Assessment of the maintenance cost and analysis of availability measures in a finite life cycle for a system subject to competing failures

Abstract

This paper deals with the assessment of the maintenance cost and the performance of a system under a finite planning horizon. The system is subject to two dependent causes of failure: internal degradation and sudden shocks. We assume that internal degradation follows a gamma process. When the deterioration level of the degradation process exceeds a threshold, a degradation failure occurs. Sudden shocks arrive at the system following a doubly stochastic Poisson process (DSPP). A sudden shock provokes the system failure. A condition-based maintenance (CBM) with periodic inspection times is implemented. Recursive methods combining numerical integration and Monte Carlo simulation are developed to evaluate the expected cost rate and its standard deviation. Also, recursive methods to calculate the reliability, the availability and the interval reliability of the system are given. Numerical examples are provided to illustrate the analytical results.

Appendices

Appendix A

For $t<T$, $E\left[ C(t)\right]$ is given by

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\left[ C(t)\right] &{}=C_d E\left[ (t-Y) \mathbf {1}_{\{\sigma _{M_s}<Y<t,~Y<\sigma _L\}}\right] +C_d E\left[ (t-\sigma _L) \mathbf {1}_{\{\sigma _{M_s}<\sigma _L<t,~\sigma _L<Y\}}\right] \\ &{}\quad +C_d E\left[ (t-Y) \mathbf {1}_{\{Y<t,~Y<\sigma _{M_s}\}}\right] .\\ \end{aligned} \end{array} \end{aligned}$$

That is,

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\left[ C(t)\right] &{}=C_d\displaystyle \int _{0}^{t}f_{\sigma _{M_s}}(u)\int _{u}^{t}\left[ -\frac{\partial }{\partial v}I(u,v)\right] {\bar{F}}_{\sigma _{L}-\sigma _{M_s}}(v-u)(t-v)~dv~du\\ &{}\quad +C_d\displaystyle \int _{0}^{t}f_{\sigma _{M_s}}(u)\int _{u}^{t}I(u,v)f_{\sigma _{L}-\sigma _{M_s}}(v-u)(t-v)~\mathrm{d}v~\mathrm{d}u\\ &{}\quad +C_d\displaystyle \int _{0}^{t}f_1(u){\bar{F}}_{\sigma _{M_s}}(u)(t-u)\mathrm{d}u.\\ \end{aligned} \end{array} \end{aligned}$$

For $t\ge T$, $E\left[ C(t)\right]$ is conditioned to D

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\left[ C(t)\right]&=E\left[ C(t),D\le t\right] +E\left[ C(t),D> t\right] . \end{aligned} \end{array} \end{aligned}$$

Thus, if $D>t$

$$\begin{aligned} \begin{array}{l} \begin{aligned} E \left[ C(t),D> t\right] &{}=\lfloor t/T\rfloor C_I\Bigg (1-\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }P_{D}(kT)\Bigg ) \\ &{}\quad +C_dE\left[ W(\lfloor t/T\rfloor T,t)\right] \Bigg (1-\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }P_{D}(kT)\Bigg ). \end{aligned} \end{array}\end{aligned}$$

If $D\le t$, $E\left[ C(t)\right]$ can be split into two terms: the cost in the first renewal cycle $\left( C(D)\right)$ and the cost in the remaining time horizon $\left( C(D,t)\right)$. Since C(D) and C(D, t) are independent, we get

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\left[ C(t),D\le t\right]&=E\left[ C(D),D\le t\right] +E\left[ C(D,t),D\le t\right] . \end{aligned} \end{array} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\left[ C(D),D\le t\right] &{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }E\left[ C(D),D=kT\right] \\ &{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor } \left( C_c+C_I(k-1)+C_d E\left[ W((k-1)T,kT)\right] \right) P_{D,c}(kT) \\ &{}\quad +\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }\left( C_p+C_I(k-1)\right) P_{D,p}(kT). \end{aligned} \end{array} \end{aligned}$$

Since C(D, t) is stochastically the same as $C(t-D)$,

$$\begin{aligned} E\left[ C(D,t),D=kT\right] =E\left[ C(t-kT)\right] P_{D}(kT). \end{aligned}$$

Hence, $E\left[ C(t)\right]$ verifies the following recursive equation:

$$\begin{aligned}E\left[ C(t)\right] =\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }E\left[ C(t-kT)\right] P_{D}(kT)+G(t),\end{aligned}$$

being

$$\begin{aligned} \begin{array}{l} \begin{aligned} G(t)&{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }\Big (C_p+C_I(k-1)\Big )P_{D,p}(kT)\\ &{}\quad +\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }\Big (C_c+C_I(k-1)+C_dE\left[ W((k-1)T,kT)\right] \Big )P_{D,c}(kT)\\ &{}\quad + \left( \lfloor t/T\rfloor C_I +C_dE\left[ W(\lfloor t/T\rfloor T,t)\right] \right) \Big (1-\sum _{k=1}^{\lfloor t/T\rfloor } P_{D}(kT)\Big ),\\ \end{aligned} \end{array} \end{aligned}$$

and the result holds.

Appendix B

For $t<T$, the expected square cost, $E\left[ C(t)^2\right]$, is given by

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\left[ C(t)^2\right] &{}=C_d^2 E\left[ (t-Y)^2 \mathbf {1}_{\{\sigma _{M_s}<Y<t,~Y<\sigma _L\}}\right] +C_d^2 E\left[ (t-\sigma _L)^2 \mathbf {1}_{\{\sigma _{M_s}<\sigma _L<t,~\sigma _L<Y\}}\right] \\ &{}\quad +C_d^2 E\left[ (t-Y)^2 \mathbf {1}_{\{Y<t,~Y<\sigma _{M_s}\}}\right] .\\ \end{aligned} \end{array} \end{aligned}$$

That is,

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\left[ C(t)\right] =&{}C_d^2\displaystyle \int _{0}^{t}f_{\sigma _{M_s}}(u)\int _{u}^{t}\left[ -\frac{\partial }{\partial v}I(u,v)\right] {\bar{F}}_{\sigma _{L}-\sigma _{M_s}}(v-u)(t-v)^2~\mathrm{d}v~\mathrm{d}u\\ &{}\quad +C_d^2\displaystyle \int _{0}^{t}f_{\sigma _{M_s}}(u)\int _{u}^{t}I(u,v)f_{\sigma _{L}-\sigma _{M_s}}(v-u)(t-v)^2~\mathrm{d}v~\mathrm{d}u\\ &{}\quad + C_d^2\displaystyle \int _{0}^{t}f_1(u){\bar{F}}_{\sigma _{M_s}}(u)(t-u)^2\mathrm{d}u.\\ \end{aligned} \end{array} \end{aligned}$$

For $t\ge T$, $E\Big [C(t)^2\Big ]$ is conditioned to D

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\left[ C(t)^2\right]&=E\left[ C(t)^2,D\le t\right] +E\left[ C(t)^2,D> t\right] . \end{aligned} \end{array} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{array}{l} \begin{aligned} E&\left[ C(t)^2,D> t\right] =\Big (\lfloor t/T\rfloor C_I +C_dE\left[ W_{T}^{M}(\lfloor t/T\rfloor T,t)\right] \Big )^2\Bigg (1-\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }P_{D}(kT)\Bigg ), \end{aligned} \end{array} \end{aligned}$$

where $P_{D}$ is given by (9). On the other hand,

$$\begin{aligned} \begin{array}{l} \begin{aligned} E&\left[ C(t)^2,D\le t\right] =E\left[ \left( C(D)+C(D,t)\right) ^2,D\le t\right] . \end{aligned} \end{array} \end{aligned}$$

Developing the expression

$$\begin{aligned} \begin{array}{l} \begin{aligned} E&{}\left[ \left( C(D)+C(D,t)\right) ^2,D\le t\right] =E\left[ C(D)^2,D\le t\right] +E\left[ C(D,t)^2,D\le t\right] \\ +&{}E\left[ 2~ C(D)C(D,t),D\le t\right] . \end{aligned} \end{array} \end{aligned}$$

Thus,

$$\begin{aligned} \begin{array}{l} \begin{aligned} E\Big [C(D)^2,D\le t\Big ] &{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }E\left[ C(D)^2,D=kT\right] \\ &{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }\left( C_c+C_I(k-1)+C_dE\left[ W((k-1)T,kT)\right] \right) ^2P_{D,c}(kT) \\ &{}\quad +\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }\left( C_c+C_I(k-1)\right) ^2P_{D,p}(kT). \end{aligned} \end{array} \end{aligned}$$

Following the same reasoning as in “Appendix A”,

$$\begin{aligned}&E \left[ C(D)C(D,t),D\le t\right] \\&\quad =\sum _{k=1}^{\lfloor t/T \rfloor }\left( C_c+C_I(k-1)\right) E\left[ C(t-kT)\right] P_{D,c}(kT)\\&\qquad +\sum _{k=1}^{\lfloor t/T \rfloor }C_dE\left[ W((k-1)T,kT)\right] E\left[ C(t-kT)\right] P_{D,c}(kT)\\&\qquad +\sum _{k=1}^{\lfloor t/T \rfloor }\left( C_p+C_I(k-1)\right) E\left[ C(t-kT)\right] P_{D,p}(kT). \end{aligned}$$

Hence, $E\left[ C(t)^2\right]$ verifies the following recursive equation:

$$\begin{aligned} E\left[ C(t)^2\right] =\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }E\left[ C(t-kT)^2\right] P_{D}(kT)+H(t), \end{aligned}$$

being

$$\begin{aligned} H(t)&=\displaystyle \sum _{k=1}^{\lfloor t/T \rfloor } \left( C_p+C_I(k-1)\right) ^2P_{D,p}(kT)\\&\quad +\displaystyle \sum _{k=1}^{\lfloor t/T \rfloor } \left( C_c+C_I(k-1)+C_dE\left[ W((k-1)T,kT)\right] \right) ^2P_{D,c}(kT)\\&\quad +2\sum _{k=1}^{\lfloor t/T \rfloor }\left( C_c+C_I(k-1)\right) E\left[ C(t-kT)\right] P_{D,c}(kT)\\&\quad +2\sum _{k=1}^{\lfloor t/T \rfloor }C_dE\left[ W((k-1)T,kT)\right] E\left[ C(t-kT)\right] P_{D,c}(kT)\\&\quad +2\sum _{k=1}^{\lfloor t/T \rfloor }\left( C_p+C_I(k-1)\right) E\left[ C(t-kT)\right] P_{D,p}(kT)\\&\quad +\Big (\lfloor t/T\rfloor C_I +C_dE\left[ W(\lfloor t/T\rfloor T,t)\right] \Big )^2\Bigg (1-\sum _{k=1}^{\lfloor t/T\rfloor }P_{D}(kT)\Bigg ), \end{aligned}$$

and the result holds.

Appendix C

For $t<T$, A(t) is given by

$$\begin{aligned} \begin{array}{l} \begin{aligned} A(t)&{}= P\left[ t<\sigma _{M_s},~Y>t\right] +P\left[ \sigma _{M_s}<t<\sigma _L,~Y>t\right] \\ &{}={\bar{F}}_{\sigma _{M_s}}(t){\bar{F}}_{1}(t)+\displaystyle \int _{0}^{t}f_{\sigma _{M_s}}(u){\bar{F}}_{\sigma _{L}-\sigma _{M_s}}(t-u)I(u,t)\mathrm{d}u.\\ \end{aligned} \end{array} \end{aligned}$$

For $t\ge T$, A(t) is conditioned to the time to the first renewal

$$\begin{aligned} \begin{array}{l} \begin{aligned} A(t)&{}= \displaystyle \sum _{j=0}^{\infty }\mathbf {1}_{\{R_{j}\le t<R_{j+1}\}}\Big [P\left[ O(t)<L,~Y>(t-R_j),~D\le t\right] \\ &{}\quad +P\left[ O(t)<L,~Y>(t-R_j),~D> t\right] \Big ]. \end{aligned} \end{array} \end{aligned}$$

If $D>t$

$$\begin{aligned} \begin{array}{l} \begin{aligned} A(t)&{}= \Big [P\left[ t<\sigma _{M},~Y>t\right] \\ &{}+P\left[ \lfloor t/T\rfloor T<\sigma _{M}<\sigma _{M_s}< t<\sigma _{L},~Y>t\right] \\ &{}\quad +P\left[ \lfloor t/T\rfloor T<\sigma _{M}< t<\sigma _{M_s},~Y>t\right] \Big ]\mathbf {1}_{\left\{ M\le M_s\right\} }\\ &{}\quad + \Big [P\left[ t<\sigma _{M_s},~Y>t\right] \\ &{}\quad +P\left[ \sigma _{M_s}< \lfloor t/T\rfloor T<\sigma _{M}< t<\sigma _{L},~Y>t\right] \\ &{}\quad +P\left[ \sigma _{M_s}<\lfloor t/T\rfloor T< t<\sigma _{M},~Y>t\right] \\ &{}\quad +P\left[ \lfloor t/T\rfloor T<\sigma _{M_s}< t<\sigma _{L},~Y>t\right] \Big ]\mathbf {1}_{\left\{ M> M_s\right\} }. \end{aligned} \end{array} \end{aligned}$$

That is

$$\begin{aligned} \begin{array}{l} \begin{aligned} A(t)&{}= \Big [{\bar{F}}_{\sigma _{M}}(t){\bar{F}}_{1}(t)\\ &{}\quad +\displaystyle \int _{\lfloor t/T \rfloor T}^t f_{\sigma _{M}}(u)\int _{u}^{t}f_{\sigma _{M_s}-\sigma _{M}}(v-u){\bar{F}}_{\sigma _{L}-\sigma _{M_s}}(t-v)I(v,t)~dv~du\\ &{}+\displaystyle \int _{\lfloor t/T \rfloor T}^t f_{\sigma _{M}}(u){\bar{F}}_{\sigma _{M_s}-\sigma _{M}}(t-u){\bar{F}}_{1}(t)~du\Big ]\mathbf {1}_{\left\{ M\le M_s\right\} } +\Big [{\bar{F}}_{\sigma _{M_s}}(t){\bar{F}}_{1}(t)\\ &{}+\displaystyle \int _{0}^{\lfloor t/T \rfloor T} f_{\sigma _{M_s}}(u)\int _{\lfloor t/T \rfloor T}^{t}f_{\sigma _{M}-\sigma _{M_s}}(v-u){\bar{F}}_{\sigma _{L}-\sigma _{M}}(t-v)I(u,t)~dv~du\\ &{}+\displaystyle \int _{0}^{\lfloor t/T \rfloor T} f_{\sigma _{M_s}}(u){\bar{F}}_{\sigma _{M}-\sigma _{M_s}}(t-u)I(u,t)~du\\ &{}+\displaystyle \int _{\lfloor t/T \rfloor T}^t f_{\sigma _{M_s}}(u){\bar{F}}_{\sigma _{L}-\sigma _{M_s}}(t-u)I(u,t)~du\Big ]\mathbf {1}_{\left\{ M> M_s\right\} }\\ &{}=J_{T,1}^M(t)\mathbf {1}_{\left\{ M\le M_s\right\} }+J_{T,2}^M(t)\mathbf {1}_{\left\{ M> M_s\right\} }. \end{aligned} \end{array} \end{aligned}$$

If $D\le t$,

$$\begin{aligned} \begin{array}{l} \begin{aligned} \displaystyle \sum _{j=0}^{\infty }&{}\mathbf {1}_{\{R_{j}\le t<R_{j+1}\}}P\left[ O(t)<L,~Y>(t-R_j),~D\le t\right] \\ =\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }&{}P_{D}(kT)\Bigg [ \sum _{j=0}^{\infty }\mathbf {1}_{\{R_{j}\le t<R_{j+1}\}}P\left[ O(t-kT)<L,~Y>(t-kT-R_j)\right] \Bigg ]\\ =\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }&{}A(t-kT)P_{D}(kT). \end{aligned} \end{array} \end{aligned}$$

Then, for $t\ge T$, A(t) verifies the following recursive equation:

$$\begin{aligned} \begin{array}{l} \begin{aligned} A(t)&=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }A(t-kT)P_{D}(kT)+J_{1}(t)\mathbf {1}_{\left\{ M\le M_s\right\} }+J_{2}(t)\mathbf {1}_{\left\{ M> M_s\right\} }, \end{aligned} \end{array} \end{aligned}$$

and the result holds.

Appendix D

For $t<T$, there is no maintenance action on [0, t]; hence, R(t) is equal to A(t).

For $t\ge T$, R(t) is conditioned to the time of the first replacement

$$\begin{aligned} \begin{array}{l} \begin{aligned} R(t)&{}=P\left[ O(u)<L,~\forall u\in (0,t],~N_s(0,t)=0,~D\le t\right] \\ &{}\quad +P\left[ O(u)<L,~\forall u\in (0,t],~N_s(0,t)=0,~D> t\right] . \end{aligned} \end{array} \end{aligned}$$

If $D>t$

$$\begin{aligned} R(t)= & {} J_{1}(t)\mathbf {1}_{\left\{ M\le M_s\right\} }+J_{2}(t)\mathbf {1}_{\left\{ M> M_s\right\} }. \end{aligned}$$

If $D\le t$,

$$\begin{aligned} \begin{array}{l} \begin{aligned} R(t)&{}=P\left[ O(u)<L,~\forall u\in (0,t],N_s(0,t)=0,~D\le t\right] \\ &{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }P_{D,p}(kT)P\left[ O(u)<L,\forall u\in (0,t-kT],N_s(0,t-kT)=0\right] \\ &{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }R(t-kT)P_{D,p}(kT). \end{aligned} \end{array} \end{aligned}$$

Then, for $t\ge T$, R(t) verifies the following recursive equation:

$$\begin{aligned} \begin{array}{l} \begin{aligned} R(t)&=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }R(t-kT)P_{D,p}(kT) +J_{1}(t)\mathbf {1}_{\left\{ M\le M_s\right\} }+J_{2}(t)\mathbf {1}_{\left\{ M> M_s\right\} }, \end{aligned} \end{array} \end{aligned}$$

and the result holds.

Appendix E

For $(t+s)<T$, there is no maintenance action on $[0, t+s]$; hence, $IR(t,t+s)$ is equal to $R(t+s)$. For $t+s\ge T$, $IR(t,t+s)$ is conditioned to the time of the first replacement

$$\begin{aligned} \begin{array}{l} \begin{aligned} IR(t,t+s)&{}=P\left[ O(u)<L,\forall u\in (t,t+s], N_s(t,t+s)=0,~D\le t\right] \\ &{}\quad +P\left[ O(u)<L,~\forall u\in (t,t+s],\, N_s(t,t+s)=0,~t<D< t+s\right] \\ &{}\quad +P\left[ O(u)<L,~\forall u\in (t,t+s],\, N_s(t,t+s)=0,~D\ge t+s\right] . \end{aligned} \end{array} \end{aligned}$$

If $D\ge t+s$

$$\begin{aligned} \begin{array}{l} \begin{aligned} IR(t,t+s)&{}=A(t+s)\\ &{}=J_{1}(t+s)\mathbf {1}_{\left\{ M\le M_s\right\} }+J_{2}(t+s)\mathbf {1}_{\left\{ M> M_s\right\} }. \end{aligned} \end{array} \end{aligned}$$

If $t<D<t+s$

$$\begin{aligned} IR(t,t+s)&=P\left[ O(u)<L,~\forall u\in (t,t+s],\, N_s(t,t+s)=0,~t<D<t+s\right] \\&=\displaystyle \sum _{k=\lfloor t/T\rfloor +1}^{\lfloor (t+s)/T\rfloor }P_{D,p}(kT)P\left[ O(u-kT)<L, \, \forall u\in \right. \\&\quad \left. (0,t+s-kT],N_s(0,t+s-kT)=0\right] \\&=\displaystyle \sum _{k=\lfloor t/T\rfloor +1}^{\lfloor (t+s)/T\rfloor }R(t+s-kT)P_{D,p}(kT). \end{aligned}$$

If $D\le t$

$$\begin{aligned} \begin{array}{l} \begin{aligned} IR(t,t+s)&{}=P\left[ O(u)<L,\forall u\in (t,t+s],~N_s(t,t+s)=0,~D\le t\right] \\ &{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }P_{D}(kT)P\left[ O(u-kT)<L,\forall u\in (t-kT,t+s-kT],\right. \\ &{}\quad \left. ~N_s(t-kT,t+s-kT)=0\right] \\ &{}=\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }IR(t-kT,t+s-kT)P_{D}(kT). \end{aligned} \end{array} \end{aligned}$$

Then, for $t+s\ge T$, $IR(t,t+s)$ verifies the following recursive equation:

$$\begin{aligned} \begin{array}{l} \begin{aligned} IR(t,t+s)&{}=\displaystyle \sum _{k=\lfloor t/T\rfloor +1}^{\lfloor (t+s)/T\rfloor }R(t+s-kT)P_{D,p}(kT)\\ &{}\quad +\displaystyle \sum _{k=1}^{\lfloor t/T\rfloor }IR(t-kT,t+s-kT)P_{D}(kT)\\ &{}\quad +J_{1}(t+s)\mathbf {1}_{\left\{ M\le M_s\right\} }+J_{2}(t+s)\mathbf {1}_{\left\{ M> M_s\right\} }, \end{aligned} \end{array} \end{aligned}$$

and the result holds.