3.1. Allocation of two relevations in cases: (i) $\mathcal{A}_i=\mathcal{A}_j$ and (iii) $\mathcal{A}_i \subseteq \mathcal{A}_j$

In this subsection, we discuss the optimal allocation policies of one and two relevations for a series system and later generalize them to a coherent system in the sense of usual stochastic order and hazard rate order, for the cases: (i) $\mathcal{A}_i=\mathcal{A}_j$ and (iii) $\mathcal{A}_i \subseteq \mathcal{A}_j$ .

Suppose that a series system has two components with lifetimes $T_{1}$ and $T_{2}$ , and two relevations are available having lifetimes $S_{1}$ and $S_{2}$ . However, due to time and budget issues, suppose one can perform only one relevation at a time to the series system. That is, we are interested in the comparison between $U_1=\min \left\{T_1 \# S_1, T_2\right\}$ and $U_2=\min \left\{T_1, T_2 \# S_2\right\}$ , where the distribution function, survival function, and density function of a random variable $U_{1}$ $(U_{2})$ are $F_{U_{1}}$ $(F_{U_{2}})$ , $\overline{F}_{U_{1}}$ $(\overline{F}_{U_{2}})$ , and $f_{U_{1}}$ $(f_{U_{2}})$ , respectively. Suppose that the component lifetimes and the available relevation lifetimes are mutually independent. Belzunce et al. [Reference Belzunce, Martínez and Ruiz10] proved that if $T_{1} \leq_{hr} T_{2}$ and $S_{1} \geq_{hr} S_{2}$ then $U_{1} \geq_{st} U_{2}$ . One natural question arises that under what sufficient conditions $U_{1} \geq_{st} U_{2}$ can be strengthened to $U_{1} \geq_{hr} U_{2}$ ? To answer this question, we provide the following result on the optimal allocation policy of two relevations (performed one at a time) for hazard rate order in the series system.

Proof. First, we prove that $U_1 \geq_{hr} U_{2}$ under the conditions ‘ $S_{1} \geq_{hr} T_{1}$ ’, ‘ $T_{1} \leq_{hr} T_{2}$ ’, and ‘ $S_{1} \geq_{hr} S_{2}$ ’. To prove $U_1 \geq_{hr} U_{2}$ , it suffices to show that $\displaystyle\frac{\overline{F}_{U_1}(t)}{\overline{F}_{U_2}(t)}$ is increasing in $t>0$ . We have

(2) \begin{align}\overline{F}_{U_1}(t)=\left(\overline{F}_{T_1}(t)+\overline{F}_{S_1}(t) \int_0^t \frac{{f}_{T_1}(u)}{\overline{F}_{S_1}(u)} \,\textrm{d} u\right) \overline{F}_{T_2}(t), \quad t > 0, \end{align}

and

(3) \begin{align}\overline{F}_{U_2}(t)=\left(\overline{F}_{T_2}(t)+\overline{F}_{S_2}(t) \int_0^t \frac{{f}_{T_2}(u)}{\overline{F}_{S_2}(u)} \,\textrm{d} u\right) \overline{F}_{T_1}(t), \quad t > 0. \end{align}

From (2) and (3), it can be seen that

(4) \begin{align}\displaystyle\frac{\overline{F}_{U_1}(t)}{\overline{F}_{U_2}(t)}=\frac{\overline{F}_{T_1}(t)\;\overline{F}_{T_2}(t)+\overline{F}_{T_2}(t)\;\overline{F}_{S_1}(t) \displaystyle \int_0^t \frac{{f}_{T_1}(u)}{\overline{F}_{S_1}(u)} \,\textrm{d} u }{\overline{F}_{T_1}(t)\;\overline{F}_{T_2}(t)+\overline{F}_{T_1}(t)\;\overline{F}_{S_2}(t) \displaystyle \int_0^t \frac{{f}_{T_2}(u)}{\overline{F}_{S_2}(u)} \,\textrm{d} u} = \frac{g(t)+g_1(t)}{g(t)+g_2(t)}, \quad t > 0, \end{align}

where

\begin{align*}g(t)=\overline{F}_{T_1}(t) \overline{F}_{T_2}(t), g_1(t)=\overline{F}_{T_2}(t) \overline{F}_{S_1}(t) \int_0^t \frac{{f}_{T_1}(u)}{\overline{F}_{S_1}(u)} \,\textrm{d} u,\end{align*}

and

\begin{align*}g_2(t)=\overline{F}_{T_1}(t) \overline{F}_{S_2}(t) \int_0^t \frac{{f}_{T_2}(u)}{\overline{F}_{S_2}(u)} \,\textrm{d} u,\quad t>0.\end{align*}

To prove that ${\overline{F}_{U_1}(t)}/{\overline{F}_{U_2}(t)}$ is increasing in $t>0$ , it suffices to show that

(5) \begin{align}\frac{g(t)+g_1(t)}{g(t)+g_2(t)}\quad \text{is increasing in } t>0. \end{align}

Now, we employ Lemma 1 to show that (5) holds true. It is evident that $g_1(t)$ , $g_2(t) \geq 0$ , and $g(t) \geq 0$ , for all $t>0$ . Further, we have

(6) \begin{align}\frac{g_1(t)}{g_2(t)} &= \frac{\overline{F}_{T_2}(t)}{\overline{F}_{T_1}(t)}\cdot \frac{\overline{F}_{S_1}(t) \displaystyle \int_0^t \frac{{f}_{T_1}(u)}{\overline{F}_{S_1}(u)}\,\textrm{d}u}{\overline{F}_{S_2}(t) \displaystyle \int_0^t \frac{{f}_{T_2}(u)}{\overline{F}_{S_2}(u)}\,\textrm{d}u} \nonumber \\[5pt] &= \frac{\overline{F}_{T_2}(t)}{\overline{F}_{T_1}(t)}\cdot \frac{\displaystyle \int_0^t \frac{\overline{F}_{S_1}(t)}{\overline{F}_{S_1}(u)} {f}_{T_1}(u) \,\textrm{d}u}{\displaystyle \int_0^t \frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_2}(u)} {f}_{T_2}(u) \,\textrm{d}u} \nonumber \\[5pt] &= \phi_1(t)\;\phi_2(t),\end{align}

where $\phi_1(t) = {\overline{F}_{T_2}(t)}/{\overline{F}_{T_1}(t)}$ and

\begin{align*}\phi_2(t)= \frac{\int_0^t \frac{\overline{F}_{S_1}(t)}{\overline{F}_{S_1}(u)} {f}_{T_1}(u) \,\textrm{d}u}{\int_0^t \frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_2}(u)} {f}_{T_2}(u) \,\textrm{d}u}.\end{align*}

If $T_{1} \leq_{hr} T_{2}$ then $\phi_{1}(t)$ is increasing in $t>0$ . To prove $\phi_2(t)$ is increasing, it is equivalent to prove that ${1}/{\phi_2(t)}$ is decreasing for $t>0$ . Now,

\begin{align*}\displaystyle\frac{1}{\phi_2(t)}= \frac{\displaystyle \int_0^t \frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_2}(u)} {f}_{T_2}(u) \,\textrm{d}u}{\displaystyle \int_0^t \frac{\overline{F}_{S_1}(t)}{\overline{F}_{S_1}(u)} {f}_{T_1}(u) \,\textrm{d}u} = \frac{\displaystyle \int_0^t {h_{2}(t, u)}{{g}_{2}(u)} \,\textrm{d}u}{\displaystyle \int_0^t {h_{1}(t, u)}{{g}_{1}(u)} \,\textrm{d}u},\end{align*}

where, $h_{i}(t, u) = {\overline{F}_{S_i}(t)}/{\overline{F}_{S_i}(u)}$ and ${g}_{i}(u) = {f}_{T_i}(u)$ , for $i=1, 2$ . On employing Lemma 2 and using the conditions ‘ $T_{1} \leq_{hr} T_{2}$ ’ and ‘ $S_{2} \leq_{hr} S_{1}$ ’, we get ${1}/{\phi_2(t)}$ is decreasing in $t>0$ . Thus, from (6), we find that

(7) \begin{align}\frac{g_1(t)}{g_2(t)} \quad\text {is increasing in } t>0 . \end{align}

Now, we have to show that ${g(t)}/({g_2(t)-g_1(t)})$ is increasing in $t>0$ . We have

(8) \begin{align}\displaystyle\frac{g(t)}{g_2(t)-g_1(t)} & =\frac{\overline{F}_{T_1}(t) \overline{F}_{T_2}(t)}{\overline{F}_{T_1}(t) \overline{F}_{S_2}(t) \displaystyle \int_0^t \frac{{f}_{T_2}(u)}{\overline{F}_{S_2}(u)} \,\textrm{d} u-\overline{F}_{T_2}(t) \overline{F}_{S_1}(t) \displaystyle \int_0^t \frac{{f}_{T_1}(u)}{\overline{F}_{S_1}(u)} \,\textrm{d} u} \nonumber \\[5pt] & =\frac{1}{\displaystyle \int_0^t \frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_2}(u)} \frac{f_{T_{2}}(u)}{\overline{F}_{T_{2}}(t)} \,\textrm{d} u-\displaystyle \int_0^t \frac{\overline{F}_{S_1}(t)}{\overline{F}_{S_1}(u)} \frac{f_{T_{1}}(u)}{\overline{F}_{T_1}(t)} \,\textrm{d} u},\quad t>0. \end{align}

Now, we can see that

(9) \begin{align}& \frac{\partial}{\partial t}\left(\int_0^t \frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_2}(u)} \frac{{f}_{T_2}(u)}{\overline{F}_{T_2}(t)} \,\textrm{d} u-\int_0^t \frac{\overline{F}_{S_1}(t)}{\overline{F}_{S_1}(u)} \frac{{f}_{T_1}(u)}{\overline{F}_{T_1}(t)} \,\textrm{d} u\right) \nonumber \\[5pt] = & \left(\frac{{f}_{T_2}(t)}{\overline{F}_{T_2}(t)}-\frac{{f}_{T_1}(t)}{\overline{F}_{T_1}(t)}\right) + \int_0^t \frac{\partial}{\partial t}\left(\frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_2}(u)} \frac{{f}_{T_2}(u)}{\overline{F}_{T_2}(t)}-\frac{\overline{F}_{S_1}(t)}{\overline{F}_{S_1}(u)} \frac{{f}_{T_1}(u)}{\overline{F}_{T_1}(t)}\right) \,\textrm{d} u \nonumber\\[5pt] = & \left(\frac{{f}_{T_2}(t)}{\overline{F}_{T_2}(t)}-\frac{{f}_{T_1}(t)}{\overline{F}_{T_1}(t)}\right) + \int_0^t \frac{\partial}{\partial t}\left\{\left(\frac{\overline{F}_{S_1}(t)}{\overline{F}_{T_1}(t)}\right)\left(\frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_1}(t)} \frac{\overline{F}_{T_1}(t)}{\overline{F}_{T_2}(t)} \frac{{f}_{T_2}(u)}{\overline{F}_{S_2}(u)}-\frac{{f}_{T_1}(u)}{\overline{F}_{S_1}(u)}\right)\right\} \,\textrm{d} u \nonumber \\[5pt] \leq & \int_0^t \frac{\partial}{\partial t}\left\{\left(\frac{\overline{F}_{S_1}(t)}{\overline{F}_{T_1}(t)}\right)\left(\frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_1}(t)} \frac{\overline{F}_{T_1}(t)}{\overline{F}_{T_2}(t)} \frac{{f}_{T_2}(u)}{\overline{F}_{S_2}(u)}-\frac{{f}_{T_1}(u)}{\overline{F}_{S_1}(u)}\right)\right\} \,\textrm{d} u \nonumber \\[5pt] \leq & 0,\quad \text {for all } t>0, \end{align}

where the second last inequality follows from the condition ‘ $T_{1} \leq_{hr} T_{2}$ ’, and, the last inequality follows from the conditions ‘ $T_{1} \leq_{hr} T_{2}$ ’, ‘ $S_{1} \geq_{hr} S_{2}$ ’, and ‘ $S_{1} \geq_{hr} T_{1}$ ’, since these conditions imply that ${\overline{F}_{S_1}(t)}/{\overline{F}_{T_1}(t)}$ is increasing, ${\overline{F}_{S_2}(t)}/{\overline{F}_{S_1}(t)}$ and ${\overline{F}_{T_1}(t)}/{\overline{F}_{T_2}(t)}$ are decreasing in $t>0$ , and

\begin{align*}\frac{\overline{F}_{S_2}(t)}{\overline{F}_{S_1}(t)} \frac{\overline{F}_{T_1}(t)}{\overline{F}_{T_2}(t)} \frac{{f}_{T_2}(u)}{\overline{F}_{S_2}(u)}-\frac{{f}_{T_1}(u)}{\overline{F}_{S_1}(u)} \leq 0\quad \text{for all }t>0.\end{align*}

Therefore, we get that

(10) \begin{align}\frac{g(t)}{g_2(t)-g_1(t)}\quad \text {is increasing in } t>0 . \end{align}

Now, by using Lemma 1, and from (7) and (10), we get that (4) is increasing (i.e. $({g(t)+g_1(t)})/({g(t)+g_2(t)})$ is increasing in $t>0$ ), thereby implying that ${\overline{F}_{U_1}(t)}/{\overline{F}_{U_2}(t)}$ is increasing in $t>0$ , i.e. $U_1 \geq_{hr} U_{2}$ . Thus, $U_1 \geq_{hr} U_{2}$ under the conditions ‘ $S_{1} \geq_{hr} T_{1}$ ’, ‘ $T_{1} \leq_{hr} T_{2}$ ’, and ‘ $S_{1} \geq_{hr} S_{2}$ ’. Similarly, we can prove $U_1 \geq_{hr} U_{2}$ under the conditions ‘ $S_{2} \geq_{hr} T_{2}$ ’, ‘ $T_{1} \leq_{hr} T_{2}$ ’, and ‘ $S_{1} \geq_{hr} S_{2}$ ’.

In a similar vein, we can prove that $U_{1} \leq_{hr} U_{2}$ under the assumptions ‘ $S_{1} \geq_{hr} T_{1}$ (or $S_{2} \geq_{hr} T_{2}$ )’, ‘ $T_{1} \geq_{hr} T_{2}$ ’, and ‘ $S_{1} \leq_{hr} S_{2}$ ’ hold true. Hence, the result is proven.

The following example illustrates the application of Theorem 1. Recall that T is said to be Weibull random variable with shape parameter $\alpha(>0)$ and scale parameter $\lambda(>0)$ , denoted by $T \sim W(\alpha, \lambda)$ , if its survival function is given by $\overline{F}_{T} (t;\;\alpha, \lambda) =\textrm{e}^{-(\lambda t)^\alpha}$ , $t>0$ .

Note that Belzunce et al. [Reference Belzunce, Martínez and Ruiz10] established $U_{1} \geq_{st} U_{2}$ under the conditions $T_{1} \leq_{hr} T_{2}$ and $S_{1} \geq_{hr} S_{2}$ . As we can see from Theorem 1 that when we move from a weaker stochastic ordering (i.e. usual stochastic order) to a stronger stochastic ordering (i.e. hazard rate order), an extra condition ‘ $S_{1} \geq_{hr} T_{1}$ or $S_{2} \geq_{hr} T_{2}$ ’ is needed for the result to hold true. The following counterexample shows that if we relax the condition ‘ $S_1 \geq_{hr} T_1$ ’ in Theorem 1, the result $U_{1}$ $\ngeq_{hr}$ $U_{2}$ no longer holds true.

Figure 2.

Schematic diagrams of systems with lifetimes $U_{1}$ (left) and $U_{2}$ (right) corresponding to Theorem 1.

Figure 3.

Plot of the ratio of survival functions of $U_{1}$ and $U_{2}$ (i.e. ${\overline{F}_{U_{1}}(t)}/{\overline{F}_{U_{2}}(t)}$ ).

With regards to relevation allocation to a coherent system, Belzunce et al. [Reference Belzunce, Martínez and Ruiz10] investigated the optimal allocation policies for two relevation (performed one at a time) to a coherent system for case (i) $\mathcal{A}_i=\mathcal{A}_j$ in the sense of usual stochastic order. They demonstrated that for components i and j in common path sets, the relevation should be applied to the component with the higher hazard rate. Further, they called for future research on comparing system lifetimes in terms of the hazard rate order after allocating a single relevation to a coherent system. To address this, we provide the following result, which employs Theorem 1 above on series systems. For this purpose, consider an n-component coherent system with components’ lifetimes $T_1, T_2, \ldots, T_n$ , having structure function $\phi$ , and minimal path sets $P_1, P_2, \ldots, P_k$ . The system lifetime can be expressed as

\begin{align*}\tau(\boldsymbol{T})=\tau\left(T_1, \ldots, T_n\right)=\max _{\mathcal{P} \in \mathcal{A}} \min _{j \in \mathcal{P}} T_j.\end{align*}

In addition, let $\boldsymbol{T}^{(i, j)} = T_{1}, \ldots, T_{i-1}, T_{i+1}, \ldots, T_{j-1}, T_{j+1}, \ldots, T_{n}$ , and denote the lifetime of the two coherent systems by $W_1\sim \tau\left(T_i \# S_1, T_j, \boldsymbol{T}^{(i, j)}\right) $ and $W_2\sim \tau\left(T_i, T_j \# S_2, \boldsymbol{T}^{(i, j)}\right) $ after performing one relevation at a time.

Theorem 2. Let $W_1\sim \tau\left(T_i \# S_1, T_j, \boldsymbol{T}^{(i, j)}\right) $ and $W_2\sim \tau\left(T_i, T_j \# S_2, \boldsymbol{T}^{(i, j)}\right) $ . Consider two locations $i, j \in \{1, \ldots , n\}$ such that $i<j$ and $\{i, j\} \subseteq P_r $ , for $r= i_1, \ldots , i_r$ , where $\left\{ i_1, \ldots , i_r \right\} \subseteq \left\{ 1, \ldots , k \right\}$ , i.e. $\mathcal{A}_i=\mathcal{A}_j$ . Suppose

\begin{align*} \frac{\mathrm{P}\left(\underset{1\leq l \leq k, l \notin \{i_1, \ldots, i_r \} }{\max} \underset{m \in P_l}{\min} \left\{T_m\right\} >t \right)}{\mathrm{P}\left( \underset{s \in \{ i_1, \ldots, i_r\} }{\max} \underset{m \in P_s, m\neq i, j}{\min} \left\{T_m\right\} > t, \underset{1\leq l \leq k, l \notin \{i_1, \ldots, i_r \} }{\max} \underset{m \in P_l}{\min} \left\{T_m\right\} \leq t \right)} \end{align*}

is decreasing in $t>0$ . If $T_i \leq_{h r} (\!\geq_{hr}\!) \;T_j$ , $S_1 \geq_{h r} (\!\leq_{hr}\!) \;S_2$ , and $S_1 \geq_{h r} T_i$ (or $S_2 \geq_{h r} T_j$ ) then $W_1$ $\geq_{hr}$ $(\!\leq_{hr}\!)$ $W_2$ .

Proof. To prove the result, let us consider two series systems having lifetime $U_1=\min \{T_i \# S_1,\ T_j\}$ and $U_2=\min \left\{T_i, T_j \# S_2\right\}$ . First, we prove that $W_1 \geq_{hr} W_2$ under the assumptions $T_i \leq_{h r} T_j$ , $S_1 \geq_{h r} S_2$ , and $S_1 \geq_{h r} T_i$ (or $S_2 \geq_{h r} T_j$ ). Then, for some minimal path set $P_r$ such that $\{i, j\} \subseteq P_r$ , the lifetimes of systems $W_{1}$ and $W_{2}$ can be expressed as

\begin{align*}\begin{aligned}W_1 & =\max\left\{\min \left\{ \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}, T_i \# S_1, T_j \right\}, \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \right\},\end{aligned}\end{align*}

and

\begin{align*}\begin{aligned}W_2 & =\max\left\{\min \left\{ \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}, T_i , T_j\# S_2 \right\}, \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \right\}.\end{aligned}\end{align*}

Thus, we get

\begin{align*} \overline{F}_{W_{1} }(t) &= 1- \mathrm{P}\bigg( \min \left\{ \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}, T_i \# S_1, T_j\right\} \leq t,\\[5pt] &\hspace{7.4cm}\max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \leq t \bigg)\\[5pt] &= \mathrm{P}\left( U_1 > t\right) \mathrm{P}\left( \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}>t , \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \leq t\right)\\[5pt] &\quad+\mathrm{P}\left( \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} > t\right), \quad t>0,\end{align*}

and

\begin{align*} \overline{F}_{W_{2} }(t)&=\mathrm{P}\left( U_2 > t\right) \mathrm{P}\left( \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}>t , \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \leq t\right) \\[5pt] &\quad+\mathrm{P}\left( \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} > t\right), \quad t>0.\end{align*}

Now,

\begin{align*} \frac{\overline{F}_{W_{1} }(t)}{\overline{F}_{W_{2} }(t)} = \frac{g(t) + g_{1}(t)}{g(t) + g_{2}(t)}, \quad t>0,\end{align*}

where

\begin{align*}g(t) = \mathrm{P}\left( \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} > t\right),\end{align*}

\begin{align*}g_1(t) = \mathrm{P}\left( U_1 > t\right) \mathrm{P}\left( \max\limits_{s \in \{i_1, \ldots, i_r\}}\min\limits_{m \in P_s, m \neq i, j}\left\{T_m\right\}>t , \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} \leq t\right) \end{align*}

and

\begin{align*} g_2(t)= \mathrm{P}\left( U_2 > t\right) \mathrm{P}\left( \max\limits_{s \in \{i_1, \ldots, i_r\}}\min\limits_{m \in P_s, m \neq i, j}\left\{T_m\right\}>t , \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} \leq t\right), \quad t>0. \end{align*}

Now, from Theorem 1, we get that $g_1(t)/g_2(t)$ is increasing in $t>0$ , since ‘ $T_{i} \leq_{hr} T_{j}$ ’ and ‘ $S_{2} \leq_{hr} S_{1}$ ’. Further, we have

\begin{align*} \frac{g(t)}{g_2(t)- g_1(t)} &= \frac{\mathrm{P}\left( \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} > t\right)}{\mathrm{P}\left( \max\limits_{s \in \{i_1, \ldots, i_r\}}\min\limits_{m \in P_s, m \neq i, j}\left\{T_m\right\}>t , \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} \leq t\right)} \\[5pt] &\quad\times \frac{1}{\mathrm{P}\left( U_2 > t \right) - \mathrm{P}\left( U_1 > t \right)},\quad t>0. \end{align*}

Now, from the condition of Theorem 2, we get that

\begin{align*}\frac{\mathrm{P}\left( \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} > t\right)}{\mathrm{P}\left( \max\limits_{s \in \{i_1, \ldots, i_r\}}\min\limits_{m \in P_s, m \neq i, j}\left\{T_m\right\}>t, \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} \leq t\right)}\end{align*}

is decreasing in $t>0$ and from the proof of Theorem 1, we get ${1}/({\mathrm{P}( U_2 > t ) - \mathrm{P}( U_1 > t )})$ is increasing in $t>0$ under the assumptions $T_i \leq_{h r} T_j$ , $S_1 \geq_{h r} S_2$ , and $S_1 \geq_{h r} T_i$ (or $S_2 \geq_{h r} T_j$ ). By combining these two facts, we get ${g(t)}/({g_2(t)- g_1(t)})$ is increasing in $t>0$ . Hence, $W_1 \geq_{hr} W_2$ .

Similarly, we can prove that $W_1 \leq_{hr} W_2$ under the assumptions $T_i \geq_{h r} T_j$ , $S_1 \leq_{h r} S_2$ , and rest of the conditions are same. Hence, the result is proved.

To illustrate the result of Theorem 2, we provide an example below.

Example 4. Consider a coherent system made up of four components $C_{1}$ , $C_{2}$ , $C_{3}$ , and $C_{4}$ (illustrated in Figure 4). Let the first two components $C_{1}$ and $C_{2}$ have lifetimes $T_{1}$ and $T_{2}$ , respectively, with $T_{i} \sim W(\alpha, \lambda_{i})$ , where $ \alpha > 0$ , $\lambda_{i} > 0$ , $i=1,2$ , and the other two components $C_{3}$ and $C_{4}$ have lifetimes $T_{3}$ and $T_{4}$ , respectively, such that $T_{1} \overset{d}{=} T_{4}$ and $T_{2} \overset{d}{=} T_{3}$ . Further, suppose two relevations $R_{1}$ and $R_{2}$ having lifetimes $S_{1}\sim W(\alpha, \beta_{1})$ and $S_{2}\sim W(\alpha, \beta_{2})$ , where $ \alpha > 0$ , $\beta _{i} > 0$ , $i=1,2$ , are available for allocation. Assume that relevation $R_{1}$ is performed to component $C_{1}$ and $R_{2}$ is performed to component $C_{2}$ , then the resultant systems have lifetimes $W_1$ and $W_2$ , respectively.

Figure 4.

Schematic diagrams of systems with lifetimes $W_{1}$ (left) and $W_{2}$ (right) corresponding to Example 2.

Note that the given coherent system has minimal path sets $P_{1} = \{{1}, {2}, {3}\}$ and $P_{2} = \{4\}$ . Thus, we can say that $\mathcal{A}_1=\mathcal{A}_2$ (i.e. components 1 and 2 are included in the same minimal path sets). Further, it is easy to see that when $\lambda_{2}$ $\leq$ $\lambda_{1}$ , then $T_{1}$ $\leq_{hr}$ $T_{2}$ . In addition, if $\beta_{1}$ $\leq$ $\beta_{2}$ and $\beta_{1}$ $\leq$ $\lambda_{1}$ , then $S_{2}$ $\leq_{hr}$ $S_{1}$ and $T_{1}$ $\leq_{hr}$ $S_{1}$ hold true. Further, it can be seen that

\begin{align*} \frac{\mathrm{P}\left( \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} > t\right)}{\mathrm{P}\left( \max\limits_{s \in \{i_1, \ldots, i_r\}}\min\limits_{m \in P_s, m \neq i, j}\left\{T_m\right\}>t, \max\limits_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min\limits_{m \in P_l}\left\{T_m\right\} \leq t\right)}&=\frac{\mathrm{P}(T_{4}>t)}{\mathrm{P}(T_{3}>t)\; \mathrm{P}(T_{4}\leq t)} \end{align*}

is decreasing in $t>0$ , if $\lambda_{2}$ $\leq$ $\lambda_{1}$ (i.e. $T_{4} \leq_{hr} T_{3}$ ). Thus, the conditions of Theorem 2 are satisfied, i.e. $W_1 \geq_{hr} W_2$ .

Up to now, we considered the optimal allocation policy for two relevations (performed one at a time) to series and general coherent systems. Moving forward, suppose there is no budget constraint, and we can afford to allocate two relevations, with lifetimes $S_1$ and $S_2$ , simultaneously to a system. It is, therefore, of interest to determine the optimal allocation policy for two relevation allocations simultaneously to a system. In this regard, Zhang and Zhang [Reference Zhang and Zhang34, Lemma 1], proved that if $T_1 \leq_{lr} T_2$ and $S_{1} \geq_{hr} S_{2}$ then $V^{\prime}_{1} \geq_{s t} V^{\prime}_{2}$ , where $V^{\prime}_{1}=\min \left\{T_1 \# S_1, T_2 \# S_2\right\}$ and $V_{2}^{\prime}=\min \left\{T_1 \# S_2, T_2 \# S_1\right\}$ . As they gave only one set of sufficient conditions for $V_{1}^{\prime} \geq_{s t} V^{\prime}_{2}$ to hold true, we provide additional sufficient conditions, in the following theorem, for the same result to hold true.

If we assume that S follows an exponential distribution in the relevation survival formula (i.e. (1)), then the standby redundancy allocation becomes a special case of the relevation allocation. In light of this, one can observe that the first set of conditions, i.e. $T_1$ $\leq_{lr}$ $(\geq_{lr})$ $T_2$ and $S_{1} \geq_{hr} S_{2}$ , is same as that in [Reference Kumar and Naqvi21, Theorem 3.1], and the second set of conditions, i.e. $T_1$ $\geq_{lr}$ $(\leq_{lr})$ $T_2$ and $S_{1} \leq_{hr} S_{2}$ , is same as that in [Reference Kumar and Naqvi21, Theorem 3.2]. This highlights that in this case, if we shift from standby redundancy allocation to relevation allocation, the conditions remain the same.

The following counterexample shows that if we relax the likelihood ratio condition $ T_1 \leq_{lr} (\geq_{lr}) T_2 $ to a weaker stochastic order in Theorem 3, then the result no longer holds.

Counterexample 2. Let $T_{1}$ and $T_{2}$ be independent random variables with distribution functions

\begin{align*}F_{T_{1}} (t) = \begin{cases} {\displaystyle 1-\exp\left(-at-\frac{bt^{3}}{3}\right)}, & \textit{if}\;\, \text{t $\geq$ 0,}\\[5pt] 0, & \textit{otherwise,} \end{cases}\end{align*}

and

\begin{align*}F_{T_{2}} (t) = \begin{cases}{\displaystyle \frac{1}{2}\Bigg[1+\textit{erf}\Bigg(\frac{\log(t)}{\sqrt{2}}\Bigg)\Bigg]}, & \textit{if}\;\, \text{t $>$ 0,}\\[5pt] 0, & \textit{otherwise,} \end{cases}\end{align*}

where $a, b > 0$ . From Figure 5(a), it can be seen that $F_{T_{2}} \leq F_{T_{1}}$ , for $a = 0.8, b = 3$ , implying that $T_{1}$ $\leq_{st}$ $T_{2}$ . It is also easy to see from Figure 5(b) (and (c)) that ${F_{T_{2}}}/{F_{T_{1}}}$ (and $f_{T_2} / f_{T_1}$ ) is not monotonically increasing or decreasing for $a = 0.8$ and $b = 3$ (i.e. neither $T_{1}$ $\nleq_{rh}$ $T_{2}$ nor $T_{1}$ $\nleq_{lr}$ $T_{2}$ ).

Figure 5.

(a) Plot of the distribution functions of $T_{1}$ and $T_{2}$ (i.e. ${{F}_{T_{1}}(t)} \;\text{and}\;{{F}_{T_{2}}(t)}$ ). (b) Plot of the ratio of the distribution functions of $T_{2}$ and $T_{1}$ (i.e. ${{F}_{T_{2}}(t)}/{{F}_{T_{1}}(t)}$ ). (c) Plot of the ratio of the density functions of $T_{2}$ and $T_{1}$ (i.e. ${{f}_{T_{2}}(t)}/{{f}_{T_{1}}(t)}$ ). (d) Plot of the difference of the survival functions of $V_{1}'$ and $V_{2}'$ (i.e. ${\overline{F}_{V_{1}'}(t)} - {\overline{F}_{V_{2}'}(t)}$ ).

Now, suppose that $S_{1}$ and $S_{2}$ are independent random variables following Gamma and Weibull distribution with density functions

\begin{align*}f_{S_{1}} (t) = \begin{cases}{\displaystyle\frac{1}{\Gamma{(\alpha_{1})}} {t^{\alpha_{1} - 1}} {\textrm{e}^{-t}}}, & \textit{if}\;\, \text{t $>$ 0,}\\[5pt] 0, & \textit{otherwise,} \end{cases} \end{align*}

and

\begin{align*} f_{S_{2}} (t) = \begin{cases} {\displaystyle \alpha_{2} t^{\alpha_{2}-1} \textrm{e}^{-t^{\alpha_{2}}}}, & \textit{if}\;\, \text{t $\geq$ 0,}\\[5pt] 0, & \textit{otherwise,} \end{cases} \end{align*}

respectively, where, $ \alpha_{1} > 0, \alpha_{2} > 0$ . One can easily prove that $S_{1}$ $\geq_{hr}$ $S_{2}$ if $\alpha_{1} = 2$ and $\alpha_{2} = 1$ . However, it is evident from Figure 5(d) that the result of Theorem 3 does not hold, i.e. if $T_{1}$ $\leq_{st}$ $T_{2}$ and $S_{1}$ $\geq_{hr}$ $S_{2}$ , it does not imply that $V^{\prime}_{1}$ $\leq_{st}$ $V^{\prime}_{2}$ . Hence, if we relax the condition $T_{1}$ $\leq_{lr}$ $(\geq_{lr})$ $T_{2}$ to $T_{1}$ $\leq_{st}$ $(\!\geq_{st}\!)$ $T_{2}$ in Theorem 3, then the result (i.e. $V^{\prime}_{1}$ $\leq_{st}$ $(\!\geq_{st}\!)$ $V^{\prime}_{2}$ ) no longer holds even if the other condition (i.e. $S_{1}$ $\geq_{hr}$ $S_{2}$ ) remains the same.

With regards to the allocation of two relevations to a coherent system simultaneously, Yan et al. [Reference Yan, Zhang and Zhang31] provided the result for case (i) $\mathcal{A}_i = \mathcal{A}_j$ . However, Zhang and Zhang [Reference Zhang and Zhang34] pointed out an error in Theorem 1 of [Reference Yan, Zhang and Zhang31] for a two-component series system. Consequently, Theorem 4 in [Reference Yan, Zhang and Zhang31] for two relevations allocation to a coherent system for case (i) $\mathcal{A}_i = \mathcal{A}_j$ is also incorrect. In the following, we present the corrected version for two relevations allocation to a coherent system in case (i) $\mathcal{A}_i = \mathcal{A}_j$ . For this purpose, we consider an n-component coherent system with components lifetimes $T_1, T_2, \ldots, T_n$ , having structure function $\phi$ , and minimal path sets $P_1, P_2, \ldots, P_k$ , where k is the total number of minimal path sets. The system lifetime can be expressed as

\begin{align*}\tau(\boldsymbol{T})=\tau\left(T_1, \ldots, T_n\right)=\min _{\mathcal{P} \in \mathcal{A}} \max _{j \in \mathcal{P}} T_j.\end{align*}

In addition, let $\boldsymbol{T}^{(i, j)} = T_{1}, \ldots, T_{i-1}, T_{i+1}, \ldots, T_{j-1}, T_{j+1}, \ldots, T_{n}$ and we denote the lifetime of the two coherent systems by $Z_1\sim \tau\left(T_i \# S_1, T_j \# S_2, \boldsymbol{T}^{(i, j)}\right) $ and $Z_2\sim \tau\left(T_i \# S_2, T_j \# S_1, \boldsymbol{T}^{(i, j)}\right) $ after performing two relevations.

Proof. To prove the result, let us consider two series systems having ${V_1}'=\min \left\{T_i \# S_1, T_j \# S_2\right\}$ and ${V_2}'=\min \left\{T_i \# S_2, T_j \# S_1\right\}$ . Now, for some minimal path set $P_r$ such that $\{i, j\} \subseteq P_r$ , the lifetimes of systems $Z_{1}$ and $Z_{2}$ can be expressed as

\begin{align*}Z_1 & =\max\left\{\min \left\{ \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}, T_i \# S_1, T_j \# S_2 \right\}, \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \right\},\end{align*}

and

\begin{align*}Z_2 & =\max\left\{\min \left\{ \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}, T_i \# S_2 , T_j\# S_1 \right\}, \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \right\}.\end{align*}

Thus, we get

\begin{align*} \overline{F}_{Z_{1} }(t) &=1- \mathrm{P}\bigg( \min \left\{ \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}, T_i \# S_1, T_j \# S_2\right\} \leq t,\\[5pt] &\quad \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \leq t\bigg) \\[5pt] &= \mathrm{P}\left( {V_1}' > t\right) \mathrm{P}\left( \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}>t , \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \leq t\right)\\[5pt] &\quad +\mathrm{P}\left( \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} > t\right),\end{align*}

and

\begin{align*} \overline{F}_{Z_{2} }(t) &= \mathrm{P}\left( {V_2}' > t\right) \mathrm{P}\left( \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}>t , \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \leq t\right)\\[5pt] &\quad +\mathrm{P}\left( \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} > t\right), \quad t>0.\end{align*}

Now,

(11) \begin{equation} \overline{F}_{Z_{1} }(t) - \overline{F}_{Z_{2} }(t) = \Big( \mathrm{P}\left( {V_1}' > t\right) - \mathrm{P}\left( {V_2}' > t\right)\Big) \times \delta(t),\end{equation}

where

\begin{eqnarray*} \delta(t)= \mathrm{P}\left( \max_{s \in \{i_1, \ldots, i_r\}}\min _{m \in P_s, m \neq i, j}\left\{T_m\right\}>t , \max_{1\leq l \leq k, l \notin \{i_1, \ldots, i_r\} }\min _{m \in P_l}\left\{T_m\right\} \leq t\right).\end{eqnarray*}

Thus, from (11), it can be seen that to prove $Z_1 \geq_{s t} (\!\leq_{st}\!) Z_2$ , it is equivalent to prove ${V_1}' \geq_{s t} (\!\leq_{st}\!)\; {V_2}'$ . Hence, from Theorem 3, the result is proved.

Up to now, we focused on the first case, i.e. $\mathcal{A}_i=\mathcal{A}_j$ . We now turn our attention to case (iii) $\mathcal{A}_i \subseteq \mathcal{A}_j$ . In the situation where only one relevation out of the two can be allocated to a coherent system, Wu et al. [Reference Wu, Ding, Zhang and Zhao30, Theorem 2.1] showed that $W_{1}$ $\leq_{st}$ $W_{2}$ under the conditions $T_{i} \geq_{hr} T_{j}$ and $S_{1} \leq_{hr} S_{2}$ . Further, when the two relevations can be allocated simultaneously to a coherent system, Zhang and Zhang [Reference Zhang and Zhang34, Theorem 1] proved $Z_1 \geq_{st} Z_2$ when $T_i \geq_{lr} T_j$ and $S_1 \leq_{hr} S_2$ . Now, we provide additional set of sufficient conditions for $Z_1 \geq_{s t} (\!\leq_{st}\!) \;Z_2$ in case (iii) $\mathcal{A}_i \subseteq \mathcal{A}_j$ .

The following counterexample shows that if we relax the likelihood ratio condition $ T_1 \leq_{lr} (\geq_{lr}) T_2 $ to a weaker stochastic order in Theorem 5, then the result no longer holds.

Figure 6.

Plot of the difference of survival functions of $Z_{1}$ and $Z_{2}$ (i.e. $\displaystyle{\overline{F}_{Z_{1}}(t)} -{\overline{F}_{Z_{2}}(t)}$ ).

It is interesting to explore the problem of allocating two relevations simultaneously to a coherent system in the sense of hazard rate order. This continues to be an open problem requiring further exploration.

3.2. Allocation of two relevations in cases: (ii) $\mathcal{S}_i=\mathcal{S}_j$ and (iv) $\mathcal{S}_i \subseteq \mathcal{S}_j$

In this subsection, we discuss the optimal allocation policies of two relevations (performed simultaneously) for a parallel system and later generalize them to a coherent system for the cases: (ii) $\mathcal{S}_i=\mathcal{S}_j$ and (iv) $\mathcal{S}_i \subseteq \mathcal{S}_j$ .

Belzunce et al. [Reference Belzunce, Martínez and Ruiz10] studied the optimal allocation policy for two relevations (performed one at a time) in a two-component series system and later generalized to a coherent system in case (ii) $\mathcal{S}_i=\mathcal{S}_j$ . They made the stochastic comparison in the sense of the usual stochastic order and demonstrated that for components i and j in common cut sets, the relevation should be done on the component with the higher reversed hazard rate. Later, Wu et al. [Reference Wu, Ding, Zhang and Zhao30] established the stochastic comparison for two relevations (performed one at a time) in the sense of the usual stochastic order for case (iv) $\mathcal{S}_i \subseteq \mathcal{S}_j$ . Recently, Zhang and Zhang [Reference Zhang and Zhang34] studied the optimal allocation policy for two relevations (performed both at a time) in the sense of a usual stochastic order for case (iv) $\mathcal{S}_i \subseteq \mathcal{S}_j$ . In their study on the allocation of two relevations to a coherent system, they proved that $ \tau\left(T_i \# S_1, T_j \# S_2, \boldsymbol{T}^{(i, j)}\right) \leq_{st} \tau\left(T_i \# S_2, T_j \# S_1, \boldsymbol{T}^{(i, j)}\right) $ if $\mathcal{S}_i \subseteq \mathcal{S}_j$ , $S_{1} \geq_{hr} S_{2}$ , and $F_{T_{i} \# S_{1}}(t) F_{T_{j} \# S_{2}}(t) \geq F_{T_{i} \# S_{2}}(t) F_{T_{j} \# S_{1}}(t)$ , for all $t>0$ (see [Reference Zhang and Zhang34, Theorem 2]). The authors further pointed out that it is of interest to find sufficient conditions for the assumption $F_{T_{i} \# S_{1}}(t) F_{T_{j} \# S_{2}}(t) \geq F_{T_{i} \# S_{2}}(t) F_{T_{j} \# S_{1}}(t)$ to hold true. They conjectured that it might require $T_{i} \leq_{hr} T_{j}$ or $T_{i} \leq_{rh} T_{j}$ . To inquire into this conjecture, we first provide the following theorem for two-component parallel system and then investigate the case for a coherent system.

In the above result, it is easy to establish reversed hazard rate ordering between $T_{1}$ and $T_{2}$ . To investigate the other sufficient condition ‘ $1-(\overline{F}_{S_{i}}(t)/\overline{F}_{S_{i}}(u))$ is $R R_2\;(T P_2)$ in $(i, u) \in\{1,2\} \times(0, t]$ ’, we illustrate in the following example that it is satisfied by some common family of distributions, such as Weibull and Pareto.

Now, we are ready to present the results for the allocation of two relevations simultaneously to a coherent system for case (ii) $\mathcal{S}_i=\mathcal{S}_j$ and case (iv) $\mathcal{S}_i \subseteq \mathcal{S}_j$ . It is to be noted that for case (ii) $\mathcal{S}_i=\mathcal{S}_j$ , although a result was provided in [Reference Belzunce, Martínez and Ruiz10] on optimal allocation policy for two relevations (performed one at a time), no result exists in the literature for two relevations (performed at the same time) allocation to a coherent system. To fill this gap in the literature, we provide the following result, which employs Theorem 6 and uses the notation from Subsection 3.1.

Theorem 7. Let $Z_1\sim \tau\left(T_i \# S_1, T_j \# S_2, \boldsymbol{T}^{(i, j)}\right) $ and $Z_2\sim \tau\left(T_i \# S_2, T_j \# S_1, \boldsymbol{T}^{(i, j)}\right) $ . Consider two locations $i, j \in \{1, \ldots , n\}$ such that $i<j$ and $\{i, j\} \subseteq C_r $ , for $r= i_1, \ldots , i_r$ , where $\left\{ i_1, \ldots , i_r \right\} \subseteq \left\{ 1, \ldots , m \right\}$ , i.e. $\mathcal{S}_i=\mathcal{S}_j$ . If any of the following sets of conditions are satisfied:

1. (i) $1-(\overline{F}_{S_{i}}(t)/\overline{F}_{S_{i}}(u))$ is $R R_2$ in $(i, u) \in\{1,2\} \times(0, t]$ and $T_i \leq_{r h} (\geq_{r h})\; T_j$ ;

2. (ii) $1-(\overline{F}_{S_{i}}(t)/\overline{F}_{S_{i}}(u))$ is $T P_2$ in $(i, u) \in\{1,2\} \times(0, t]$ and $T_i \geq_{r h} (\leq_{r h})\; T_j$ ;

then

\begin{align*}Z_1 \geq_{s t} (\!\leq_{st}\!)\; Z_2.\end{align*}

Proof. The idea of the proof is borrowed from [Reference Belzunce, Martínez and Ruiz10, Theorem 2]. First, we prove (i). To prove $Z_1 \geq_{st} Z_2$ under the assumptions ‘ $1-{\overline{F}_{S_{i}}(t)}/{\overline{F}_{S_{i}}(u)}$ is $R R_2$ in $(i, u) \in\{1,2\} \times(0, t]$ ’ and ‘ $T_i \leq_{rh} T_j$ ’, let us consider two parallel systems having lifetime $V_1=\max \left\{T_i \# S_1, T_j \# S_2 \right\}$ and $V_2=\max \left\{T_i \# S_2, T_j \# S_1\right\}$ . Then, for some minimal cut set $C_r$ such that $\{i, j\} \subseteq C_r$ , the survival functions of systems $Z_{1}$ and $Z_{2}$ can be expressed as

\begin{align*} \overline F_{Z_1}(t) &= \mathrm{P}\bigg( \min\limits_{1 \leq l \leq m, l \notin \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_l}\left\{T_k\right\} \Big \} > t, \min\limits_{s \in \{i_1, i_2, \ldots, i_r\} } \Big\{ \max \Big\{\max _{k \in C_s, k \neq i, j}\left\{T_k\right\}, T_i \# S_1,\\[5pt] & \quad T_j \# S_2\Big\} \Big\} >t \bigg) \\[5pt] &=\mathrm{P}\bigg( \min\limits_{1 \leq l \leq m, l \notin \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_l}\left\{T_k\right\} \Big \} > t, \max\Big\{ \min\limits_{s \in \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_s, k \neq i, j}\left\{T_k\right\} \Big\}, V_1 \Big\} >t \bigg) \\[5pt] &= \bigg(1- F_{V_1}(t) \; \mathrm{P}\bigg( \min\limits_{s \in \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_s, k \neq i, j}\left\{T_k\right\} \Big\} \leq t \; \big| \\[5pt] & \hspace{+1cm} \min\limits_{1 \leq l \leq m, l \notin \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_l}\left\{T_k\right\} \Big \} > t \bigg) \bigg) \; \mathrm{P}\bigg( \min\limits_{1 \leq l \leq m, l \notin \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_l}\left\{T_k\right\} \Big \} > t \bigg) , \end{align*}

and

\begin{align*} \overline F_{Z_2}(t) &= \mathrm{P}\bigg( \min\limits_{1 \leq l \leq m, l \notin \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_l}\left\{T_k\right\} \Big \} > t, \max\Big\{ \min\limits_{s \in \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_s, k \neq i, j}\left\{T_k\right\} \Big\}, V_1 \Big\} >t \bigg) \\[5pt]&= \bigg(1- F_{V_2}(t) \; \mathrm{P}\bigg( \min\limits_{s \in \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_s, k \neq i, j}\left\{T_k\right\} \Big\} \leq t \; \big| \\[5pt] &\quad \min\limits_{1 \leq l \leq m, l \notin \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_l}\left\{T_k\right\} \Big \} > t \bigg) \bigg) \; \mathrm{P}\bigg( \min\limits_{1 \leq l \leq m, l \notin \{i_1, i_2, \ldots, i_r\} } \Big\{ \max _{k \in C_l}\left\{T_k\right\} \Big \} > t \bigg). \end{align*}

Thus, to prove $Z_1 \geq_{st} Z_2$ , it suffices to show that $V_1 \geq_{st} V_2$ , which we obtain, by using conditions ‘ $1-{\overline{F}_{S_{i}}(t)}/{\overline{F}_{S_{i}}(u)}$ is $R R_2$ in $(i, u) \in\{1,2\} \times(0, t]$ ’ and ‘ $T_i \leq_{rh} T_j$ ’, from Theorem 6.

In a similar vein we can prove that $Z_1 \leq_{st} Z_2$ under the assumptions ‘ $1-\frac{\overline{F}_{S_{i}}(t)}{\overline{F}_{S_{i}}(u)}$ is $RR_2$ in $(i, u) \in\{1,2\} \times(0, t]$ ’ and ‘ $T_i \leq_{rh} T_j$ ’. Hence, the results of Theorem 7 of (i) are proved.

The proof of (ii) of Theorem 7 is analogous to the proof of part (i), hence it is omitted.

To illustrate the application of Theorem 7, we present the following example. Recall that T is said to be a truncated normal random variable with mean zero, variance $\sigma(>0)$ , and parameter $\mu=1/2$ , denoted by $T \sim TN( \mu, \sigma)$ , if its density function is given by

\begin{align*}\overline{F}_{T} (t;\; \mu, \sigma) = \frac{1}{\mu \sigma \sqrt{2 \pi}} {\exp\left(-\displaystyle \frac{t^{2}}{2 {\sigma^2}}\right)},\quad t>0.\end{align*}

Figure 7.

Schematic diagrams of systems with lifetimes $Z_{1}$ (left) and $Z_{2}$ (right) corresponding to Example 4.

With regards to case (iv) $\mathcal{S}_i \subseteq \mathcal{S}_j$ , Wu et al. [Reference Wu, Ding, Zhang and Zhao30] established the usual stochastic order for two relevations allocation (performed one at a time). Their result was further extended by [Reference Zhang and Zhang34, Theorem 2] by establishing the usual stochastic order for two relevations allocation (performed both at the same time). As mentioned previously (Theorem 6), we explore the conjecture of [Reference Zhang and Zhang34, Theorem 2] by employing Theorem 6 and establishing the usual stochastic order for two relevations allocation (performed both at the same time) to a coherent system.

As observed, one of the conditions of Theorem 8, i.e. $S_{1} \geq_{hr} S_{2}$ , is the same as a condition in [Reference Zhang and Zhang34, Theorem 2], where the authors conjectured that $T_{i} \leq_{rh} T_{j}$ might be required to prove $F_{T_{i} \# S_{1}}(t) F_{T_{j} \# S_{2}}(t) \geq F_{T_{i} \# S_{2}}(t) F_{T_{j} \# S_{1}}(t)$ , for $t>0$ . We have found in Theorem 8 that to prove this conjecture we require $T_{i} \leq_{rh} T_{j}$ , as well as $1-(\overline{F}_{S_{i}}(t)/\overline{F}_{S_{i}}(u))$ is $TP_2$ in $(i, u) \in\{1,2\} \times(0, t]$ . Moreover, our proposed Theorems 7 and 8 contribute to the literature by adding some new results on coherent systems, and also, collectively summarizing all the existing results in a comprehensive manner.

The following example illustrates the application of Theorem 8.

We provide the following counterexample, which shows that the conditions of Theorem 8 are not sufficient to establish hazard rate order (or reversed hazard rate order), i.e. $Z_1 \geq_{hr} (\!\leq_{hr}\!)\; Z_2$ (or $Z_1 \geq_{rh} (\leq_{rh})\; Z_2$ ). Furthermore, we also show that the conjecture proposed in [Reference Zhang and Zhang34, Theorem 2] (i.e. stating that $T_{i} \leq_{rh} T_{j}$ might required to prove $F_{T_{i} \# S_{1}}(t) F_{T_{j} \# S_{2}}(t) \geq F_{T_{i} \# S_{2}}(t) F_{T_{j} \# S_{1}}(t)$ , for $t>0$ ), may not hold true for a coherent system.

Counterexample 4. Under the setup of Example 7, assume that $\sigma_{1}=2$ , $\sigma_{2}=20$ , $\sigma_{3}=10$ , $\mu_{1}=\mu_{2}=\mu_{3}=0.5$ , $\alpha_{1}=\alpha_{2}=2$ , $\lambda_{1}=1$ , and $\lambda_{2}=2$ . Then, the conditions of Theorem 8 are satisfied. However, from Figure 8(a) we can see that the ratio of distribution functions of $Z_{1}$ and $Z_{2}$ (i.e. ${F_{Z_{1}}}/{F_{Z_{2}}}$ ) is neither increasing nor decreasing. Hence, we can conclude that $Z_1 \ngeq_{rh} (\nleq_{rh})\;Z_2$ . Similarly, we can see from Figure 8(b) that $Z_1 \ngeq_{hr} (\nleq_{hr})\;Z_2$ , when $\sigma_{1}=2$ , $\sigma_{2}=20$ , $\sigma_{3}=20$ , $\mu_{1}=\mu_{2}=\mu_{3}=0.5$ , $\alpha_{1}=\alpha_{2}=2$ , $\lambda_{1}=1$ , and $\lambda_{2}=2$ (i.e. conditions of Theorem 8 are satisfied). Thus, the conditions of Theorem 8 are not sufficient to establish reversed hazard rate and hazard rate orders between $Z_{1}$ and $Z_{2}$ .

Figure 8.

(a) Plot of the ratio of distribution functions of $Z_{1}$ and $Z_{2}$ (i.e. ${{F}_{Z_{1}}(t)}/{{F}_{Z_{2}}(t)}$ ). (b) Plot of the ratio of survival functions of $Z_{1}$ and $Z_{2}$ (i.e. ${\overline{F}_{Z_{1}}(t)}/{\overline{F}_{Z_{2}}(t)}$ ). (c) Plot of the difference of survival functions of $Z_{1}$ and $Z_{2}$ (i.e. $\overline{F}_{Z_{1}}(t) - \overline{F}_{Z_{2}}(t)$ ).

Moreover, it is interesting to see that if condition ‘ $1-\overline{F}_{S_{i}}(t)/\overline{F}_{S_{i}}(u)$ is $R R_2$ (or $TP_2)$ in $(i, u) \in\{1,2\} \times(0, t]$ ’ is relaxed in Theorem 8, then will the result hold or not $?$ It is easy to see that if $\sigma_{1}=2$ , $\sigma_{2}=10$ , $\sigma_{3}=20$ , $\mu_{1}=\mu_{2}=\mu_{3}=0.5$ , $\alpha_{1}=5$ , $\alpha_{2}=10$ , and $\lambda_{1}=\lambda_{2}=1$ , then $T_{1}$ $\leq_{rh}$ $T_{3}$ and $S_{1}$ $\geq_{hr}$ $S_{2}$ hold but the condition ‘ $1-\overline{F}_{S_{i}}(t)/\overline{F}_{S_{i}}(u)$ is $R R_2$ (or $TP_2)$ in $(i, u) \in\{1,2\} \times(0, t]$ ’ does not hold. Thus, we can see from Figure 8(c) that if $T_{1}$ $\leq_{rh}$ $T_{3}$ and $S_{1}$ $\geq_{hr}$ $S_{2}$ hold but ‘ $1-\overline{F}_{S_{i}}(t)/\overline{F}_{S_{i}}(u)$ is $R R_2$ (or $TP_2)$ in $(i, u) \in\{1,2\} \times(0, t]$ ’ does not hold, then neither $Z_{1} \ngeq_{st} Z_{2}$ nor $Z_{1} \nleq_{st} Z_{2}$ . Hence, the above conjecture given by [Reference Zhang and Zhang34] may not be true for a general coherent system.

From the discussion, it is clear that exploring the problem of simultaneously allocating two relevations to a coherent system in terms of reversed hazard rate order is intriguing. This remains an open problem that warrants further investigation.

4.1. Simulation study

4.1.1. An empirical data analysis at component level:

In this subsection, we conduct a simulation study to investigate the implications and performance of Theorem 2. To achieve this, we analyze and assess the behavior of the following systems:

\begin{align*}W_{1} = \max\{\min\{T_{1} \# S_{1}, T_{2}, T_{3}\}, T_{4}\}\quad \text{and} \quad W_{2} = \max\{\min\{T_{1}, T_{2} \# S_{2}, T_{3}\}, T_{4}\}.\end{align*}

We begin with the assumption that the random variables $ T_{1} $ and $ T_{2} $ follow Weibull distributions with the distribution functions:

\begin{align*}F_{T_{1}} (t;\; \alpha, \lambda_{1}) = \begin{cases} 1- {\textrm{e}^{-{(\lambda_{1}t)}^{\alpha}}}, & \text{if t $\geq$ 0,}\\[5pt] 0, & \text{otherwise,} \end{cases} \end{align*}

and

\begin{align*} F_{T_{2}} (t;\; \alpha, \lambda_{2}) = \begin{cases} 1- {\textrm{e}^{-{(\lambda_{2}t)}^{\alpha}}}, & \text{if t $\geq$ 0,}\\[5pt] 0, & \text{otherwise,} \end{cases} \end{align*}

where $ \alpha > 0$ , $\lambda_{i} > 0$ , $i=1,2$ . In addition, assume that $T_{1} \overset{d}{=} T_{4}$ and $T_{2} \overset{d}{=} T_{3}$ . Further, let two random variables $S_{1}$ and $S_{2}$ that follow Weibull distributions with distribution functions,

\begin{align*}F_{S_{1}} (t;\; \alpha, \beta_{1}) = \begin{cases} 1- {\textrm{e}^{-{(\beta_{1}t)}^{\alpha}}}, & \text{if t $\geq$ 0,}\\[5pt] 0, & \text{otherwise,} \end{cases} \end{align*}

and

\begin{align*} F_{S_{2}} (t;\; \alpha, \beta_{2}) = \begin{cases} 1- {\textrm{e}^{-{(\beta_{2}t)}^{\alpha}}}, & \text{if t $\geq$ 0,}\\[5pt] 0, & \text{otherwise,} \end{cases} \end{align*}

where $ \alpha > 0$ , $\beta _{i} > 0$ , $i=1,2$ . First, we fix the value of $\alpha$ as 2. We now follow a step-by-step procedure.

Table 1.

Goodness-of-fit test.

Table 1 Long description

A table with four rows and three columns presents the results of a goodness-of-fit test. The columns are labeled Parameter, Estimate, Statistic, and p-value. The rows list the parameters lambda 1, lambda 2, beta 1, and beta 2 along with their corresponding estimates, statistics, and p-values. Row 1: lambda 1, 0.3992, 0.1152, 0.5155. Row 2: lambda 2, 0.1993, 0.1149, 0.5168. Row 3: beta 1, 0.2991, 0.1363, 0.4340. Row 4: beta 2, 0.5960, 0.3880, 0.0778.

Step 1. Simulate random numbers from the distribution functions $F_{T_{i}}$ for $i=1, 2, 3, 4,$ and $G_{S_{i}}$ for $i=1, 2$ . We generate 10,000 random numbers from $F_{T_{1}}(t;\; 2, 0.4)$ , $F_{T_{2}}(t;\; 2, 0.2)$ , $F_{T_{3}}(t;\; 2, 0.2)$ , $F_{T_{4}}(t;\; 2,\ 0.4)$ , $G_{S_{1}}(t;\; 2, 0.3)$ , and $G_{S_{2}}(t;\; 2, 0.6)$ .

Step 2. Use the simulated data to obtain an estimate (see, [Reference Zhu, Byrd, Lu and Nocedal36]) of parameters ( $\hat{\lambda}_1, \hat{\lambda}_2, \hat{\beta}_1$ and $\hat{\beta}_2$ ). The Cramér–von Mises statistic is used in the context of goodness-of-fit tests [Reference Fermanian14, Reference Genest and Rémillard15, Reference You, Fang and Li32, Reference Zhang and Zhang33]. It measures the discrepancy between an empirical distribution function (EDF) $ K_n(v) $ and a theoretical distribution function $ K_{\theta_n}(v) $ . The statistic is given by

\[S_n(K) = n \int_{0}^{1} \left( K_n(v) - K_{\theta_n}(v) \right)^2 \, \textrm{d} K_{\theta_n}(v)\]

where $ K_n(v) $ is the EDF based on the sample data and $ K_{\theta_n}(v) $ is the theoretical cumulative distribution function (CDF) under the null hypothesis, parameterized by $\theta_n$ . The p-values of the goodness-of-fit test are obtained as follows.

• • From the estimated parameters ( $\hat{\lambda}_1, \hat{\lambda}_2, \hat{\beta}_1$ and $\hat{\beta}_2$ ), we obtain estimated $F_{T_{i}}$ for $i=1, 2, 3, 4,$ and $G_{S_{i}}$ for $i=1, 2$ .

• • For each $F_{T_{i}}$ , $i=1, 2, 3, 4,$ and $G_{S_{i}}$ , $i=1, 2$ , generate random observations and obtain Cramer–von Mises statistic $S_n(K)$ . Repeat this process 10,000 times to obtain 10,000 test statistics.

• • Estimate the p-value by calculating the percentage of the above statistics that exceed the value obtained from the simulated data.

The p-values in Table 1 are relatively high, indicating that the simulated data aligns well with the model. From Table 1, it can be seen that the conditions of our Theorem 2 (i.e. $\hat{\lambda}_{2} \leq \hat{\lambda}_{1}$ , $\hat{\beta}_{1} \leq \hat{\beta}_{2}$ , and $\hat{\beta}_{1} \leq \hat{\lambda}_{1}$ , imply ‘ $T_{1} \leq_{hr} T_{2}$ ’, ‘ $S_{2} \leq_{hr} S_{1}$ ’, and ‘ $T_{1} \leq_{hr} S_{1}$ ’, respectively) are satisfied. In addition, from Example 2, we can see that since $\hat{\lambda}_{2} \leq \hat{\lambda}_{1}$ , i.e. $T_{4} \leq_{hr} T_{3}$ implies that condition

\begin{align*}\frac{\mathrm{P}\left(\underset{1\leq l \leq k, l \notin \{i_1, \ldots, i_r \} }{\max} \underset{m \in P_l}{\min} \left\{T_m\right\} >t \right)}{\mathrm{P}\left( \underset{s \in \{ i_1, \ldots, i_r\} }{\max} \underset{m \in P_s, m\neq i, j}{\min} \left\{T_m\right\} > t, \underset{1\leq l \leq k, l \notin \{i_1, \ldots, i_r \} }{\max} \underset{m \in P_l}{\min} \left\{T_m\right\} \leq t \right)}\end{align*}

is decreasing in $t>0$ , is satisfied. Thus, all conditions of Theorem 2 are satisfied.

Therefore, we can conclude that $ W_{1} \geq_{hr} W_{2} $ . To illustrate this, we plot the survival functions of $W_{1}$ and $W_{2}$ , and also plot the ratio of survival functions $W_{1}$ and $W_{2}$ $ ({\rm i.e.}\ {\overline{F}_{W_{1}}(t)}/{\overline{F}_{W_{2}}(t)} ) $ using the estimated parameters, as shown in Figure 9(a), and (b), respectively. The plot demonstrates that $ {\overline{F}_{W_{1}}(t)}/{\overline{F}_{W_{2}}(t)} $ is increasing in $ t > 0 $ , implies that $ W_{1} \geq_{hr} W_{2} $ .

Figure 9.

(a) Plot of the survival functions of $W_{1}$ and $W_{2}$ $ \left({\rm i.e.}\ {\overline{F}_{W_{1}}(t)} \;\text{and}\;{\overline{F}_{W_{2}}(t)} \right) $ . (b) Plot of the ratio of survival functions of $W_{1}$ and $W_{2}$ $ ({\rm i.e.}\ {\overline{F}_{W_{1}}(t)}/{\overline{F}_{W_{2}}(t)} ) $ .

4.1.2. An empirical data analysis at system level:

This subsection presents a simulation study to illustrate the result of Theorem 1. Consider two systems defined by

\[U_1 = \min\{T_1 \# S_1, T_2\}, \quad \text{and} \quad U_2 = \min\{T_2 \# S_2, T_1\},\]

where $T_1$ and $T_2$ are component lifetimes, and $S_1$ and $S_2$ represent the lifetimes of two available relevations.

Assume that the lifetimes $T_1$ , $T_2$ , $S_1$ , and $S_2$ are mutually independent and exponentially distributed. Let $T_1 \sim \text{Exp}(\lambda_1)$ , $T_2 \sim \text{Exp}(\lambda_2)$ , $S_1 \sim \text{Exp}(\beta_1)$ , and $S_2 \sim \text{Exp}(\beta_2)$ , with survival functions

\begin{align*}& \bar{F}_{T_1}(t) = \textrm{e}^{-\lambda_1 t}, \quad \bar{F}_{T_2}(t) = \textrm{e}^{-\lambda_2 t}, \quad\bar{F}_{S_1}(t) = \textrm{e}^{-\beta_1 t}, \\[5pt] &\bar{F}_{S_2}(t) = \textrm{e}^{-\beta_2 t}, \quad \text{where} \ \lambda_1, \lambda_2, \beta_1, \ \text{and} \ \beta_2 >0.\end{align*}

If $\lambda_{1} \geq \lambda_{2}$ , $\beta_{1} \leq \beta_{2}$ , and $\beta_{1} \leq \lambda_{1}$ then the conditions $T_{1} \leq_{hr} T_{2}$ , $S_{1} \geq_{hr} S_{2}$ , and $S_{1} \geq_{hr} T_{1}$ hold true, respectively. Consequently, the conditions of Theorem 1 are satisfied. Thus, the values of the parameters are fixed (such that the conditions of Theorem 1 are satisfied) as

\[\lambda_1 = 0.07, \quad \lambda_2 = 0.05, \quad \beta_1 = 0.03, \quad \beta_2 = 0.04.\]

To examine the behavior of the systems $U_1$ and $U_2$ , we simulate 1,000 observations from each distribution. The samples are generated using the inverse transform method based on the derived forms of the CDFs for each system. The Kaplan–Meier estimator (Kaplan and Meier [Reference Kaplan and Meier18]) is used to estimate the survival functions of $U_1$ and $U_2$ . Figure 10(a) illustrates that the system $U_1$ tends to exhibit higher survival probabilities than $U_2$ across time. Moreover, the ratio of the survival functions plotted in Figure 10(b) shows a non-decreasing trend. This empirical observation provides numerical support for the hazard rate ordering $ U_1 \geq_{hr} U_2 $ , thereby validating the theoretical result established in Theorem 1.

Figure 10.

(a) Plot of the survival functions of $U_{1}$ and $U_{2}$ $ \left({\rm i.e.}\ {\overline{F}_{U_{1}}(t)} \;\text{and}\;{\overline{F}_{U_{2}}(t)} \right) $ . (b) Plot of the ratio of survival functions of $U_{1}$ and $U_{2}$ $ ({\rm i.e.}\ {\overline{F}_{U_{1}}(t)}/{\overline{F}_{U_{2}}(t)} ) $ .

4.2. A real data analysis

A cable is usually made up of several wires and, over time, the tensile strength of a cable generally diminishes due to the wear and aging of its internal parts. As a result, improving the tensile strength of cables is an essential and fascinating challenge, with the goal of maximizing their reliability and maintaining their operational status for as long as possible. To address this issue, a common strategy is to introduce redundancies, such as relevation, at the component level.

Based on the data recorded by Hand et al. [Reference Hand, Daly, McConway, Lunn and Ostrowski17], the tensile strengths of wires (refer to Table 2) were examined in a study that included a collection of cables taken from various high-voltage electricity transmission networks. A cable typically comprises multiple wires, and the strength of the structured cable can be viewed as a k-out-of-n system, with each wire regarded as an individual component. There are nine cables, and each cable contains twelve wires. This dataset has been widely used in various research papers on reliability optimization, including [Reference Guo, Zhang, Zhang and Zhao16, Reference Panja, Kundu and Pradhan26, Reference You, Fang and Li32, Reference Zhang and Zhang33]. Now, let us consider a cable composed of two wires connected in series, namely Wire 7 (lifetime is denoted by $T_{1}$ ) and Wire 5 (lifetime is denoted by $T_{2}$ ), and two available relevations to perform, namely Wire 4 (lifetime is denoted by $S_{1}$ ) and Wire 6 (lifetime is denoted by $S_{2}$ ). In addition, assume that either the relevation $S_{1}$ is performed with $T_{1}$ or $S_{2}$ is performed with $T_{2}$ , which gives a resultant system has lifetime $U_{1}$ or $U_{2}$ , i.e. $U_1=\min \left\{T_1 \# S_1, T_2\right\}$ and $U_2=$ $\min \left\{T_1, T_2 \# S_2\right\}$ . Thus, we want to make a comparison between $U_1=\min \left\{T_1 \# S_1, T_2\right\}$ and $U_2=$ $\min \left\{T_1, T_2 \# S_2\right\}$ .

To start, we use Gamma, Log-Normal, Exponential, Weibull, and Pareto distributions to model the tensile strengths of these four wires (Wires 4, 5, 6, and 7). We estimated the parameters of these four wires based on the method of maximum-likelihood estimation [Reference Zhu, Byrd, Lu and Nocedal36], wherein the values of the criterion such as the Akaike information criterion (AIC; cf. [Reference Akaike1]) and Bayesian information criterion (BIC; cf. [Reference Schwarz28]) are also estimated. It can be seen from Table 3 that the best-fitted distribution is Log-Normal for Wire 5, Wire 6, and Wire 7. Overall, it should be also noted that the best-fit distribution is Log-Normal (since AIC and BIC of Log-Normal are lower compared with other distributions) of all wires. Thus, we can conclude that the best-fitted distribution is Log-Normal for all four wires. From Table 3, for Log-Normal distribution, it can be seen that the estimated standard deviations are approximately the same ( $\approx 0.02$ , say $\hat{\sigma}$ ) for $T_{1}$ , $T_{2}$ , $S_{1}$ , and $S_{2}$ . Now, suppose, the density functions of the Log-Normal distribution of $T_{1}$ , $T_{2}$ , $S_{1}$ , and $S_{2}$ are denoted as $f_{T_{1}}(t;\; \mu_{1}, \sigma)$ , $f_{T_{2}}(t;\; \mu_{2}, \sigma)$ , $f_{S_{1}}(t;\; \mu_{3}, \sigma)$ , and $f_{S_{2}}(t;\; \mu_{4}, \sigma)$ , respectively, where $\sigma > 0$ , $\mu_{{i}'s} > 0$ , for $i=1, 2, 3, 4$ , and $t>0$ . Suppose the maximum-likelihood estimates of $\mu_{i}$ and $\sigma$ are $\hat{\mu}_{i}$ and $\hat{\sigma}$ , respectively, for Log-normal distribution. Since, it can be seen from Table 3 that $\hat{\mu}_{1} \leq \hat{\mu}_{2}$ , $\hat{\mu}_{4} \leq \hat{\mu}_{3}$ , and $\hat{\mu}_{1} \leq \hat{\mu}_{3}$ , which imply ‘ $T_{1} \leq_{hr} T_{2}$ ’, ‘ $S_{2} \leq_{hr} S_{1}$ ’, and ‘ $T_{1} \leq_{hr} S_{1}$ ’, respectively. Hence, we can see that the conditions of our Theorem 1 are satisfied for the real data set. It can be seen from Figure 11 that the ratio of survival functions of $U_{1}$ and $U_{2}$ (i.e. ${\overline{F}_{U_{1}}}/{\overline{F}_{U_{2}}}$ ) is increasing. Thus, $U_{1} \geq_{hr} U_{2}$ .

Table 2.

Dataset for wire measurements.

Table 2 Long description

A table comparing the tensile strength of wires, with four rows and seven columns. The rows are labeled as Wire 4, Wire 5, Wire 6, and Wire 7. The columns represent different measurements of tensile strength. Row 1: Wire 4, 338, 348, 328, 350, 340, 336, 341, 345, 348. Row 2: Wire 5, 330, 337, 338, 335, 350, 339, 336, 350, 355. Row 3: Wire 6, 334, 328, 332, 332, 346, 340, 342, 348, 351. Row 4: Wire 7, 335, 328, 335, 328, 345, 342, 347, 341, 333. The table provides detailed tensile strength values for each wire, allowing for comparison across different measurements.

Table 3.

Estimate of parameters of four wires.

Table 3 Long description

A table comparing estimated parameters of four wires using different distributions. The table has 25 rows and 5 columns. The columns are labeled Wire, Distribution, MLE, AIC, and BIC. The rows are labeled with different wires (Wire 4, Wire 5, Wire 6, Wire 7) and distributions (Gamma, LNorm, Exp, Weibull, Pareto). Each row provides the maximum-likelihood estimates (MLE), Akaike information criterion (AIC), and Bayesian information criterion (BIC) values for the respective wire and distribution. Row 1: Wire 4, Gamma, shape = 2321.367, rate = 6.7967, 63.73970, 64.13415. Row 2: Wire 4, LNorm, meanlog = 5.8333, sdlog = 0.0195, 63.70862, 64.10307. Row 3: Wire 4, Exp, rate = 0.0029, 125.00319, 125.20041. Row 4: Wire 4, Weibull, shape = 64.8032, scale = 344.6259, 62.57137, 62.96582. Row 5: Wire 4, Pareto, shape = 5725063.5725, scale = 1952741996.3093, 127.00320, 127.39765. Row 6: Wire 5, Gamma, shape = 1869.757, rate = 5.482, 66.79049, 67.18494. Row 7: Wire 5, LNorm, meanlog = 5.8319, sdlog = 0.0232, 66.75677, 67.15122. Row 8: Wire 5, Exp, rate = 0.0029, 124.97975, 125.17697. Row 9: Wire 5, Weibull, shape = 43.4577, scale = 345.1539, 68.51233, 68.90678. Row 10: Wire 5, Pareto, shape = 4217915.3987, scale = 1439034532.0315, 126.97975, 127.37420. Row 11: Wire 6, Gamma, shape = 1712.0152, rate = 5.0488, 66.25260, 66.64705. Row 12: Wire 6, LNorm, meanlog = 5.8264, sdlog = 0.0226, 66.16431, 66.55876. Row 13: Wire 6, Exp, rate = 0.0029, 124.87980, 125.07702. Row 14: Wire 6, Weibull, shape = 49.1605, scale = 342.9849, 66.74408, 67.13853. Row 15: Wire 6, Pareto, shape = 4778084.8466, scale = 1620415101.2033, 126.87980, 127.27425. Row 16: Wire 7, Gamma, shape = 2232.0617, rate = 6.6202, 63.58830, 63.98275. Row 17: Wire 7, LNorm, meanlog = 5.8202, sdlog = 0.0196, 63.48042, 63.87486. Row 18: Wire 7, Exp, rate = 0.003, 124.76743, 124.96465. Row 19: Wire 7, Weibull, shape = 57.3725, scale = 340.3374, 63.90417, 64.29862. Row 20: Wire 7, Pareto, shape = 2982905.8774, scale = 1005566065.5832, 126.76743, 127.16188.

4.3. Minimal repair policy

In reliability engineering, the minimal repair policy is a highly cost-effective approach, frequently employed to extend the lifetime of a system. In minimal repair, a failed component is replaced with a redundant one that has the same age and distribution as the original component at the time of failure. Suppose that the survival functions of random variables T and S are the same (say $\overline{F}$ ), then relevation is reduced to minimal repair policy, as follows:

\begin{align*}\overline{F}_{T \# S}(t)=\overline{F}(t)+\overline{F}(t) \int_0^t \frac{1}{\overline{F}(x)}\, \mathrm{d} F(x) = \overline{F}(t) [1 - \ln(\overline{F}(t))].\end{align*}

Barlow and Hunter [Reference Barlow and Hunter4] first introduced the notion of minimal repair. Note that the minimal repair is a special case of relevation [Reference Yan, Zhang and Zhang31] and is also an alternate to standby redundancy allocation. Minimal repair is employed in various real-world situations. For instance, in the case of heart failure caused by coronary occlusion, cardiopulmonary resuscitation (CPR) can revive the patient. If the revival occurs without significant damage to vital organs, CPR can be considered a minimal repair [Reference Balaban and Singpurwalla6]. For more details on minimal repair, the reader may refer to [Reference Ascher2, Reference Arriaza, Navarro and Suárez3, Reference Belzunce, Martínez and Ruiz10, Reference Navarro, Arriaza and Suárez24, Reference Yan, Zhang and Zhang31, Reference Zhang and Zhang34, Reference Zhang and Zhao35] and others.

Belzunce et al. [Reference Belzunce, Martínez and Ruiz10] studied the optimal allocation policy for two minimal repairs (performed one at a time) to a two-component series system. They proved that $U_{1} \geq_{st} U_{2}$ , if $T_{1} \leq_{st} T_{2}$ and $T_{i} \overset{d}{=} S_{i}$ for $i=1, 2$ (see [Reference Belzunce, Martínez and Ruiz10, Lemma 2]). Later, Yan et al. [Reference Yan, Zhang and Zhang31] provided a necessary and sufficient condition to prove $U_{1} \geq_{hr} U_{2}$ for a two-component series system. They established that $U_{1} \geq_{hr} U_{2}$ if and only if $ {r_{T_1}(t)}/({1-\ln(\overline{F}_{T_1}(t))}) \geq {r_{T_2}(t)}/({1-\ln(\overline{F}_{T_2}(t))})$ , where $r_{T_i}$ is the hazard rate of random variable $T_i$ for $i=1, 2$ . Moreover, in Theorem 1, we studied the optimal allocation of two relevations (performed one at a time) in the two-component series system, in the sense of hazard rate order. As we know, minimal repair is a special case of relevation, in light of this, it is interesting to determine the sufficient conditions in the case of minimal repair. The following theorem provides sufficient conditions to establish $U_{1}\geq_{hr}(\!\leq_{hr}\!)\;U_{2}$ in case of the minimal repair policy (i.e. $T_{i} \overset{d}{=} S_{i}$ for $i=1, 2$ ).

Figure 11.

Plot of the ratio of survival functions of $U_{1}$ and $U_{2}$ (i.e. ${\overline{F}_{U_{1}}(t)}/{\overline{F}_{U_{2}}(t)}$ ).

Proof. First, we prove that $U_1 \geq_{hr} U_{2}$ under the conditions ${H_{T_{1}}}/{H_{T_{2}}}$ is increasing, and $T_{1} \leq_{hr} T_{2}$ . To prove $U_1 \geq_{hr} U_{2}$ , it suffices to show that ${\overline{F}_{U_1}(t)}/{\overline{F}_{U_2}(t)}$ is increasing in $t>0$ . We have

(12) \begin{align}\overline{F}_{U_1}(t)=\left(\overline{F}_{T_1}(t)+\overline{F}_{T_1}(t) \int_0^t \frac{{f}_{T_1}(u)}{\overline{F}_{T_1}(u)} \,\textrm{d} u\right) \overline{F}_{T_2}(t) = \overline{F}_{T_{1}}(t) \big[1 - \ln(\overline{F}_{T_{1}}(t))\big] \overline{F}_{T_{2}}(t) , \quad t > 0, \end{align}

and

(13) \begin{align}\overline{F}_{U_2}(t)=\left(\overline{F}_{T_2}(t)+\overline{F}_{T_2}(t) \int_0^t \frac{{f}_{T_2}(u)}{\overline{F}_{T_2}(u)} \,\textrm{d} u\right) \overline{F}_{T_1}(t) = \overline{F}_{T_{2}}(t) \big[1 - \ln(\overline{F}_{T_{2}}(t))\big] \overline{F}_{T_{1}}(t) , \quad t > 0. \end{align}

From (12) and (13), it can be seen that

(14) \begin{align}\displaystyle\frac{\overline{F}_{U_1}(t)}{\overline{F}_{U_2}(t)}= \frac{1 - \ln(\overline{F}_{T_{1}}(t))}{1 - \ln(\overline{F}_{T_{2}}(t))} = \frac{g(t)+g_1(t)}{g(t)+g_2(t)}, \quad t > 0, \end{align}

where $g(t)=1, g_1(t)= - \ln(\overline{F}_{T_{1}}(t))$ and $g_2(t)= - \ln(\overline{F}_{T_{2}}(t))$ . To prove ${\overline{F}_{U_1}(t)}/{\overline{F}_{U_2}(t)}$ is increasing in $t>0$ , it suffices to show that

(15) \begin{align}\frac{g(t)+g_1(t)}{g(t)+g_2(t)} \quad \text{is increasing in } t>0. \end{align}

Now, ${{g_1}(t)}/{{g_2}(t)} = {- \ln(\overline{F}_{T_{1}}(t))}/{- \ln(\overline{F}_{T_{2}}(t))} = {H_{T_{1}}(t)}/{H_{T_{2}}(t)}$ is increasing in $t>0$ , where $H_{T_{1}}$ and $H_{T_{2}}$ are cumulative hazard rates of random variables $T_{1}$ and $T_{2}$ , respectively. Further, to prove ${g(t)}/({g_{2}(t) - g_{1}(t)}) = {1}/{\ln{\Big({\overline{F}_{T_1}(t)}/{\overline{F}_{T_2}(t)}\Big)}}$ is increasing, it suffices to show that $\ln{({\overline{F}_{T_1}(t)}/{\overline{F}_{T_2}(t)})}$ is decreasing in $t>0$ :

\begin{align*} \Bigg[\displaystyle \ln{\Bigg(\frac{\overline{F}_{T_1}(t)}{\overline{F}_{T_2}(t)}\Bigg)} \Bigg]' &\overset{sgn}{=} - \frac{f_{T_{1}}(t)}{\overline{F}_{T_1}(t)} + \frac{f_{T_{2}}(t)}{\overline{F}_{T_2}(t)} \\[5pt] & = \frac{f_{T_{2}}(t)}{\overline{F}_{T_2}(t)} - \frac{f_{T_{1}}(t)}{\overline{F}_{T_1}(t)} \\[5pt] & \leq 0.\end{align*}

The last inequality is due to $T_{1} \leq_{hr} T_{2}$ (i.e. ${f_{T_{2}}(t)}/{\overline{F}_{T_2}(t)} \leq {f_{T_{1}}(t)}/{\overline{F}_{T_1}(t)}$ for all $t>0$ ). Thus, ${g(t)}/({g_{2}(t) - g_{1}(t)})$ is increasing in $t>0$ . From Lemma 1, it is easy to see that $({g(t)+g_1(t)})/({g(t)+g_2(t)})$ is increasing in $t>0$ , since ${g_1(t)}/{g_2(t)}$ and ${g(t)}/({g_{2}(t) - g_{1}(t)})$ are increasing in $t>0$ . Hence, $U_{1} \geq_{hr} U_{2}$ holds true under the assumptions ${H_{T_{1}}}/{H_{T_{2}}}$ is increasing in $t>0$ , and $T_{1} \leq_{hr} T_{2}$ .

Similarly, one can prove that $U_{1} \leq_{hr} U_{2}$ holds true under the assumptions ${H_{T_{1}}}/{H_{T_{2}}}$ is decreasing in $t>0$ , and $T_{1} \geq_{hr} T_{2}$ .

Note that the conditions in Theorem 9 are easy to prove and satisfied by some general families of distribution. For instance, see the following examples which satisfy by some common families of distributions, such as Weibull and Pareto.

Note that [Reference Belzunce, Martínez and Ruiz10, Lemma 2] showed that $ U_{1} \geq_{st} U_{2} $ if $T_{1} \leq_{st} T_{2}$ in a two-component series system under the minimal repair policy, i.e. $T_{i} \overset{d}{=} S_{i}$ for $i=1, 2$ . In Theorem 9, we showed that $ U_{1} \geq_{hr} (\!\leq_{hr}\!)\; U_{2} $ if $T_{1} \leq_{hr} (\!\geq_{hr}\!)\;T_{2}$ and ${H_{T_{1}}}/{H_{T_{2}}}$ is increasing (decreasing) in $t>0$ . Thus, it is evident that when we move from a weaker stochastic ordering, i.e. $ U_{1} \geq_{st} U_{2} $ to a stronger stochastic ordering, i.e. $ U_{1} \geq_{hr} U_{2} $ , $T_{1} \leq_{st} T_{2}$ gets strengthen to $T_{1} \leq_{hr} T_{2}$ and an additional condition ${H_{T_{1}}(t)}/{H_{T_{2}}(t)}$ is increasing in $t>0$ , is also needed for the result to hold true. This highlights the trade-off between the set of sufficient conditions when we move from a weaker to a stronger stochastic ordering.

The following example illustrates the application of Theorem 9.

The following counterexample shows that the condition ${H_{T_{1}}(t)}/{H_{T_{2}}(t)}$ is increasing in $t>0$ in Theorem 9, is not sufficient to prove $U_1 \geq_{hr} U_2$ .

In Theorem 9, we established the result for minimal repairs in a series system, in the sense of hazard rate order. In light of this, it is interesting to determine sufficient conditions for minimal repairs in a coherent system in the sense of hazard rate order. In this regard, Yan et al. [Reference Yan, Zhang and Zhang31] provide an ‘if and only if’ condition to establish $ W_{1} \geq_{hr} W_{2} $ for minimal repair policy for a coherent system. In the following theorem, we give sufficient conditions to establish hazard rate order for a coherent system in the case of minimal repairs.

Figure 12.

Plot of the ratio of survival functions of $U_{1}$ and $U_{2}$ (i.e. ${\overline{F}_{U_{1}}(t)}/{\overline{F}_{U_{2}}(t)})$ .

Theorem 10. Let $W_1\sim \tau\left(T_i \# T_i, T_j, \boldsymbol{T}^{(i, j)}\right) $ and $W_2\sim \tau\left(T_i, T_j \# T_j, \boldsymbol{T}^{(i, j)}\right) $ . Consider two locations $i, j \in \{1, \ldots , n\}$ such that $i<j$ and $\{i, j\} \subseteq P_r $ , for $r= i_1, \ldots , i_r$ , where $\left\{ i_1, \ldots , i_r \right\} \subseteq \left\{ 1, \ldots , k \right\}$ , i.e. $\mathcal{A}_i=\mathcal{A}_j$ . Suppose

\begin{align*} \frac{\mathrm{P}\left(\underset{1\leq l \leq k, l \notin \{i_1, \ldots, i_r \} }{\max} \underset{m \in P_l}{\min} \left\{T_m\right\} >t \right)}{\mathrm{P}\left( \underset{s \in \{ i_1, \ldots, i_r\} }{\max} \underset{m \in P_s, m\neq i, j}{\min} \left\{T_m\right\} > t, \underset{1\leq l \leq k, l \notin \{i_1, \ldots, i_r \} }{\max} \underset{m \in P_l}{\min} \left\{T_m\right\} \leq t \right)} \end{align*}

is decreasing in $t>0$ . If $\overline{F}_{U_2}(t) - \overline{F}_{U_1}(t)$ is decreasing (increasing) and negative (positive) in $t>0$ then $W_1 \geq_{hr} (\!\leq_{hr}\!) W_2$ .

It is worth mentioning here that the condition $\overline{F}_{U_2}(t) - \overline{F}_{U_1}(t)$ is decreasing (increasing) and negative (positive) in $t>0$ implies that ${\overline{F}_{U_1}(t)}/{\overline{F}_{U_2}(t)}$ is increasing (decreasing) in $t>0$ , i.e. $U_1 \geq_{hr} (\!\leq_{hr}\!) U_2$ (the result established in Theorem 9).

Note that establishing the stochastic comparison between $ V_{1}^{\prime} $ and $ V^{\prime}_{2} $ (for the series system with two revelations), $ V_{1} $ and $ V_{2} $ (for the parallel system with two revelations), and $ Z_{1} $ and $ Z_{2} $ (for a coherent system with two revelations) under a minimal repair policy is trivial, since both systems have same lifetimes.

Up to now, we focused on the optimal allocation policy for two minimal repairs (performed one at a time). In real-life situations, reliability engineers might face the problem of where to allocate multiple minimal repairs to a series or coherent system. This has caught the attention of the researchers over the past decade. In this regard, Arriaza et al. [Reference Arriaza, Navarro and Suárez3], Navarro et al. [Reference Navarro, Arriaza and Suárez24], and Navarro and Fernández-Martnez [Reference Navarro and Fernández-Martnez25] studied the allocation of r number of minimal repairs to the same components in coherent systems by using redundancy distortion function approach. Moreover, the majority of the results focused either on identical components or on the same number of minimal repairs performed to series and coherent systems. In light of this, we focus on the optimal allocation policy for multiple minimal repairs (not necessarily the same number of repairs) on a single node (or component) within coherent systems with nonidentical components. To begin with, we develop the result of a series system and then later generalize it to a coherent system. Suppose $T_{i}\; \#^r\; T_{i}$ denote the lifetime of the component i (having lifetime $T_{i}$ ) after allocating $r\,(>0)$ minimal repairs. Then, the survival function of $T_{i}\; \#^r \;T_{i}$ is given by

\begin{align*}\displaystyle \mathrm{P}(T_{i}\; \#^r\; T_{i} > t) = \displaystyle \overline{F}_{T_{i}}(t) \sum_{k=0}^{r} \frac{1}{k!} [-\ln \overline{F}_{T_{i}}(t)]^k,\end{align*}

where $\overline{F}_{T_{i}}$ is the survival function of $T_{i}$ (see, [Reference Navarro and Fernández-Martnez25]). Suppose, we have a two-component series system with component lifetimes $T_{1}$ and $T_{2}$ . Furthermore, assume that either $k_1$ minimal repairs are performed on $T_{1}$ or $k_2$ are on $T_{2}$ . That is, the resultant series systems have lifetimes $U_1^{*}=\min \left\{T_1\; \#^{k_1}\; T_1, T_2\right\}$ and $U_2^{*}=$ $\min \left\{T_1, T_2\; \#^{k_2}\; T_2\right\}$ . In addition, suppose the system consists of nonidentical components. With these assumptions, we are now ready to state the following theorem for the series system.