## 3 Methodology
### Reliability of Standby Redundancy  
A two-component standby system succeeds in two conditions: (1) when the active component, the sensing subsystem and the switch do not fail; (2) the active component fails before the end of the mission, the sensing and switching subsystems do not fail, and the standby component not having already failed succeeds for the remainder of the mission. The reliability block diagram of such a standby system is given in Figure 8.  
Figure 7: Load-sharing with two redundant components (\(k_{1}+k_{2}=1\)). Assuming that the failures of different components and the failures due to different failure modes are independent, the probability of the first condition is expressed mathematically by  
\[R_{1}^{sb}(t)=R_{1E}(t)\cdot R_{SE}(t)\cdot R_{SWO}(t) \tag{20}\]  
where \(R_{1E}(t)\) is the reliability of component 1 in the energized mode at time \(t\) during the whole mission, \(R_{SE}(t)\) means the reliability of the sensing subsystem at time \(t\), and \(R_{SWO}(t)\) represents the switch does not fail open from 0 to \(t\). For the second condition, component 1 fails in a time interval \((t_{1},t_{1}+\Delta t_{1})\) before time \(t\), and the sensing subsystem does not fail by \(t_{1}\). The switching subsystem functions in the quiescent mode by \(t_{1}\), and the switch successfully switches in the standby component in the energized mode with a one-cycle operation, at which time the standby component does not fail by \(t_{1}\) in quiescent mode. Following this, the standby component will function successfully for the remainder of the mission in the energized mode. The switch does not fail open until the end of the mission. Therefore, the probability of the second condition can be derived as follows  
\[R_{2}^{sb}(t)=\int_{t_{1}=0}^{t}f_{1}(t_{1})\cdot R_{SE}(t_{1})\cdot R_{SWQ}(t _{1})\cdot R_{SWO}(t_{1},t-t_{1})\cdot R_{SWE}(1\;\text{cycle})\cdot R_{2Q}(t_ {1})\cdot R_{2E}(t_{1e},t-t_{1})dt_{1} \tag{21}\]  
where \(t_{1}\) is a variable quantity (\(0\leq t_{1}\leq t\)). Assuming that all components have constant failure rates and the lifetime of each component follows an Exponential distribution. From the formula \(R_{2Q}(t_{1})=R_{2E}(t_{1e})\), taking the natural logarithm of both sides of Equation \(e^{-\lambda_{2Q}t_{1}}=e^{-\lambda_{2E}t_{1e}}\), the equivalent time \(t_{1e}=\frac{\lambda_{2Q}t_{1}}{\lambda_{2E}}\) can be obtained. Finally, the reliability of standby systems can be expressed as  
\[R_{sys}^{sb}(t)= R_{1}^{sb}(t)+R_{2}^{sb}(t)\] \[= e^{-(\lambda_{1}+\lambda_{SE}+\lambda_{SWO})t}+\lambda_{1}e^{-( \lambda_{SWE}+\lambda_{SWO}t+\lambda_{2E}t)}\int_{0}^{t}e^{-\left(\lambda_{1} +\lambda_{SE}+\lambda_{SWQ}-\lambda_{SWO}+\lambda_{2Q}-\lambda_{2E}\right)t_ {1}}dt_{1} \tag{22}\]  
In addition, for a standby system with \(n\) identical components where one component is functioning and \((n-1)\) components are in standby, the reliability of the standby system can be calculated by  
\[R_{sys}^{sb}(t)=P(N_{f}\leq n-1)=e^{-\lambda t}\left[1+\lambda t+\frac{( \lambda t)^{2}}{2!}+\cdots+\frac{(\lambda t)^{n-1}}{(n-1)!}\right] \tag{23}\]  
where \(N_{f}\) is the number of components that fail. If the lifetime of components in a standby system follows a two-parameter Weibull distribution with parameters \(\lambda\) and \(\gamma\), the equivalent time \(t_{1e}=\frac{1}{\lambda_{2E}}\left(\lambda_{2Q}t_{1}\right)^{\gamma_{2Q}/ \gamma_{2E}}\) and the \(i\)th component reliability \(R_{i}(t)=e^{-(\lambda_{i}t)^{\gamma_{i}}}\) can be obtained. The reliability formula of the standby system can also be derived by the proposed method from Equations (20) and (21)  
Figure 8: Two-unit standby system with sensing and switching subsystems.