Single inspection

Consider a coherent system with lifetime T, as described in the introduction. The system begins to operate at time ${t=0}$ and each component is subject to failure over time. Assume that the operator inspects the system at time t and he/she observes that at least k components have already failed before t, but the system is still working. As we mentioned in the previous section, the probability of the number of failed components up to time t is given by ${p_{k,n}^{t}(i)}$ in (2). Asadi and Berred [Reference Asadi and Berred2] explored several properties of the above conditional probability ${p^t_{k,n}(i)}$ in the case that ${k=0}$ . Eryilmaz [Reference Eryilmaz10] considered the number of failed components for a coherent system whose component lifetimes are exchangeable. Ashrafi and Asadi [Reference Ashrafi and Asadi3] studied the number of failed components in three-state networks under different conditions and applied their results for age-replacement of three-state networks. The conditional probability ${p^t_{k,n}(i)}$ can be represented as follows:

(4) \begin{eqnarray}p^t_{k,n}(i)&\,=\,&\mathbb{P}(N_t=i\mid X_{k:n}\leq t<T)\nonumber\\[3pt]&\,=\,&\frac{\bar{S}_{i}\binom{n}{i}\phi^i(t)}{\sum_{j=k}^{n-1}\bar{S}_j\binom{n}{j}\phi^j(t)}, \ \ \ i=k,...,n-1,\end{eqnarray}

where ${\bar{S}_i=\sum_{j=i+1}^{n}s_j}$ and ${\phi(t)=F(t)/\bar{F}(t)}$ , provided that ${\bar{F}(t)=1-F(t)>0}$ ; see Appendix A for the proof. The expectation of ${(N_t\mid X_{k:n}\leq t<T)}$ can be calculated as follows:

(5) \begin{eqnarray}\mathbb{E}(N_t\mid X_{k:n}\leq t<T)=\frac{n\sum_{i=k}^{n-1}\bar{S}_{i}\binom{n-1}{i-1}\phi^i(t)}{\sum_{j=k}^{n-1}\bar{S}_{j}\binom{n}{j}\phi^j(t)}, \ \ \ k=0,1,..., n-1.\end{eqnarray}

One can easily verify that the common reliability function ${\bar{F}(t)}$ of the components can be recovered from two successive values ${p^t_{k,n}(i)}$ and ${p^t_{k,n}(i+1)}$ as follows:

\[ \bar{F}(t)=\left(1+\frac{i+1}{n-i}\left(\frac{s_{i+1}}{\bar{S}_{i+1}}+1\right) \frac{p^t_{k,n}(i+1)}{p^t_{k,n}(i)}\right)^{-1}, \quad t>0.\]

Figure 1:

The bridge system.

Example 2.1. In Figure 1, a bridge system consisting of five identical components is pictured. We assume that the component lifetimes are i.i.d. random variables having Weibull distribution with CDF ${F(t)=1-\exp\{-t^{2}\}}$ , ${t\geq0}$ . The system signature can be computed as ${\textbf{{s}}=(0,0.2,0.6,0.2,0)}$ . Then we have ${\phi(t)=e^{t^2}-1}$ and

(6) \begin{eqnarray} p^t_{k,5}(i)=\frac{\bar{S}_{i}\binom{5}{i}(e^{t^2}-1)^i}{\sum_{j=k}^{4}\bar{S}_j\binom{5}{j}(e^{t^2}-1)^j}, \ \ \ 0\leq k\leq i\leq 4. \end{eqnarray}

As can be seen in Figure 2(a), ${p^t_{1,5}(1)}$ is decreasing in t, ${p^t_{1,5}(2)}$ increases for a period of time and then decreases, and ${p^t_{1,5}(3)}$ is increasing in t (see Remark 2.2). It should be mentioned that in this system, ${p^t_{1,5}(4)=0}$ , since for the component with lifetime ${X_{5:5}}$ , ${\mathbb{P}(T=X_{5:5})=s_5=0}$ ; i.e., the component with lifetime ${X_{5:5}}$ will never entail the failure of the system. Also, using Equation (5), we obtain

(7) \begin{eqnarray}H_{k,n}^t\,:\!=\mathbb{E}(N_t\mid X_{k:n}\leq t<T)=\frac{5\sum_{i=k}^{4}\bar{S}_{i}\binom{4}{i-1}(e^{t^2}-1)^i}{\sum_{j=k}^{4}\bar{S}_{j}\binom{5}{j}(e^{t^2}-1)^j},\ \ \ 0\leq k\leq 4.\end{eqnarray}

Figure 2(b) shows the graph of ${H_{k,n}^t}$ as a function of t for ${k=0,1,2}$ . It is seen that ${H_{k,n}^t}$ is increasing in k and t. We have proved the same result for a general system structure and an arbitrary baseline distribution in Corollary 2.1.

Figure 2:

(a) The plots of ${p^t_{k,n}(i)}$ for ${i=1, 2,3}$ and ${k=1}$ in Example 2.1. (b) The plots of ${H_{k,n}^t}$ for ${k=0, 1,2}$ in Example 2.1.

The following theorem gives some stochastic properties of ${(N_t\mid X_{k:n}\leq t<T)}$ . Before giving the theorem, we mention the well-known result that, if X and Y are two nonnegative random variables with probability density functions ${f_1}$ and ${f_2}$ , respectively, then the order ${X\leq_{lr}Y}$ is equivalent to saying that ${f_m(t)}$ is TP ${_2}$ in ${(m,t)\in \{1,2\}\times [0,\infty)}$ (see Shaked and Shanthikumar [Reference Shaked and Shanthikumar33]).

Proof. See Appendix B. □

It is known that the likelihood ratio order between two random variables implies the usual stochastic order between them. Using this fact and Theorem 2.1, one can verify the behavior of ${\mathbb{E}(N_t\mid X_{k:n}\leq t<T)}$ in terms of k and t, as shown in the following corollary.

We now provide a comparison between two different coherent systems based on the number of failed components in each system.

Theorem 2.2. Suppose that two coherent systems with, respectively, orders n and ${n+1}$ have signature vectors ${\textbf{{s}}^{(1)}=(s_1, s_2,... , s_n)}$ and ${\textbf{{s}}^{(2)}=(p_1, p_2,... , p_{n+1})}$ . Let the component lifetimes of the first system be ${X_1,X_2,...,X_n}$ , and let those of the second system be ${Y_1,Y_2,...,Y_{n+1}}$ , where in the two systems the components are independent and have a common distribution function F. Denote by ${N_t}$ and ${N_t^*}$ the number of failed components of the first and second systems, respectively, at time t. If ${\textbf{{s}}^{(1)}\leq _{\rm hr}\textbf{{s}}^{(2)}}$ , then

\[(N_t^*\mid Y_{k:n+1}\leq t<T_2)\geq_{\rm lr}(N_t\mid X_{k:n}\leq t<T_1),\ \ \ t\geq 0,\]

where ${T_1}$ and ${T_2}$ denote the lifetimes of the system with order n and the system with order ${n+1}$ , respectively.

Proof. See Appendix C. □

The next result gives a likelihood ratio order comparison on the failed components between two coherent systems.

Theorem 2.3. Assume that ${\mathcal{S}_{1}}$ and ${\mathcal{S}_{2}}$ are two coherent systems with i.i.d. component lifetimes ${X_1,X_2,...,X_n}$ and ${Y_1,Y_2,...,Y_n}$ whose CDFs are F and G, respectively. Let ${N_t}$ and ${N_t^*}$ be the numbers of failed components in ${\mathcal{S}_{1}}$ and ${\mathcal{S}_{2}}$ , respectively, on [0, t]. Further, suppose that the lifetime of the system ${\mathcal{S}_{1}}$ ( ${\mathcal{S}_{2}}$ ) is denoted by ${T_1}$ ( ${T_2}$ ) and the corresponding signature vector is denoted by ${\textbf{{s}}^{(1)}}$ ( ${\textbf{{s}}^{(2)}}$ ). If ${{X_1}\leq _{\rm st}{Y_1}}$ and ${\textbf{{s}}^{(1)}\geq _{\rm hr}\textbf{{s}}^{(2)}}$ , then

\[(N_t\mid X_{k:n}\leq t<T_1)\geq_{\rm lr}(N_t^*\mid Y_{k:n}\leq t<T_2),\ \ \ t\geq 0.\]

Proof. See Appendix D. □

Double inspection

Consider again a coherent system with n i.i.d. components and suppose that the system starts working at time ${t=0}$ . We assume that the system is monitored by the operator at two time instances ${t_1}$ and ${t_2}$ (with ${t_1<t_2}$ ). This method of inspection is known in the literature as double monitoring. Some recent references in this regard are Zhang and Meeker [Reference Zhang and Meeker43], Parvardeh et al. [Reference Parvardeh, Balakrishnan and Arshadipour26], and Navarro and Calì [Reference Navarro and Calì22]. Suppose that the number of failed components up to time ${t_1}$ is k, and at time ${t_2}$ the system is still operating. Under these circumstances, we intend to study the probability of the number of failed components in the system at time ${t_2}$ . The probability mass function of this random variable, ${p^{t_1,t_2}_{k,n}(i)}$ , can be written as

(8) \begin{eqnarray} p^{t_1,t_2}_{k,n}(i)&\,=\,&\mathbb{P}(N_{t_2}=i\mid X_{k:n}\leq t_1<X_{k+1:n}, T>t_2)\nonumber\\[3pt]&\,=\,&\dfrac{\bar{S}_i\binom{n}{i}\binom{i}{k}\left(\frac{\bar{F}(t_1)}{\bar{F}(t_2)}-1\right)^{i-k}}{\sum_{j=k}^{n-1}\bar{S}_j\binom{n}{j}\binom{j}{k}\left(\frac{\bar{F}(t_1)}{\bar{F}(t_2)}-1\right)^{j-k}}, \ \ i=k,..., n-1.\end{eqnarray}

For the proof, see Appendix E. From (8), the mean number of failed components up to time ${t_2}$ can be represented as

(9) \begin{eqnarray}\varphi(t_1,t_2)&\,=\,&\mathbb{E}(N_{t_2}\mid X_{k:n}\leq t_1<X_{k+1:n}, T>t_2)\nonumber\\[3pt]&\,=\,&\dfrac{n\sum_{i=k}^{n-1}\bar{S}_i\binom{n-1}{i-1}\binom{i}{k}\left(\frac{\bar{F}(t_1)}{\bar{F}(t_2)}-1\right)^{i}}{\sum_{j=k}^{n-1}\bar{S}_j\binom{n}{j}\binom{j}{k}\left(\frac{\bar{F}(t_1)}{\bar{F}(t_2)}-1\right)^{j}}.\end{eqnarray}

The next theorem reveals some stochastic properties of ${(N_{t_2}\mid X_{k:n}\leq t_1<X_{k+1:n}, T>t_2)}$ in terms of k, ${t_1}$ , and ${t_2}$ .

Proof. See Appendix F. □

As the likelihood ratio order is a subclass of the usual stochastic order, we get the following corollary from Theorem 2.4.

The next theorem provides a comparison between the failed components in two coherent systems in terms of likelihood ratio order.

Theorem 2.5. Let ${\mathcal{S}_{1}}$ and ${\mathcal{S}_{2}}$ denote two coherent systems with i.i.d. component lifetimes ${X_1,X_2,...,X_n}$ and ${Y_1,Y_2,...,Y_n}$ whose CDFs are F and G, respectively. Let ${N_t}$ and ${N_t^*}$ be the number of failed components of ${\mathcal{S}_{1}}$ and ${\mathcal{S}_{2}}$ , respectively, on [0, t]. Further, suppose the lifetime of the system ${\mathcal{S}_{1}}$ ( ${\mathcal{S}_{2}}$ ) is ${T_1}$ ( ${T_2}$ ) with signature vector ${\textbf{{s}}^{(1)}}$ ( ${\textbf{{s}}^{(2)}}$ ). If ${{X_1}\leq _{\rm hr}{Y_1}}$ and ${\textbf{{s}}^{(2)}\leq _{\rm hr}\textbf{{s}}^{(1)}}$ , then

\[(N_{t_2}\mid X_{k:n}\leq t_1<X_{k+1:n}, T_1>t_2)\geq_{\rm lr}(N_{t_2}^*\mid Y_{k:n}\leq t_1<Y_{k+1:n}, T_2>t_2),\ \ \ 0\leq t_1 < t_2.\]

Proof. See Appendix G. □

Strategy I

Assume that a coherent system begins to operate at time 0. A minimal repair has been performed on each component of the system that fails in the interval ${(0,\tau)}$ . Thus, we can assume that the system, consisting of n unfailed components with age ${\tau}$ , is alive at ${\tau}$ . Here, ${\tau}$ is a predetermined constant which may be considered, for example, as a guarantee time of the system. We assume that, after ${\tau}$ , a warning lamp is installed on the system that turns on at the time of the kth component failure, where k is predetermined (for a realistic example where warning lamps are employed in a system, we refer to Shimizu and Kawai [Reference Shimizu and Kawai35], where the authors consider a warning lamp in a vehicle’s electronic power steering system (EPS) which will operate in case of failure of components of the EPS; see also Khaledi and Shaked [Reference Khaledi and Shaked16]).

The operator decides to perform CM on the whole system at a cost of ${c_{cms}}$ once the system fails in the interval ${(\tau,t)}$ , or he/she decides to perform a PM action when the total operating time reaches t if the lamp turns on, whichever occurs first. More precisely, the operator decides to perform PM on all operating components together with CM on the failed ones at a cost of ${c_{pm}}$ for PM and a cost of ${c_{cm}}$ for CM, when the total operating time reaches t, provided that the warning light has lit up before or at time t. On the other hand, if the system is alive at t and the lamp does not turn on, he/she performs a perfect PM on the entire system with a cost of ${c_{pms}}$ . It should be mentioned here that, in the above situation, the operator knows only whether the light is on or off. In other words, the warning light might have been turned on at a time before t; the operator does not know the exact time of the kth failure in the interval ${(\tau,t)}$ . Two different cases of renewal cycle for Strategy ${\textbf{I}}$ are shown in Figure 3. The upper axis in Figure 3 shows the case where system failure has not occurred up to time t; that is, the system age reaches t. The lower axis depicts the case where the system fails before the age reaches t.

Figure 3:

Maintenance Strategy ${\textbf{I}}$ .

It is evident that in the above policy, the inspection of the system is done in the interval ${(0,\tau)}$ , where the components of the system have been monitored continuously, and at the single time instant t (see Pham and Wang [Reference Pham and Wang27]). As mentioned in [Reference Pham and Wang27], a justification for the first part is that in the interval ${(0,\tau)}$ the components are ‘young’ and hence a minor repair is adequate. Thus, before ${\tau}$ , only minimal repairs, which require little time and have low cost, are carried out. After the system reaches age ${\tau}$ , as the cost of continuous monitoring may be substantial and minimal repair may not be reasonable because of the increased failure rate of the components, the operator does not monitor the system; instead, an alarm is installed which operates at the time of the kth failure. Using this information we obtain the optimal time of PM under the following settings on the cost function.

To evaluate the cost in interval ${(0,\tau)}$ , we shall utilize a cost function used by Sheu [Reference Sheu34]; see also Pham and Wang [Reference Pham and Wang27]. For each component, let the cost of the ith minimal repair depend on the deterministic part ${a_1(t,i)}$ and the age-dependent random part ${a_2(t)}$ . Note that ${a_1(t,i)}$ depends on the number i of minimal repairs performed on the component and the age t of that component. Two parts are now linked by a positive function h. In fact, the required cost of the ith minimal repair at age t for each component is ${h(a_1(t,i),a_2(t))}$ , where h is a continuous nondecreasing function of t, and is a nondecreasing function of i. Thus, the expected cost of minimal repairs for the whole system in a renewal period is

\[c^*_{min}=n \mathbb{E}\left[\sum_{i=1}^{N(\tau)}h(a_1(S_i,i),a_2(S_i))\right]\!,\]

where ${N(\tau)}$ denotes the total number of minimal repairs during the time interval ${(0,\tau)}$ , and ${S_1,S_2,...}$ are the successive failure times of each component at which minimal repairs have been performed. It is known that for the minimal repair the failure times follow a nonhomogeneous Poisson process with rate r(t) (see Barlow and Hunter [Reference Barlow and Hunter6] or Gertsbakh [Reference Gertsbakh14]). Note that r(t) is the failure rate of a component lifetime. It has been proved by Sheu [Reference Sheu34] that

(10) \begin{eqnarray}c^*_{min}=n\int_{0}^{\tau}\nu(y)r(y)dy,\end{eqnarray}

where ${\nu(y)=\mathbb{E}_{N(y)}\mathbb{E}_{a_2(y)}[h(a_1(y,N(y)+1),a_2(y))]}$ ; see also Pham and Wang [Reference Pham and Wang27]. A special case considered in the literature is that in which ${h(a_1(t,i),a_2(t))}$ is assumed to be a constant ${c_{min}}$ (see Barlow and Hunter [Reference Barlow and Hunter6] and Tahara and Nishida [Reference Tahara and Nishida37]). We assume that this cost includes the cost of monitoring and the cost of repair. Thus ${c^*_{min}}$ reduces to

\[c^*_{\min}=n c_{\min}H(\tau),\]

where ${H(\tau)=\int_{0}^{\tau}r(t)dt}$ is known as the mean value function of the failure process.

In this strategy, we consider t as a decision variable and ${\tau}$ as a predetermined time instant. By using the renewal reward theorem (see, e.g., Ross [Reference Ross30], p. 52), the average cost of system maintenance per unit time is then defined as the ratio of the average cost of the system maintenance per renewal cycle to the expected duration of a renewal cycle. In other words,

(11) \begin{eqnarray}\eta_I(t)&\,=\,&\frac{n \int_{0}^{\tau}\nu(y)r(y)dy+F_{\tau}(t-\tau)c_{cms}+c_{pms}P_{1,k}(\tau, t)}{\tau+\mathbb{E}(\!\min\!(t-\tau,T_\tau))}\nonumber\\[3pt]&&+\,\frac{P_{2,k}(\tau ,t)\left[(c_{cm}-c_{pm})\mathbb{E}(N_{t,\tau}\mid (X_{\tau})_{k:n}\leq t<T_{\tau})+nc_{pm}\right]}{\tau+\mathbb{E}(\!\min\!(t-\tau,T_\tau))},\end{eqnarray}

where ${F_{\tau}(\!\cdot\!)}$ denotes the CDF of the lifetime of an n-component system where each component’s age is ${\tau}$ , ${\mathbb{E}(N_{t,\tau}\mid (X_{\tau})_{k:n}\leq t<T_{\tau})}$ is the expectation of the number of failed components of the live system at time t with at least k failed components when all components are functioning at time ${\tau}$ (for ${\tau<t}$ ),

\[P_{1,k}(\tau, t)=\mathbb{P}(X_{k:n}>t, T>t\mid X_{1:n}>\tau),\]

and

\[P_{2,k}(\tau ,t)=\mathbb{P}(X_{k:n}\leq t<T\mid X_{1:n}>\tau).\]

Note that ${(X_{\tau})_{k:n}}$ is the kth order statistic from the CDF ${F(t|\tau)=1-\bar{F}(t)/\bar{F}(\tau)}$ , ${t>\tau}$ . We can easily show that

(12) \begin{eqnarray}\quad \ \ F_{\tau}(t-\tau)=1-\sum_{j=0}^{n-1}\bar{S}_{j}\binom{n}{j}\left(1-\frac{\bar{F}(t)}{\bar{F}(\tau)}\right)^{j}\left(\frac{\bar{F}(t)}{\bar{F}(\tau)}\right)^{n-j},\end{eqnarray}

\begin{eqnarray*}P_{1,k}(\tau, t)=\sum_{i=0}^{k-1}\bar{S}_{i}\binom{n}{i}\left(1-\frac{\bar{F}(t)}{\bar{F}(\tau)}\right)^i\left(\frac{\bar{F}(t)}{\bar{F}(\tau)}\right)^{n-i},\end{eqnarray*}

and

\begin{eqnarray*}P_{2,k}(\tau, t)=\sum_{i=k}^{n-1}\bar{S}_{i}\binom{n}{i}\left(1-\frac{\bar{F}(t)}{\bar{F}(\tau)}\right)^i\left(\frac{\bar{F}(t)}{\bar{F}(\tau)}\right)^{n-i}.\end{eqnarray*}

Also, from Equation (5), we can obtain

\begin{eqnarray*}\mathbb{E}(N_t\mid X_{k:n}\leq t<T)=\frac{n\sum_{i=k}^{n-1}\bar{S}_{i}\binom{n-1}{i-1}\phi^i(t)}{\sum_{j=k}^{n-1}\bar{S}_{j}\binom{n}{j}\phi^j(t)}.\end{eqnarray*}

By replacing ${\phi(t)}$ with ${\left(\frac{\bar{F}(\tau)}{\bar{F}(t)}-1\right)}$ , we may obtain the corresponding formula for ${\mathbb{E}(N_{t,\tau}\mid (X_{\tau})_{k:n}\leq t<T_{\tau})}$ . Also, it can easily be shown that

\begin{eqnarray*}\mathbb{E}(\!\min (t-\tau, T_{\tau}))&\,=\,&\int_{0}^{t-\tau}[1-F_{\tau}(x)]dx\nonumber\\[3pt]&\,=\,&\frac{1}{\bar{F}^{n}(\tau)}\sum_{j=0}^{n-1}\bar{S}_{j}\binom{n}{j}\int_{0}^{t-\tau}(\bar{F}(\tau)-\bar{F}(x+\tau))^j(\bar{F}(x+\tau))^{n-j}dx.\end{eqnarray*}

In the following proposition, in the case that the decision variable in ${\eta_I(t)}$ is t, we verify the existence of the optimal value ${t^*}$ minimizing ${\eta_I(t)}$ .

Proposition 3.1. Let r(t) be the failure rate of components and ${\eta_I(t)}$ be as given in Equation (11). If ${\lim_{t\rightarrow \infty}r(t)}$ is finite and

(13) \begin{eqnarray} \lim_{t\rightarrow \infty}r(t)>\frac{c_{cms}+n\int_{0}^{\tau}\nu(y)r(y)dy}{(n-i^{*}+1)(c_{cms}-n c_{pm}-(i^{*}-1)(c_{cm}-c_{pm}))(\tau+\mu_{\tau})},\end{eqnarray}

then there exists a finite ${t^*}$ which satisfies ${\frac{d}{dt}\eta_I(t)\mid_{t=t^*}=0}$ and minimizes ${\eta_I(t)}$ , where ${\mu_{\tau}}$ is the expectation of the lifetime of an n-component system whose components have age ${\tau}$ , and ${i^{*}}$ is defined in Remark 2.2.

Proof. See Appendix H. □

It should be pointed out that, in the policy described above, we assume that all maintenance actions take negligible times. Now, suppose that minimal repair also takes negligible time, but PM combined with CM takes ${w_1}$ time units, CM on the whole system at time t takes ${w_2}$ time units, and PM on the whole system at time t takes ${w_3}$ time units. In the literature, a well-known criterion in the maintenance of systems is stationary availability. The stationary availability is defined as the ratio of the average time that the system is in a functioning state to the average length of a cycle. The stationary availability for Strategy ${\textbf{I}}$ is then given by

\begin{eqnarray*}A_I(t)=\frac{\tau+\mathbb{E}(\!\min\!(t-\tau,T_\tau))}{\tau+\mathbb{E}(\!\min\!(t-\tau,T_\tau))+w_1 P_{2,k}(\tau, t)+w_2 F_{\tau}(t-\tau)+w_3P_{1,k}(\tau, t)}.\end{eqnarray*}

Example 3.1. Assume that the bridge system in Figure 1 has component lifetimes which are independent Weibull random variables with CDF ${{F}(t)=1-\exp\{-t^{2}\}}$ , ${t\geq0}$ . It is known that the system signature is ${(0, 0.2, 0.6, 0.2, 0)}$ (see, e.g., Samaniego [Reference Samaniego32]). Let ${c_{min}=0.5}$ , ${c_{cms}=25}$ , ${c_{pms}=4}$ , ${c_{cm}=2}$ , and ${c_{pm}=1}$ . In Figure 4, the graphs of ${\eta_I(t)}$ are presented for different values of k and ${\tau}$ .

Figure 4:

The expected cost of the system maintenance per unit time in Example 3.1: (a) ${\tau=0.1}$ , ${k=1,2,3}$ from top to bottom; (b) ${k=0}$ , ${\tau=0.1,0.3,0.5}$ from top to bottom.

In order to investigate the robustness of Strategy ${\textbf{I}}$ with respect to the model parameters ${c_{pms}}$ , ${c_{cms}}$ , ${{c}_{min}}$ , ${{c}_{cm}}$ , and ${{c}_{pm}}$ , we have provided some numerical results in Tables 1 and 2. As the tables show, when ${c_{cms}}$ increases, ${t^*}$ decreases and the operator should perform preventive action sooner. On the other hand, when ${c_{pms}}$ gets larger, ${t^*}$ increases too; i.e., the larger the cost of system PM, the later the time of performing system PM. When ${{c}_{min}}$ gets larger, the optimal time of PM gets larger too, and when ${{c}_{cm}}$ increases, as expected, ${t^*}$ decreases. Hence, according to this example, it is evident that the model is robust in terms of ${c_{cms}}$ , ${c_{pms}}$ , ${{c}_{min}}$ , and ${{c}_{cm}}$ . In the same manner one can see that the model is robust in terms of the other parameters ${{c}_{pm}}$ and ${\tau}$ .

Table 1:

Optimal maintenance time for Strategy ${\textbf{I}}$ .

Table 2:

Optimal maintenance time for Strategy ${\textbf{I}}$ .

The optimal times ${t^*}$ that minimize the average cost per unit of time and ${\eta_I(t^*)}$ are presented in Table 3 for several time instants ${\tau}$ . Note that if we consider ${\tau}$ and t as two decision variables, then the optimal value for the pair ${(\tau, t)}$ is ${(1.25, 1.25787)}$ , which results in the minimum maintenance cost ${6.30534}$ . The optimal values of ${(\tau,t)}$ are also tabulated in Table 4 for different values of ${c_{cms}}$ and ${c_{pms}}$ .

Table 3:

Optimal maintenance times for Strategy ${\textbf{I}}$ with ${k=2}$ , ${c_{min}=0.5}$ , ${c_{cms}=25}$ , ${c_{pms}=4}$ , ${c_{cm}=2}$ , and ${c_{pm}=1}$ .

Table 4:

Bivariate optimal maintenance times for Strategy ${\textbf{I}}$ .

Figure 5 depicts the three-dimensional plots of the cost function in terms of ${(\tau,t)}$ for ${k=0, 2}$ , ${c_{min}=0.5}$ , ${c_{cms}=25}$ , ${c_{pms}=4}$ , ${c_{cm}=2}$ , and ${c_{pm}=1}$ .

Figure 5:

The three-dimensional plot of cost function in Example 3.1: (a) ${k=0}$ , (b) ${k=2}$ .

Figure 6 depicts the plots of ${A_I(t)}$ for ${w_{1}=0.08}$ , ${w_{2}=0.2}$ , and ${w_{3}=0.02}$ , and for several values of k and ${\tau}$ . As the plots show, the system availability first increases to attain its maximum and then decreases.

Figure 6:

The stationary availability in Example 3.1: (a) ${\tau=0.1}$ , ${k=1,2,3}$ from top to bottom; (b) ${k=0}$ , ${\tau=0.1,0.3,0.5}$ from top to bottom.

Now, assume that ${h(a_1(t,i),a_2(t))=a_1(t)+a_2(t)}$ , where ${a_1(t)=3t}$ and ${a_2(t)}$ follows the normal distribution with mean 1. Then, by (10), ${c^*_{min}=n(2\tau^3+\tau^2)}$ . The expected maintenance cost ${\eta_I(t)}$ is presented in Figure 7 for several values of k and ${\tau}$ .

Figure 7:

The expected cost of the system maintenance per unit time in Example 3.1 with ${h(a_1(t,i),a_2(t))=3t+a_2(t)}$ where ${a_2(t)}$ follows the normal distribution with mean 1: (a) ${\tau=0.1}$ , ${k=1,2,3}$ from top to bottom; (b) ${k=0}$ , ${\tau=0.1,0.3,0.5}$ from top to bottom.

The special case of this policy in which there is no minimal repair, i.e. ${\tau=0}$ , may be of interest. In this particular case, the average cost of the system maintenance per unit of time can be reduced to

\begin{eqnarray*}\eta_{I}(t)&\,=\,&\frac{c_{cms}F_T(t)+c_{pms}\mathbb{P}(T>t,X_{k:n}>t)}{\mathbb{E}(\!\min\!(t,T))}\\[4pt]&&+\, \frac{\mathbb{P}(X_{k:n}\leq t<T)[(c_{cm}-c_{pm})\mathbb{E}(N_t\mid X_{k:n}\leq t<T)+n c_{pm}]}{\mathbb{E}(\!\min\!(t,T))},\end{eqnarray*}

where

\begin{eqnarray*}\mathbb{P}(X_{k:n}\leq t<T)=\sum_{i=k}^{n-1}\bar{S}_i\binom{n}{i}F^i(t)\bar{F}^{n-i}(t),\end{eqnarray*}

\begin{eqnarray*}\mathbb{P}(T>t,X_{k:n}>t)=\sum_{i=0}^{k-1}\bar{S}_{i}\binom{n}{i}F^i(t)\bar{F}^{n-i}(t),\end{eqnarray*}

and ${\mathbb{E}(N_t\mid X_{k:n}\leq t<T)}$ is defined in (5).

Also, for this particular case where ${\tau=0}$ , the stationary availability may be written as

\begin{eqnarray*}A_{I}(t)=\frac{\mathbb{E}(\!\min\!(t,T))}{\mathbb{E}(\!\min\!(t,T))+w_1\mathbb{P}(X_{k:n}\leq t<T)+w_{2}F_T(t)+w_3 \mathbb{P}(T>t,X_{k:n}>t)}.\end{eqnarray*}

Now we consider the following example.

An analysis of the results of Table 5 indicates that the maintenance strategy ${\textbf{I}}$ with ${\tau=0}$ is robust. It is seen that when ${{c}_{cms}}$ increases, the optimal time of PM decreases, as expected. Also, when ${{c}_{pms}}$ gets larger, the optimal time of PM gets larger, too. One can easily see that, based on this example, the model is robust in terms of the costs ${{c}_{pm}}$ and ${{c}_{cm}}$ .

Table 5:

Optimal maintenance time for Strategy ${\textbf{I}}$ with ${\tau=0}$ .

Figure 8:

(a) The average maintenance cost per unit time in Example 3.2; (b) the stationary availability in Example 3.2.

We should mention here that the only difference between the maintenance schedules given in this example and those of Example 3.1 is that, in Example 3.1, the operator performs minimal repair in a time interval at the beginning of the system operation, while in the present example there is no minimal repair. By considering the same initial values for these two maintenance policies, one can make a comparison based on the expected cost or the availability criterion. For example, let ${c_{min}=0.5}$ , ${c_{cms}=25}$ , ${c_{pms}=4}$ , ${c_{cm}=2}$ , ${c_{pm}=1}$ , ${w_{1}=0.08}$ , ${w_{2}=0.2}$ , and ${w_{3}=0.02}$ . If ${k=1}$ and ${\tau=0.1}$ , the optimal values of ${\eta_I}$ and ${A_I}$ are 15.0488 and 0.8790, respectively, whereas if we do not perform minimal repair, the optimal values of ${\eta_I}$ and ${A_I}$ are 15.4200 and 0.8727, respectively. Therefore, based on these observations, the operator prefers to perform minimal repair at the starting point of the maintenance. One can show that the situation is reversed if ${k=3}$ , in which case no minimal repair is preferred.

Strategy II

Assume that a new coherent system begins to operate at time 0. Suppose that the system has been inspected at two times ${t_1}$ and ${t_2}$ , with ${t_1<t_2}$ . If the system fails before ${t_1}$ , then the operator performs CM on the entire system with a cost of ${c_{cms}}$ at the time of the system failure. He/she performs the same action if the system fails during the time interval ${(t_1,t_2)}$ . On the other hand, if the system is functioning at ${t_2}$ , the operator decides among three different actions:

1. (a) If the number of components that have failed by ${t_1}$ , namely ${N_{t_1}}$ , is at most ${(k_1-1)}$ , the operator performs PM on the whole system with a cost of ${c_{pms}}$ .

2. (b) If ${k_1\leq N_{t_1}\leq k_2}$ , then he/she decides to perform PM on all operating components of the system together with CM on all failed ones, at a cost of ${c_{pm}}$ for PM and a cost of ${c_{cm}}$ for CM.

3. (c) If ${N_{t_1}}$ is at least ${(k_2+1)}$ , then the operator decides to perform a more rigid PM on the system (than in Case (a)) at a cost of ${c^*_{pms}}$ .

In this strategy, we assume that ${t_2}$ is the decision variable, while ${t_1}$ , ${k_1}$ , and ${k_2}$ are fixed constants. Three different cases of renewal cycle for Strategy ${\textbf{II}}$ are shown in Figure 9. The upper axis in Figure 9 depicts the case in which system failure has not occurred up to time ${t_2}$ ; that is, the age of the system reaches ${t_2}$ . The middle axis shows the case in which the system is alive at the inspection time ${t_1}$ and fails before attaining the age ${t_2}$ . The lower axis depicts the case where the system fails before its age reaches the time of inspection ${t_1}$ .

Figure 9:

Maintenance Strategy ${\textbf{II}}$ .

The average cost of the system maintenance per unit of time is

(14) \begin{eqnarray}\eta_{II}(t_2)&\,=\,&\frac{D(t_2)}{\mathbb{E}(\!\min\!(t_2,T))},\end{eqnarray}

where

\begin{eqnarray*}D(t_2)&\,=\,&c_{cms}\mathbb{P}(T\leq t_2)+c_{pms}\mathbb{P}(T>t_2,N_{t_1}\leq k_1-1)\\[3pt]&&+\, [(c_{cm}-c_{pm})\mathbb{E}(N_{t_2}\mid k_1\leq N_{t_1}\leq k_2,T>t_2)+n c_{pm}]\\[3pt]& &\times\ \mathbb{P}(T>t_2,k_1\leq N_{t_1}\leq k_2)+c^*_{pms}\mathbb{P}(T>t_2,N_{t_1}\geq k_2+1),\end{eqnarray*}

and

\begin{eqnarray*}\mathbb{E}(\!\min\!(t_2,T))&\,=\,&\int_{0}^{t_2}\sum_{j=0}^{n-1}\bar{S}_j\binom{n}{j}F^j(t)\bar{F}^{n-j}(t) dt.\end{eqnarray*}

In the special case where ${k_1=k_2=k}$ , ${D(t_2)}$ may be reduced to

\begin{eqnarray*}D(t_2)&\,=\,&c_{cms}\mathbb{P}(T\leq t_2)+c_{pms}\mathbb{P}(T>t_2,N_{t_1}\leq k-1)\\[3pt]&&+\, [(c_{cm}-c_{pm})\varphi(t_1,t_2)+n c_{pm}]\mathbb{P}(T>t_2, N_{t_1}= k)+c^*_{pms}\mathbb{P}(T>t_2,N_{t_1}\geq k+1),\end{eqnarray*}

where, from (9),

\begin{eqnarray*}\varphi(t_1,t_2)=\frac{n\sum_{i=k}^{n-1}\bar{S}_i\binom{n-1}{i-1}\binom{i}{k}\left(\frac{\bar{F}(t_1)}{\bar{F}(t_2)}-1\right)^{i}}{\sum_{j=k}^{n-1}\bar{S}_j\binom{n}{j}\binom{j}{k}\left(\frac{\bar{F}(t_1)}{\bar{F}(t_2)}-1\right)^{j}}.\end{eqnarray*}

Also,

\begin{eqnarray*}&&\mathbb{P}(T>t_2,N_{t_1}\leq k-1)=\sum_{i=1}^{k}s_i\sum_{j=0}^{i-1}\binom{n}{j}F^j(t_2)\bar{F}^{n-j}(t_2)\\[3pt]&&\ \ \ \ \ \ +\sum_{i=k+1}^{n}s_i\sum_{m=n-i+1}^{n}\sum_{l=\max(m,n-k+1)}^{n}\binom{n}{l}\binom{ l}{m}F^{n-l}(t_1)(\bar{F}(t_1)-\bar{F}(t_2))^{l-m}\bar{F}^{m}(t_2)\end{eqnarray*}

and

\begin{eqnarray*}\mathbb{P}(T>t_2,N_{t_1}\geq k+1)&\,=\,&\sum_{i=k+2}^{n}s_i\sum_{j=k+1}^{i-1}\sum_{m=n-i+1}^{n-k-1}\binom{n}{m}\binom{ n-m}{j}F^{j}(t_1)\\[3pt]&\times&(\bar{F}(t_1)-\bar{F}(t_2))^{n-j-m}\bar{F}^{m}(t_2).\end{eqnarray*}

On the other hand,

\begin{eqnarray*}\mathbb{P}(T>t_2,N_{t_1}=k)&\,=\,&\sum_{i=k+1}^{n}s_i\sum_{j=k}^{i-1}\binom{n}{j}\binom{j}{k}F^k(t_1)\bar{F}^{n-j}(t_2)(\bar{F}(t_1)-\bar{F}(t_2))^{j-k}\\[3pt]&\,=\,&\sum_{j=k}^{n-1}\bar{S}_j \binom{n}{j}\binom{j}{k}F^k(t_1)\bar{F}^{n-j}(t_2)(\bar{F}(t_1)-\bar{F}(t_2))^{j-k}.\end{eqnarray*}

Also, we may obtain

\begin{eqnarray*}\mathbb{P}(T<t_2)=1-\sum_{j=0}^{n-1}\bar{S}_j\binom{n}{j}F^{j}(t_2)\bar{F}^{n-j}(t_2).\end{eqnarray*}

The aim here is to minimize ${\eta_{II}(t_2)}$ with respect to the decision variable ${t_2}$ ; that is, we should find the possible value ${t_2}$ , if it exists, such that

\[\eta_{II}(t_2^*)=\min_{t_2>t_1}\eta_{II}(t_2).\]

Now, let us assume that PM combined with CM takes ${w_1}$ time units, CM on the whole system takes ${w_2}$ time units, PM on the whole system takes ${w_3}$ time units, and the rigid PM on the system takes ${w_4}$ time units. The stationary availability for Strategy ${\textbf{II}}$ is given by

\begin{eqnarray*}A_{II}(t_2)=\frac{\mathbb{E}(\!\min\!(t_2,T))}{\mathbb{E}(\!\min\!(t_2,T))+B(t_2)},\nonumber\end{eqnarray*}

where

\begin{eqnarray*}B(t_2)&\,=\,&w_1 \mathbb{P}(T>t_2,k_1\leq N_{t_1}\leq k_2)+w_{2}\mathbb{P}(T\leq t_2)\\[3pt]&&+\,w_{3}\mathbb{P}(T>t_2,N_{t_1}\leq k_1-1)+w_{4}\mathbb{P}(T>t_2,N_{t_1}\geq k_2+1).\end{eqnarray*}

Example 3.3. Let us look at again the bridge system in Example 2.1, where the component lifetimes are i.i.d. with Weibull distribution, with reliability function ${\bar{F}(t)=\exp\{-t^{2}\}}$ , ${t\geq0}$ . Figure 10(a) shows the plot of ${\eta_{II}(t_2)}$ for ${t_1=0.5}$ , ${c^*_{pms}=20}$ , ${c_{cms}=20}$ , ${c_{pms}=5}$ , ${c_{cm}=2}$ , and ${c_{pm}=1}$ , and different values ${k = 1,2,3}$ . Figure 10(b) depicts the plots of ${A_{II}(t)}$ for ${k=2}$ , ${t_1=0.5}$ , ${w_{1}=0.04}$ , ${w_{2}=0.05}$ , ${w_{3}=0.02}$ , and ${w_{4}=0.06}$ . It can be seen that the system availability first increases to arrives at its maximum and then decreases. The availability attains its maximum at ${t_2^*=0.719917}$ and ${A_{II}(t_2^*)=0.952746}$ .

Figure 10:

(a) The average maintenance cost per unit time in Example 3.3; (b) the stationary availability for Strategy ${\textbf{II}}$ in Example 3.3.

An analysis of the results in Table 6 indicates that the maintenance strategy ${\textbf{II}}$ is robust. It is seen that when ${{c}_{cms}}$ increases (or ${c_{pm}}$ decreases), the optimal time of PM decreases, as expected. This robustness is true also in terms of ${{c}^*_{pms}}$ . That is, when ${{c}^*_{pms}}$ gets larger, the optimal time of PM gets larger, too. One can easily see that, based on this example, the model is robust in terms of the costs ${{c}_{pms}}$ and ${{c}_{cm}}$ .

Table 6:

Optimal maintenance time for Strategy ${\textbf{II}}$ with ${t_1=0.5}$ , ${k=1}$ .

Table 7:

Bivariate optimal maintenance times for Strategy ${\textbf{II}}$ .

Considering ${t_1}$ and ${t_2}$ as two decision variables, the three-dimensional plot of the cost function is depicted in Figure 11. The optimal values of ${(t_1,t_2)}$ are also tabulated in Table 7 for different values of ${c_{cms}}$ and ${c^*_{pms}}$ .

Figure 11:

The three-dimensional plot of the cost function in Example 3.3.

Recalling Example 3.2, it is interesting to compare Strategies I and II for the bridge system. To do so, let ${\tau=0.1}$ , ${t_1=0.5}$ , ${c_{min}=0.5}$ , ${c_{cms}=25}$ , ${c_{pms}=4}$ , ${c_{cm}=2}$ , ${c_{pm}=1}$ , ${c^*_{pms}=20}$ , ${w_{1}=0.08}$ , ${w_{2}=0.2}$ , ${w_{3}=0.02}$ , and ${w_{4}=0.06}$ . If ${k=2}$ in Strategy ${\textbf{I}}$ and ${k_1=k_2=1}$ in Strategy ${\textbf{II}}$ , then the optimal values of ${\eta_I}$ and ${\eta_{II}}$ are 13.1557 and 20.6165, and the optimal values of ${A_I}$ and ${A_{II}}$ are 0.9250 and 0.8728, respectively. This means that based on both the expected maintenance cost and the stationary availability criteria, Strategy ${\textbf{I}}$ is preferred.