This paper considers an $M/G/1$ repairable queueing system with $N$-policy and delayed uninterrupted single vacation, in which the repair facility subject to breakdowns and then replaced during the repair facility busy period. By the renewal process theory, the total probability decomposition technique and the Laplace transform tool, some reliability indices of the service station and the repair facility are discussed, such as the transient-state and steady-state unavailability, and the expected failure number during $(0,t]$, etc., and the parameter sensitivity analysis is carried out on the steady-state unavailability and the steady-state failure frequency.
He Yaxing, Tang Yinghui, Liu Qionglin. Analysis of ${M/G/1}$ Repairable Queueing System with a Replaceable Repair Facility, ${N}$-policy and Delayed Uninterrupted Single Vacation[J]. Acta Mathematica Scientia, 2023, 43(2): 625-645
其中 $y(s)=\int_{0}^{\infty }{{{e}^{-st}}{\rm d}Y(t)},w(s)=\int_{0}^{\infty }{{{e}^{-st}}{\rm d}W(t)},$${{W}^{\left( k \right)}}\left( t \right)$表示自身$k$重卷积, $k\ge 1,$ 且${{W}^{\left( 0 \right)}}\left( t \right)=1,t\ge 0$, 以下符号同理.
令${{\tilde{\chi }}_{n}}$ 表示第$n$个顾客的“广义服务时间”, ${{\tilde{G}}_{n}}(t)=P\left\{ {{{\tilde{\chi }}}_{n}}\le t \right\}$, 如果把服务台的“广义修理时间”看成服务台失效后的修理时间, 可得
$\begin{matrix}\tilde{G}(t)\triangleq{{\tilde{G}}_{n}}(t)=P\left\{ {{{\tilde{\chi }}}_{n}}\le t \right\}=\sum\limits_{k=0}^{\infty }{\int_{0}^{t}{{{{\tilde{Y}}}^{\left( k \right)}}\left( t-x \right){{e}^{-\alpha x}}\frac{{{\left( \alpha x \right)}^{k}}}{k!}{\rm d}G(x)}},\end{matrix}$
对(2.4)式作$LS$变换可得
$\begin{matrix}\tilde{g}(s)=\int_{0}^{\infty }{{{e}^{-st}}{\rm d}\tilde{G}(t)}=\sum\limits_{k=0}^{\infty }{{{\left( \tilde{y}\left( s \right) \right)}^{k}}\int_{0}^{\infty }{{{e}^{-\left( s+\alpha \right)t}}\frac{{{\left( \alpha t \right)}^{k}}}{k!}{\rm d}G(t)}}=g(s+\alpha -\alpha \tilde{y}(s)),\end{matrix}$
符号$b$通常用来表示在经典的$M/G/1$排队系统中, 从一个顾客开始的服务台忙期长度, 有分布$P\left\{ {{b}}\le t \right\}={{B}}\left( t \right)$, ${{b}^{\left\langle i \right\rangle }}$则表示从$i$个顾客开始的服务台忙期长度, 有分布$P\left\{ {{b}^{\left\langle i \right\rangle }}\le t \right\}={{B}^{\left( i \right)}}\left( t \right)$, 用$\tilde{b}$表示该可修排队系统从一个顾客开始的服务台的“广义忙期”长度, 其分布$\tilde{B}\left( t \right)=P\left\{ \tilde{b}\le t \right\}$, ${{\tilde{b}}^{\left\langle i \right\rangle }}$表示从$i$个顾客开始的服务台的“广义忙期”长度, 有分布${{\tilde{B}}^{\left( i \right)}}\left( t \right)=P\left\{ {{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\}, t\ge 0,i\ge 1.$ 类似于文献[39]中有关可修排队系统的讨论, 我们有如下引理.
引理 2.2[39] 对$\mathfrak{R}e \left( s \right)>0$, $\tilde{b}(s)$是方程$z=\tilde{g}\left( s+\lambda -\lambda z \right)$在$\left| z \right|<1$内的唯一根, 且
$\tilde{B}\left( t \right)=\sum\limits_{k=1}^{\infty }{\int_{0}^{t}{\frac{{{\left( \lambda x \right)}^{k-1}}}{k!}{{e}^{-\lambda x}}{\rm d}}}{{\tilde{G}}^{\left( k \right)}}\left( x \right),t\ge 0, $
$\tilde{\Phi }\left( t \right)=P\left\{ \mbox{时刻t服务台失效} \right\}, {{\tilde{\varphi }}^{*}}\left( s \right)=\int_{0}^{\infty }{{{e}^{-st}}\tilde{\Phi }\left( t \right){\rm d}t},$
$\tilde{M}\left( t \right)=E\left\{ \left( 0,t \right]\mbox{内系统失效的次数} \right\},\tilde{m}\left( s \right)=\int_{0}^{\infty }{{{e}^{-st}}{\rm d}}\tilde{M}\left( t \right).$
$\begin{matrix} {{\tilde{\varphi }}^{*}}\left( s \right)=\frac{\alpha \left[ 1-\tilde{y}\left( s \right) \right]}{s\left[ s+\alpha -\alpha \tilde{y}\left( s \right) \right]},\tilde{m}\left( s \right)=\frac{\alpha }{s+\alpha -\alpha \tilde{y}\left( s \right)}, \end{matrix}$
且平稳结果
$\begin{matrix}\lim\limits_{t\to \infty }\,\tilde{\Phi }\left( t \right)=\lim\limits_{s\to {{0}^{+}}}\,s{{\tilde{\varphi }}^{*}}\left( s \right)=\frac{\alpha \beta \left( 1+\nu \gamma \right)}{1+\alpha \beta \left( 1+\nu \gamma \right)},\lim\limits_{t\to \infty }\,\frac{\tilde{M}\left( t \right)}{t}=\lim\limits_{s\to {{0}^{+}}}\,s\tilde{m}\left( s \right)=\frac{\alpha }{1+\alpha \beta \left( 1+\nu \gamma \right)}. \end{matrix}$
${{\varphi }_{0}}^{*}\left( s \right)={{\tilde{\varphi }}^{*}}\left( s \right)f\left( s \right)\left\{ 1-\frac{\tilde{b}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},$
${{\varphi }_{i}}^{*}\left( s \right)={{\tilde{\varphi }}^{*}}\left( s \right)\left\{ 1-\frac{{{{\tilde{b}}}^{i}}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},i\ge 1,$
$A(s)=\int_{0}^{\infty }{{{e}^{-st}}V(t){\rm d}{{F}^{\left( N \right)}}(t)},\tilde{C}\left( s \right)=\sum\limits_{m=0}^{N-1 }\int_{0}^{\infty }{{{{\tilde{b}}}^{m}}\left( s \right)\frac{{{\left( \lambda t \right)}^{m}}}{m!}{{e}^{-\left( s+\lambda \right)t}}{\rm d}V\left( t \right)},$
$D(s)=\int_{0}^{\infty }{{{e}^{-st}}{{F}^{(N)}}\left( t \right){\rm d}V\left( t \right)},\xi \left( s \right)=f\left( s \right)\left[ 1-h\left( s+\lambda \right) \right]+h\left( s+\lambda \right)\left[ A\left( s \right)+D\left( s \right) \right],$
$\tilde{\Delta }\left( s \right)=1-\left[ 1-h\left( s+\lambda \right) \right]f\left( s \right)\tilde{b}\left( s \right)-h\left( s+\lambda \right)[ {{{\tilde{b}}}^{N}}\left( s \right)A\left( s \right)+v( s+\lambda -\lambda \tilde{b}\left( s \right) )-\tilde{C}\left( s \right) ].$
由于服务台的寿命服从负指数分布, 根据负指数分布的无记忆性, 用${{\tilde{b}}^{\left\langle i \right\rangle }}$对引理3.1中的$\tilde{\Phi }\left( t \right)$作全概率分解, 得
$\begin{matrix} \tilde{\Phi }\left( t \right)&=&P\left\{ \mbox{时刻t服务台失效} \right\} \\ &=&P\left\{ 0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }},\mbox{时刻t服务台失效} \right\}+P\left\{ {{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t,\mbox{时刻t服务台失效}\right\} \\ &=&{{S}_{i}}\left( t \right)+\int_{0}^{t}{\tilde{\Phi }\left( t-x \right){\rm d}{{{\tilde{B}}}^{\left( i \right)}}\left( x \right)}, \end{matrix}$
则如下得(3.7)式
$\begin{matrix} {{S}_{i}}\left( t \right)=\tilde{\Phi }\left( t \right)-\int_{0}^{t}{\tilde{\Phi }\left( t-x \right){\rm d}{{{\tilde{B}}}^{\left( i \right)}}\left( x \right)}, \end{matrix}$
将(3.7)式代入(3.4)式, 得
$\begin{matrix} {{\Phi }_{i}}\left( t \right)&=&\tilde{\Phi }\left( t \right)-\int_{0}^{t}{\tilde{\Phi }\left( t-x \right){\rm d}{{{\tilde{B}}}^{\left( i \right)}}\left( x \right)}+\int_{0}^{t}{\int_{0}^{t-x}{{{\Phi }_{1}}(t-x-y)\bar{H}(y){\rm d}{{F}^{\left( N \right)}}(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}} \\ &&+\int_{0}^{t}{\int_{0}^{t-x}{\int_{0}^{t-x-y}{{{\Phi }_{N}}(t-x-y-z)\bar{F}(y)V(z){\rm d}{{F}^{\left( N \right)}}(z){\rm d}H(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}}} \\ &&+\sum\limits_{n=N}^{\infty }{\int_{0}^{t}{\int_{0}^{t-x}{\int_{0}^{t-x-y}{{{\Phi }_{n}}(t-x-y-z)\bar{F}(y)\frac{{{(\lambda z)}^{n}}}{n!}{{e}^{-\lambda z}}{\rm d}V(z){\rm d}H(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}}}}, \end{matrix} $
对(3.8)式与(3.5)式作$L$变换, 有
$\begin{matrix} {{\varphi }_{i}}^{*}\left( s \right)&=&{{\tilde{\varphi }}^{*}}\left( s \right)\left[ 1-{{{\tilde{b}}}^{i}}\left( s \right) \right]+{{\varphi }_{1}}^{*}\left( s \right){{\tilde{b}}^{i}}\left( s \right)\int_{0}^{\infty }{{{e}^{-st}}\bar{H}\left( t \right){\rm d}F\left( t \right)} \\ &&+{{\varphi }_{N}}^{*}\left( s \right){{\tilde{b}}^{i}}\left( s \right)\left[ \int_{0}^{\infty }{{{e}^{-st}}V\left( t \right){\rm d}{{F}^{\left( N \right)}}\left( t \right)} \right]\left[ \int_{0}^{\infty }{{{e}^{-st}}\bar{F}\left( t \right){\rm d}H\left( t \right)} \right] \\ &&+\sum\limits_{n=N}^{\infty }{{{\varphi }_{n}}^{*}\left( s \right)}{{\tilde{b}}^{i}}\left( s \right)\left[ \int_{0}^{\infty }{{{e}^{-\left( s+\lambda \right)t}}\frac{{{\left( \lambda t \right)}^{n}}}{n!}{\rm d}V\left( t \right)} \right]\left[ \int_{0}^{\infty }{{{e}^{-st}}\bar{F}\left( t \right){\rm d}H\left( t \right)} \right], \end{matrix} $
${{\varphi }_{0}}^{*}\left( s \right)={{\varphi }_{1}}^{*}\left( s \right)f\left( s \right), $
令(3.9)式中$i=1$, 将得到的${{\varphi }_{1}}^{*}\left( s \right)$代入(3.10)式中得
$\begin{matrix} {{\varphi }_{0}}^{*}\left( s \right)&=&{{\tilde{\varphi }}^{*}}\left( s \right)f\left( s \right)[ 1-\tilde{b}\left( s \right) ]+{{\varphi }_{1}}^{*}\left( s \right)f\left( s \right)\tilde{b}\left( s \right)\int_{0}^{\infty }{{{e}^{-st}}\bar{H}\left( t \right){\rm d}F\left( t \right)} \\ &&+{{\varphi }_{N}}^{*}\left( s \right)f\left( s \right)\tilde{b}\left( s \right)\left[ \int_{0}^{\infty }{{{e}^{-st}}V\left( t \right){\rm d}{{F}^{\left( N \right)}}\left( t \right)} \right]\left[ \int_{0}^{\infty }{{{e}^{-st}}\bar{F}\left( t \right){\rm d}H\left( t \right)} \right] \\ &&+\sum\limits_{n=N}^{\infty }{{{\varphi }_{n}}^{*}\left( s \right)}f\left( s \right)\tilde{b}\left( s \right)\left[ \int_{0}^{\infty }{{{e}^{-\left( s+\lambda \right)t}}\frac{{{\left( \lambda t \right)}^{n}}}{n!}{\rm d}V\left( t \right)} \right]\left[ \int_{0}^{\infty }{{{e}^{-st}}\bar{F}\left( t \right){\rm d}H\left( t \right)} \right], \end{matrix} $
由(3.9)与(3.11)式可得到${{\varphi }_{0}}^{*}\left( s \right)$与${{\varphi }_{i}}^{*}\left( s \right)$关系式为
$\begin{matrix} {{\varphi }_{i}}^{*}\left( s \right)={{\tilde{\varphi }}^{*}}\left( s \right)[ 1-{{{\tilde{b}}}^{i-1}}\left( s \right)]+\frac{{{{\tilde{b}}}^{i-1}}\left( s \right)}{f\left( s \right)}{{\varphi }_{0}}^{*}\left( s \right),i\ge 1, \end{matrix}$
$\begin{matrix}\lim\limits_{t\to \infty }{{\Phi }_{i}}\left( t \right)&=&\lim\limits_{s\to {{0}^{+}}}\,s{{\varphi }_{i}}^{*}\left( s \right) \\&=&\lim\limits_{s\to {{0}^{+}}}\,s{{\tilde{\varphi }}^{*}}\left( s \right)\left\{ 1-\frac{{{{\tilde{b}}}^{i}}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\} \\&=&\frac{\alpha \beta \left( 1+\nu \gamma \right)}{1+\alpha \beta \left( 1+\nu \gamma \right)}\cdot \lim\limits_{s\to {{0}^{+}}}\,\left\{ 1-\frac{{{{\tilde{b}}}^{i}}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},\end{matrix}$
经过简单的计算可知需对(3.13)式运用洛必达法则, 并且注意到
$\lim\limits_{s\to {{0}^{+}}}{{\left[ 1-\xi \left( s \right) \right]}^{\prime }}=\frac{1}{\lambda }[ 1-h\left( \lambda \right) ]+h\left( \lambda \right)\left[ \int_{0}^{\infty }{tV\left( t \right){\rm d}{{F}^{\left( N \right)}}\left( t \right)}+\int_{0}^{\infty }{t{{F}^{\left( N \right)}}\left( t \right){\rm d}V\left( t \right)} \right],$
${{M}_{i}}\left( t \right)=E\left\{\mbox{(0,t]内服务台失效次数},\left| N\left( 0 \right)=i \right. \right\}, {{m}_{i}}\left( s \right)=\int_{0}^{\infty }{{{e}^{-st}}{\rm d}{{M}_{i}}}\left( t \right),i\ge 0.$
定理 3.2 对$\mathfrak{R}e \left( s \right)>0$,有
${{m}_{0}}\left( s \right)=\tilde{m}\left( s \right)f\left( s \right)\left\{ 1-\frac{\tilde{b}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},$
${{m}_{i}}\left( s \right)=\tilde{m}\left( s \right)\left\{ 1-\frac{{{{\tilde{b}}}^{i}}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},i\ge 1,$
$\begin{matrix}{{M}_{i}}\left( t \right)&=&E\left\{\mbox{(0,t]内服务台失效次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\}{+}E\left\{\mbox{(0,t] 内服务台失效次数}, {{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&=&E\left\{\mbox{(0,t]内服务台失效次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\}+E\left\{ (0,{{{\tilde{b}}}^{\left\langle i \right\rangle }} ]\mbox{内服务台失效次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&&+\int_{0}^{t}{\int_{0}^{t-x}{{{M}_{1}}(t-x-y)\bar{H}(y){\rm d}F(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}} \\&&+\int_{0}^{t}{\int_{0}^{t-x}{\int_{0}^{t-x-y}{{{M}_{N}}(t-x-y-z)\bar{F}(y)V(z){\rm d}{{F}^{\left( N \right)}}(z){\rm d}H(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}}} \\&&+\sum\limits_{n=N}^{\infty }{\int_{0}^{t}{\int_{0}^{t-x}{\int_{0}^{t-x-y}{{{M}_{n}}(t-x-y-z)\bar{F}(y)\frac{{{(\lambda z)}^{n}}}{n!}{{e}^{-\lambda z}}{\rm d}V(z){\rm d}H(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}}}} \\&=&{{\Gamma }_{i}}\left( t \right)+{{L}_{i}}\left( t \right)+\int_{0}^{t}{\int_{0}^{t-x}{{{M}_{N}}(t-x-y)V(y){\rm d}{{F}^{\left( N \right)}}(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}} \\&&+\sum\limits_{n=N}^{\infty }{\int_{0}^{t}{\int_{0}^{t-x}{{{M}_{n}}(t-x-y)\frac{{{(\lambda y)}^{n}}}{n!}{{e}^{-\lambda y}}{\rm d}V(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}}},\end{matrix}$
${{M}_{0}}\left( t \right)=\int_{0}^{t}{{{M}_{1}}\left( t-x \right){\rm d}F\left( x \right)},$
其中
$\begin{matrix}&&{{\Gamma }_{i}}\left( t \right)=E\left\{ \mbox{(0,t]内服务台失效次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\}\\&&{{L}_{i}}\left( t \right)=E\left\{ (0,{{{\tilde{b}}}^{\left\langle i \right\rangle }} ]\mbox{内服务台失效次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\},i\ge 1,\end{matrix}$
用${{\tilde{b}}^{\left\langle i \right\rangle }}$对引理3.1中的$\tilde{M}\left( t \right)$作全概率分解, 得
$\begin{matrix}\tilde{M}\left( t \right)&=&E\left\{ \mbox{(0,t]内服务台失效次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\}+E\left\{\mbox{(0,t] 内服务台失效次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&=&E\left\{ \mbox{(0,t]内服务台失效次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\}+E\left\{ (0,{{{\tilde{b}}}^{\left\langle i \right\rangle }} ]\mbox{内服务台失效次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&&+E\left\{ ( {{{\tilde{b}}}^{\left\langle i \right\rangle }},t ]\mbox{内服务台失效次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&=&{{\Gamma }_{i}}\left( t \right)+{{L}_{i}}\left( t \right)+\int_{0}^{t}{\tilde{M}\left( t-x \right){\rm d}{{{\tilde{B}}}^{\left( i \right)}}\left( x \right)},\end{matrix}$
于是可得如下(3.22)式
$\begin{matrix}{{\Gamma }_{i}}\left( t \right)+{{L}_{i}}\left( t \right)=\tilde{M}\left( t \right)-\int_{0}^{t}{\tilde{M}\left( t-x \right){\rm d}{{{\tilde{B}}}^{\left( i \right)}}\left( x \right)},\end{matrix}$
$\begin{matrix}{{m}_{i}}\left( s \right)&=&\tilde{m}\left( s \right)\left[ 1-{{{\tilde{b}}}^{i}}\left( s \right) \right]+{{m}_{1}}\left( s \right){{\tilde{b}}^{i}}\left( s \right)\int_{0}^{\infty }{{{e}^{-st}}\bar{H}\left( t \right){\rm d}F\left( t \right)} \\&&+{{m}_{N}}\left( s \right){{\tilde{b}}^{i}}\left( s \right)\left[ \int_{0}^{\infty }{{{e}^{-st}}V\left( t \right){\rm d}{{F}^{\left( N \right)}}\left( t \right)} \right]\left[ \int_{0}^{\infty }{{{e}^{-st}}\bar{F}\left( t \right){\rm d}H\left( t \right)} \right] \\&&+\sum\limits_{n=N}^{\infty }{{{m}_{n}}\left( s \right)}{{\tilde{b}}^{i}}\left( s \right)\left[ \int_{0}^{\infty }{{{e}^{-\left( s+\lambda \right)t}}\frac{{{\left( \lambda t \right)}^{n}}}{n!}{\rm d}V\left( t \right)} \right]\left[ \int_{0}^{\infty }{{{e}^{-st}}\bar{F}\left( t \right){\rm d}H\left( t \right)} \right],\\ \end{matrix}$
${{m}_{0}}\left( s \right)={{m}_{1}}\left( s \right)f\left( s \right),$
令(3.23)式中$i=1$, 将得到的${{m}_{1}}\left( s \right)$代入(3.24)式中得
$\begin{matrix}{{m}_{0}}\left( s \right)&=&\tilde{m}\left( s \right)f\left( s \right)[ 1-\tilde{b}\left( s \right) ]+{{m}_{1}}\left( s \right)f\left( s \right)\tilde{b}\left( s \right)\int_{0}^{\infty }{{{e}^{-st}}\bar{H}\left( t \right){\rm d}F\left( t \right)} \\&&+{{m}_{N}}\left( s \right)f\left( s \right)\tilde{b}\left( s \right)\left[ \int_{0}^{\infty }{{{e}^{-st}}V\left( t \right){\rm d}{{F}^{\left( N \right)}}\left( t \right)} \right]\left[ \int_{0}^{\infty }{{{e}^{-st}}\bar{F}\left( t \right){\rm d}H\left( t \right)} \right] \\&&+\sum\limits_{n=N}^{\infty }{{{m}_{n}}\left( s \right)}f\left( s \right)\tilde{b}\left( s \right)\left[ \int_{0}^{\infty }{{{e}^{-\left( s+\lambda \right)t}}\frac{{{\left( \lambda t \right)}^{n}}}{n!}{\rm d}V\left( t \right)} \right]\left[ \int_{0}^{\infty }{{{e}^{-st}}\bar{F}\left( t \right){\rm d}H\left( t \right)} \right],\end{matrix}$
由(3.23)与(3.25)式得${{m}_{i}}\left( s \right)$与${{m}_{0}}\left( s \right)$的关系式为
$\begin{matrix}{{m}_{i}}\left( s \right)=\tilde{m}\left( s \right)[ 1-{{{\tilde{b}}}^{i-1}}\left( s \right)]+\frac{{{{\tilde{b}}}^{i-1}}\left( s \right)}{f\left( s \right)}{{m}_{0}}\left( s \right),i\ge 1,\end{matrix}$
再将(3.25)式与(3.26)式进行简单的计算化简整理可得(3.16)式, 将(3.16)式代入(3.26)式得(3.17)式, 再根据$\lim\limits_{t\to \infty }\,\frac{{{M}_{i}}\left( t \right)}{t}=\lim\limits_{s\to {{0}^{+}}}\,s{{m}_{i}}\left( s \right)$, 类似定理3.1的证明可得(3.18)式, 证毕.
$\xi \left( s \right)=f\left( s \right)\left[ 1-h\left( s+\lambda \right) \right]+h\left( s+\lambda \right)\left[ A\left( s \right)+D\left( s \right) \right], $
$C\left( \alpha \right)=\sum_{m=0}^{N-1}\int_{0}^{\infty }{{{b}^{m}}\left( \alpha \right)\frac{{{\left( \lambda t \right)}^{m}}}{m!}{{e}^{-\lambda t}}{\rm d}V\left( t \right)},$
$C\left( s+\alpha \right)=\sum_{m=0}^{N-1}\int_{0}^{\infty }{{{b}^{m}}\left( s+\alpha \right)\frac{{{\left( \lambda t \right)}^{m}}}{m!}{{e}^{-\left( s+\lambda \right)t}}{\rm d}V\left( t \right)},$
而平均首次失效时间为$\int_{0}^{\infty }{t{\rm d}}{{\Psi }_{i}}\left( t \right)=-\frac{{\rm d}{{\psi }_{i}}\left( s \right)}{{\rm d}s}\left| _{s=0} \right.,i\ge 1,$并注意到
$\lim\limits_{s\to {{0}^{+}}}\,\left[ 1-\xi \left( s \right) \right]=\lim\limits_{s\to {{0}^{+}}}\,1-f\left( s \right)\left[ 1-h\left( s+\lambda \right) \right]-h\left( s+\lambda \right)\left[ A\left( s \right)+D\left( s \right) \right]=0,$
$\lim\limits_{s\to {{0}^{+}}}\,{{\left[ 1-\xi \left( s \right) \right]}^{\prime }}=\frac{1}{\lambda }\left[ 1-h\left( \lambda \right) \right]+h\left( \lambda \right)\left[ \int_{0}^{\infty }{tV\left( t \right){\rm d}{{F}^{\left( N \right)}}\left( t \right)}+\int_{0}^{\infty }{t{{F}^{\left( N \right)}}\left( t \right){\rm d}V\left( t \right)} \right],$
$\Pi \left( t \right)=P\left\{\mbox{时刻t修理设备处于更换状态} \right\}, {{\pi }^{*}}\left( s \right)=\int_{0}^{\infty }{{{e}^{-st}}\Pi \left( t \right){\rm d}t}, $
$\Sigma \left( t \right)=E\left\{ \mbox{修理设备在( 0,t]内更换的次数} \right\}, \sigma \left( s \right)=\int_{0}^{\infty }{{{e}^{-st}}{\rm d}\Sigma }\left( t \right).$
$\begin{matrix}{{\pi }^{*}}\left( s \right)=\frac{\nu \left( {1}-w\left( s \right) \right)}{s\left( s+\nu -w\left( s \right) \right)},{ }{{\sigma }^{*}}\left( s \right)=\frac{\nu }{s+\nu -w\left( s \right)},\end{matrix}$
平稳结果为
$\begin{matrix}\lim\limits_{t\to \infty }\,\Pi \left( t \right)=\lim\limits_{s\to {{0}^{+}}}\,s{{\pi }^{*}}\left( s \right)=\frac{\nu \gamma }{1+\nu \gamma },{ }\lim\limits_{t\to \infty }\,\frac{\Sigma \left( t \right)}{t}=\lim\limits_{s\to {{0}^{+}}}\,s\sigma \left( s \right)=\frac{\nu }{1+\nu \gamma }.\end{matrix}$
$\tilde{S}\left( t \right)=\Pi \left( t \right)+\int_{0}^{t}{\Pi \left( t-x \right){\rm d}\tilde{Y}\left( x \right)},$
$\tilde{K}\left( t \right)+\tilde{L}\left( t \right)=\Sigma \left( t-x \right)-\int_{0}^{t}{\Sigma \left( t-x \right)}{\rm d}\tilde{Y}\left( x \right),$
其中$\Pi \left( t \right)$与$\Sigma \left( t \right)$由引理4.1给出.
$\begin{matrix}{{a}^{*}}\left( s \right)={{\pi }^{*}}\left( s \right)\frac{\left( 1-\tilde{y}\left( s \right) \right)x\left( s \right)}{1-x\left( s \right)\tilde{y}\left( s \right)}=\frac{\nu \left( {1}-w\left( s \right) \right)}{s\left( s+\nu -w\left( s \right) \right)}\cdot \frac{\alpha \left( 1-y\left( s+\nu -\nu w\left( s \right) \right) \right)}{s+\alpha -\alpha y\left( s+\nu -\nu w\left( s \right) \right)},\end{matrix}$
$\begin{matrix}A\left( t \right)&=&\sum\limits_{m=1}^{\infty }{P\left\{ \sum\limits_{l=1}^{m-1}{( {{X}_{l}}+{{{\tilde{Y}}}_{l}} )+{{X}_{m}}\le t<\sum\limits_{l=1}^{m}{( {{X}_{l}}+{{{\tilde{Y}}}_{l}}),\mbox{时刻t修理设备处于更换状态}}} \right\}} \\&=&\sum\limits_{m=1}^{\infty }{\int_{{0}}^{t}{P\left\{ {{{\tilde{Y}}}_{m}}>t-x,\mbox{时刻t-x修理设备处于更换状态} \right\}d\left[ {{X}^{m}}*{{{\tilde{Y}}}^{\left( m-1 \right)}}\left( x \right) \right]}} \\&=&\sum\limits_{m=1}^{\infty }{\int_{{0}}^{t}{\tilde{S}\left( t-x \right)}}d\left[ {{X}^{m}}*{{{\tilde{Y}}}^{\left( m-1 \right)}}\left( x \right) \right],\end{matrix}$
其中$\tilde{S}\left( t \right)$由引理4.2给出.
对(4.9)式作$L$变换, 并结合引理4.1与引理4.2即可得(4.7)式, 再由
$\begin{matrix}\lim\limits_{t\to \infty }\,A\left( t \right)=\lim\limits_{t\to \infty }\,\lim\limits_{s\to {{0}^{+}}}\,\int_{0}^{t}{{{e}^{-sx}}{\rm d}}A\left( x \right)=\lim\limits_{s\to {{0}^{+}}}\,\int_{0}^{t}{{{e}^{-sx}}{\rm d}}A\left( x \right)=\lim\limits_{s\to {{0}^{+}}}\,s{{a}^{*}}\left( s \right),\end{matrix}$
$\begin{matrix}d\left( s \right)=\sigma \left( s \right)\frac{\left( 1-\tilde{y}\left( s \right) \right)x\left( s \right)}{1-x\left( s \right)\tilde{y}\left( s \right)}=\frac{\nu }{s+\nu -w\left( s \right)}\cdot \frac{\alpha \left( 1-y\left( s+\nu -\nu w\left( s \right) \right) \right)}{s+\alpha -\alpha y\left( s+\nu -\nu w\left( s \right) \right)},\end{matrix}$
$\begin{matrix}R\left( t \right)=&&E\left\{ ( {{{\tilde{Y}}}_{k}},{{X}_{k}} ),\left( 0,t \right]\mbox{内修理设备的更换次数},{{{\tilde{Y}}}_{1}}>t\ge 0 \right\}\\ &&+E\left\{ ( {{{\tilde{Y}}}_{k}},{{X}_{k}} ),( 0,{{{\tilde{Y}}}_{1}}]\mbox{内修理设备的更换次数},{{{\tilde{Y}}}_{1}}\le t \right\} \\&&+E\left\{ ( {{{\tilde{Y}}}_{k}},{{X}_{k}} ),( {{{\tilde{Y}}}_{1}},t ]\mbox{内修理设备的更换次数},{{{\tilde{Y}}}_{1}}\le t \right\} \\&=&E\left\{(0,t]\mbox{内修理设备的更换次数},{{{\tilde{Y}}}_{1}}>t\ge 0 \right\}\\&&+E\left\{ ( 0,{{{\tilde{Y}}}_{1}}]\mbox{内修理设备的更换次数},{{{\tilde{Y}}}_{1}}\le t \right\}+\int_{0}^{t}{D\left( t-x \right){\rm d}}P\left\{ \tilde{Y}\le x \right\} \\&=&\tilde{K}\left( t \right)+\tilde{L}\left( t \right)+\int_{0}^{t}{D\left( t-x \right){\rm d}}P\left\{ \tilde{Y}\le x \right\},\end{matrix}$
其中$\tilde{K}\left( t \right)+\tilde{L}\left( t \right)$由上面的引理4.2给出, 对(4.12)与(4.13)式作$LS$换, 结合引理4.1和引理4.2即可解得(4.10)式, 再根据Tauber定理得$\lim\limits_{t\to \infty }\,\frac{D\left( t \right)}{t}=\lim\limits_{s\to {{0}^{+}}}\,sd\left( s \right)$, 于是使用洛必达法则可得(4.11)式. 证毕.
4.2 修理设备的不可用度
对$t\ge 0$, 令
${{\Pi }_{i}}\left( t \right)=P\left\{ \mbox{时刻t修理设备处于更换状态}\left| N\left( 0 \right)=i \right. \right\},{{\pi }_{i}}^{*}\left( s \right)=\int_{0}^{\infty }{{{e}^{-st}}{{\Pi }_{i}}\left( t \right){\rm d}t},i\ge 0.$
定理 4.1 对$\mathfrak{R}e \left( s \right)>0$, 有
$\begin{matrix}{{\pi }_{0}}^{*}\left( s \right)={{a}^{*}}\left( s \right)f\left( s \right)\left\{ 1-\frac{\tilde{b}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},\end{matrix}$
$\begin{matrix}{{\pi }_{i}}^{*}\left( s \right)={{a}^{*}}\left( s \right)\left\{ 1-\frac{{{{\tilde{b}}}^{i}}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},i\ge 1,\end{matrix}$
${{\Pi }_{0}}\left( t \right)=\int_{0}^{t}{{{\Pi }_{1}}\left( t-x \right){\rm d}F\left( x \right)},$
其中${{\tilde{\Lambda }}_{i}}\left( t \right)=P\left\{ 0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }},\mbox{时刻$t$处于修理设备“广义忙期”且处于更换状态} \right\}$, 且
$\begin{matrix}{{\tilde{\Lambda }}_{i}}\left( t \right)=A\left( t \right)-\int_{0}^{t}{A\left( t-x \right)}{\rm d}{{\tilde{B}}^{\left( i \right)}}\left( x \right),i\ge 1,\end{matrix}$
其中$A(t)$由引理4.3给出, 下面验证(4.19)式.
用${{\tilde{b}}^{\left\langle i \right\rangle }}$对$A(t)$ 进行全概率分解, 可得
$\begin{matrix}A\left( t \right)&=&P\left\{ \mbox{时刻t处于修理设备“广义忙期”且处于更换状态} \right\} \\&=&P\left\{ 0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }},\mbox{时刻t处于修理设备“广义忙期”且处于更换状态} \right\} \\&&+P\left\{ {{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t,\mbox{时刻t处于修理设备“广义忙期”且处于更换状态} \right\} \\&=&{{\tilde{\Lambda }}_{i}}\left( t \right)+\int_{0}^{t}{A\left( t-x \right)}{\rm d}{{\tilde{B}}^{(i)}}\left( x \right),\end{matrix}$
再将(4.19)式代入(4.17)式, 对代入后的(4.17), (4.18)式作$L$变换, 同定理3.1的方法化简整理得(4.14), (4.15)式, 再根据$\lim\limits_{t\to \infty }\,{{\Pi }_{i}}\left( t \right)=\lim\limits_{s\to {{0}^{+}}}\,s{{\pi }_{i}}^{*}\left( s \right)$, 仿照定理3.1的证明使用洛必达法则可得(4.16)式, 证毕.
4.3 修理设备在$\left( 0,t \right]$ 内的平均更换次数
对$t\ge 0$, 令
${{Z}_{i}}\left( t \right)=E\left\{\mbox{修理设备在 \left( 0,t \right]内的平均更换次数}\left| N\left( 0 \right)=i \right. \right\}, {{z}_{i}}\left( s \right)=\int_{0}^{\infty }{{{e}^{-st}}{\rm d}{{Z}_{i}}}\left( t \right).$
定理 4.2 对$\mathfrak{R}e \left( s \right)>0$, 有
${{z}_{0}}\left( s \right)=d\left( s \right)f\left( s \right)\left\{ 1-\frac{\tilde{b}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},$
${{z}_{i}}\left( s \right)=d\left( s \right)\left\{ 1-\frac{{{{\tilde{b}}}^{i}}\left( s \right)\left[ 1-\xi \left( s \right) \right]}{\tilde{\Delta }\left( s \right)} \right\},i\ge 1,$
$\begin{matrix}{{Z}_{i}}\left( t \right)&=&E\left\{ \mbox{修理设备在}\left( 0,t \right]\mbox{内的更换次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\} \\&&+E\left\{ \mbox{修理设备在}\left( 0,t \right]\mbox{内的更换次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&=&E\left\{\mbox{修理设备在}\left( 0,t \right]\mbox{内的更换次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\} \\&&+E\left\{ \mbox{修理设备在}( 0,{{{\tilde{b}}}^{\left\langle i \right\rangle }}]\mbox{内的更换次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&&+E\left\{ \mbox{修理设备在}( {{{\tilde{b}}}^{\left\langle i \right\rangle }}+{{{\hat{\tau }}}_{1}},t ]\mbox{内的更换次数},0<{{{\hat{\tau }}}_{1}}\le H,{{{\tilde{b}}}^{\left\langle i \right\rangle }}+{{{\hat{\tau }}}_{1}}\le t \right\} \\&& +E\left\{\mbox{修理设备在}( {{{\tilde{b}}}^{\left\langle i \right\rangle }}+H+{{{\hat{\hat{\tau }}}}_{1}}+{{l}_{N-1}},t ]\mbox{内的更换次数},\right. \\ && \left. {{{\hat{\tau }}}_{1}}>H,V<{{{\hat{\hat{\tau }}}}_{1}}+{{l}_{N-1}},{{{\tilde{b}}}^{\left\langle i \right\rangle }}+H+{{{\hat{\hat{\tau }}}}_{1}}+{{l}_{N-1}}\le t \right\} \\ &&+\sum\limits_{n=N}^{\infty }{E\left\{ \mbox{修理设备在}( {{{\tilde{b}}}^{\left\langle i \right\rangle }}+H+{{{\hat{\hat{\tau }}}}_{1}}+{{l}_{N-1}},t ]\mbox{内的更换次数}, \right.} \\ & & \left. {{{\hat{\tau }}}_{1}}>H,{{{\hat{\hat{\tau }}}}_{1}}+{{l}_{n-1}}\le V<{{{\hat{\hat{\tau }}}}_{1}}+{{l}_{n}},{{b}^{\left\langle i \right\rangle }}+H+V\le t \right\} \\&=&{{C}_{i}}\left( t \right)+{{\Omega }_{i}}\left( t \right)+\int_{0}^{t}{\int_{0}^{t-x}{{{Z}_{1}}(t-x-y)\bar{H}(y){\rm d}F(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}} \\&&+\int_{0}^{t}{\int_{0}^{t-x}{\int_{0}^{t-x-y}{{{Z}_{N}}(t-x-y-z)\bar{F}(y)V(z){\rm d}{{F}^{\left( N \right)}}(z){\rm d}H(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}}} \\&&+\sum\limits_{n=N}^{\infty }{\int_{0}^{t}{\int_{0}^{t-x}{\int_{0}^{t-x-y}{{{Z}_{n}}(t-x-y-z)\bar{F}(y)\frac{{{(\lambda z)}^{n}}}{n!}{{e}^{-\lambda z}}{\rm d}V(z){\rm d}H(y){\rm d}{{{\tilde{B}}}^{(i)}}(x)}}}},\end{matrix}$
${{Z}_{0}}\left( t \right)=\int_{0}^{t}{{{Z}_{1}}\left( t-x \right){\rm d}F\left( x \right)},$
其中
${{C}_{i}}\left( t \right)=E\left\{\mbox{修理设备在$\left( 0,t \right]$内的更换次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\},$
${{\Omega }_{i}}\left( t \right)=E\left\{ \mbox{修理设备在$( 0,{{{\tilde{b}}}^{\left\langle i \right\rangle }}]$内的更换次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\},$
与定理4.1同理, 用${{\tilde{b}}^{\left\langle i \right\rangle }}$对引理4.4中的$D(t)$作全概率分解, 得
$\begin{matrix}D\left( t \right)&=&E\left\{ \mbox{修理设备在$\left( 0,t \right]$内的更换次数}\right\} \\&=&E\left\{ \mbox{修理设备在$\left( 0,t \right]$内的更换次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\} \\&&+E\left\{\mbox{修理设备在$\left( 0,t \right]$内的更换次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&=&E\left\{ \mbox{修理设备在$\left( 0,t \right]$内的更换次数},0\le t<{{{\tilde{b}}}^{\left\langle i \right\rangle }} \right\} \\&&+E\left\{ \mbox{修理设备在$( 0,{{{\tilde{b}}}^{\left\langle i \right\rangle }}]$内的更换次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&&+E\left\{ \mbox{修理设备在$( {{{\tilde{b}}}^{\left\langle i \right\rangle }},t]$内的更换次数},{{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\} \\&=&{{C}_{i}}\left( t \right)+{{\Omega }_{i}}\left( t \right)+\int_{0}^{t}{D\left( t-x \right){\rm d}{{{\tilde{B}}}^{\left( i \right)}}\left( x \right)},\end{matrix}$
于是
$\begin{matrix}{{C}_{i}}\left( t \right)+{{\Omega }_{i}}\left( t \right)=D\left( t \right)-\int_{0}^{t}{D\left( t-x \right){\rm d}{{{\tilde{B}}}^{\left( i \right)}}\left( x \right)},\end{matrix}$
将(4.27)式代入(4.24)式, 再对代入后的(4.24)与(4.25)式作$LS$变换, 经过化简整理得 (4.21) 与(4.22) 式, 再根据$\lim\limits_{t\to \infty }\,\frac{{{Z}_{i}}\left( t \right)}{t}=\lim\limits_{s\to {{0}^{+}}}\,s{{z}_{i}}\left( s \right)$,同理定理3.1的证明用洛必达法则可得(4.23)式, 证毕.
A number of situations is considered where the single service station is incapacitated from time to time to render service to the incoming (stationary) Poisson stream of customers. Five models are presented Model. A deals with the situation in which the service interruption is brought about by a Poisson process homogeneous in time. The preemptive priority queuing model is shown to be a special case. Model B is concerned with the case in which breakdowns occur only while the station is giving service. In Model C it is assumed that the repair process cannot start without customers at the station. Model D describes a situation in which the repair of the station starts at the initiative of a customer who wishes to improve the standard of service. In Model E the assumption is that the station can break down only while no customer is being serviced. In all models it is assumed that service times and repair times possess arbitrary distribution functions each having a density and finite second moment. The expected queue lengths and related operating characteristics of the various systems are derived using relatively simple mathematical methods.
In many queueing systems, the service process is subject to interruptions resulting from breakdowns, scheduled off-periods or the arrival of customers with preemptive priority. We consider a single server, first-come, first-served queueing system that alternates between periods when service is available (“on-periods”) and periods when the server is unavailable (“off-periods”). We consider the case of Poisson arrivals and general service, on- and off-time distributions and derive bounds and approximations for the mean waiting time, probability of delay and steady-state distribution of the number in system. These results are exact for the case of exponentially distributed on-times. Computational results are reported.
Reliability analysis of discrete-time $Ge{{o}^{({{\lambda }_{1}},{{\lambda }_{2}})}}/G/1$ repairable queue with Bernoulli feedback and Min$(N,D)$-policy
Journal of Systems Science and Mathematical Sciences, 2016, 36(11): 2070-2086
... 符号$b$通常用来表示在经典的$M/G/1$排队系统中, 从一个顾客开始的服务台忙期长度, 有分布$P\left\{ {{b}}\le t \right\}={{B}}\left( t \right)$, ${{b}^{\left\langle i \right\rangle }}$则表示从$i$个顾客开始的服务台忙期长度, 有分布$P\left\{ {{b}^{\left\langle i \right\rangle }}\le t \right\}={{B}^{\left( i \right)}}\left( t \right)$, 用$\tilde{b}$表示该可修排队系统从一个顾客开始的服务台的“广义忙期”长度, 其分布$\tilde{B}\left( t \right)=P\left\{ \tilde{b}\le t \right\}$, ${{\tilde{b}}^{\left\langle i \right\rangle }}$表示从$i$个顾客开始的服务台的“广义忙期”长度, 有分布${{\tilde{B}}^{\left( i \right)}}\left( t \right)=P\left\{ {{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\}, t\ge 0,i\ge 1.$ 类似于文献[39]中有关可修排队系统的讨论, 我们有如下引理. ...
... 引理 2.2[39] 对$\mathfrak{R}e \left( s \right)>0$, $\tilde{b}(s)$是方程$z=\tilde{g}\left( s+\lambda -\lambda z \right)$在$\left| z \right|<1$内的唯一根, 且 ...
... 符号$b$通常用来表示在经典的$M/G/1$排队系统中, 从一个顾客开始的服务台忙期长度, 有分布$P\left\{ {{b}}\le t \right\}={{B}}\left( t \right)$, ${{b}^{\left\langle i \right\rangle }}$则表示从$i$个顾客开始的服务台忙期长度, 有分布$P\left\{ {{b}^{\left\langle i \right\rangle }}\le t \right\}={{B}^{\left( i \right)}}\left( t \right)$, 用$\tilde{b}$表示该可修排队系统从一个顾客开始的服务台的“广义忙期”长度, 其分布$\tilde{B}\left( t \right)=P\left\{ \tilde{b}\le t \right\}$, ${{\tilde{b}}^{\left\langle i \right\rangle }}$表示从$i$个顾客开始的服务台的“广义忙期”长度, 有分布${{\tilde{B}}^{\left( i \right)}}\left( t \right)=P\left\{ {{{\tilde{b}}}^{\left\langle i \right\rangle }}\le t \right\}, t\ge 0,i\ge 1.$ 类似于文献[39]中有关可修排队系统的讨论, 我们有如下引理. ...
... 引理 2.2[39] 对$\mathfrak{R}e \left( s \right)>0$, $\tilde{b}(s)$是方程$z=\tilde{g}\left( s+\lambda -\lambda z \right)$在$\left| z \right|<1$内的唯一根, 且 ...
