# Fuzzy Evaluation and Improvement Decision-Making Model for Machining Operation Performance
## 5 Numerical Example  
This paper takes a numerical example to explain the fuzzy evaluation and improvement decision-making model proposed in this section. Suppose that the required performance value of the operation performance index \(\theta_{OPI}\) is 3.5 (\(V=3.5\)). Then, this paper considers the problem of the hypothesis test as follows:  
null hypothesis \(H_{0}:\theta_{OPI}=3.5\)  
versus  
alternative hypothesis \(H_{1}:\theta_{OPI}\
eq 3.5\).  
_Let \(T_{1},\ldots,T_{j},\ldots,T_{30}\)_ be a random sample with sample size n =30 and \(t_{j}\) be the observed value of random variable \(T_{j}\), then we can calculate that the value of \(\bar{t}\) is  
\[\bar{t}=\frac{\sum_{j=1}^{30}t_{j}}{30}=2.2 \tag{32}\]  
The administrator sets the minimum required value as \(L=\)1 for the time interval between machine failures. Then, the observed value of test statistical \(\theta_{OPI}^{*}\) is  
\[\theta_{O}^{*}=\frac{\bar{t}}{L}=\frac{2.2}{1}=2.2. \tag{33}\]  
According to Equations (17) and (18), we have \(\theta_{OPI}(\alpha)=66/G_{1-\alpha/2}(30,1)\) and \(\theta_{OPI}(\alpha)=66/G_{\alpha/2}(30,1)\) with \(\theta_{O}^{*}=2.2\). To use the EXCEL software to calculate the two function values corresponding to \(\alpha\) from 0.01 to 1.00 to find the data of membership function. Then, the values of \(\theta_{0L}\), \(\theta_{0M}\), and \(\theta_{0R}\) can be calculated as follows:  
\[\theta_{0L}=\frac{30}{G_{0.995}(30,1)}\theta_{0}^{*}=\frac{30\times 2.2}{G_{0.995 }(30,1)}=1.436 \tag{34}\]  
\[\theta_{0M}=\frac{30}{G_{0.5}(30,1)}\theta_{0}^{*}=\frac{30\times 2.2}{G_{0.5}(30,1 )}=2.225, \tag{35}\]  
and  
\[\theta_{0R}=\frac{30}{G_{0.005}(30,1)}\theta_{0}^{*}=\frac{30\times 1.3}{G_{0.0 05}(30,1)}=3.715. \tag{36}\]  
Then, the triangular-shaped fuzzy number of \(\widetilde{\theta}_{OPI}\) is \(\widetilde{\theta}_{OPI}=\Delta(1.436,\,2.225,\,3.715)\) and the membership function of fuzzy number \(\widetilde{\theta}_{OPI}\) can be shown as follows:  
\[\eta(x)=\left\{\begin{array}{ll}0,\ x<1.436\\ \alpha_{1},\ 1.436\leq x<2.225\\ 1,\ x=2.225\\ \alpha_{2},\ 2.225<x\leq 3.715\\ 0,\ 3.715<x\end{array}\right. \tag{37}\]  
where \(\alpha_{1}\) and \(\alpha_{2}\) are determined by \(G_{1-\alpha_{1}/2}(30,1)x=66\) and \(G_{\alpha_{2}/2}(30,1)x=66\). Based on the above-stated, the value of \(\theta_{0M}\) is 2.225, the value of V is 3.5, and the value of \(\theta_{0R}\) is 3.175. Then, we have \(\theta_{0M}\leq V\leq\theta_{0R}\). According to Equation (31), the value of \(d_{R}/d_{T}\) is shown as follows:  
\[d_{R}/d_{T}=\frac{3.715-3.5}{3.715-2.225}=0.144 \tag{38}\]  
Based on the rules of the fuzzy evaluation and improvement decision-making and based on \(\phi=0.15\), since \(d_{R}/d_{T}<\phi=0.15\) and \(\theta_{0M}\leq V\leq\theta_{0R}\), then we can reject \(H_{0}\) and assume that \(\theta_{OPI}<V\). Thus, the machining operation performance must be advanced. The value of upper confidence limit of index \(\theta_{OPI}\) is 3.715, which is greater than 3.5. Therefore, the statistical test does not reject \(H_{0}\). According to Equation (33), the index has an estimated value of 2.2, much smaller than 3.5. Clearly, fuzzy test methods are better than statistical test methods without missing opportunities for improvement [28; 29].  
In addition, enterprises can use the above cases to learn the method of the model proposed in this article, and at the same time, they can apply this method to actual cases. In addition to grasping the opportunity to improve quality, they can also achieve the effect of energy saving and carbon reduction.