## 3 IA for Two-Unit Parallel System with Three States
### Mathematical Model of IA  
According to [22], for establishing the Markov model of IA for the two-unit parallel system with three states, the system must meet the following assumptions. **Assumption 3**. : _Each unit has its own maintenance equipment, when the unit is down, its maintenance equipment will repair the unit immediately._  
**Assumption 4**. : _Each unit i has the working time \(X_{i}\), the delay-repair time \(W_{i}\) and repair time \(Y_{i}\)\((i=1,2)\), which are subject to the following exponential functions_  
\[X_{i} \sim F_{i}(t) =1-e^{-\lambda_{i}t}, \tag{29}\] \[W_{i} \sim W_{i}(t) =1-e^{-v_{i}t},\] (30) \[Y_{i} \sim G_{i}(t) =1-e^{-\mu_{i}t}, \tag{31}\]  
_where \(\lambda_{i}\) is the fault rate, \(\mu_{i}\) delay-repair rate and \(v_{i}\) repair rate._  
**Assumption 5**. : _After maintenance of the unit, it will be as new and we suppose that at the beginning all units are new._  
Under these assumptions, we can get all states of the system:  
* unit 1 and unit 2 are up. * unit 1 is up, unit 2 is delay-repair. * unit 1 is delay-repair, unit 2 is up. * unit 1 is up, unit 2 is repaired. * unit 1 is repaired, unit 2 is up. * unit 1 is delay-repair, unit 2 is delay-repair. * unit 1 is repaired, unit 2 is repaired. The state transition diagram is shown in Figure 4.  
According to [22], for this Markov process, we have the state transfer probability function as follows  
\[\left\{ \begin{array}{l}P_{00}(\Delta t)=1-(\lambda_{1}+\lambda_{2})\Delta t +o(\Delta t),P_{01}(\Delta t)=\lambda_{2}\Delta t+o(\Delta t),P_{02}(\Delta t) =\lambda_{1}\Delta t+o(\Delta t),\\ P_{11}(\Delta t)=1-(\lambda_{1}+v_{2})\Delta t+o(\Delta t),P_{13}(\Delta t)=v_ {2}\Delta t+o(\Delta t),P_{15}(\Delta t)=\lambda_{1}\Delta t+o(\Delta t),\\ P_{22}(\Delta t)=1-(\lambda_{2}+v_{1})\Delta t+o(\Delta t),P_{24}(\Delta t)=v_ {1}\Delta t+o(\Delta t),P_{25}(\Delta t)=\lambda_{2}\Delta t+o(\Delta t),\\ P_{30}(\Delta t)=\mu_{2}\Delta t+o(\Delta t),P_{33}(\Delta t)=1-(\lambda_{1}+ \mu_{2})\Delta t+o(\Delta t),P_{36}(\Delta t)=\lambda_{1}\Delta t+o(\Delta t), \\ P_{40}(\Delta t)=\mu_{1}\Delta t+o(\Delta t),P_{44}(\Delta t)=1-(\lambda_{2}+ \mu_{1})\Delta t+o(\Delta t),P_{47}(\Delta t)=\lambda_{2}\Delta t+o(\Delta t), \\ P_{55}(\Delta t)=1-(v_{1}+v_{2})\Delta t+o(\Delta t),P_{56}(\Delta t)=v_{2} \Delta t+o(\Delta t),P_{57}(\Delta t)=v_{1}\Delta t+o(\Delta t),\\ P_{62}(\Delta t)=u_{2}\Delta t+o(\Delta t),P_{66}(\Delta t)=1-(v_{1}+u_{2}) \Delta t+o(\Delta t),P_{68}(\Delta t)=v_{1}\Delta t+o(\Delta t),\\ P_{71}(\Delta t)=u_{1}\Delta t+o(\Delta t),P_{77}(\Delta t)=1-(v_{2}+u_{1}) \Delta t+o(\Delta t),P_{78}(\Delta t)=v_{2}\Delta t+o(\Delta t),\\ P_{83}(\Delta t)=u_{1}\Delta t+o(\Delta t),P_{84}(\Delta t)=u_{2}\Delta t+o( \Delta t),P_{88}(\Delta t)=1-(u_{1}+u_{2})\Delta t+o(\Delta t),\end{array} \right. \tag{32}\]  
and the elements \(m_{ij}\) of the transfer rate matrix \(M\) satisfies  
\[\left\{ \begin{array}{l}\lim_{\Delta t\to 0}\frac{P_{ij}(\Delta t)}{ \Delta t}=m_{ij}\ \ \ \ \ i\
eq j,\\ \lim_{\Delta t\to 0}-\frac{1-P_{ii}(\Delta t)}{\Delta t}=m_{ii}.\end{array} \right. \tag{33}\]So the transfer rate matrix \(M\) is as follows  
\[M=-\left[ \begin{array}{ccccccccc}\lambda_{1}&-\lambda_{2}&-\lambda_{1}&0&0&0&0&0 &0\\ 0&\lambda_{1}+v_{2}&0&-v_{2}&0&-\lambda_{1}&0&0&0\\ 0&0&\lambda_{2}+v_{1}&0&-v_{1}&-\lambda_{2}&0&0&0\\ -u_{2}&0&0&\lambda_{1}+u_{2}&0&0&-\lambda_{1}&0&0\\ -u_{1}&0&0&0&\lambda_{2}+u_{1}&0&0&-\lambda_{2}&0\\ 0&0&0&0&v_{1}+v_{2}&-v_{2}&-v_{1}&0\\ 0&0&-u_{2}&0&0&0&v_{1}+u_{2}&0&-v_{1}\\ 0&-u_{1}&0&0&0&0&0&v_{+}u_{1}&-v_{2}\\ 0&0&0&-u_{1}&-u_{2}&0&0&0&u_{1}+u_{2}\end{array} \right] \tag{34}\]  
Let \(P(t)=\left(P_{0}(t),P_{1}(t),P_{2}(t),P_{3}(t),P_{4}(t),P_{5}(t),P_{6}(t),P_{7 }(t),P_{8}(t)\right)\), where \(P_{i}\left(t\right),\ i=0,1,2,3\), denote the probability that state \(i\) occur at time \(t\). So the ordinary differential equation of \(P(t)\) can be obtained as follows  
\[\left\{ \begin{array}{l}P^{\prime}\left(t\right)=P\left(t\right)M,\\ P\left(0\right)=(1,0,0,0,0,0,0,0).\end{array} \right.