# Research on Reliability Assessment of Mechanical Equipment Based on the Performance-Feature Model
## 4 Case Study
### Reliability Evaluation of Tools  
The PFCs obtained are regarded as complete failure characteristic data, and the tool was evaluated according to the reliability evaluation method of the exponential distribution. The reliability function of the model A is shown below:  
\[R(c)=R_{G}(Gnt)=e^{-Gn/15.2544} \tag{14}\]  
Figure 4 shows a reliability curve that follows an exponential distribution.  
When the reliability is equal to 0.5, the corresponding eigenvalue is defined as the median eigenvalue \(Gn_{R=0.5}\), and when the reliability is equal to \(e^{-1}\), the corresponding eigenvalue is defined as the reliable eigenvalue \(Gn_{R=\mathrm{e}^{-1}}\).It can be obtained by Formula (14) that,  
\[Gn_{R=0.5}=10.5735\]  
\begin{table}
\begin{tabular}{c c c c c c} \hline \(i\) & \(x_{i}\) & \(F_{0}(x_{i})\) & \((i-1)/n\) & \(i/n\) & \(\delta_{i}\) \\ \hline
**1** & 4.26 & 0.4963 & 0 & 0.111 & 0.4963 \\
**2** & 4.55 & 0.5193 & 0.111 & 0.222 & 0.4083 \\
**3** & 5.11 & 0.5607 & 0.222 & 0.333 & 0.3387 \\
**4** & 5.31 & 0.5746 & 0.333 & 0.444 & 0.2416 \\
**5** & 6.13 & 0.6272 & 0.444 & 0.556 & 0.1832 \\
**6** & 6.14 & 0.6278 & 0.556 & 0.667 & 0.0718 \\
**7** & 6.49 & 0.6482 & 0.667 & 0.778 & 0.1298 \\
**8** & 7.14 & 0.6832 & 0.778 & 0.889 & 0.2058 \\
**9** & 10.79 & 0.8239 & 0.889 & 1 & 0.1761 \\ \hline \end{tabular}
\end{table}
Table 6: KS test of Logarithmic distribution.  
Figure 4: Reliability curve of exponential distribution.  
\[Gn_{R=\text{e}^{-1}}=15.2544\]  
The number of cuts of tool wear limit, marked as \(N_{t}\), and the number of cuts based on the characteristic value were found, then the number of cuts based on the median eigenvalue was marked as \(N_{0.5}\), and the number of cuts based on the natural eigenvalue was recorded as \(N_{e}\). The relative error between actual cutting times and estimated cutting times was calculated and used to validate the model and the method. The results are shown in Table 7.  
The estimated cutting times were higher than the test times, so the error was larger.  
The reliability function of the model B is shown below:  
\[R(c)=R_{L}(\text{Gn})=e^{-Gn/6.212} \tag{15}\]  
Figure 5 shows a reliability curve that follows an exponential distribution.  
It can be obtained by Formula (15) that,  
\[Gn_{R=0.5}=4.306\]  
\[Gn_{R=e^{-1}}=6.212\]  
The results of relative error and the number of cuts are shown in Table 8.  
\begin{table}
\begin{tabular}{c c c c c c c} \hline \hline
**Sample Number** & **1** & **2** & **3** & **4** & **5** & **6** \\ \hline \(N_{t}\) & 285 & 274 & 276 & 285 & 236 & 230 \\ \(N_{0.5}\) & \(>\)315 & \(>\)315 & \(>\)315 & 274 & 274 & 274 \\ \(N_{e}\) & \(>\)315 & \(>\)315 & \(>\)315 & \(>\)315 & \(>\)315 & \(>\)315 \\ \hline \hline \end{tabular}
\end{table}
Table 7: The number of cuts corresponding to the failure threshold.  
Figure 5: Reliability curve of exponential distribution.  
The relative error was concentrated between one and five percent, and the verification result was good, so compared with model A, model B was superior and more suitable to evaluate tool reliability.