A New Smoothing Conjugate Gradient Method for Solving Nonlinear Nonsmooth Complementarity Problems

Abstract

In this paper, by using the smoothing Fischer-Burmeister function, we present a new smoothing conjugate gradient method for solving the nonlinear nonsmooth complementarity problems. The line search which we used guarantees the descent of the method. Under suitable conditions, the new smoothing conjugate gradient method is proved globally convergent. Finally, preliminary numerical experiments show that the new method is efficient.

1. Introduction

We consider the nonlinear nonsmooth complementarity problem, which is to find a vector in $R^n$ satisfying the conditions

$$x\ge 0,F(x)\ge 0,x^{T}F(x)=0$$

(1)

where $F:R^n\to R^n$ is a locally Lipschitz continuous function. If F is continuously differentiable, then Problem (1) is called the nonlinear complementarity problems NCP(F). As we all know, Equation (1) is a very useful general mathematics model, which is closely related to the mathematical programming, variational inequalities, fixed point problems and mixed strategy problems (such as [1,2,3,4,5,6,7,8,9,10,11,12,13]). The methods for solving NCP(F) are classified into three categories: nonsmooth Newton methods, Jacobian smoothing methods and smoothing methods (see [14,15,16,17,18,19]). Conjugate gradient methods are widely and increasingly used for solving unconstrained optimization problem, especially in large-scale cases. There are few scholar has investigated the problem how to use the conjugate gradient method to solve NCP(F) (such as [10,20]). Moreover, in these papers, F is required to be a continuous differentiable $P_0+R_0$ function. In this paper, we present a new smoothing conjugate gradient method for solving Equation (1), where F is only required to be a locally Lipschitz continuous function.

In this paper, we also define the generalized gradient of F at x is

$$\partial F(x)=conv\{\underset{x_k⟶x,x_k\in D_F}{lim}▽F(x_k)\}$$

where $\prime \prime conv\prime \prime$ denotes the convex hull of a set, $D_F$ denotes the set of points at which F is differentiable (see [21]). In the following, we introduce the definition of the smoothing function.

Definition 1 

(see [22]) Let $F:R^n⟶R^n$ be a locally Lipschitz continuous function. We call $\tilde{F}:R^n\times R_{+}⟶R^n$ is a smoothing function of F, if $\tilde{F}(x,\mu )$ is continuously differentiable in $R^n$ for any fixed $\mu >0$, and

$$\underset{\mu ⟶0}{lim}\tilde{F}(x,\mu )=F(x)$$

for any fixed $x\in R^n$. If

$$\underset{x_k⟶x,\mu \downarrow 0}{lim}▽\tilde{F}(x_k,\mu )\in \partial F(x)$$

for any $x_k\in R^n$, we say F satisfies gradient consistency property.

In the following sections of our paper, we also use the Fischer-Burmeister function (see [23]) and the smoothing Fischer-Burmeister function. (1) The Fischer-Burmeister function

$$\phi (a,b)=\sqrt{a^2+b^2}-a-b,{(a,b)}^T\in R^2$$

where $\phi :R^2⟶R$. From the definition of $\phi$, we know that it is twice continuously differentiable besides ${(0,0)}^T$. Moreover, it is a complementarity function, which satisfies

$$\phi (a,b)=0⟺a\ge 0,b\ge 0,ab=0$$

Denote

$$H(x)=\left(\begin{array}{c}\hfill \phi (x_1,F_1(x))\hfill \\ \hfill ⋮\hfill \\ \hfill \phi (x_n,F_n(x))\hfill \end{array}\right)$$

It is obvious that $H(x)$ is zero at a point x if and only if x is a solution of Equation (1). Then Equation (1) can be transformed into the following unconstrained optimization problem

$$min\Psi (x)=\frac{1}{2}{\parallel H(x)\parallel }^2$$

We know that the optimal value of $\Psi$ is zero, and $\Psi$ is called the value function of Equation (1).

(2) The smoothing Fischer-Burmeister function

$${\phi }_{\mu }(a,b)=\sqrt{a^2+b^2+\mu }-a-b$$

where $\phi :R^3⟶R$ and $\mu >0.$

Let

$$H_{\mu }(x)=\left(\begin{array}{c}\hfill {\phi }_{\mu }(x_1,{\tilde{F}}_1(x,\mu ))\hfill \\ \hfill ⋮\hfill \\ \hfill {\phi }_{\mu }(x_n,{\tilde{F}}_n(x,\mu ))\hfill \end{array}\right)$$

$${\Psi }_{\mu }(x)=\frac{1}{2}{\parallel H_{\mu }(x)\parallel }^2$$

where ${\tilde{F}}_i(x,\mu )$ is the smoothing function of $F_i(x),i=1,\cdots ,n$.

The rest of this work is organized as follows. In Section 2, we describe the new smoothing conjugate gradient method for the solution of Problem (1). It is shown that this method has global convergence properties under fairly mild assumptions. In Section 3, preliminary numerical results and some discussions for this method are presented.

2. The New Smoothing Conjugate Gradient Method and its Global Convergence

The new smoothing conjugate gradient direction is defined as

$$d_k=\left\{\begin{array}{cc}-\nabla {\Psi }_{{\mu }_0}(x_0),\hfill & k=0\hfill \\ -\nabla {\Psi }_{{\mu }_{k-1}}(x_k)+{\beta }_k{d}_{k-1},\hfill & k\ge 1\hfill \end{array}\right.$$

(2)

where ${\beta }_k$ is a scalar. Here, we use ${\beta }_k$ (see [24]) which is defined as

$${\beta }_k^{DY}=\frac{\parallel \nabla {\Psi }_{{\mu }_{k-1}}(x_k){\parallel }^2}{d_{k-1}^T{y}_{k-1}}$$

(3)

where $y_{k-1}=\nabla {\Psi }_{{\mu }_{k-1}}(x_k)-\nabla {\Psi }_{{\mu }_{k-2}}(x_{k-1})$. When $k=1$, we set ${\mu }_{k-2}={\mu }_0$. The line search is Armijo-type line search (see [25]), where ${\alpha }_k={\eta }^{m_k},0<\eta <1,m_k$ is the smallest nonnegative integer satisfies

$${\Psi }_{{\mu }_k}(x_k+{\alpha }_k{d}_k)\le {\Psi }_{{\mu }_k}(x_k)+\delta {\alpha }_k{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k$$

(4)

$${(\nabla {\Psi }_{{\mu }_k}(x_k+{\alpha }_k{d}_k))}^T{d}_{k+1}\le -\sigma {\parallel \nabla {\Psi }_{{\mu }_k}(x_k+{\alpha }_k{d}_k)\parallel }^2,0<\sigma \le 1,0<\delta <1$$

(5)

Then, we give the new smoothing conjugate gradient method for solving Equation (1).

Algorithm 1: Smoothing Conjugate Gradient Method

(S.0) Choose $x_0\in R^n,\epsilon >0,{\mu }_0>0,m>0,\sigma ,\delta ,m_1\in (0,1),$$d_0=-\nabla {\Psi }_{{\mu }_0}(x_0)$. Set $k=0$.

(S.1) If $\Psi (x_k)\le \epsilon$, then stop, otherwise go to Step 2.

(S.2) Compute ${\alpha }_k$ by Equations (4) and (5), where $d_{k+1}=-\nabla {\Psi }_{{\mu }_k}(x_k+{\alpha }_k{d}_k)+{\beta }_{k+1}{d}_k$ and ${\beta }_{k+1}$ is given by Equation (3). Let $x_{k+1}=x_k+{\alpha }_k{d}_k$.

(S.3) If $\parallel \nabla {\Psi }_{{\mu }_k}(x_{k+1})\parallel \ge m{\mu }_k$, then set ${\mu }_{k+1}={\mu }_k$, otherwise set ${\mu }_{k+1}=m_1{\mu }_k$.

(S.4) Let $k=k+1$, go back to Step 1.

Algorithm 2: Algorithm Framework of Algorithm 1

PROGRAM ALGORITHM

INITIALIZE $x_0\in R^n,\epsilon >0,{\mu }_0>0,m>0,m_1\in (0,1)$;

Set $k=0$ and $d_0=-\nabla {\Psi }_{{\mu }_0}(x_0)$.

WHILE the termination condition

$\Psi (x_k)\le \epsilon$ is not met

Find step sizes ${\alpha }_k$;

Set $x_{k+1}=x_k+{\alpha }_k{d}_k$

Evaluate $\nabla {\Psi }_{{\mu }_k}(x_{k+1})$ and $d_{k+1}$;

IF $\parallel \nabla {\Psi }_{{\mu }_k}(x_{k+1})\parallel \ge m{\mu }_k$ THEN

Set ${\mu }_{k+1}={\mu }_k$;

ELSE

Set ${\mu }_{k+1}=m_1{\mu }_k$;

END IF

Set $k=k+1$;

END WHILE

RETURN final solution $x_k$;

END ALGORITHM

In the following, we will give the analysis about the global convergence of Algorithm 1. (The Algorithm 2 is the algorithm framework of Algorithm 1.) Before doing this work, we need the following basic assumptions.

Assumption 1.

(i) For any $x\in R^n$, $0<\mu \le {\mu }_0$, the level set $L_{\mu }(c)=\{x\in R^n|{\Psi }_{\mu }(x)\le c\}$ is bounded.

(ii) $\nabla {\Psi }_{{\mu }_k}(x_{k+1})$ is Lipschitz continuous, that is, there exists a constant $L>0$ such that

$$\parallel \nabla {\Psi }_{{\mu }_k}(x_{k+1})-\nabla {\Psi }_{{\mu }_{k-1}}(x_k)\parallel \le L\parallel x_{k+1}-x_k\parallel ,\forall x_{k+1},x_k\in L_{\mu }(c)$$

Lemma 1. 

Suppose that $\left\{d_k\right\}$ is an infinite sequence of directions generated by Algorithm 1, then

$$-{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k\ge \overline{c}{\parallel \nabla {\Psi }_{{\mu }_{k-1}}(x_k)\parallel }^2,\forall k\ge 0,\overline{c}<\sigma$$

(6)

Proof 

$\mathrm{If}$ $k=0$, by Equation (2) and $\overline{c}<1$, we can know that Equation (6) holds. If $k>0$, by Equation (5) and $\overline{c}<\sigma$, we can conclude that Equation (6) holds.

Lemma 2. 

Suppose that Assumption 1 holds. Then, there exists ${\alpha }_k>0$ for every k, and

$${\alpha }_k\ge \omega \frac{|{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k|}{\parallel d_k{\parallel }^2}$$

(7)

with ω is a positive constant.

Proof 

$\mathrm{From}$ Step 0 of Algorithm 1, we know that $d_0=-\nabla {\Psi }_{{\mu }_0}(x_0)$, i.e., $d_0$ is a descent direction. Suppose that $d_k$ is satisfied

$${(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k\le -\sigma {\parallel \nabla {\Psi }_{{\mu }_{k-1}}(x_k)\parallel }^2\le 0$$

(8)

for any ${\tilde{\alpha }}_k$. We denote

$${\tilde{x}}_{k+1}=x_k+{\tilde{\alpha }}_k{d}_k$$

$${\tilde{d}}_{k+1}=-\nabla {\Psi }_{{\mu }_k}({\tilde{x}}_{k+1})+{\tilde{\beta }}_{k+1}{d}_k$$

(9)

By

$$\begin{gathered} {(\nabla {\Psi }_{{\mu }_k}({\tilde{x}}_{k+1}))}^T{\tilde{d}}_{k+1}=-\parallel \nabla {\Psi }_{{\mu }_k}({\tilde{x}}_{k+1}){\parallel }^2+{\tilde{\beta }}_{k+1}[{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k+ \\ {(\nabla {\Psi }_{{\mu }_k}({\tilde{x}}_{k+1})-\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k] \end{gathered}$$

(10)

We know that ${\beta }_k$ in Equation (3) is equivalent to (see [24])

$${\beta }_k=\frac{{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k}{{(\nabla {\Psi }_{{\mu }_{k-2}}(x_{k-1}))}^T{d}_{k-1}}>0$$

(11)

Since Assumption 1, Equations (10) and (11) yield

$${(\nabla {\Psi }_{{\mu }_k}({\tilde{x}}_{k+1}))}^T{\tilde{d}}_{k+1}\le -\sigma {\parallel \nabla {\Psi }_{{\mu }_k}({\tilde{x}}_{k+1})\parallel }^2,{\tilde{\alpha }}_k\in (0,\frac{|{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k|}{L\parallel d_k{\parallel }^2})$$

(12)

by Mean Value Theorem, we have

$$\begin{array}{cc}\hfill {\Psi }_{{\mu }_k}({\tilde{x}}_{k+1})-{\Psi }_{{\mu }_k}(x_k)& ={\int }_0^1{\tilde{\alpha }}_k{(\nabla {\Psi }_{{\mu }_k}(x_k+t{\tilde{\alpha }}_k{d}_k))}^T{d}_{k}dt\hfill \\ \hfill ={\tilde{\alpha }}_k{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T& d_k+{\int }_0^1{\tilde{\alpha }}_k{[\nabla {\Psi }_{{\mu }_k}(x_k+t{\tilde{\alpha }}_k{d}_k)-\nabla {\Psi }_{{\mu }_{k-1}}(x_k)]}^T{d}_{k}dt\hfill \\ & \le {\tilde{\alpha }}_k{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k+{\int }_0^{1}L{\tilde{\alpha }}_k^2{\parallel d_k\parallel }^{2}tdt\hfill \\ & \le {\tilde{\alpha }}_k{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k+\frac{1}{2}L{\tilde{\alpha }}_k^2{\parallel d_k\parallel }^2\hfill \end{array}$$

Then, we obtain that

$$\begin{gathered} {\Psi }_{{\mu }_k}({\tilde{x}}_{k+1})-{\Psi }_{{\mu }_k}(x_k)\le \delta {\tilde{\alpha }}_k{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k, \\ \forall {\tilde{\alpha }}_k\in (0,\frac{2(1-\delta )}{L}\frac{|{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k|}{\parallel d_k{\parallel }^2}) \end{gathered}$$

(13)

By Equations (12) and (13), we know that Equations (4) and (5) determine a positive stepsize ${\alpha }_k$. And there must exists a constant $\xi \in (0,1)$ yields

$$\xi ·\frac{|{(\nabla {\Psi }_{{\mu }_{k-1}}(x_k))}^T{d}_k|}{L\parallel d_k{\parallel }^2}<1$$

Denote $\omega =min\{\frac{\xi }{L},\frac{2\xi (1-\delta )}{L}\}$, then Equation (7) holds. And Equation (5) implies that Equation (8) holds for $k+1$. Hence, the proof is completed.

Theorem 1. 

Suppose that for any fixed $\mu >0$, ${\Psi }_{\mu }$ satisfies Assumption 1, then the infinite sequence $\left\{x_k\right\}$ generated by Algorithm 1 satisfies

$$\underset{k⟶\infty }{lim}{\mu }_k=0,\underset{k⟶\infty }{lim\; inf}\parallel \nabla {\Psi }_{{\mu }_{k-1}}(x_k)\parallel =0$$

(14)

Proof 

$\mathrm{Denote}$ $K=\left\{k\right|{\mu }_{k+1}=m_1{\mu }_k\}$, we first show that K is an infinite set.

If K is a finite set, there exists an integer $\overline{k}$ such that

$$\parallel \nabla {\Psi }_{{\mu }_{k-1}}(x_k)\parallel \ge m{\mu }_{k-1}$$

for all $k>\overline{k}$. We also have ${\mu }_k={\mu }_{\overline{k}}=:\overline{\mu }$ for all $k>\overline{k}$ and

$$\underset{k⟶\infty }{lim\; inf}\parallel \nabla {\Psi }_{\overline{\mu }}(x_k)\parallel >0$$

(15)

In the following, we will proof

$$\underset{k⟶\infty }{lim\; inf}\parallel \nabla {\Psi }_{\overline{\mu }}(x_k)\parallel =0$$

(16)

By Lemma 1 and Assumption 1, we know that $\left\{{\Psi }_{\overline{\mu }}(x_k)\right\}$ is a monotone decreasing sequence and the limit of $\left\{{\Psi }_{\overline{\mu }}(x_k)\right\}$ is exist. Summing Equation (7), we get

$$\sum _{k\ge \overline{k}+1}\frac{{(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k{)}^2}{\parallel d_k{\parallel }^2}<\infty$$

(17)

Due to Equation (2), we also have

$$d_k+\nabla {\Psi }_{\overline{\mu }}(x_k)={\beta }_k{d}_{k-1},\forall k\ge \overline{k}+1$$

(18)

Square both sides of Equation (18), we get

$$\parallel d_k{\parallel }^2={({\beta }_k)}^2\parallel d_{k-1}{\parallel }^2-2{(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k-{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k)\parallel }^2$$

(19)

Divided both sides of Equation (19) by ${({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2$, we have

$$\begin{array}{cc}\hfill \frac{\parallel d_k{\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}& =\frac{{({\beta }_k)}^2{\parallel d_{k-1}\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}-\frac{2{(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k}{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}-\frac{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k){\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}\hfill \\ & =\frac{{({\beta }_k)}^2{\parallel d_{k-1}\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}-{(\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k)\parallel }+\frac{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k)\parallel }{{(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k})}^2+\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k){\parallel }^2}\hfill \\ & \le \frac{{({\beta }_k)}^2{\parallel d_{k-1}\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}+\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k){\parallel }^2}\hfill \\ & =\frac{\parallel d_{k-1}{\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_{k-1}))}^T{d}_{k-1})}^2}+\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k){\parallel }^2}\hfill \\ & \le \frac{\parallel d_{k-2}{\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_{k-2}))}^T{d}_{k-2})}^2}+\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_{k-1}){\parallel }^2}+\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k){\parallel }^2}\hfill \\ & \le \cdots \hfill \\ & \le \frac{\parallel d_{\overline{k}+1}{\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_{\overline{k}+1}))}^T{d}_{\overline{k}+1})}^2}+\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_{\overline{k}+2}){\parallel }^2}+\cdots +\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_k){\parallel }^2}\hfill \end{array}$$

Denote

$$\frac{\parallel d_{\overline{k}+1}{\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_{\overline{k}+1}))}^T{d}_{\overline{k}+1})}^2}=\lambda >0$$

Then

$$\frac{\parallel d_k{\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}\le \lambda +\sum _{i=\overline{k}+2}^k\frac{1}{\parallel \nabla {\Psi }_{\overline{\mu }}(x_i){\parallel }^2}$$

(20)

If Equation (16) is not hold, there exists $\gamma >0$ such that

$$\parallel \nabla {\Psi }_{\overline{\mu }}(x_k)\parallel \ge \gamma ,\forall k>\overline{k}+1$$

(21)

We obtain from Equations (20) and (21) that

$$\frac{\parallel d_k{\parallel }^2}{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}\le \frac{\lambda {\gamma }^2+k-\overline{k}-1}{{\gamma }^2}$$

Because of

$$\frac{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}{\parallel d_k{\parallel }^2}\ge \frac{{\gamma }^2}{\lambda {\gamma }^2+k-\overline{k}-1}$$

provies

$$\sum _{k\ge \overline{k}+1}\frac{{({(\nabla {\Psi }_{\overline{\mu }}(x_k))}^T{d}_k)}^2}{\parallel d_k{\parallel }^2}=+\infty$$

which leads to a contradiction with Equation (17). This show that Equation (16) holds. There are conflicts between Equations (16) and (15). This show that K must be an infinite set and

$$\underset{k⟶\infty }{lim}{\mu }_k=0$$

(22)

Then, we can assume that $K=\{k_0,k_1,...\}$ with $k_0<k_1<...$ Hence, we get

$$\underset{i⟶\infty }{lim}\parallel \nabla {\Psi }_{{\mu }_{k_i}}(x_{k_i+1})\parallel \le m\underset{i⟶\infty }{lim}{\mu }_{k_i}=0$$

and completes the proof.

3. Numerical Tests

In this section, we intend to test the efficiency of Algorithm 1 by numerical experiments. We use Algorithm 1 to solve eleven examples, some of them are proposed the first time, some of them are modified by the examples of the references (such as [26,27]).

The smoothing function of F as ${\tilde{F}}_i(x,\mu )=\sqrt{F_i{(x)}^2+\mu }$ is used in solving Examples 1–4. From Example 5 to Example 11, the smoothing function of F is defined by (see [26]). $\tilde{F}(x,\mu )=\mu ln{\sum }_{i=1}^{m}exp(\frac{f_i(x)}{\mu }),$ where $F(x)=max\{f_1(x),\cdots ,f_m(x)\}$, $i=1,\cdots m$.

Throughout the experiments, we set $\sigma ={10}^{-2},m=1.5,m_1=0.5.$ In Examples 1–3 and Examples 5–8, we set $\epsilon ={10}^{-4},\delta ={10}^{-3},\eta =0.4,{\mu }_0=0.2$. Example 4, in which we set parameters $\epsilon ={10}^{-3},\delta ={10}^{-2},\eta =0.1,{\mu }_0=0.02$. In the case of Examples 9–11, we set $\epsilon ={10}^{-2},\delta ={10}^{-3},\eta =0.4,{\mu }_0=0.2$. We choose $\Psi (x)\le \epsilon$ as the termination criterion. Our numerical results are summaried in Table 1, Table 2, Table 3, Table 4, Table 5, Table 6, Table 7, Table 8, Table 9, Table 10 and Table 11, where all components of $x_0$ are randomly selected from 0 to 10. We randomly generate 10 initial points, then implement Algorithm 1 to solve the test problem for each initial point. By the numerical of results of Examples 10–11, we can see that Algorithm 1 is suitably to solve the large scale problems.

Example 1. 

We consider Equation (1), where F is defined by $F(x)=|2x-1|$. The exact solutions of this problem are 0 and $0.5$.

Table 1. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
0.9713	5.053658e−1	5.636783e−5	1
1.7119	4.977618e−1	9.929266e−6	11
2.7850	−1.295343e−2	8.495830e−5	8
3.1710	5.422178e−3	1.461954e−5	8
4.0014	5.562478e−3	1.538368e−5	8
5.4688	−7.521520e−3	2.849662e−5	7
6.5574	5.926470e−3	1.745635e−5	10
7.9221	1.276205e−2	8.037197e−5	7
8.4913	−1.994188e−3	1.992344e−6	7
9.3399	1.723553e−3	1.482749e−6	7

Table 1. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
0.9713	5.053658e−1	5.636783e−5	1
1.7119	4.977618e−1	9.929266e−6	11
2.7850	−1.295343e−2	8.495830e−5	8
3.1710	5.422178e−3	1.461954e−5	8
4.0014	5.562478e−3	1.538368e−5	8
5.4688	−7.521520e−3	2.849662e−5	7
6.5574	5.926470e−3	1.745635e−5	10
7.9221	1.276205e−2	8.037197e−5	7
8.4913	−1.994188e−3	1.992344e−6	7
9.3399	1.723553e−3	1.482749e−6	7

Example 2. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill |2x_1-1|\hfill \\ \hfill |4x_2+x_1-\frac{1}{2}|\hfill \end{array}\right).$ There are three exact solutions as ${(\frac{1}{2},0)}^T,{(0,\frac{1}{8})}^T$ and ${(0,0)}^T$.

Table 2. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
${(4.6939,0.1190)}^T$	${(-0.0058,0.0031)}^T$	2.162427e−5	7
${(5.2853,1.6565)}^T$	${(0.0077,0.1233)}^T$	2.974655e−5	13
${(9.9613,0.7818)}^T$	${(0.0030,0.1248)}^T$	7.106107e−6	5
${(4.9836,9.5974))}^T$	${(0.4979,-0.0097)}^T$	6.744680e−5	12
${(1.4495,8.5303)}^T$	${(0.4978,0.0080)}^T$	3.327241e−5	13
${(0.4965,9.0272)}^T$	${(0.0045,0.1221)}^T$	3.348105e−5	15
${(9.1065,1.8185)}^T$	${(-0.0045,0.1296)}^T$	9.857281e−5	6
${(4.0391,0.9645)}^T$	${(0.0087,0.1247)}^T$	6.417282e−5	10
${(7.7571,4.8679)}^T$	${(0.0045,-0.0043)}^T$	1.946526e−5	13
${(7.0605,0.3183)}^T$	${(0.0086,0.122)}^T$	4.037476e−5	8

Table 2. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
${(4.6939,0.1190)}^T$	${(-0.0058,0.0031)}^T$	2.162427e−5	7
${(5.2853,1.6565)}^T$	${(0.0077,0.1233)}^T$	2.974655e−5	13
${(9.9613,0.7818)}^T$	${(0.0030,0.1248)}^T$	7.106107e−6	5
${(4.9836,9.5974))}^T$	${(0.4979,-0.0097)}^T$	6.744680e−5	12
${(1.4495,8.5303)}^T$	${(0.4978,0.0080)}^T$	3.327241e−5	13
${(0.4965,9.0272)}^T$	${(0.0045,0.1221)}^T$	3.348105e−5	15
${(9.1065,1.8185)}^T$	${(-0.0045,0.1296)}^T$	9.857281e−5	6
${(4.0391,0.9645)}^T$	${(0.0087,0.1247)}^T$	6.417282e−5	10
${(7.7571,4.8679)}^T$	${(0.0045,-0.0043)}^T$	1.946526e−5	13
${(7.0605,0.3183)}^T$	${(0.0086,0.122)}^T$	4.037476e−5	8

Example 3. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill |5x_1+x_2-x_3|\hfill \\ \hfill x_1^2+4x_2-x_3-2\hfill \\ \hfill 5x_2^2-6x_1-2x_3\hfill \end{array}\right).$ ${(0,\frac{1}{2},0)}^T$ is one of the exact solutions of this problem.

Table 3. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
${(1.9175,7.3843,2.4285)}^T$	${(0.0087,0.5021,-0.0026)}^T$	9.785244e−5	21
${(1.1921,9.3983,6.4555)}^T$	${(0.0047,0.5019,0.0102)}^T$	6.577107e−5	25
${(1.8687,4.8976,4.4559)}^T$	${(-0.0055,0.4974,-0.0109)}^T$	7.552798e−5	19
${(2.7029,2.0846,5.6498)}^T$	${(0.0099,0.4998,0.0028)}^T$	5.759549e−5	26
${(7.2866,7.3784,0.6340)}^T$	${(0.0099,0.5003,-0.0063)}^T$	9.612693e−5	36
${(1.2991,5.6882,4.6939)}^T$	${(-0.0115,0.5009,0.0065)}^T$	9.216436e−5	31
${(5.3834,9.9613,0.7818)}^T$	${(0.0022,0.4952,-0.0083)}^T$	9.798255e−5	26
${(9.5613,5.7521,0.5978)}^T$	${(0.0089,0.5029,0.0074)}^T$	7.570081e−5	28
${(7.7571,4.8679,4.3586)}^T$	${(0.0048,0.5025,-0.0008)}^T$	6.774521e−5	24
${(3.8827,5.5178,2.2895)}^T$	${(-0.0047,0.4981,-0.0111)}^T$	7.903206e-5	25

Table 3. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
${(1.9175,7.3843,2.4285)}^T$	${(0.0087,0.5021,-0.0026)}^T$	9.785244e−5	21
${(1.1921,9.3983,6.4555)}^T$	${(0.0047,0.5019,0.0102)}^T$	6.577107e−5	25
${(1.8687,4.8976,4.4559)}^T$	${(-0.0055,0.4974,-0.0109)}^T$	7.552798e−5	19
${(2.7029,2.0846,5.6498)}^T$	${(0.0099,0.4998,0.0028)}^T$	5.759549e−5	26
${(7.2866,7.3784,0.6340)}^T$	${(0.0099,0.5003,-0.0063)}^T$	9.612693e−5	36
${(1.2991,5.6882,4.6939)}^T$	${(-0.0115,0.5009,0.0065)}^T$	9.216436e−5	31
${(5.3834,9.9613,0.7818)}^T$	${(0.0022,0.4952,-0.0083)}^T$	9.798255e−5	26
${(9.5613,5.7521,0.5978)}^T$	${(0.0089,0.5029,0.0074)}^T$	7.570081e−5	28
${(7.7571,4.8679,4.3586)}^T$	${(0.0048,0.5025,-0.0008)}^T$	6.774521e−5	24
${(3.8827,5.5178,2.2895)}^T$	${(-0.0047,0.4981,-0.0111)}^T$	7.903206e-5	25

Example 4. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill |2x_1-x_2+3x_3+2x_4-6|\hfill \\ \hfill 3x_1-3x_2+3x_3+2x_4-5\hfill \\ \hfill 3x_1-x_2-x_3+2x_4-3\hfill \\ \hfill 3x_1-x_2+3x_3-x_4-4\hfill \end{array}\right).$ ${(\frac{31}{13},\frac{22}{13},0,\frac{19}{13})}^T$, ${(\frac{7}{4},0,0,\frac{5}{4})}^T$, ${(0,0,\frac{11}{5},\frac{13}{5})}^T$, ${(3,0,0,0)}^T$ are four of the exact solutions of this problem.

Table 4. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
${(5.6743,9.6878,8.2450,9.5961)}^T$	${(0.0062,-0.0249,2.1692,2.5652)}^T$	4.436641e-4	21
${(0.1485,1.5669,4.7157,5.4299)}^T$	${(0.0009,0.0320,2.2173,2.6291)}^T$	5.987181e−4	37
${(0.5969,6.5803,8.8964,1.0963)}^T$	${(3.0522,0.0031,-0.0202,-0.0151)}^T$	3.790585e−4	23
${(8.7494,1.2100,8.5635,8.9978)}^T$	${(0.0296,-0.0290,2.1300,2.5000)}^T$	9.638295e−4	17
${(7.7836,0.6937,2.7878,3.7937)}^T$	${(3.0003,0.0096,0.0154,-0.0158)}^T$	3.052065e−4	13
${(0.6837,0.8497,0.6834,4.0982)}^T$	${(3.0055,0.0333,-0.0050,0.0131)}^T$	7.098848e−4	21
${(7.6034,5.8410,4.0295,5.1004)}^T$	${(2.4048,1.7076,-0.0096,1.4753)}^T$	4.242339e−4	25
${(9.8754,9.2271,5.6426,4.3146)}^T$	${(3.0166,0.0148,-0.0271,0.0280)}^T$	8.913504e−4	20
${(8.5061,1.4453,3.7049,6.2239)}^T$	${(2.9635,0.0040,0.0176,0.0220)}^T$	5.981987e−4	26
${(2.7744,0.0611,3.7471,4.3693)}^T$	${(0.0132,-0.0399,2.1529,2.5310)}^T$	9.794091e−4	21

Table 4. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
${(5.6743,9.6878,8.2450,9.5961)}^T$	${(0.0062,-0.0249,2.1692,2.5652)}^T$	4.436641e-4	21
${(0.1485,1.5669,4.7157,5.4299)}^T$	${(0.0009,0.0320,2.2173,2.6291)}^T$	5.987181e−4	37
${(0.5969,6.5803,8.8964,1.0963)}^T$	${(3.0522,0.0031,-0.0202,-0.0151)}^T$	3.790585e−4	23
${(8.7494,1.2100,8.5635,8.9978)}^T$	${(0.0296,-0.0290,2.1300,2.5000)}^T$	9.638295e−4	17
${(7.7836,0.6937,2.7878,3.7937)}^T$	${(3.0003,0.0096,0.0154,-0.0158)}^T$	3.052065e−4	13
${(0.6837,0.8497,0.6834,4.0982)}^T$	${(3.0055,0.0333,-0.0050,0.0131)}^T$	7.098848e−4	21
${(7.6034,5.8410,4.0295,5.1004)}^T$	${(2.4048,1.7076,-0.0096,1.4753)}^T$	4.242339e−4	25
${(9.8754,9.2271,5.6426,4.3146)}^T$	${(3.0166,0.0148,-0.0271,0.0280)}^T$	8.913504e−4	20
${(8.5061,1.4453,3.7049,6.2239)}^T$	${(2.9635,0.0040,0.0176,0.0220)}^T$	5.981987e−4	26
${(2.7744,0.0611,3.7471,4.3693)}^T$	${(0.0132,-0.0399,2.1529,2.5310)}^T$	9.794091e−4	21

Example 5. 

We consider Equation (1), where $F=max\{(x-2),(2x-5)\}$. There are two exact solutions as 0 and 2.

Table 5. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
0.2922	2.0024	2.787626e−6	5
1.7071	1.9894	5.621467e−5	3
2.2766	2.0075	2.836408e−5	3
3.1110	2.0001	2.938429e−9	1
4.3570	2.0101	5.061011e−5	4
5.7853	2.0109	5.937701e−5	5
6.2406	1.9871	8.325445e−5	6
7.1122	2.0116	6.635145e−5	3
8.8517	1.9970	4.557770e−6	6
9.7975	1.9928	2.607803e−5	4

Table 5. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
0.2922	2.0024	2.787626e−6	5
1.7071	1.9894	5.621467e−5	3
2.2766	2.0075	2.836408e−5	3
3.1110	2.0001	2.938429e−9	1
4.3570	2.0101	5.061011e−5	4
5.7853	2.0109	5.937701e−5	5
6.2406	1.9871	8.325445e−5	6
7.1122	2.0116	6.635145e−5	3
8.8517	1.9970	4.557770e−6	6
9.7975	1.9928	2.607803e−5	4

Example 6. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill f_1(x)\hfill \\ \hfill f_2(x)\hfill \\ \hfill f_3(x)\hfill \\ \hfill f_4(x)\hfill \end{array}\right).$ $f_i(x)=max\{x_1^2,x_2^2,x_3^2,x_4^2\}$, $i=1,2,3,4.$ The exact solution of this problem is ${(0,0,0,0)}^T$.

Table 6. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
${(7.4003,2.3483,7.3496,9.7060)}^T$	${(0.0825,0.0842,0.0825,0.0830)}^T$	9.228501e−5	29
${(1.3393,0.3089,9.3914,3.0131)}^T$	${(0.0858,0.0629,0.0484,0.0414)}^T$	9.446089e−5	22
${(7.3434,0.5133,0.7289,0.8853)}^T$	${(0.0852,0.0817,0.0801,0.0607)}^T$	9.546345e−5	36
${(6.7865,4.9518,1.8971,4.9501)}^T$	${(0.0431,0.0800,0.0733,0.0801)}^T$	7.428734e−5	34
${(1.4761,0.5497,8.5071,5.6056)}^T$	${(0.0774,0.0691,0.0717,0.0655)}^T$	6.591548e−5	39
${(0.5670,5.2189,3.3585,1.7567)}^T$	${(0.0244,0.0421,0.0575,0.0604)}^T$	2.433055e−5	25
${(7.6903,5.8145,9.2831,5.8009)}^T$	${(0.0477,0.0739,0.0745,0.0766)}^T$	6.280631e−5	32
${(6.9475,7.5810,4.3264,6.5550)}^T$	${(0.0846,0.0819,0.0566,0.0850)}^T$	9.476538e−5	21
${(2.8785,4.1452,4.6484,7.6396)}^T$	${(0.0637,0.0802,0.0742,0.0012)}^T$	5.739573e−5	33
${(2.9735,0.6205,2.9824,0.4635)}^T$	${(0.0854,0.0727,0.0831,0.0542)}^T$	9.575717e−5	22

Table 6. Number of iterations and the final value of $\Psi (x^{*}).$

x0	x*	$\Psi$(x*)	k
${(7.4003,2.3483,7.3496,9.7060)}^T$	${(0.0825,0.0842,0.0825,0.0830)}^T$	9.228501e−5	29
${(1.3393,0.3089,9.3914,3.0131)}^T$	${(0.0858,0.0629,0.0484,0.0414)}^T$	9.446089e−5	22
${(7.3434,0.5133,0.7289,0.8853)}^T$	${(0.0852,0.0817,0.0801,0.0607)}^T$	9.546345e−5	36
${(6.7865,4.9518,1.8971,4.9501)}^T$	${(0.0431,0.0800,0.0733,0.0801)}^T$	7.428734e−5	34
${(1.4761,0.5497,8.5071,5.6056)}^T$	${(0.0774,0.0691,0.0717,0.0655)}^T$	6.591548e−5	39
${(0.5670,5.2189,3.3585,1.7567)}^T$	${(0.0244,0.0421,0.0575,0.0604)}^T$	2.433055e−5	25
${(7.6903,5.8145,9.2831,5.8009)}^T$	${(0.0477,0.0739,0.0745,0.0766)}^T$	6.280631e−5	32
${(6.9475,7.5810,4.3264,6.5550)}^T$	${(0.0846,0.0819,0.0566,0.0850)}^T$	9.476538e−5	21
${(2.8785,4.1452,4.6484,7.6396)}^T$	${(0.0637,0.0802,0.0742,0.0012)}^T$	5.739573e−5	33
${(2.9735,0.6205,2.9824,0.4635)}^T$	${(0.0854,0.0727,0.0831,0.0542)}^T$	9.575717e−5	22

Example 7. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill f_1(x)\hfill \\ \hfill ⋮\hfill \\ \hfill f_{10}(x)\hfill \end{array}\right)$ with $f_i(x)=max\{x_1^2,\cdots ,x_{10}^2\}$, $i=1,\cdots ,10.$ The exact solution of this problem is ${(0,0,0,0,0,0,0,0,0,0)}^T$.

Table 7. Number of iterations and the final value of $\Psi (x^{*}).$

x0	$\Psi$(x*)	k
${(8.2408,8.2798,2.9337,3.0937,5.2303,3.2530,8.3184,8.1029,5.5700,2.6296)}^T$	9.719070e−5	27
${(9.5089,4.4396,0.6002,8.6675,6.3119,3.5507,9.9700,2.2417,6.5245,6.0499)}^T$	9.957464e−5	45
${(4.1705,9.7179,9.8797,8.6415,3.8888,4.5474,2.4669,7.8442,8.8284,9.1371)}^T$	8.965459e−5	39
${(8.3975,3.7172,8.2822,1.7652,1.2952,8.7988,0.4408,6.8672,7.3377,4.3717)}^T$	9.644608e−5	47
${(9.7209,0.3146,8.3540,8.3571,0.4986,5.4589,9.4317,3.2147,8.0647,6.0140)}^T$	9.737485e−5	37
${(8.3336,4.0363,3.9018,3.6045,1.4026,2.6013,0.8682,4.2940,2.5728,2.9756)}^T$	9.240212e−5	47
${(4.8267,3.7601,5.2378,2.6487,0.6836,4.3633,1.7385,0.2611,9.5468,4.3060)}^T$	8.801643e−5	44
${(0.5398,0.2062,6.8148,5.9863,1.1403,7.9625,6.1785,0.7021,0.6928,1.3601)}^T$	8.946151e−5	44
${(5.7099,1.6977,1.4766,4.7608,9.0810,5.5218,0.3294,0.5386,8.0506,4.5137)}^T$	8.815143e−5	45
${(2.1647,7.8620,7.2309,2.7884,5.8243,4.2101,0.9207,0.2403,4.9115,2.7827)}^T$	6.697806e−5	39

Table 7. Number of iterations and the final value of $\Psi (x^{*}).$

x0	$\Psi$(x*)	k
${(8.2408,8.2798,2.9337,3.0937,5.2303,3.2530,8.3184,8.1029,5.5700,2.6296)}^T$	9.719070e−5	27
${(9.5089,4.4396,0.6002,8.6675,6.3119,3.5507,9.9700,2.2417,6.5245,6.0499)}^T$	9.957464e−5	45
${(4.1705,9.7179,9.8797,8.6415,3.8888,4.5474,2.4669,7.8442,8.8284,9.1371)}^T$	8.965459e−5	39
${(8.3975,3.7172,8.2822,1.7652,1.2952,8.7988,0.4408,6.8672,7.3377,4.3717)}^T$	9.644608e−5	47
${(9.7209,0.3146,8.3540,8.3571,0.4986,5.4589,9.4317,3.2147,8.0647,6.0140)}^T$	9.737485e−5	37
${(8.3336,4.0363,3.9018,3.6045,1.4026,2.6013,0.8682,4.2940,2.5728,2.9756)}^T$	9.240212e−5	47
${(4.8267,3.7601,5.2378,2.6487,0.6836,4.3633,1.7385,0.2611,9.5468,4.3060)}^T$	8.801643e−5	44
${(0.5398,0.2062,6.8148,5.9863,1.1403,7.9625,6.1785,0.7021,0.6928,1.3601)}^T$	8.946151e−5	44
${(5.7099,1.6977,1.4766,4.7608,9.0810,5.5218,0.3294,0.5386,8.0506,4.5137)}^T$	8.815143e−5	45
${(2.1647,7.8620,7.2309,2.7884,5.8243,4.2101,0.9207,0.2403,4.9115,2.7827)}^T$	6.697806e−5	39

Example 8. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill f_1(x)\hfill \\ \hfill ⋮\hfill \\ \hfill f_4(x)\hfill \end{array}\right)$ with $f_i(x)={\sum }_{j=1}^{4}max\{-x_j-x_{j+1},-x_j-x_{j+1}+(x_j^2+x_{j+1}^2+1)\}$, $i=1,2,3,4.$ The exact solution of this problem is ${(0,0,0,0)}^T$.

Table 8. Number of iterations and the final value of $\Psi (x^{*}).$

x0	$\Psi$(x*)	k
${(4.1131,8.2898,9.3511,3.9907)}^T$	3.670149e−5	4
${(0.5221,5.7119,7.4767,3.2024)}^T$	4.216994e−5	4
${(5.4000,2.2106,0.9595,0.6017)}^T$	6.167554e−5	7
${(6.6015,0.5231,5.5683,7.1203)}^T$	3.838925e−5	4
${(1.6924,2.5845,1.9791,6.0569)}^T$	6.272257e−5	6
${(3.3969,1.9786,5.0683,9.5076)}^T$	7.097729e−5	5
${(4.2175,4.1131,9.5914,7.5025)}^T$	2.693701e−5	4
${(8.8728,0.5585,1.3822,8.6306)}^T$	9.021922e−5	7
${(9.8100,2.3352,0.9623,3.8458)}^T$	4.687797e−5	5
${(9.6426,6.7115,2.9917,5.3113)}^T$	8.657057e−5	6

Table 8. Number of iterations and the final value of $\Psi (x^{*}).$

x0	$\Psi$(x*)	k
${(4.1131,8.2898,9.3511,3.9907)}^T$	3.670149e−5	4
${(0.5221,5.7119,7.4767,3.2024)}^T$	4.216994e−5	4
${(5.4000,2.2106,0.9595,0.6017)}^T$	6.167554e−5	7
${(6.6015,0.5231,5.5683,7.1203)}^T$	3.838925e−5	4
${(1.6924,2.5845,1.9791,6.0569)}^T$	6.272257e−5	6
${(3.3969,1.9786,5.0683,9.5076)}^T$	7.097729e−5	5
${(4.2175,4.1131,9.5914,7.5025)}^T$	2.693701e−5	4
${(8.8728,0.5585,1.3822,8.6306)}^T$	9.021922e−5	7
${(9.8100,2.3352,0.9623,3.8458)}^T$	4.687797e−5	5
${(9.6426,6.7115,2.9917,5.3113)}^T$	8.657057e−5	6

Example 9. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill f_1(x)\hfill \\ \hfill ⋮\hfill \\ \hfill f_4(x)\hfill \end{array}\right)$ with $f_i(x)={\sum }_{j=1}^{4}max\{-x_j-x_{j+1},-x_j-x_{j+1}+(x_j^2+x_{j+1}^2-1)\}$, $i=1,2,3,4.$ The exact solution of this problem is ${(0,0,0,0)}^T$.

Table 9. Number of iterations and the final value of $\Psi (x^{*}).$

x0	$\Psi$(x*)	k
${(1.5290,1.5254,1.5555,0.8957)}^T$	9.777886e−3	8
${(4.5442,6.6890,8.3130,7.9024)}^T$	3.912481e−3	5
${(9.0150,3.1834,5.9708,2.9780)}^T$	9.081688e−3	3
${(3.1781,9.8445,5.4825,7.4925)}^T$	6.868711e−3	7
${(8.4185,1.6689,9.0310,1.0512)}^T$	5.318627e−3	4
${(7.4509,7.2937,7.1747,1.3343)}^T$	7.203761e−3	9
${(4.4579,5.0879,5.3049,8.5972)}^T$	9.500345e−3	4
${(6.7772,8.0584,5.3124,9.5590)}^T$	9.421194e−3	4
${(0.6668,5.4152,2.8166,4.8090)}^T$	6.718722e−3	7
${(6.8486,2.0826,6.0816,3.2618)}^T$	3.494877e−3	4

Table 9. Number of iterations and the final value of $\Psi (x^{*}).$

x0	$\Psi$(x*)	k
${(1.5290,1.5254,1.5555,0.8957)}^T$	9.777886e−3	8
${(4.5442,6.6890,8.3130,7.9024)}^T$	3.912481e−3	5
${(9.0150,3.1834,5.9708,2.9780)}^T$	9.081688e−3	3
${(3.1781,9.8445,5.4825,7.4925)}^T$	6.868711e−3	7
${(8.4185,1.6689,9.0310,1.0512)}^T$	5.318627e−3	4
${(7.4509,7.2937,7.1747,1.3343)}^T$	7.203761e−3	9
${(4.4579,5.0879,5.3049,8.5972)}^T$	9.500345e−3	4
${(6.7772,8.0584,5.3124,9.5590)}^T$	9.421194e−3	4
${(0.6668,5.4152,2.8166,4.8090)}^T$	6.718722e−3	7
${(6.8486,2.0826,6.0816,3.2618)}^T$	3.494877e−3	4

Example 10. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill f_1(x)\hfill \\ \hfill ⋮\hfill \\ \hfill f_n(x)\hfill \end{array}\right)$ with $f_i(x)=max\{x_1^2-6x_1,\cdots ,x_n^2-6x_n\}$, $i=1,\cdots ,n.$ n represents the problem dimension. The solution is $x^{*}={(\lambda \cdots \lambda )}^T$ (λ is no more than 6). In this problem, we intend to check the efficiency of Algorithm 1 with the dimension of test problem is $50,100,$ and 200. We randomly selected ten initial values when $n=50,$ $n=100$ and $n=200$.

Table 10. Number of iterations, the final value of $\Psi (x^{*})$ and dimension of the test problem.

n = 50		n = 100		n = 200
$\Psi (x^{*})$	k	$\Psi (x^{*})$	k	$\Psi (x^{*})$	k
1.625691e−3	9	9.444914e−3	11	9.897292e−3	15
4.082584e−3	7	5.358975e−5	9	3.937758e−4	5
6.082289e−3	7	4.734809e−3	9	5.800944e−3	16
2.042082e−3	9	3.249863e−3	6	3.289200e−3	11
3.765484e−3	9	6.587880e−3	10	4.674659e−3	10
7.553578e−3	13	2.632872e−3	10	1.450852e−3	13
4.208302e−4	14	4.177174e−3	3	9.461359e−3	16
4.250316e−3	9	9.744427e−3	7	3.778464e−3	15
2.634965e−5	10	5.854241e−6	10	1.501579e−3	8
3.445498e−3	11	4.209193e−3	6	1.984871e−3	25

Table 10. Number of iterations, the final value of $\Psi (x^{*})$ and dimension of the test problem.

n = 50		n = 100		n = 200
$\Psi (x^{*})$	k	$\Psi (x^{*})$	k	$\Psi (x^{*})$	k
1.625691e−3	9	9.444914e−3	11	9.897292e−3	15
4.082584e−3	7	5.358975e−5	9	3.937758e−4	5
6.082289e−3	7	4.734809e−3	9	5.800944e−3	16
2.042082e−3	9	3.249863e−3	6	3.289200e−3	11
3.765484e−3	9	6.587880e−3	10	4.674659e−3	10
7.553578e−3	13	2.632872e−3	10	1.450852e−3	13
4.208302e−4	14	4.177174e−3	3	9.461359e−3	16
4.250316e−3	9	9.744427e−3	7	3.778464e−3	15
2.634965e−5	10	5.854241e−6	10	1.501579e−3	8
3.445498e−3	11	4.209193e−3	6	1.984871e−3	25

Example 11. 

We consider Equation (1), where $F=\left(\begin{array}{c}\hfill f_1(x)\hfill \\ \hfill ⋮\hfill \\ \hfill f_n(x)\hfill \end{array}\right)$ with $f_i(x)=max\{x_1^2,\cdots ,x_n^2\}$, $i=1,\cdots ,n.$ The problem has only unique solution $x^{*}={(0,\cdots ,0)}^T$. We randomly selected ten initial values when $n=100,$$n=200$ and $n=500$.

Table 11. Number of iterations, the final value of $\Psi (x^{*})$ and dimension of the test problem.

n = 100		n = 200		n = 500
$\Psi (x^{*})$	k	$\Psi (x^{*})$	k	$\Psi (x^{*})$	k
9.152621e−03	17	9.040255e−03	9	7.682471e−3	14
4.383679e−3	15	6.976857e−3	9	8.861191e−3	15
5.172738e−3	15	6.902897e−3	10	8.892858e−3	12
5.796109e−3	12	7.686345e−3	12	9.210427e−3	14
7.613768e−3	16	8.400876e−3	10	9.843579e−3	10
5.398565e−3	12	8.066523e−3	10	9.717126e−3	13
3.403516e−3	15	9.097423e−3	12	8.999900e−3	15
8.701785e−3	13	7.208014e−3	11	9.970099e−3	12
8.302172e−3	11	7.822304e−3	13	9.391355e−3	15
6.610621e−3	13	7.278306e-3	9	9.624919e−3	10

Table 11. Number of iterations, the final value of $\Psi (x^{*})$ and dimension of the test problem.

n = 100		n = 200		n = 500
$\Psi (x^{*})$	k	$\Psi (x^{*})$	k	$\Psi (x^{*})$	k
9.152621e−03	17	9.040255e−03	9	7.682471e−3	14
4.383679e−3	15	6.976857e−3	9	8.861191e−3	15
5.172738e−3	15	6.902897e−3	10	8.892858e−3	12
5.796109e−3	12	7.686345e−3	12	9.210427e−3	14
7.613768e−3	16	8.400876e−3	10	9.843579e−3	10
5.398565e−3	12	8.066523e−3	10	9.717126e−3	13
3.403516e−3	15	9.097423e−3	12	8.999900e−3	15
8.701785e−3	13	7.208014e−3	11	9.970099e−3	12
8.302172e−3	11	7.822304e−3	13	9.391355e−3	15
6.610621e−3	13	7.278306e-3	9	9.624919e−3	10

4. Conclusions

In this paper, we have presented a new smoothing conjugate gradient method for the nonlinear nonsmooth complementarity problems. The method is based on a smoothing Fischer-Burmeister function and Armijo-type line search. With careful analysis, we are able to show that our method is globally convergent. Numerical tests illustrate that the method can efficiently solve the given test problems, therefor the new method is promising. We might consider more effective ways of choosing smoothing functions and line search methods for our method. This remains under investigation.