# Classification of Systems and Maintenance Models
## 4 Classification of Maintenance Models
### Mean Times for the System to Stay in Different States
}^{5}E\big{(}T_{i}^{s}\big{)}}\sum\limits_{k=0}^{\infty} \int\limits_{0}^{\tau-\rho}\int\limits_{k\tau+\pi+\rho}^{\infty}P_{TP}(\tau, \,\ldots,k\tau\mid\theta)\omega(\theta)d\theta d\tau \tag{94}\]  
To prove (94), we express the probability \(A_{ma}(k\tau,\,\rho)\) using the renewal density function and then proceed to the limit. \[A_{ma}(\rho)=\lim\limits_{k\to\infty}A_{ma}(k\tau,\rho) \tag{95}\]  
Since system recovery is only possible at discrete moments of time \(k\tau\) (\(k=0,\,1,\,2,\,\ldots\,\) ), we express the renewal density function through the \(\delta\)-function:  
\[h(x)=\sum\limits_{j=0}^{k}P_{R}(j\tau)\delta(x-j\tau) \tag{96}\]  
Using (96), we present (92) in the integral form. \[A_{ma}(k\tau,\rho)=\int\limits_{0}^{k\tau}P(k\tau-x+\rho)h(x)dx \tag{97}\]  
Furthermore, because the function \(P(k\tau-x+\rho)\) is not negative, it has limited variation on the semi-axis (0, \(\infty\)) and satisfies the following inequality:  
\[\int\limits_{0}^{\infty}P(t)dt\leq\int\limits_{0}^{\infty}[1-F(t)]dt<\infty \tag{98}\]  
Subsequently, according to [PERSON]'s theorem in the case of a lattice random variable [95], we have:  
\[\lim\limits_{k\to\infty}\int\limits_{0}^{k\tau}P(k\tau-x+\rho)h(x)dx=\frac{ \tau}{\alpha}\sum\limits_{k=0}^{\infty}P(k\tau+\rho) \tag{99}\]  
where \(F(t)\) is the unreliability function and \(\alpha\) is the average time between system recoveries. From (93), it follows that:  
\[P(k\tau+\rho)=\frac{1}{\tau-\rho}\int\limits_{0}^{\tau-\rho}\int\limits_{k\tau+ \tau+\rho}^{\infty}P_{TP}(\tau,\,\ldots,k\tau|\theta)\omega(\vartheta)d\vartheta d\pi \tag{100}\]  
Since the system is used in intermittent mode (\(l=2\)):  
\[\alpha=\sum\limits_{ \begin{array}{c}i=1\\ i\
eq 3\end{array} }^{5}E(T_{i}^{s}) \tag{101}\]  
Substituting (100) and (101) into (99), we obtain (94). For subgroup \(I_{1311}\), the achieved availability \(A_{a}^{r}\) is the maintenance effectiveness indicator. We determined the achieved availability using the well-known properties of regenerative processes:  
\[A_{a}^{r}=E(T_{1}^{s})\Bigg{/}\sum\limits_{i=1}^{5}E(T_{i}^{s}) \tag{102}\]  
For subgroup \(I_{1321}\), the maintenance effectiveness indicator can be either AOPUT or AMCUT, which we determined as follows:  
\[W_{1}^{r}=\Big{[}C_{prof}E(T_{1}^{s})-C_{uf}E(T_{2}^{s})-(C_{ins}+C_{dt})E(T_{3 }^{s})\,-\,\,(C_{PR}+C_{dt})E(T_{4}^{s})-(C_{CR}+C_{dt})E(T_{5}^{s})\Bigg{]} \Bigg{/}\sum\limits_{i=1}^{5}E(T_{i}^{s}) \tag{103}\]  
\[V_{1}^{r}=\Big{[}C_{uf}E(T_{2}^{s})+(C_{ins}+C_{dt})E(T_{3}^{s})\,+(C_{PR}+C_{ dt})E(T_{4}^{s})\,+(C_{CR}+C_{dt})E(T_{5}^{s})\Big{]}\Bigg{/}\sum\limits_{i=1}^{5}E(T _{i}^{s}) \tag{104}\]  
where \(C_{prof}\) is the average profit from using the system per unit time and \(C_{uf}\) is the average loss due to the system being in a hidden failure state per unit time. For subgroup \(I_{1412}\), we determined the inherent availability as follows:  
\[A_{i}^{r}=\left\{ \begin{array}{c}E(T_{1}^{s})\Bigg{/}\sum\limits_{i=1}^{5} E(T_{i}^{r}),\text{if only CBM inspections can be carried out during time intervals when the system is not used}\\ i\
eq 3\\ E(T_{1}^{s})\Bigg{/}\sum\limits_{i=1}^{5}E(T_{i}^{s}),\text{if both CBM inspections and preventive and corrective repair can be carried out during}\\ \text{time intervals when the system is not used}\end{array} \right. \tag{105}\]  
For subgroup \(I_{1422}\), the maintenance effectiveness indicators were AOPUT and AMCUT; however, because the mode of operation was intermittent, \(C_{dt}=0\) for the cost component associated with scheduled CBM inspections. \[W_{2}^{r}=\frac{C_{prof}E\big{(}T_{1}^{s}\big{)}-C_{uf}E(T_{2}^{s})-C_{ins}E \big{(}T_{3}^{s}\big{)}-(C_{PR}+C_{dt})E\big{(}T_{4}^{s}\big{)}-(C_{PR}+C_{dt })E\big{(}T_{5}^{s}\big{)}}{\sum\limits_{ \begin{array}{c}i=1\\ i\
eq 3\end{array} }^{5}E\big{(}T_{i}^{s}\big{)}} \tag{106}\]  
\[V_{2}^{r}=\Big{[}C_{uf}E(T_{2}^{s})+C_{ins}E(T_{3}^{s})\,+(C_{PR}+C_{dt})E(T_{ 4}^{s})+(C_{PR}+C_{dt})E(T_{5}^{s})\Bigg{]}\Bigg{/}\sum\limits_{ \begin{array}{c}i=1\\ i\
eq 3\end{array}