A nonmonotone trust-region method for generalized Nash equilibrium and related problems with strong convergence properties

Abstract

The generalized Nash equilibrium problem (GNEP) is often difficult to solve by Newton-type methods since the problem tends to have locally nonunique solutions. Here we take an existing trust-region method which is known to be locally fast convergent under a relatively mild error bound condition, and modify this method by a nonmonotone strategy in order to obtain a more reliable and efficient solver. The nonmonotone trust-region method inherits the nice local convergence properties of its monotone counterpart and is also shown to have the same global convergence properties. Numerical results indicate that the nonmonotone trust-region method is significantly better than the monotone version, and is at least competitive to an existing software applied to the same reformulation used within our trust-region framework. Additional tests on quasi-variational inequalities (QVI) are also presented to validate efficiency of the proposed extension.

A Proofs of Lemma 1 and Proposition 1

We first recall some elementary properties of the projection operator.

Lemma 3

The following statements hold:

(a) :

$\big ( P_{\varOmega } (z) - z)^T \big ( P_{\varOmega } (z) - x \big ) \le 0 \quad \forall x \in \varOmega , \ \forall z \in \mathbb {R}^n$;

(b) :

$\Vert P_{\varOmega }(x_2) - P_{\varOmega }(x_1) \Vert \le \Vert x_2 - x_1 \Vert \quad \forall x_1,x_2 \in \mathbb {R}^n.$

(c) :

Given $x, d \in \mathbb {R}^n$, the function

$$\begin{aligned} \theta (t) := \frac{\big \Vert P_{\varOmega } (x+ t d) - x \big \Vert }{t}, \quad t > 0, \end{aligned}$$

is nonincreasing.

The first two properties in Lemma 3 are a well-known characterization of the projection (see [4]), whereas the third property was shown, e.g., in [3] in the context of a suitable globalization of a projected gradient method.

Proof of Lemma 1

Let $k \in \mathbb {N}$ be fixed, choose $\Delta > 0$, and define

$$\begin{aligned} z_k := x_k + d_k^G (\Delta ) = x_k - \frac{\Delta }{\Delta _{\max }} \gamma _k \nabla \varPsi (x_k) \end{aligned}$$

for the sake of notational convenience. Then an elementary calculation yields

$$\begin{aligned} \nabla \varPsi (x_k)^T \bar{d}_k^G ( \Delta )= & {} \nabla \varPsi (x_k)^T \big ( P_{\varOmega } (z_k) - x_k \big ) \\= & {} \frac{\Delta _{\max }}{\Delta \cdot \gamma _k} \big ( x_k - z_k \big )^T\big ( P_{\varOmega }(z_k) - x_k \big ) \\= & {} \frac{\Delta _{\max }}{\Delta \cdot \gamma _k} \big ( P_{\varOmega } (z_k)- z_k \big )^T \big ( P_{\varOmega } (z_k) - x_k \big ) \\&+\,\,\frac{\Delta _{\max }}{\Delta \cdot \gamma _k} \big ( x_k - P_{\varOmega } (z_k) \big )^T \big ( P_{\varOmega } (z_k) - x_k \big ) \\\le & {} -\,\,\frac{\Delta _{\max }}{\Delta \cdot \gamma _k} \big \Vert \bar{d}_k^G ( \Delta ) \big \Vert ^2, \end{aligned}$$

where the inequality follows from Lemma 3(a), the definition of $\bar{d}_k^G (\Delta )$, and the feasibility of $x_k$. On the other hand, Lemma 3(c) with $d := -\, \nabla \varPsi (x_k)$ implies that

$$\begin{aligned} \frac{\big \Vert \bar{d}_k^G (\Delta ) \big \Vert }{\Delta }= & {} \frac{\left\| P_{\varOmega } \big ( x_k - \frac{\Delta }{\Delta _{\max }} \gamma _k \nabla \varPsi (x_k) \big ) - x_k \right\| }{\Delta } \\\ge & {} \frac{\left\| P_{\varOmega } \big (x_k - \gamma _k \nabla \varPsi (x_k) \big ) - x_k \right\| }{\Delta _{\max }} \\= & {} \frac{\left\| \bar{d}_k^G (\Delta _{\max }) \right\| }{\Delta _{\max }} \end{aligned}$$

holds for all $0 < \Delta \le \Delta _{\max }$. Combining the last two inequalities yields the assertion. $\square$

Proof of Proposition 1

Since $x_k\rightarrow x^*$ for $k\in K$ and $k\rightarrow \infty$, the continuity of $F'$ implies that there is a constant $b_1$ such that $\Vert F'(x_k)\Vert \le b_1$ for all $k \in K$. Using (4), (5), and Lemma 3(b), we therefore obtain for all $k\in K$

$$\begin{aligned} \begin{aligned} \big \Vert F'(x_k) \bar{d}_k^G(\Delta ) \big \Vert&= \big \Vert F'(x_k) \big ( P_{\varOmega }\big [x_k + d_k^G(\Delta ) \big ] - x_k\big ) \big \Vert \\&\le \big \Vert F'(x_k) \big \Vert \big \Vert x_k + d_k^G(\Delta ) - x_k \big \Vert \\&\le \frac{\Delta \gamma _k}{\Delta _{max} } \big \Vert F'(x_k) \big \Vert \big \Vert \nabla \varPsi (x_k) \big \Vert \\&\le b_1 \Delta , \end{aligned} \end{aligned}$$

(27)

where the last inequality follows from the definition of $\gamma _k$ in (S.2).

Since $x^*$ is not a stationary point by assumption, we can follow the argument from the first part of the proof of Theorem 1 in order to see that there is a constant $b > 0$ such that

$$\begin{aligned} \big \Vert \bar{d}_k^G(\Delta _{max}) \big \Vert \ge b \quad \forall k\in K, k \ge \hat{k}. \end{aligned}$$

(28)

Let

$$\begin{aligned} \tilde{\Delta } = \text {min} \bigg \{ \Delta _{max}, \frac{(1- \sigma )b^2}{b_1^2 \Delta _{max}} \bigg \}. \end{aligned}$$

(29)

We first prove that (6) holds for all $k\in K, k \ge \hat{k}$ and all $\Delta \in (0, \tilde{\Delta }]$. From the definition of $\bar{d}_k(\Delta )$, we get that

$$\begin{aligned} \begin{aligned} Pred_k(\Delta )&= \frac{1}{2} \big \Vert F(x_k) + F'(x_k) \bar{d}_k (\Delta ) \big \Vert ^2- \varPsi (x_k)\\&\le \frac{1}{2} \big \Vert F(x_k) + F'(x_k) \bar{d}_k^G (\Delta ) \big \Vert ^2 - \varPsi (x_k)\\&= \nabla \varPsi (x_k)^T \bar{d}_k^G (\Delta ) + \frac{1}{2} \big \Vert F'(x_k)\bar{d}_k^G (\Delta ) \big \Vert ^2 \\&= \sigma \nabla \varPsi (x_k)^T \bar{d}_k^G (\Delta ) + (1-\sigma ) \nabla \varPsi (x_k)^T\bar{d}_k^G (\Delta ) + \frac{1}{2} \big \Vert F'(x) \bar{d}_k^G (\Delta ) \big \Vert ^2 \\&\le \sigma \nabla \varPsi (x_k)^T\bar{d}_k^G (\Delta ) - (1-\sigma ) \Big (\frac{\Delta }{\Delta _{max} \gamma _k} \Big ) \big \Vert \bar{d}_k^G (\Delta _{max}) \big \Vert ^2 + \frac{1}{2}b_1^2\Delta ^2\\&\le \sigma \nabla \varPsi (x_k)^T\bar{d}_k^G (\Delta ) - b_1^2 \Delta \tilde{\Delta } + \frac{1}{2}b_1^2\Delta ^2\\&\le \sigma \nabla \varPsi (x_k)^T\bar{d}_k^G (\Delta ), \end{aligned} \end{aligned}$$

where the second inequality follows directly from Lemma 1 and (27), the third inequality follows from (28) and (29) and recalling that $0 < \gamma _k\le 1$, while the last inequality holds since $\Delta \le \tilde{\Delta }.$

In order to prove that (7) holds for $k\in K$ and k sufficiently large and for $\Delta$ belonging to an interval $(0,\bar{\Delta }]$, we will first show that

$$\begin{aligned} - Pred_k(\Delta ) \ge \beta \Delta , \end{aligned}$$

(30)

and

$$\begin{aligned} Ared_k(\Delta ) - Pred_k(\Delta ) \le c_1 \Delta ^2 \end{aligned}$$

(31)

hold for suitable constants $\beta > 0$ and $c_1 > 0$

First we show that (30) holds. To this aim, taking $\Delta \in (0,\tilde{\Delta }]$, using Lemma 1 and (27), we can write

$$\begin{aligned} \begin{aligned} \frac{1}{2} \big \Vert F(x_k) + F'(x_k)\bar{d}_k^G(\Delta ) \big \Vert ^2&= \frac{1}{2} \Vert F(x_k) \Vert ^2 + \nabla \varPsi (x_k)^T \bar{d}_k^G(\Delta ) + \frac{1}{2} \big \Vert F'(x_k) \bar{d}_k^G(\Delta ) \big \Vert ^2\\&\le \varPsi (x_k) - \Big ( \frac{\Delta }{\gamma _k \Delta _{max}} \Big ) \big \Vert \bar{d}_k^G(\Delta _{max}) \big \Vert ^2 + \frac{1}{2} b_1^2 \Delta ^2\\&\le \varPsi (x_k) - \Big ( \frac{\Delta }{\gamma _k \Delta _{max}} \Big ) \big \Vert \bar{d}_k^G(\Delta _{max}) \big \Vert ^2 + \frac{1}{2}\Delta \frac{b^2(1-\sigma )}{\Delta _{max}} \\&\le \varPsi (x_k) - \Big ( \frac{\Delta }{\gamma _k \Delta _{max}} \Big ) \big \Vert \bar{d}_k^G(\Delta _{max}) \big \Vert ^2 + \frac{1}{2}\Delta \frac{\big \Vert \bar{d}_k^G(\Delta _{max}) \big \Vert ^2}{\gamma _k \Delta _{max}}\\&= \varPsi (x_k) - \Big ( \frac{\Delta }{2 \gamma _k \Delta _{max}} \Big ) \big \Vert \bar{d}_k^G(\Delta _{max}) \big \Vert ^2, \end{aligned} \end{aligned}$$

where the second inequality follows from (29), and the third holds recalling that $(1- \sigma )< 1, \gamma _k \le 1$, and (28). Consequently, we have

$$\begin{aligned} Pred_k(\Delta )\le & {} \frac{1}{2} \big \Vert F(x_k) + F'(x_k)\bar{d}_k^G(\Delta ) \big \Vert ^2 - \varPsi (x_k) \\\le & {} - \,\bigg ( \frac{\Delta }{2 \gamma _k \Delta _{max}} \bigg ) \big \Vert \bar{d}_k^G(\Delta _{max}) \big \Vert ^2 < 0, \end{aligned}$$

where the first inequality follows from the definitions of $\bar{d}_k$ and $t^*(\Delta )$ in (S.4). Thus, using (28) and recalling again that $\gamma _k \le 1$, we obtain that there exists $\beta > 0$ such that (30) is satisfied for all $k\in K, k \ge \hat{k}$ and all $\Delta \in (0, \tilde{\Delta }]$.

Now we prove (31). From Lemma 3(b), recalling the definitions of $d_k^G(\Delta )$, $\bar{d}_k^G(\Delta )$ and $\gamma _k$, we have

$$\begin{aligned} \big \Vert \bar{d}_k^G(\Delta ) \big \Vert \le \big \Vert d_k^G(\Delta ) \big \Vert \le \Delta , \quad \forall \Delta \in (0, \Delta _{max}]. \end{aligned}$$

From the definition of $\bar{d}_k^{tr}(\Delta )$, using Lemma 3(b) again, and recalling that $d_k^{tr}(\Delta )$ is the trust-region solution, we have

$$\begin{aligned} \big \Vert \bar{d}_k^{tr}(\Delta ) \big \Vert \le \big \Vert d_k^{tr}(\Delta ) \big \Vert \le \Delta , \quad \forall \Delta \in (0, \Delta _{max}]. \end{aligned}$$

Consequently, from the last two inequalities we get $\Vert \bar{d}_k(\Delta ) \Vert \le \Delta$. Since $F'$ is locally Lipschitzian, it is globally Lipschitz on compact sets. Consequently, $\nabla \varPsi$ is also globally Lipschitz on compact sets. Since $x_k \rightarrow x^*$ for $k \in K$ and $\bar{d}_k(\Delta )$ is bounded for all $\Delta \in (0, \Delta _{\max } ]$, we can apply the Mean Value Theorem and obtain the existence of suitable numbers $\theta _k \in (0,1)$ and a Lipschitz constant $L > 0$ such that

$$\begin{aligned} \begin{aligned}&\varPsi \big ( x_k +\bar{d}_k(\Delta ) \big ) - \varPsi (x_k) - \nabla \varPsi (x_k)^T\bar{d}_k(\Delta )\\&\quad = \nabla \varPsi \big (x_k +\theta _k\bar{d}_k(\Delta ) \big )^T \bar{d}_k(\Delta )-\nabla \varPsi (x_k)^T\bar{d}_k(\Delta ) \\&\quad \le L \Delta \Vert \bar{d}_k(\Delta ) \Vert \end{aligned} \end{aligned}$$

for all $k \in K$ and all $\Delta \in (0, \Delta _{\max }]$, where the last inequality takes into account the Cauchy–Schwarz inequality. Hence, we can write

$$\begin{aligned} \begin{aligned} Ared_k(\Delta ) - Pred_k(\Delta )&= \varPsi \big ( x_k +\bar{d}_k(\Delta ) \big ) - \frac{1}{2} \big \Vert F(x_k) + F'(x_k)\bar{d}_k(\Delta ) \big \Vert ^2 \\&= \varPsi \big ( x_k +\bar{d}_k(\Delta ) \big ) - \varPsi (x_k) - \nabla \varPsi (x_k)^T\bar{d}_k(\Delta )\\&\quad -\,\,\frac{1}{2} \bar{d}_k(\Delta )^T F'(x_k)^T F'(x_k) \bar{d}_k(\Delta ) \\&\le L\Delta \big \Vert \bar{d}_k(\Delta ) \big \Vert - \frac{1}{2} \bar{d}_k(\Delta )^T F'(x_k)^T F'(x_k) \bar{d}_k(\Delta )\\&\le L \Delta ^2 + c_2 \big \Vert \bar{d}_k(\Delta ) \big \Vert ^2\\&\le c_1 \Delta ^2 \end{aligned} \end{aligned}$$

for suitable constants $c_1, c_2 > 0$.

Finally, exploiting (30) and (31), it follows that there exists $\bar{\Delta } > 0$ such that

$$\begin{aligned} \hat{r}_k =1 - \frac{Ared_k(\Delta ) - Pred_k(\Delta )}{- Pred_k(\Delta )} \ge \rho _1, \quad \forall k\in K, k \ge \hat{k} \text { and } \forall \Delta \in (0, \bar{\Delta }]. \end{aligned}$$

This concludes the proof. $\square$