An algorithm for solving a multi-valued variational inequality

Abstract

We propose a new extragradient method for solving a multi-valued variational inequality. It is showed that the method converges globally to a solution of the multi-valued variational inequality, provided the multi-valued mapping is continuous with nonempty compact convex values. Preliminary computational experience is also reported.

MSC: 47H04, 47H10, 47J20, 47J25.

1 Introduction

We consider the following multi-valued variational inequality, denoted by $MVI(F,C)$: to find $x^{\ast }\in C$ and $\xi \in F(x^{\ast })$ such that

$$〈\xi ,y-x^{\ast }〉\ge 0,\mathrm{\forall }y\in C,$$

(1.1)

where C is a nonempty closed convex set in ${\mathbb{R}}^n$, F is a multi-valued mapping from C into ${\mathbb{R}}^n$ with nonempty values, and $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$ denote the inner product and the norm in ${\mathbb{R}}^n$, respectively.

Extragradient-type algorithms have been extensively studied in the literature; see [1–3]. Various algorithms for solving the multi-valued variational inequality have been extensively studied in the literature [4–15]. The well-known proximal point algorithm [12] requires the multi-valued mapping F to be monotone. [11] proposes a projection algorithm for solving the multi-valued variational inequality with a pseudomonotone mapping. In [11], choosing $u_i\in F(x_i)$ needs solving a single-valued variational inequality; see the expression (2.1) in [11]. [6] presents a double projection algorithm, which is an improvement of [11], so that $u_i\in F(x_i)$ can be taken arbitrarily. In [6], however, choosing the hyperplane needs computing the supremum and hence is computationally expensive. To overcome this difficulty, [7] introduces an extragradient algorithm for solving the multi-valued variational inequality in which computing the supremum is avoided. In this paper, we present a new extragradient method for solving the multi-valued variational inequality. In our method, $u_i\in F(x_i)$ can be taken arbitrarily. Moreover, the main difference of our method from those of [6, 7, 11] is the procedure of Armijo-type linesearch. We also present numerical tests to compare our Algorithm 2.2 with those in [6, 11].

This paper is organized as follows. In Section 2, we present the algorithm details. We prove the preliminary results for convergence analysis in Section 3. Numerical results are reported in the last section.

2 Algorithms

Let us recall the definition of a continuous multi-valued mapping. F is said to be upper semicontinuous at $x\in C$ if for every open set V containing $F(x)$, there is an open set U containing x such that $F(y)\subset V$ for all $y\in C\cap U$. F is said to be lower semicontinuous at $x\in C$ if given any sequence $x_k$ converging to x and any $y\in F(x)$, there exists a sequence $y_k\in F(x_k)$ that converges to y. F is said to be continuous at $x\in C$ if it is both upper semicontinuous and lower semicontinuous at x. If F is single-valued, then both upper semicontinuity and lower semicontinuity reduce to the continuity of F.

F is called pseudomonotone on C in the sense of Karamardian [16] if for any $x,y\in C$,

$$〈v,x-y〉\ge 0\text{for some }v\in F(y)\Rightarrow 〈u,x-y〉\ge 0\text{ for all }u\in F(x).$$

(2.1)

Let S be the solution set of (1.1), that is, those points $x^{\ast }\in C$ satisfying (1.1). Throughout this paper, we assume that the solution set S of problem (1.1) is nonempty and F is continuous on C with nonempty compact convex values satisfying the following property:

$$〈\zeta ,y-x〉\ge 0,\mathrm{\forall }y\in C,\mathrm{\forall }\zeta \in F(y),\mathrm{\forall }x\in S.$$

(2.2)

The property (2.2) holds if F is pseudomonotone on C.

Let $P_C$ denote the projector onto C, and let $\mu >0$ be a parameter.

Proposition 2.1 $x\in C$ and $\xi \in F(x)$ solve problem (1.1) if and only if

$$r_{\mu }(x,\xi ):=x-P_C(x-\mu \xi )=0.$$

Algorithm 2.2 Choose $x_0\in C$ and two parameters $\gamma ,\sigma \in (0,1)$. Set $i=0$.

Step 1. Choose $u_i\in F(x_i)$ and let $k_i$ be the smallest nonnegative integer satisfying

$$v_i\in F\left(P_C(x_i-{\gamma }^{k_i}{u}_i)\right),$$

(2.3)

$${\gamma }^{k_i}〈u_i-v_i,r_{{\gamma }^{k_i}}(x_i,u_i)〉\le \sigma {\parallel r_{{\gamma }^{k_i}}(x_i,u_i)\parallel }^2.$$

(2.4)

Set ${\eta }_i={\gamma }^{k_i}$. If $r_{{\eta }_i}(x_i,u_i)=0$, stop.

Step 2. Compute $x_{i+1}:=P_C(x_i-{\alpha }_i{d}_i)$, where

$$d_i=r_{{\eta }_i}(x_i,u_i)+{\eta }_i{v}_i,$$

(2.5)

$${\alpha }_i=\frac{(1-\sigma ){\parallel r_{{\eta }_i}(x_i,u_i)\parallel }^2}{{\parallel d_i\parallel }^2}.$$

(2.6)

Let $i:=i+1$ and go to Step 1.

Remark 2.3 Let us compare the above algorithm with those in [6, 7, 11]. First, Aimijo-type linesearch procedures in the four algorithms are different. [6, 7, 11] use different procedures which replace (2.4) by the following ones:

$$〈v_i,r_{\mu }(x_i,u_i)〉\ge \sigma {\parallel r_{\mu }(x_i,u_i)\parallel }^2\text{or}〈u_i-v_i,r_{\mu }(x_i,u_i)〉\le \sigma {\parallel r_{\mu }(x_i,u_i)\parallel }^2,$$

where μ is required to be strictly less than 1 or $1/\sigma$, and $v_i\in F(x_i-{\gamma }^{k_i}{r}_{\mu }(x_i,u_i))$. In our algorithm, μ can change according to the value of ${\eta }_i$ in each iteration and $v_i\in F(P_C(x_i-{\gamma }^{k_i}{u}_i))$. Secondly, the way to generate the next iterate is different. In [6, 11], the next iterate is a projection of the current iterate onto the intersection of the feasible set C and a hyperplane, while in our algorithm as well as in [7] the next iterate is a projection onto the feasible set C. In addition, the searching directions in [7] and our algorithm are also different.

Lemma 2.4 Let C be a closed convex subset of ${\mathbb{R}}^n$. For any $x,y\in {\mathbb{R}}^n$ and $z\in C$, the following statements hold.

(i) $〈x-P_C(x),z-P_C(x)〉\le 0$.

(ii) ${\parallel P_C(x)-P_C(y)\parallel }^2\le {\parallel x-y\parallel }^2-{\parallel P_C(x)-x+y-P_C(y)\parallel }^2$.

Proof See [17]. □

The proof of the following lemma is easy and we omit it (see Lemma 3.1 in [18] for example).

Lemma 2.5 For any $x\in {\mathbb{R}}^n$, $\xi \in F(x)$ and $\mu >0$,

$$min\{1,\mu \}\parallel r_1(x,\xi )\parallel \le \parallel r_{\mu }(x,\xi )\parallel \le max\{1,\mu \}\parallel r_1(x,\xi )\parallel .$$

We first show that Algorithm 2.2 is well defined.

Proposition 2.6 If $x_i$ is not a solution of problem (1.1), then there exists a nonnegative integer $k_i$ satisfying (2.3) and (2.4).

Proof Suppose that for all k and all $v\in F(P_C(x_i-{\gamma }^k{u}_i))$, we have

$${\gamma }^k〈u_i-v,r_{{\gamma }^k}(x_i,u_i)〉>\sigma {\parallel r_{{\gamma }^k}(x_i,u_i)\parallel }^2,$$

and hence,

$${\gamma }^k\parallel u_i-v\parallel >\sigma \parallel r_{{\gamma }^k}(x_i,u_i)\parallel .$$

Therefore,

$$\begin{array}{rl}\parallel u_i-v\parallel & >\frac{\sigma }{{\gamma }^k}\parallel r_{{\gamma }^k}(x_i,u_i)\parallel \\ \ge \frac{\sigma }{{\gamma }^k}min\{1,{\gamma }^k\}\parallel r_1(x_i,u_i)\parallel \\ =\sigma \parallel r_1(x_i,u_i)\parallel ,\end{array}$$

(2.7)

where the second inequality follows from Lemma 2.5 and the equality follows from $\gamma \in (0,1)$ and $k\ge 0$. Since $P_C(\cdot )$ is continuous and $x_i\in C$, $P_C(x_i-{\gamma }^k{u}_i)\to x_i(k\to \mathrm{\infty })$. Since F is lower semicontinuous, $u_i\in F(x_i)$ and $P_C(x_i-{\gamma }^k{u}_i)\to x_i(k\to \mathrm{\infty })$, there is $v_k\in F(P_C(x_i-{\gamma }^k{u}_i))$ such that $v_k\to u_i(k\to \mathrm{\infty })$. Therefore,

$$\parallel u_i-v_k\parallel >\sigma \parallel r_1(x_i,u_i)\parallel ,\mathrm{\forall }k.$$

(2.8)

Let $k\to \mathrm{\infty }$ in (2.8), we have

$$0=\parallel u_i-u_i\parallel \ge \sigma \parallel r_1(x_i,u_i)\parallel >0.$$

This contradiction completes the proof. □

3 Main results

Now we obtain the following auxiliary result that will be used for proving the convergence of Algorithm 2.2.

Theorem 3.1 If the assumption (2.2) holds and $x_i\notin S$, then for any $x^{\ast }\in S$,

$$〈d_i,x_i-x^{\ast }〉\ge (1-\sigma ){\parallel r_{{\eta }_i}(x_i,{\xi }_i)\parallel }^2>0.$$

(3.1)

Proof Let $x^{\ast }\in S$. Since $u_i\in F(x_i)$ and ${\eta }_i>0$, it follows from (2.2) that

$$〈{\eta }_i{u}_i,x_i-x^{\ast }〉\ge 0.$$

(3.2)

Similarly, we have

$$〈{\eta }_i{v}_i,P_C(x_i-{\eta }_i{u}_i)-x^{\ast }〉\ge 0,$$

(3.3)

because $v_i\in F(P_C(x_i-{\eta }_i{u}_i))$. Since $x^{\ast }\in C$, from Lemma 2.4(i) we have

$$〈x_i-{\eta }_i{u}_i-P_C(x_i-{\eta }_i{u}_i),P_C(x_i-{\eta }_i{u}_i)-x^{\ast }〉\ge 0.$$

(3.4)

It follows from (3.2), (3.3) and (3.4) that

$$\begin{array}{rl}〈d_i,x_i-x^{\ast }〉=& 〈r_{{\eta }_i}(x_i,u_i)+{\eta }_i{v}_i,x_i-x^{\ast }〉\\ =& 〈r_{{\eta }_i}(x_i,u_i)+{\eta }_i(v_i-u_i),x_i-x^{\ast }〉+〈{\eta }_i{u}_i,x_i-x^{\ast }〉\\ \ge & 〈r_{{\eta }_i}(x_i,u_i)+{\eta }_i(v_i-u_i),x_i-x^{\ast }〉\\ =& 〈r_{{\eta }_i}(x_i,u_i)+{\eta }_i(v_i-u_i),r_{{\eta }_i}(x_i,u_i)〉\\ +〈x_i-{\eta }_i{u}_i-P_C(x_i-{\eta }_i{u}_i),P_C(x_i-{\eta }_i{u}_i)-x^{\ast }〉\\ +〈{\eta }_i{v}_i,P_C(x_i-{\eta }_i{u}_i)-x^{\ast }〉\\ \ge & 〈r_{{\eta }_i}(x_i,u_i)+{\eta }_i(v_i-u_i),r_{{\eta }_i}(x_i,u_i)〉.\end{array}$$

(3.5)

Therefore,

$$\begin{array}{rl}〈d_i,x_i-x^{\ast }〉& \ge 〈r_{{\eta }_i}(x_i,u_i)+{\eta }_i(v_i-u_i),r_{{\eta }_i}(x_i,u_i)〉\\ ={\parallel r_{{\eta }_i}(x_i,u_i)\parallel }^2-{\eta }_i〈u_i-v_i,r_{{\eta }_i}(x_i,u_i)〉\\ \ge {\parallel r_{{\eta }_i}(x_i,u_i)\parallel }^2-\sigma {\parallel r_{{\eta }_i}(x_i,u_i)\parallel }^2\\ =(1-\sigma ){\parallel r_{{\eta }_i}(x_i,u_i)\parallel }^2,\end{array}$$

(3.6)

where the second inequality follows from (2.4). This completes the proof. □

Theorem 3.2 If $F:C\to 2^{{\mathbb{R}}^n}$ is continuous with nonempty compact convex values on C and the assumption (2.2) holds, then the sequence $\{x_i\}$ generated by Algorithm  2.2 converges to a solution $\overline{x}$ of (1.1).

Proof Let $x^{\ast }\in S$. It follows from Lemma 2.4(ii), Lemma 2.5, (2.5), (2.6) and (3.6) that

$$\begin{array}{rl}{\parallel x_{i+1}-x^{\ast }\parallel }^2& \le {\parallel x_i-x^{\ast }-{\alpha }_i{d}_i\parallel }^2\\ ={\parallel x_i-x^{\ast }\parallel }^2-2{\alpha }_i〈d_i,x_i-x^{\ast }〉+{\alpha }_i^2{\parallel d_i\parallel }^2\\ \le {\parallel x_i-x^{\ast }\parallel }^2-\frac{{(1-\sigma )}^2{\parallel r_{{\eta }_i}(x_i,u_i)\parallel }^4}{{\parallel d_i\parallel }^2}\\ \le {\parallel x_i-x^{\ast }\parallel }^2-\frac{{(1-\sigma )}^2{(min\{{\eta }_i,1\}\parallel r_1(x_i,u_i)\parallel )}^4}{{\parallel d_i\parallel }^2}\\ ={\parallel x_i-x^{\ast }\parallel }^2-\frac{{(1-\sigma )}^2{\eta }_i^4{\parallel r_1(x_i,u_i)\parallel }^4}{{\parallel r_{{\eta }_i}(x_i,u_i)+{\eta }_i{v}_i\parallel }^2}.\end{array}$$

(3.7)

It follows that the sequence $\{{\parallel x_{i+1}-x^{\ast }\parallel }^2\}$ is nonincreasing, and hence is a convergent sequence. Therefore, $\{x_i\}$ is bounded. Since F is continuous with compact values, Proposition 3.11 in [19] implies that $\{F(x_i):i\in N\}$ is a bounded set, and so are $\{u_i\}$, $\{r_{{\eta }_i}(x_i,u_i)\}$ and $\{v_i\}$. Thus, $\{r_{{\eta }_i}(x_i,u_i)+{\eta }_i{v}_i\}$ is bounded. Then there exists a positive number M such that

$$\parallel r_{{\eta }_i}(x_i,u_i)+{\eta }_i{v}_i\parallel \le M.$$

It follows from (3.7) that

$${\parallel x_{i+1}-x^{\ast }\parallel }^2\le {\parallel x_i-x^{\ast }\parallel }^2-{(1-\sigma )}^2{M}^{-2}{\eta }_i^4{\parallel r_1(x_i,u_i)\parallel }^4.$$

(3.8)

Therefore,

$$\underset{i\to \mathrm{\infty }}{lim}{\eta }_i\parallel r_1(x_i,u_i)\parallel =0.$$

(3.9)

By the boundedness of $\{x_i\}$, there exists a convergent subsequence $\{x_{i_j}\}$ converging to $\overline{x}$.

If $\overline{x}$ is a solution of problem (1.1), we show next that the whole sequence $\{x_i\}$ converges to $\overline{x}$. Replacing $x^{\ast }$ by $\overline{x}$ in the preceding argument, we obtain that the sequence $\{\parallel x_i-\overline{x}\parallel \}$ is nonincreasing and hence converges. Since $\overline{x}$ is an accumulation point of $\{x_i\}$, some subsequence of $\{\parallel x_i-\overline{x}\parallel \}$ converges to zero. This shows that the whole sequence $\{\parallel x_i-\overline{x}\parallel \}$ converges to zero, hence ${lim}_{i\to \mathrm{\infty }}{x}_i=\overline{x}$.

Suppose now that $\overline{x}$ is not a solution of problem (1.1). We show first that $k_i$ in Algorithm 2.2 cannot tend to ∞. Since F is continuous with compact values, Proposition 3.11 in [19] implies that $\{F(x_i):i\in N\}$ is a bounded set, and so the sequence $\{u_i\}$ is bounded. Therefore, there exists a subsequence $\{u_{i_j}\}$ converging to $\overline{u}$. Since F is upper semicontinuous with compact values, Proposition 3.7 in [19] implies that F is closed, and so $\overline{u}\in F(\overline{x})$. By the definition of $k_i$, we have

$${\gamma }^{k_i-1}〈u_i-v,r_{{\gamma }^{k_i-1}}(x_i,u_i)〉>\sigma {\parallel r_{{\gamma }^{k_i-1}}(x_i,u_i)\parallel }^2,\mathrm{\forall }v\in F\left(P_C(x_i-{\gamma }^{k_i-1}{u}_i)\right),$$

(3.10)

and hence

$${\gamma }^{k_i-1}\parallel u_i-v\parallel >\sigma \parallel r_{{\gamma }^{k_i-1}}(x_i,u_i)\parallel ,\mathrm{\forall }v\in F\left(P_C(x_i-{\gamma }^{k_i-1}{u}_i)\right).$$

(3.11)

Therefore,

$$\begin{array}{rl}\parallel u_i-v\parallel & >\frac{\sigma }{{\gamma }^{k_i-1}}\parallel r_{{\gamma }^{k_i-1}}(x_i,u_i)\parallel \\ \ge \frac{\sigma }{{\gamma }^{k_i-1}}min\{1,{\gamma }^{k_i-1}\}\parallel r_1(x_i,u_i)\parallel \\ =\sigma \parallel r_1(x_i,u_i)\parallel ,\mathrm{\forall }v\in F\left(P_C(x_i-{\gamma }^{k_i-1}{u}_i)\right),\mathrm{\forall }{k}_i\ge 1,\end{array}$$

(3.12)

where the second inequality follows from Lemma 2.5 and the equality follows from $\gamma \in (0,1)$.

If $k_{i_j}\to \mathrm{\infty }$, then $P_C(x_{i_j}-{\gamma }^{k_{i_j}-1}{u}_{i_j})\to \overline{x}$. The lower continuity of F, in turn, implies the existence of ${\overline{u}}_{i_j}\in F(P_C(x_{i_j}-{\gamma }^{k_{i_j}-1}{u}_{i_j}))$ such that ${\overline{u}}_{i_j}$ converges to $\overline{u}$. Therefore,

$$\parallel u_{i_j}-{\overline{u}}_{i_j}\parallel >\sigma \parallel r_1(x_{i_j},u_{i_j})\parallel .$$

(3.13)

Letting $j\to \mathrm{\infty }$, we obtain the contradiction

$$0\ge \sigma {\parallel r_1(\overline{x},\overline{u})\parallel }^2>0,$$

(3.14)

being $r_1(\cdot ,\cdot )$ continuous. Therefore, $\{k_i\}$ is bounded and so is $\{{\eta }_i\}$.

By the boundedness of $\{{\eta }_i\}$, it follows from (3.9) that ${lim}_{i\to \mathrm{\infty }}\parallel r_1(x_i,u_i)\parallel =0$. Since $r_1(\cdot ,\cdot )$ is continuous and the sequences $\{x_i\}$ and $\{u_i\}$ are bounded, there exists an accumulation point $(\overline{x},\overline{u})$ of $\{(x_i,u_i)\}$ such that $r_1(\overline{x},\overline{u})=0$. This implies that $\overline{x}$ solves the variational inequality (1.1). Similar to the preceding proof, we obtain that ${lim}_{i\to \mathrm{\infty }}{x}_i=\overline{x}$.

Now we provide a result on the convergence rate of the iterative sequence generated by Algorithm 2.2. To establish this result, we need a certain error bound to hold locally (see (3.15) below). The research on an error bound is a large topic in mathematical programming. One can refer to the survey [20] for the roles played by error bounds in the convergence analysis of iterative algorithms; more recent developments on this topic are included in Chapter 6 in [21]. A condition similar to (3.15) has also been used in [22] (see expression (10) therein) to analyze the convergence rate in a very general framework.

For any $\delta >0$, define

$$P(\delta ):=\{(x,\xi )\in C\times {\mathbb{R}}^n:\xi \in F(x),\parallel r_1(x,\xi )\parallel \le \delta \}.$$

We say that F is Lipschitz continuous on C if there exists a constant $L>0$ such that, for all $x,y\in C$, $H(F(x),F(y))\le L\parallel x-y\parallel$, where H denotes the Hausdorff metric. □

Theorem 3.3 In addition to the assumptions in Theorem  3.2, if F is Lipschitz continuous with modulus $L>0$ and if there exist positive constants c and δ such that

$$dist(x,S)\le c\parallel r_1(x,\xi )\parallel ,\mathrm{\forall }(x,\xi )\in P(\delta ),$$

(3.15)

then there is a constant $\alpha >0$ such that for sufficiently large i,

$$dist(x_i,S)\le \frac{1}{\sqrt{\alpha i+{dist}^{-2}(x_0,S)}}.$$

Proof Put $\eta :=min\{1/2,L^{-1}\gamma \sigma \}$. We first prove that ${\eta }_i>\eta$ for all i. By the construction of ${\eta }_i$, we have ${\eta }_i\in (0,1]$. If ${\eta }_i=1$, then clearly ${\eta }_i>\frac{1}{2}\ge \eta$. Now we assume that ${\eta }_i<1$. Since ${\eta }_i={\gamma }^{k_i}$, it follows that the nonnegative integer $k_i\ge 1$. Thus the construction of $k_i$ implies that

$${\gamma }^{k_i-1}〈u_i-v,r_{{\gamma }^{k_i-1}}(x_i,u_i)〉>\sigma {\parallel r_{{\gamma }^{k_i-1}}(x_i,u_i)\parallel }^2,\mathrm{\forall }v\in F\left(P_C(x_i-{\gamma }^{k_i-1}{u}_i)\right),$$

(3.16)

and hence, as $k_i\ge 1$,

$$\parallel u_i-v\parallel >\frac{\sigma }{{\gamma }^{k_i-1}}\parallel r_{{\gamma }^{k_i-1}}(x_i,u_i)\parallel ,\mathrm{\forall }v\in F\left(P_C(x_i-{\gamma }^{k_i-1}{u}_i)\right).$$

Since $u_i\in F(x_i)$ and F is compact-valued, the definition of the Hausdorff metric implies the existence of $v_i\in F(P_C(x_i-{\gamma }^{k_i-1}{u}_i))$ such that

$$\frac{\sigma }{{\gamma }^{k_i-1}}\parallel r_{{\gamma }^{k_i-1}}(x_i,u_i)\parallel <\parallel u_i-v_i\parallel \le H(F(x_i),F\left(P_C(x_i-{\gamma }^{k_i-1}{u}_i)\right))\le L\parallel r_{{\gamma }^{k_i-1}}(x_i,u_i)\parallel .$$

Therefore ${\eta }_i>L^{-1}\gamma \sigma \ge \eta$.

Let $x^{\ast }\in P_S(x_i)$. By (3.8) and (3.15), we obtain that for sufficiently large i,

$$\begin{array}{rl}{dist}^2(x_{i+1},S)& \le {\parallel x_{i+1}-x^{\ast }\parallel }^2\\ \le {\parallel x_i-x^{\ast }\parallel }^2-{(1-\sigma )}^2{M}^{-2}{\eta }_i^4{\parallel r_1(x_i,u_i)\parallel }^4\\ \le {\parallel x_i-x^{\ast }\parallel }^2-{(1-\sigma )}^2{M}^{-2}{\eta }^4{\parallel r_1(x_i,u_i)\parallel }^4\\ \le {dist}^2(x_i,S)-{(1-\sigma )}^2{M}^{-2}{\eta }^4{c}^{-4}{dist}^4(x_i,S),\end{array}$$

where the second inequality follows from ${\eta }_i>\eta$.

Write α for ${(1-\sigma )}^2{M}^{-2}{\eta }^4{c}^{-4}$. Applying Lemma 6 in Chapter 2 of [23], we have

$$dist(x_i,S)\le dist(x_0,S)/\sqrt{\alpha i{dist}^2(x_0,S)+1}=1/\sqrt{\alpha i+{dist}^{-2}(x_0,S)}.$$

This completes the proof. □

4 Numerical experiments

In this section, we present some numerical experiments for the proposed algorithm. The MATLAB codes are run on a PC (with CPU Intel P-T2390) under MATLAB Version 7.0.1.24704(R14) Service Pack 1. We compare the performance of our Algorithm 2.2, [6], Algorithm 1] and [11], Algorithm 1]. In Tables 1 and 2, ‘It.’ denotes the number of iteration and ‘CPU’ denotes the CPU time in seconds. The tolerance ε means when $\parallel r_{\mu }(x,\xi )\parallel \le \epsilon$, the procedure stops.

Table 1 Example 4.1

Table 2 Example 4.1

Example 4.1 Let $n=3$,

$$C:=\{x\in {\mathbb{R}}_{+}^n:\sum _{i=1}^n{x}_i=1\}$$

and $F:C\to 2^{{\mathbb{R}}^n}$ be defined by

$$F(x):=\{(t,t-x_1,t-x_2):t\in [0,1]\}.$$

Then the set C and the mapping F satisfy the assumptions of Theorem  3.2 and $(0,0,1)$ is a solution of the multi-valued variational inequality. Example  4.1 is tested in [6, 11]. We choose $\sigma =0.5$, $\gamma =0.9$ for our algorithm; $\sigma =0.1$, $\gamma =0.8$, $\mu =1$ for Algorithm  1 in [6]; $\sigma =0.9$, $\gamma =0.4$, $\mu =1$ for Algorithm  1 in [11]. We use $x_0=(0,0.5,0.5)$ as the initial point (Table 1 and Table 2).