\nu (x^*)= \sum _{i \notin \mathcal {I}_\nu (x^*)} \lambda ^*_i\nabla g^\nu _{i}(x^*) +\sum _{i\notin \mathcal {E}_\nu (x^*)} \mu ^*_i\nabla h^\nu _{i}(x^*)+ \sum _{i=1}^{n_{-\nu }} \xi _i^* \nabla x^{-\nu }_i + \alpha ^*. \end{aligned}$$

Hence, $x^*$ is a KKT point for (5) with Lagrange multiplier $(\lambda ^*,\mu ^*,\xi ^*,\alpha ^*)$. $\square$

Appendix 2: Notations

We sum up here the notations used throughout the paper. We refer the reader to the Sect. 2 for notations relative to the GNEP (Table 2).

Table 2 Notations