## 4 Numerical Results  
In this section, firstly, we make a numerical comparison between Method 1 and the scaled trust region method (STRM) in [20]. We apply Method 1 and the scaled trust region method to solve Examples 1 and 2. Secondly, we use Method 1 to solve the related refinery production problem, which also has been studied in [4; 13]. Finally, numerical results about large-scale stochastic linear complementarity problems are also presented. We implement Method 1 in MATLAB and test the method on the given test problems using the reformulation from the previous section. Additionally, all of these problems were done on a PC (Acer) with \(i\)5-3210\(M\) and RAM of 2 GB. Throughout the computational experiments, the parameters in Method 1 are taken as:  
\[\sigma=0.3,\beta=0.5,\gamma=0.5.\]  
The stopping criteria for Method 1 are \(\|\theta(z_{k})\|\leq 10^{-15}\) or \(k_{max}=5000\). The parameters in the STRM method (see [20]) are taken as:  
\[\Delta_{0}=10,\Delta_{min}=10^{-6},\rho_{1}=10^{-4},\rho_{2}=0.75,\sigma_{1}=0.5,\sigma_{2}=2,\eta=0.5.\]  
The stopping criteria for the STRM method are \(\|D_{k}\Re^{k}\|\leq 10^{-15}\) or \(k_{max}=5000\). In the tables of the numerical results, DIM denotes the dimension of the problem (the dimension of the variable x); \(x^{*}\) denotes the solution of \(\theta(x,y)=0\); In the following part of this section, we give the detailed description of the given test problems. **Example 1**: _Consider \(SACP(M(\omega),q(\omega))\) with:_  
\[M(\omega)=\left( \begin{array}{cc}1-2\omega&-1\\ 0&-\omega\end{array} \right),q(\omega)=\left( \begin{array}{c}1\\ \omega+1\end{array} \right),\]  
_where \(\Omega=\{\omega_{1},\omega_{2}\}=\{0,1\}\), and \(p_{i}=P(\omega_{i}\in\Omega)=0.5,i=1,2\)._  
Numerical results of Example 1 are given in Table 1, Figures 1 and 2, respectively. \(x_{0}\) are chosen randomly in \(\Re^{2}\); \(y_{0}\) are chosen randomly in \(\Re^{4}\) and \(\lambda=0.1\). Figure 1: Numerical results for Example 1 by Method 1. The \(x\)-axis represents the iteration step; the \(y\)-axis represents \(\theta(x,y)=\frac{1}{2}\|F(x,y)\|^{2}\). \begin{table}
\begin{tabular}{c c c c c c c} \hline  & \multicolumn{3}{c}{**Method 1**} & \multicolumn{3}{c}{**STRM**} \\ \hline \(p\) & \(\mathbf{x^{*}}\) & **Final Value** & **Iteration** & \(\mathbf{x^{*}}\) & **Final Value** & **Iteration** \\ \hline  
& \(1.0\times 10^{-7}\times(0.1087,0.0382)\) & \(2.4368\times 10^{-16}\) & 9 & \(1.0\times 10^{-15}\times(0.7305,0.4090)\) & \(5.7596\times 10^{-32}\) & 9 \\ \hline \end{tabular}
\end{table} 
Table 1: Numerical results for Example 1. Figure 2: Numerical results for Example 1 by the STRM method. The \(x\)-axis represents the iteration step; the \(y\)-axis represents \(\theta(x,y)=\frac{1}{2}\|F(x,y)\|^{2}\). In Table 4, we also give the comparison of Method 1 with fmincon. For the propose of comparison, we fixed \(p=10\) and the same initial points. From Table 4, we can see that Method 1 is also more effective than fmincon. \begin{table}
\begin{tabular}{c c c} \hline  & \(\mathbf{x^{*}}\) & **Final Value** \\ \hline Method 1 & \((0,1,1)\) & \(5.0000\times 10^{-17}\) \\ fmincon & \((0,1,1)\) & \(4.9780\times 10^{-14}\) \\ \hline \end{tabular}
\end{table} 
Table 4: Numerical results for Example 2 by Method 1 and fmincon. Figure 4: Numerical results for Example 2 by the STRM method. The \(x\)-axis represents the iteration step; the \(y\)-axis represents \(\theta(x,y)=\frac{1}{2}\|F(x,y)\|^{2}\). Figure 3: Numerical results for Example 2 by Method 1. The \(x\)-axis represents the iteration step; the \(y\)-axis represents \(\theta(x,y)=\frac{1}{2}\|F(x,y)\|^{2}\). **Example 3**. : _This example is a refinery production problem, which is also considered in [2; 13]._  
The problem is defined as:  
\[M(\omega)=\left( \begin{array}{cccc}0&0&1&-2-\omega_{1}&-3\\ 0&0&1&-6&\omega_{2}-3.4\\ -1&-1&0&0&0\\ 2+\omega_{1}&6&0&-\omega_{3}&-\omega_{3}\\ 3&3.4-\omega_{2}&0&-\omega_{4}&\omega_{4}\end{array} \right),\]  
\[q(\omega)=\left(