## 5 Case Study
### Maintenance Period of Most Critical Machines for the Automotive Part Production Line  
The statistics and analysis were performed on the failure data and potential failure data of all most critical machines, and its results were taken into Equation (14); then, the delay-time function \(F(h)\), the probability of potential failure \(\theta\) and the probability of potential failure detected in preventive maintenance \(r\) were gained, that is, \(F(t_{h})=1-\exp(-0.0098t_{h})\), \(\theta=0.8012\), and \(r=0.4145\). The integral transformation of Equation (9) was carried out and while \(t=T_{n-1}\), \(t+\Delta t=T_{n-1}+T_{m}=T_{n}\). The expected number of the machine failure in the preventive maintenance period \(T_{m}\) was expressed:  
\[EN_{f}(T_{m})=\theta T_{m}-\frac{\theta\times r\times(e^{\eta T_{m}}-1)}{\eta \times(e^{\eta T_{m}}-1+r)}. \tag{21}\] \begin{table}
\begin{tabular}{c c c} \hline
**Coefficient** & **Value** & **Corresponding Machine** \\ \hline \(a\) & \(0.996\) & Constant term \\ \(b_{1}\) & \(-1.766\) & M01 \\ \(b_{2}\) & \(-1.897\) & M02 \\ \(b_{3}\) & \(-1.051\) & M03 \\ \(b_{4}\) & \(-1.129\) & M04 \\ \(b_{5}\) & \(-3.441\) & M05 \\ \(b_{6}\) & \(-1.231\) & M06 \\ \(b_{7}\) & \(-0.953\) & M07 \\ \(b_{8}\) & \(-1.551\) & M08 \\ \(b_{9}\) & \(-0.407\) & M09 \\ \(b_{10}\) & \(-1.060\) & M10 \\ \(b_{11}\) & \(-1.263\) & M11 \\ \(b_{12}\) & \(-1.848\) & M12 \\ \(b_{13}\) & \(-0.847\) & M13 \\ \(b_{14}\) & \(-0.914\) & M14 \\ \(b_{15}\) & \(-2.286\) & M15 \\ \(b_{16}\) & \(-1.405\) & M16 \\ \(b_{17}\) & \(-1.338\) & M17 \\ \(b_{18}\) & \(-1.437\) & M18 \\ \(b_{19}\) & \(-1.875\) & M19 \\ \(b_{20}\) & \(-1.977\) & M20 \\ \hline \end{tabular}
\end{table} 
Table 3: The coefficients of regression equation. Equation (21) was taken into the decision objective Equation (4):  
\[ED(T_{m})=\frac{t_{f}EN_{f}(T_{m})+t_{p}}{T_{m}}=\frac{t_{f}\Big{[}\theta T_{m}- \frac{\theta\times r\times(\varepsilon^{\eta T_{m}}-1)}{\eta\times(\varepsilon^ {\eta T_{m}}-1+r)}\Big{]}+t_{p}}{T_{m}}. \tag{22}\]  
On the basis of the statistics and analysis of maintenance data, the average shutdown time of automatic production line caused by the most critical machines' failure was 2.4 h (\(t_{f}=2.4\) h) and the average preventive maintenance time of the most critical machines was 1.5 h (\(t_{p}=1.5\) h). Then, the parameters \(F(t_{h})\), \(\theta\), \(r\), \(t_{f}\) and \(t_{p}\) were taken into Equation (22), and the relationship between the expected shutdown time of the production line and the preventive maintenance period was obtained by MATLAB software, as shown in Figure 4.  
According to Figure 4, it can be found that the shutdown time is the shortest while the preventive maintenance period \(T_{m}\) is 7.30 days and the value of shortest shutdown time is 0.44. Considering of the actual operation of the production enterprise and simplifying the maintenance plan, the preventive maintenance period for the most critical machines (M02, M05, M15, M20) was designated as seven days. In order to ensure the reliability and operating rate of the automatic production line, the reliability of each machine in the production line is not less than 0.7, namely, the cumulative failure probability of each machine should be lower than 0.3 in this case study [25]. To the best of the authors' knowledge, there is no unified standard of minimum value of machine reliability now. Thus, in this case study, we referred to the study results of reference [25]. Thus, the Weibull distribution model of each machine and reliability of each machine were calculated when the \(T_{m}\) was 7. The results are shown in Table 4.  
It can be seen from Table 4 that the reliability of each machine in the preventive maintenance period are higher than 0.7. Therefore, while the preventive maintenance period of the most critical machine was designed as seven days, the production line had the shortest shutdown time and sufficient reliability.