## 4 Numerical Results
\begin{array}{c}c\\ -b\\ -180-\omega_{3}\\ -162-\omega_{4}\end{array} \right),\]  
where \(\omega_{1}\), \(\omega_{2}\), \(\omega_{3}\) and \(\omega_{4}\) satisfy the following distribution:  
\[dist\omega_{1}\approx u[-0.8,0.8],\]  
\[dist\omega_{2}\approx exp(\lambda=2.5),\]  
\[dist\omega_{3}\approx N(0,12),\]  
\[dist\omega_{4}\approx N(0,9).\]  
* Generate samples \(\omega_{j}^{k},j=1,2,3,4,k=1,2,\ldots,K\), respectively, from their 99% confidence intervals (except uniform distributions): \[\omega_{1}\in I_{1}=[-0.8,0.8],\] \[\omega_{2}\in I_{2}=[0.0,1.84],\] \[\omega_{3}\in I_{3}=[-30.91,30.91],\] \[\omega_{4}\in I_{4}=[-23.18,23.18],\]  
* For each \(j\), divide the \(I_{j}\) into \(m_{j}\) cells \(I_{j,i}\), \(i=1,2,\ldots,m_{j}\). * For each \((j,i)\), calculate the average \(v_{j,i}\) of \(\omega_{j}^{k}\); it belongs to \(I_{j,i}\). * For each \((j,i)\), the estimated probability of \(v_{j,i}\) is \(p_{j,i}=k_{j,i}/K\), where \(k_{j,i}\) is the number of \(\omega_{j}^{k}\in I_{j,i}\). * Let \(N=m_{1}\times m_{2}\times m_{3}\times m_{4}\), and set the joint distribution of \(\{(\omega^{\ell},p^{\ell}),\ell=1,2,\ldots,N\}\),  
\[\omega^{\ell}=\left( \begin{array}{c}v_{1,i_{1}}\\ v_{2,i_{2}}\\ v_{3,i_{3}}\\ v_{4,i_{4}}\end{array} \right),p^{\ell}=p_{1,i_{1}}p_{2,i_{2}}p_{3,i_{3}}p_{4,i_{4}}\]  
for \(i_{1}=1,\ldots,m_{1}\), \(i_{2}=1,\ldots,m_{2}\), \(i_{3}=1,\ldots,m_{3}\), \(i_{4}=1,\ldots,m_{4}\). In the following part of this section, we use Method 1 to solve the constrained optimization problem:  
\[\min_{z\geq 0}\theta(z)=\frac{1}{2}\|F(z)\|^{2},\]  
where \(z=(x,y)\),  
\[F(x,y)=\left( \begin{array}{c}\phi_{n}(x)\\ M(\omega^{\ell})x+q(\omega^{\ell})-y_{\ell},\ell=1\cdots N\end{array} \right).\]and:  
\[\phi_{n}(x)=\left( \begin{array}{c}\lambda\phi_{p}(x_{i},\bar{M}_{i}x+\bar{q}_{i }),i=1\cdots 5\\ \vdots\\ (1-\lambda)\phi_{+}(x_{i},\bar{M}_{i}x+\bar{q}_{i}),i=1\cdots 5\end{array}  \right).\]  
Now, we examine the following two conditions:  
\[Condition1:\omega_{1}\equiv 0,\omega_{2}\equiv 0,m_{3}=15,m_{4}=15.\]  
\[Condition2:m_{1}=5,m_{2}=9,m_{3}=7,m_{4}=11.\]  
The numerical results of Example 3 are given in Tables 5 and 6, where \(\Psi(x)=\frac{1}{2}\|\phi_{n}(x)\|^{2}\) is the merit function; \(\theta(x,y)=\frac{1}{2}\|F(x,y)\|^{2};k=10^{i},i=3,4,5\). \(2x_{1}+3x_{2}\) is the initial production cost; and \(\lambda=0.5\). In [13], in the case of \(\omega_{1}\equiv 0,\omega_{2}\equiv 0.4,m_{3}=15,m_{4}=15\). [PERSON] and [PERSON] get the optimal solution \((x_{1},x_{2})=(38.539,20.539)\); initial production cost \(2x_{1}+3x_{2}=138.695\). Here, by Method 1, we get the optimal solution \((x_{1},x_{2})=(41.6939,16.1036)\), and the production cost is \(2x_{1}+3x_{2}=131.6985\). **Remark 1**. : _In this paper, we use:_  
\[\omega_{i}=\left\{ \begin{array}{c}\omega_{j},i=j,\\ E(\omega_{j}),i\
eq j.\end{array} \right. \tag{14}\]  
_The computation cost of our method is greatly reduced. In fact, when we think about the general case of \(\omega_{1},\omega_{2},\omega_{3}\) and \(\omega_{4}\) varying the random distribution of discrete approximation by a 5-, 9-, 7- and 11-point distribution, respectively. This yields a joint discrete distribution of \(5\times 9\times 7\times 11=3465\) realizations. Then, \(F(z)\) is a function of 17,335 \((3465\times 5+10=17,335)\) dimensions. This is a more complex optimization problem._