A dynamic stress–strength model with stochastically decreasing strength

Abstract

We consider a dynamic stress–strength model under external shocks. The strength of the system decreases with time and the failure occurs when the strength finally vanishes. Furthermore, there is another cause of the system failure induced by an external shock process. Each shock is characterized by the corresponding stress. If the magnitude of the stress exceeds the current strength, then the system also fails. We assume that the initial strength of the system and its decreasing drift pattern are random. We derive the survival function of the system and interpret the time-dependent dynamic changes of the random quantities which govern the reliability performance of the system.

Appendices

Appendix 1

Substituting a random $R$ instead of its realization $r$ into (6) and applying the operation of conditional expectation,

$$\begin{aligned} \lambda _S(t)=\int _{0}^{\infty } {\bar{M}}(r)\lambda (t)g(r|T_S>t)dr=E[{\bar{M}}(R)|T_S>t]\lambda (t), \end{aligned}$$

where, as follows from

$$\begin{aligned} P(T_S \ge t, R>r)=\int _{r}^{\infty }\exp \{-{\bar{M}}(u)\int _{0}^{t}\lambda (x)dx\}\cdot g(u)du \end{aligned}$$

and

$$\begin{aligned} P(R>r|T_S\ge t)=\frac{\int _{r}^{\infty }\exp \{-{\bar{M}}(u)\int _{0}^{t}\lambda (x)dx\}\cdot g(u)du}{\int _{0}^{\infty }\exp \{-{\bar{M}}(u)\int _{0}^{t}\lambda (x)dx\}\cdot g(u)du}. \end{aligned}$$

Thus the conditional density of strength is given by

$$\begin{aligned} g(r|T_S>t)=\frac{\exp \{-{\bar{M}}(r)\int _{0}^{t}\lambda (x)dx\}\cdot g(r)}{\int _{0}^{\infty }\exp \{-{\bar{M}}(u)\int _{0}^{t}\lambda (x)dx\}\cdot g(u)du}. \end{aligned}$$

(19)

Observe that, for $t_1<t_2$,

$$\begin{aligned} \frac{g(r|T_S> t_1)}{g(r|T_S> t_2)}=\exp \{{\bar{M}}(r)\int _{t_1}^{t_2}\lambda (x)dx\}\frac{\int _{0}^{\infty }\exp \{-{\bar{M}}(u)\int _{0}^{t_2}\lambda (x)dx\}\cdot g(u)du}{\int _{0}^{\infty }\exp \{-{\bar{M}}(u)\int _{0}^{t_1}\lambda (x)dx\}\cdot g(u)du}, \end{aligned}$$

which is decreasing in $r$. Thus, $(R|T_S>t_1) \le _{fr} (R|T_S>t_2)$, which also implies $(R|T_S>t_1) \le _{st} (R|T_S>t_2)$. As ${\bar{M}}(r)$ is decreasing function of $r$, we have

$$\begin{aligned} E[{\bar{M}}(R)|T_S>t_1]\ge E[{\bar{M}}(R)|T_S>t_2]. \end{aligned}$$

Thus ${\tilde{p}}(t)$ is decreasing.

Alternatively, obtaining the derivative of the function ${\tilde{p}}(t)$ defined in (8) with respect to $t$, it is easy to see that

$$\begin{aligned} {\tilde{p}}'(t)&= \Big (-E[({\bar{M}}(R))^2|T_S>t]+(E[{\bar{M}}(R)|T_S>t])^2\Big )\lambda (t) \\&= -Var({\bar{M}}(R)|T_S>t)\lambda (t)<0, \end{aligned}$$

where expectation is with respect to the pdf (19).

Appendix 2 (Proof of Theorem 1)

The unconditional survival function can be obtained by taking expectation of the conditional survival function in (9) with respect to all random quantities in the conditional part. This procedure will be performed sequentially.

First, we take expectation of

$$\begin{aligned} P(T_S>t|N(s),0\le s\le t; S_1,S_2,\ldots , S_{N(t)};R=r,V=v) \end{aligned}$$

with respect to the conditional distribution of $(S_1,S_2,\ldots ,S_{N(t)}|N(s),0\le s\le t;R=r,V=v)$ to obtain $P(T_S>t|N(s),0\le s\le t;R=r,V=v)$. However, due to independence assumption,

$$\begin{aligned} (S_1,S_2,\ldots ,S_{N(t)}|N(s),0\le s\le t;R=r,V=v)=_d (S_1,S_2,\ldots ,S_{N(t)}) \end{aligned}$$

and thus,

$$\begin{aligned}&P(T_S>t|N(s),0\le s\le t;R=r,V=v)\\&\quad =I(r-vt>0)\prod \limits _{i=1}^{N(t)}M(r-vT_i). \end{aligned}$$

Again, due to the independence assumption, $P(T_S>t|R=r,V=v)$ can be obtained by taking expectation of $P(T_S>t|N(s),0\le s\le t;R=r,V=v)$ with respect to the shock process $\{N(s),0\le s \le t\}=\{T_1,T_2,\ldots ,T_{N(t)}, N(t)\}$:

$$\begin{aligned} P(T_S>t|R=r,V=v)&= E\Big [I(r-vt>0)\prod \limits _{i=1}^{N(t)}M(r-vT_i)\Big ] \nonumber \\&= E\Big [E\Big [I(r\!-\!vt>0)\prod \limits _{i=1}^{N(t)}M(r\!-\!vT_i)\Big |N(t)\Big ]\Big ].\quad \end{aligned}$$

(20)

It is well known that the joint distribution of $T_1,T_2,\ldots ,T_n$ given $N(t)=n$ is the same as the joint distribution of order statistics $T_{(1)}'\le T_{(2)}'\le \cdots \le T_{(n)}'$ of i.i.d. random variables $T_1',T_2',\ldots ,T_{n}'$, where the pdf of the common distribution of $T_j'$’s is given by $\lambda (x)/\varLambda (t)$, $0 \le x \le t$, $\varLambda (t)\equiv \int _{0}^{t}\lambda (u)du$. Therefore, it can be easily verified that

$$\begin{aligned}&E\Big [I(r-vt>0)\prod \limits _{i=1}^{N(t)}M(r-vT_i)\Big |N(t)=n\Big ]\nonumber \\&\quad =I(r-vt>0)\Big (E[M(r-vT_i')]\Big )^n \nonumber \\&\quad =I(r-vt>0)\Big ( \frac{1}{\varLambda (t)} \int _{0}^{t}M(r-vx)\lambda (x)dx\Big )^n. \end{aligned}$$

(21)

From Eqs. (20) and (21),

$$\begin{aligned}&P(T_S>t|R=r,V=v)\nonumber \\&\quad =I(r-vt>0)\sum \limits _{n=0}^{\infty } \Big ( \frac{1}{\varLambda (t)} \int _{0}^{t}M(r-vx)\lambda (x)dx\Big )^n \nonumber \\&\qquad \times \frac{ \Big (\int _{0}^{t}\lambda (u)du\Big )^n}{n!}\exp \Big \{-\int _{0}^{t}\lambda (u)du\Big \}\nonumber \\&\quad =I(r-vt>0)\exp \Big \{-\int _{0}^{t}{\bar{M}}(r-vx)\lambda (x)dx\Big \}. \end{aligned}$$

(22)

Finally, the unconditional survival function can be obtained by taking expectation of $P(T_S>t|R=r,V=v)$ in (22) with respect to the joint distribution of $(R,V)$:

$$\begin{aligned} P(T_S>t)&= \int _{0}^{\infty }\int _{0}^{\infty } I(r-vt>0)\exp \Big \{-\int _{0}^{t}{\bar{M}}(r-vx)\lambda (x)dx\Big \}h(v)g(r)dvdr\\&= \int _{0}^{\infty }\int _{0}^{r/t} \exp \Big \{-\int _{0}^{t}{\bar{M}}(r-vx)\lambda (x)dx\Big \}h(v)g(r)dvdr. \end{aligned}$$

By applying Leibniz rule, the failure rate function of a system $\lambda _S(t)$ is

$$\begin{aligned} \lambda _S(t)&= -\frac{P'(T_S>t)}{P(T_S>t)}=\frac{\int _{0}^{\infty } \frac{r}{t^2}\exp \Big \{-\int _{0}^{t}{\bar{M}}(r-\frac{r}{t}x)\lambda (x)dx \Big \}h(\frac{r}{t})g(r)dr}{\int _{0}^{\infty }\int _{0}^{r/t} \exp \Big \{-\int _{0}^{t}{\bar{M}}(r-vx)\lambda (x)dx\Big \}h(v)g(r)dvdr} \\&+\,\frac{ \lambda (t) \int _{0}^{\infty }\int _{0}^{r/t}{\bar{M}}(r-vt)\exp \Big \{-\int _{0}^{t}{\bar{M}}(r-vx)\lambda (x)dx\Big \}h(v)g(r)dvdr}{\int _{0}^{\infty }\int _{0}^{r/t} \exp \Big \{-\int _{0}^{t}{\bar{M}}(r-vx)\lambda (x)dx\Big \}h(v)g(r)dvdr}. \end{aligned}$$