Appendices
A Expected Values of Replacement Policy I
In order to obtain E(L) we required following computation,
Let \(\psi _{j}\) be the distribution function of \(({X_{j}^{(2)}-Y_j^{(1)}})\), where
$$\begin{aligned} \psi _{j}(v)= & {} P\left[ {X_{j}^{(2)}-Y_{j}^{(1)}}\le v\right] \\= & {} \int _{0}^{\infty } P\left[ X_{j}^{(2)}\le (y+v)|Y_{j}^{(1)}=y\right] g_j^{(1)}(y) dy\\= & {} \int _{0}^{\infty }F_{j}^{(2)}(y+v)g_{j}^{(1)}(y)dy, \\&~~ \hbox {where F(.) is the distribution function of working time.} \end{aligned}$$
Assume working time \(X_j^{(i)}\) and repair time \(Y_j^{(i)}\) are exponentially distributed with rate \(\lambda _i\) and \(\mu _i\) respectively, for i=1,2. So \(\psi _{j}(v)\) becomes,
$$\begin{aligned} \psi _{j}(v)= & {} \int _{0}^{\infty } \left[ 1-e^{-\lambda _{2}(y+v)}\right] \mu _{1} e^{-\mu _{1} y} dy\nonumber \\= & {} 1-\frac{\mu _{1} e^{-\lambda _{2}v}}{\mu _{1}+\lambda _{2}} \end{aligned}$$
(A.1)
Therefore, by using Eq. A.1
$$\begin{aligned} E\left[ \left( {X_{j}^{(2)}-Y_{j}^{(1)}}\right) _I{_{\left( {X_{j}^{(2)} -Y_{j}^{(1)}}>0\right) }}\right]= & {} \int _{0}^{\infty }vd\psi _{j}(v)\nonumber \\= & {} \int _{0}^{\infty }v\lambda _{2}\frac{\mu _{1} e^{-\lambda _{2}v}}{\mu _{1}+\lambda _{2}}dv\nonumber \\= & {} \frac{\mu _{1}}{\lambda _{2}(\lambda _{2}+\mu _{1})} \end{aligned}$$
(A.2)
So by using Eq. A.2, we have,
$$\begin{aligned} E\left[ \sum \limits _{j=1}^{S2}\left[ X^{(2)}_{j}-Y^{(1)}_{j}\right] _{I_{\left[ {Y}^{(1)}_{j}<{X}^{(2)}_{j}\right] }} \right]= & {} \frac{\mu _{1}}{\lambda _{2}(\lambda _{2}+\mu _{1})}E[S2]\nonumber \\= & {} \frac{\mu _{1}}{\lambda _{2}(\lambda _{2}+\mu _{1})}E\big (E[S2|K1]\big )\nonumber \\= & {} \frac{\mu _{1}}{\lambda _{2}(\lambda _{2}+\mu _{1})}E\bigg [\frac{(k-K_1)q_2}{p_2}\bigg ]\nonumber \\= & {} \frac{\mu _{1}}{\lambda _{2}(\lambda _{2}+\mu _{1})}\bigg [\frac{kq_2}{p_2}-\frac{kp_1}{1-q^{k}_{1}}\frac{q_2}{p_2}\bigg ] \end{aligned}$$
(A.3)
$$\begin{aligned} E \left[ \sum \limits _{j=1}^{S_1+K_1}{X}^{(1)}_{j}\right]= & {} \frac{1}{\lambda _{1}}E[S_1+K_1]\nonumber \\= & {} \frac{1}{\lambda _{1}} E\bigg (E[S_{1}+K_{1}|K_{1}]\bigg )\nonumber \\= & {} \frac{1}{\lambda _{1}}E\bigg (\frac{K_{1}q_{1}}{p_{1}(1-p^{K_{1}}_{1})}+K_{1}\bigg )\nonumber \\= & {} \frac{1}{\lambda _{1}}\left[ \sum \limits _{k_1=1}^{k}\frac{k_1}{(1-p^{k_1}_{1})}\frac{\left( {\begin{array}{c}k\\ k_1\end{array}}\right) p^{k_1-1}_{1}q^{k-k_1+1}_{1}}{1-q^k_{1}}+\frac{kp_{1}}{1-q^k_{1}}\right] \nonumber \\ E \left[ ({X}^{(2)}_{j})_{I_{\left[ {Y}^{(1)}_{j}< {X}^{(2)}_{j}\right] }}\right]= & {} \int _{0}^{\infty }\int _{0}^{y^{(1)}_{j}}x^{(2)}_{j} f(x^{(2)}_{j},{y^{(1)}_{j}}) dx^{(2)}_{j}dy^{(1)}_{j}\nonumber \\= & {} \int _{0}^{\infty }\int _{0}^{y^{(1)}_{j}}x^{(2)}_{j} f(x^{(2)}_{j})g({y^{(1)}_{j}}) dx^{(2)}_{j}dy^{(1)}_{j} \end{aligned}$$
(A.4)
After solving above integration by using method of integration by parts, we get
$$\begin{aligned} E\left[ ({X}^{(2)}_{j})_{I_{\left[ {Y}^{(1)}_{j}< {X}^{(2)}_{j}\right] }}\right]= & {} \frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{\lambda _{2}(\lambda _{2}+\mu _{1})^2} \end{aligned}$$
(A.5)
and
$$\begin{aligned} E \left[ ({X}^{(2)}_{j})_{I_{\left[ {Y}^{(1)}_{j}> {X}^{(2)}_{j}\right] }} \right]= & {} \frac{1}{\lambda _{2}}\left[ 1-\frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{(\lambda _{2}+\mu _{1})^2}\right] \end{aligned}$$
(A.6)
$$\begin{aligned} E \left[ \sum \limits _{j=1}^{S_2}({X}^{(2)}_{j})_{I_{ \left[ {Y}^{(1)}_{j}< {X}^{(2)}_ {j}\right] }}\right]= & {} \left[ \frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{\lambda _{2}(\lambda _{2}+\mu _{1})^2}\right] E[S_2]\nonumber \\= & {} \left[ \frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{\lambda _{2}(\lambda _{2}+\mu _{1})^2}\right] E \left( E[S_2|K_1]\right) \nonumber \\= & {} \left[ \frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{\lambda _{2}(\lambda _{2}+\mu _{1})^2}\right] E \left[ \frac{(k-K_1)q_2}{p_2}\right] \nonumber \\= & {} \left[ \frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{\lambda _{2}(\lambda _{2}+\mu _{1})^2} \right] \left[ \frac{kq_2}{p_2}- \frac{kq_1}{1-q^{k}_{1}}\frac{q_2}{p_2}\right] \nonumber \\= & {} \left[ \frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{\lambda _{2}(\lambda _{2}+\mu _{1})^2}\right] \frac{kq_2}{p_2}\left[ 1-\frac{p_1}{1-q^{k}_{1}}\right] \end{aligned}$$
(A.7)
$$\begin{aligned} E \left[ \sum \limits _{j=1}^{k-K_1}({X}^{(2)}_{j})_{I_{\left[ {Y}^{(1)}_{j}> {X}^{(2)}_{j}\right] }}\right]= & {} \left[ 1-\frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{(\lambda _{2}+\mu _{1})^2}\right] \frac{1}{\lambda _{2}}E[k-K_1]\nonumber \\= & {} \left[ 1-\frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{(\lambda _{2}+\mu _{1})^2}\right] \left[ \frac{k}{\lambda _{2}}-\frac{kp_1}{\lambda _2(1-q^{k}_{1})}\right] \nonumber \\ \end{aligned}$$
(A.8)
Let \(\phi _{j}\) be the distribution function of \(({Y_{j}^{(2)}-X_{j+1}^{(1)}})\), where
$$\begin{aligned} \phi _{j}(u)= & {} P\left[ {Y_{j}^{(2)}-X_{j+1}^{(1)}}\le u\right] \\= & {} \int _{0}^{\infty } P\left[ Y_{j}^{(2)}\le (x+u)|X_{j+1}^{(1)}=X\right] f_j^{(1)}(x) dx \\= & {} \int _{0}^{\infty }G_{j}^{(2)}(x+u)f_{j+1}^{(1)}(x)dx, \\&~~ \hbox {where G(.) is the distribution function of repair time.} \end{aligned}$$
Assume working time \(X_j^{(i)}\) and repair time \(Y_j^{(i)}\) are exponentially distributed with rate \(\lambda _i\) and \(\mu _i\) respectively, for i=1,2. So \(\phi _{j}(u)\) becomes,
$$\begin{aligned} \phi _{j}(u)= & {} \int _{0}^{\infty } [1-e^{-\mu _{2}(x+u)}]\lambda _{1} e^{-\lambda _{1} x} dx\nonumber \\= & {} 1-\frac{\lambda _{1} e^{-\mu _{2}u}}{\lambda _{1}+\mu _{2}} \end{aligned}$$
(A.9)
Therefore, by using Eq. A.9
$$\begin{aligned} E\left[ \big ({Y_{j}^{(2)}-X_{j+1}^{(1)}}\big )_I{_{\big ({Y_{j}^{(2)}-X_{j+1}^{(1)}}>0\big )}}\right]= & {} \int _{0}^{\infty }ud\phi _{j}(u)\nonumber \\= & {} \int _{0}^{\infty }u\mu _{2}\frac{\lambda _{1} e^{-\mu _{2}u}}{\lambda _{1}+\mu _{2}}du\nonumber \\= & {} \frac{\lambda _{1}}{\mu _{2}(\mu _{2}+\lambda _{1})} \end{aligned}$$
(A.10)
Similarly,
$$\begin{aligned} E \left[ \left( {Y_{j}^{(1)}-X_{j+1}^{(2)}}\right) _I{_{ \left( {Y_{j}^{(2)}-X_{j+1}^{(1)}}>0\right) }} \right]= & {} \frac{\lambda _{2}}{\mu _{1}(\mu _{1}+\lambda _{2})} \end{aligned}$$
(A.11)
$$\begin{aligned} E \left[ \sum \limits _{j=1}^{K_1-1} \left[ {Y}^{(2)}_{j}-{X}^{(1)}_{j+1}\right] _ {I_{\left[ {Y}^{(2)}_{j}>{X}^{(1)}_{j+1}\right] }}\right]= & {} \frac{\lambda _{1}}{\mu _{2}(\lambda _{1}+\mu _{2})}E(K_1-1)\nonumber \\= & {} \frac{\lambda _{1}}{\mu _{2}(\lambda _{1}+\mu _{2})}\left[ \frac{kp_1}{1-q^{k}_{1}}-1\right] \end{aligned}$$
(A.12)
$$\begin{aligned} E\left[ \sum \limits _{j=1}^{k-K_1} \left[ {Y}^{(1)}_{j}-{X}^{(2)}_{j}\right] _ {I_{\left[ {Y}^{(1)}_{j}>{X}^{(2)}_{j}\right] }}\right]= & {} \frac{\lambda _{2}}{\mu _{1}(\lambda _{2}+\mu _{1})} \left[ k-\frac{kp_1}{1-q^{k}_{1}}\right] \nonumber \\ \end{aligned}$$
(A.13)
Now,
$$\begin{aligned} E\left[ \sum \limits _{j=1}^{S_1+K_1}a^{(j-1)}{X}^{(1)}_{j}\right]= & {} \frac{1}{\lambda _{1}}E\bigg [\frac{1-a^{S_1+K_1}}{1-a}\bigg ]\nonumber \\= & {} \frac{1}{\lambda _{1}{(1-a)}}E\bigg (E\bigg [{1-a^{S_1+K_1}}|K1\bigg ]\bigg )\nonumber \\= & {} \frac{1}{\lambda _{1}{(1-a)}}E \left[ 1-\frac{\sum \nolimits _{s_1=1}^{\infty }a^{s_1+K_1}\left( {\begin{array}{c}s_1+K_1-1\\ K1-1\end{array}}\right) p^{K1}_{1}q^{s_1}_{1}}{(1-p^{K_1})}\right] \nonumber \\= & {} \frac{1}{\lambda _{1}{(1-a)}} \left[ 1-E\bigg (\frac{(ap_{1})^{K_1}}{(1-p^{K_1}_{1})}\right. \nonumber \\&\left. \times \,\bigg (\sum \limits _{s_1=1}^{\infty }(aq_1)^{s_1}\left( {\begin{array}{c}s_1+K_1-1\\ K_1-1\end{array}}\right) \bigg )\bigg )\right] \nonumber \\= & {} \frac{1}{\lambda _{1}{(1-a)}}\bigg [1-E\bigg (\frac{(ap_1)^{K_{1}}}{(1-p^{K_1}_{1})}\bigg (\frac{1}{(1-aq_1)^K_{1}}-1\bigg )\bigg )\bigg ]\nonumber \\= & {} \frac{1}{\lambda _{1}(1-a)} \left[ 1-\sum \limits _{k_1=1}^{k}\frac{(ap_{1})^{k_{1}}}{(1-p^{k_1}_{1})}\bigg [\frac{1}{(1-aq_{1})^{k_1}}-1\right] \nonumber \\&\times \,\frac{\left( {\begin{array}{c}k\\ k_1\end{array}}\right) p^{k_1}_{1}q^{k-k_1}_{1}}{1-q^{k}_{1}}\bigg ] \end{aligned}$$
(A.14)
$$\begin{aligned} E\left[ \sum \limits _{j=1}^{k-K1}a^{(j-1)}({X}^{(2)}_{j})_ {I_{[{Y}^{(1)}_{j}> {X}^{(2)}_{j}}}]\right]= & {} \bigg [1-\frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{(\lambda _{2}+\mu _{1})^2} \bigg ]\frac{1}{\lambda _{2}(1-a)}E[1-a^{k-K1}]\nonumber \\= & {} \bigg [1-\frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{(\lambda _{2}+\mu _{1})^2}\bigg ]\frac{1}{\lambda _{2}(1-a)}\nonumber \\&\times \,\left[ 1-\frac{\sum \nolimits _{k1=1}^{k}{a^{k-k1}\left( {\begin{array}{c}k\\ k1\end{array}}\right) p^{k_1}_{1}q^{k-k1}_{1}}}{1-q^{k}_{1}}\right] \nonumber \\= & {} \bigg [1-\frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{(\lambda _{2}+\mu _{1})^2}\bigg ]\frac{1}{\lambda _{2}(1-a)}\nonumber \\&\times \,\left[ 1-\frac{(aq_1+p_1)^{k}-(aq_1)^k}{1-q^{k}_{1}}\right] \nonumber \\ \end{aligned}$$
(A.15)
$$\begin{aligned} E\left[ \sum \limits _{j=1}^{S_1+K_1-1}b^{(j-1)}{Y}^{(1)}_{j}\right]= & {} \frac{1}{\mu _{1}}E\bigg [\frac{{b^{S_1+K_1-1}-1}}{b-1}\bigg ]\nonumber \\= & {} \frac{1}{\mu _{1}{(b-1)}}E\left( E\big [{b^{S_1+K_1-1}}-1|K1\big ]\right) \nonumber \\= & {} \frac{1}{\mu _{1}{(b-1)}}E \left[ \frac{\sum \nolimits _{s_1=1}^{\infty }b^{s_1+K_1-1}\left( {\begin{array}{c}s_1+K_1-1\\ K1-1\end{array}}\right) p^{K1}_{1}q^{s_1}_{1}}{(1-p^{K_1})}-1\right] \nonumber \\= & {} \frac{1}{\mu _{1}{(b-1)}}E \left[ \frac{(bp_1)^{K_1}}{(1-p^{K_1}_{1})b}\sum \limits _{s_1=1}^{\infty }(bq_1)^{s_1}\left( {\begin{array}{c}s_1+K_1-1\\ K_1-1\end{array}}\right) -1\right] \nonumber \\= & {} \frac{1}{\mu _{1}{(b-1)}}E\left[ \bigg (\frac{(bp_1)^{K_1}}{(1-p^{K_1}_{1})b}\bigg [\frac{1}{(1-bq_1)^K_1}- 1\bigg ]\bigg )-1\right] \nonumber \\= & {} \frac{1}{\lambda _{1}(b-1)}\left[ \sum \limits _{k_1=1}^{k} \frac{(bp_{1})^{k_1}}{(1-p^{k_1}_{1})}\right. \nonumber \\&\left. \times \,\bigg [\frac{1}{(1-bq_{1})^{k_1}} -1\bigg ]\frac{\left( {\begin{array}{c}k\\ k_1\end{array}}\right) p^{k_1}_{1}q^{k-k_1}_{1}}{1-q^{k}_{1}}-1 \right] \nonumber \\ \end{aligned}$$
(A.16)
$$\begin{aligned} E\left[ \sum \limits _{j=1}^{k-K1}b^{(j-1)}Y^{(2)}_{j}\right]= & {} \frac{1}{\mu _{2}}E\bigg [\frac{b^{k-K1}-1}{b-1}\bigg ]\nonumber \\= & {} \frac{1}{\mu _{2}(b-1)}E\left[ 1-a^{k-K1}\right] \nonumber \\= & {} \frac{1}{\mu _{2}(b-1)}\left[ \frac{\sum \nolimits _{k1=1}^{k}{b^{k-k1}\left( {\begin{array}{c}k\\ k1\end{array}}\right) p^{k_1}_{1}q^{k-k1}_{1}}}{1-q^{k}_{1}}-1\right] \nonumber \\= & {} \frac{1}{\mu _{2}(b-1)}\left[ \frac{(bq_1+p_1)^{k}-(bq_1)^k}{1-q^{k}_{1}}-1\right] \end{aligned}$$
(A.17)
B Expected Values of Replacement Policy II
Now consider,
$$\begin{aligned} E\left[ \sum \limits _{j=1}^{N}{X_j^{(1)}}\right]= & {} \frac{N}{\lambda _{1}} \end{aligned}$$
(B.1)
$$\begin{aligned} E\left[ \sum \limits _{j=1}^{N-1}{Y_j^{(1)}}\right]= & {} \frac{N-1}{\mu _{1}} \end{aligned}$$
(B.2)
From Eq. A.10,
$$\begin{aligned} E\left[ \left( {Y_{j-1}^{(2)}-X_j^{(1)}}\right) _ I{_{\left( {Y_{j-1}^{(2)}-X_j^{(1)}}>0\right) }}\right]= & {} \frac{\lambda _{1}}{\mu _{2}(\mu _{2}+\lambda _{1})} \end{aligned}$$
(B.3)
To obtain E(C), let us obtain following quantities,
By using Eq. A.2
$$\begin{aligned} E\left[ \big ({X_{j}^{(2)}-Y_j^{(1)}}\big )_I{_{ \big ({X_{j}^{(2)}-Y_j^{(1)}}>0\big )}}\right]= & {} \frac{\mu _{1}}{\lambda _{2}(\mu _{1}+\lambda _{2})} \end{aligned}$$
(B.4)
by using Eq. A.11. ew have
$$\begin{aligned} E\left[ \big ({Y_{j}^{(1)}-X_j^{(2)}}\big )_I{_{ \big ({Y_{j}^{(1)}-X_j^{(2)}}>0\big )}}\right]= & {} \frac{\lambda _{2}}{\mu _{1}(\mu _{2}+\lambda _{2})} \end{aligned}$$
(B.5)
$$\begin{aligned} E\left[ \left( {Y}^{(2)}_{j}\right) _{I_{\big ({X}^{(2)}_{j}\le {Y}^{(1)}_{j}\big )}}\right]= & {} E\big [{Y}^{(2)}_{j}\big ] P\big [{X}^{(2)}_{j}\le {Y}^{(1)}_{j}\big ]\nonumber \\= & {} \frac{1}{\mu _2}\psi _j(0)\nonumber \\= & {} \frac{\lambda _{2}}{\mu _{2}(\lambda _{2}+\mu _{1})} \end{aligned}$$
(B.6)
From Eqs. A.5 and A.6 we have obtained,
$$\begin{aligned} E\left[ \left( {X}^{(2)}_{j}\right) _{I_{ \left[ {Y}^{(1)}_{j}< {X}^{(2)}_{j}\right] }}\right]= & {} \frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{\lambda _{2}(\lambda _{2}+\mu _{1})^2} \end{aligned}$$
(B.7)
and
$$\begin{aligned} E\left[ \left( {X}^{(2)}_{j}\right) _{I_{\left[ {Y}^{(1)}_{j}> {X}^{(2)}_{j}\right] }}\right]= & {} \frac{1}{\lambda _{2}}\left[ 1-\frac{\mu _{1}(2\lambda _{2}+\mu _{1})}{(\lambda _{2}+\mu _{1})^2}\right] \end{aligned}$$
(B.8)
