## 3 Methodology
### Reliability Assessment Method for Series Systems Under No-Failure Scenarios
#### 3.1.1 All Components Are Different  
**Theorem 1.**_Assume that the series system consists of \(\,\)different components. Given the mission time \(t\), if we take_  
\[\eta(t)=\min\Bigl{(}\frac{T_{1}}{t^{\alpha_{1}}},\frac{T_{2}}{t^{\alpha_{2}}},\cdots,\frac{T_{m}}{t^{\alpha_{m}}}\Bigr{)} \tag{7}\]  
_then the LCL of the series system reliability \(R(t)\) with the confidence level \(\gamma\) can be calculated as_  
\[R_{\text{L},\gamma}(t)=\exp\Bigl{\{}-\chi_{\gamma}^{2}(2)/\bigl{[}2\eta(t) \bigr{]}\Bigr{\}} \tag{8}\]  
_which satisfies_  
\[P\bigl{[}R(t)\geq R_{\text{L},\gamma}(t)\bigr{]}\geq\gamma \tag{9}\]  
**Proof of Theorem 1.** Firstly, given the confidence level \(\gamma\), the LCL \(R_{\text{L},i,\gamma}(t)\) of the reliability \(R_{i}(t)\) for component \(i\) is obtained by Equation (6), which satisfies  
\[P\bigl{[}R_{i}(t)\geq R_{\text{L},i,\gamma}(t)\bigr{]}\geq\gamma,i=1,2,\cdots,m \tag{10}\]Substituting Equation (6) into Equation (10) yields  
\[P\left[\ln R_{i}(t)\geq-\frac{\chi_{\gamma}^{2}(2)t^{\mu_{i}}}{2T_{i}}\right]\geq \gamma,i=1,2,\cdots,m \tag{11}\]  
and  
\[P\left[\eta_{i}(t)\leq-\frac{\chi_{\gamma}^{2}(2)}{2\ln R_{i}(t)}\right]\geq \gamma,i=1,2,\cdots,m \tag{12}\]  
where \(\eta_{i}(t)=T_{i}/t^{\mu_{i}}\) is the minimum life distribution parameter.  
For a value of \(z_{i}\) taken by the random variable \(\eta_{i}(t)\), such that  
\[z_{i}=-\frac{\chi_{\gamma}^{2}(2)}{2\ln R_{i}(t)}=\frac{\ln(1-\gamma)}{\ln R_ {i}(t)},i=1,2,\cdots,m \tag{13}\]  
then the confidence level \(\gamma\) can be obtained as  
\[\gamma=1-\exp[z_{i}\ln R_{i}(t)]=1-R_{i}^{z_{i}}(t),i=1,2,\cdots,m \tag{14}\]  
Substituting Equations (13) and (14) into (12), the following inequality holds:  
\[P[\eta_{i}(t)\leq z_{i}]\geq 1-R_{i}^{z_{i}}(t),i=1,2,\cdots,m \tag{15}\]  
According to the minimum lifetime distribution theory in a series model, let \(\eta(t)=\min[\eta_{1}(t),\eta_{2}(t),\cdots,\eta_{m}(t)]\). Since \(\eta_{i}(t)\) are independent of each other, it can be deduced from Equation (15) that  
\[\begin{array}{rl}P[\eta(t)\leq z]&=1-\prod\limits_{i=1}^{m}\left\{1-P[\eta_ {i}(t)\leq z]\right\}\\ &\geq 1-\prod\limits_{i=1}^{m}R_{i}^{z}(t)=1-\left[\prod\limits_{i=1}^{m}R _{i}(t)\right]^{z}=1-R^{z}(t)\end{array} \tag{16}\]  
Let \(\gamma=1-R^{z}(t)\); then, we find  
\[z=\frac{\ln(1-\gamma)}{\ln R(t)}=-\frac{\chi_{\gamma}^{2}(2)}{2\ln R(t)}, \tag{17}\]  
and substitute it into Equation (16) to obtain  
\[P\left[\eta(t)\leq-\frac{\chi_{\gamma}^{2}(2)}{2\ln R(t)}\right]\geq\gamma \tag{18}\]  
By appropriate equivalent transformation, we have  
\[P\left(R(t)\geq\exp\left\{-\chi_{\gamma}^{2}(2)/[2\eta(t)]\right\}=R_{\text{L },\gamma}(t)\right)\geq\gamma \tag{19}\]  
Equation (19) is the definition for the LCL of \(R(t)\). \(\square\)  
Similar to the system reliability function \(R(t)\), its LCL \(R_{\text{L},\gamma}(t)\) also varies with the mission time \(t\). In other words, Equation (8) constructs an LCL curve of the reliability function at the confidence level \(\gamma\). For continuous distributions, the LCL curve of the reliability function is equivalent to that of the reliable life function at the same confidence level [30]. Figure 1 visualizes this equivalence of LCL curves, revealing the relationship between system reliability and reliable life.  
According to the above analysis, for a given reliability \(R\), the LCL of \(t_{R}\) at the confidence level \(\gamma\) can be solved inversely by Equation (8), i.e.,  
\[R=\exp\Bigl{\{}-\chi_{\gamma}^{2}(2)/\left[2\eta\bigl{(}t_{\mathrm{RL},\gamma} \bigr{)}\right]\Bigr{\}} \tag{20}\]  
Since Equation (20) is a strictly monotonic decreasing function of \(t_{\mathrm{RL},\gamma}\), it is recommended to solve it iteratively using the bisection method. In practical calculation, continuously adjust the mission time \(t\) in Equation (7) and repeat the series system reliability assessment process until \(R_{\mathrm{L},\gamma}(t)\) obtained from Equation (8) equals the given reliability \(R\). The corresponding time \(t\) is then the solution for \(t_{\mathrm{RL},\gamma}\).