Loss systems in a random environment: steady state analysis

Abstract

We consider a single server system with infinite waiting room in a random environment. The service system and the environment interact in both directions. Whenever the environment enters a prespecified subset of its state space the service process is completely blocked: Service is interrupted and newly arriving customers are lost. We prove a product-form steady state distribution of the joint queueing-environment process. A consequence is a strong insensitivity property for such systems. We discuss several applications, for example, from inventory theory and reliability theory, and show that our result extends and generalizes several theorems found in the literature, for example, of queueing-inventory processes. We investigate further classical loss systems, where, due to finite waiting room, loss of customers occurs. In connection with loss of customers due to blocking by the environment and service interruptions new phenomena arise.

Appendix: Proofs

Proof of Proposition 1

The inventory management process under $(r,S)$-policy with distribution function $B$ of the lead times fits into the definition of the environment process by setting

$$\begin{aligned} K=\{r\!+\!1,r\!+\!2,\ldots ,S\}\cup \left( \{0,1,\ldots ,r\}\times \{L,\dots ,1\}\right) ,\quad K_\mathrm{B}=\{0\}\!\times \!\{L,\dots ,1\}\,. \end{aligned}$$

The non-negative transition rates of $(X,Y)$ are, for $(n,k)\in E$,

$$\begin{aligned} \begin{array}{ll} q((n,k),(n+1,k)) = \lambda , &{} \quad k\in K_\mathrm{W},~n\ge 0,\\ q((n,k),(n-1,m)) = \mu (n)R(k,m), &{} \quad k\in K_{W},~m\in K,~n>0,\\ q((n,k),(n,m)) = v(k,m)\in \mathbb {R}_{0}^{+}, &{} \quad k\ne m,~~k,m\in K,\\ q((n,k),(i,m)) = 0, &{} \quad \text {otherwise for }(n,k)\ne (i,m)\in E, \end{array} \end{aligned}$$

$$\begin{aligned} {\begin{array}{lll}\text {where} &{} \quad R(k,k-1)= 1 &{}\quad \text {if} ~ k\in \{r+2,\ldots ,S\},\\ &{} \quad R(r+1,(r,\ell ))=b{(\ell )} &{}\quad \text {if} ~ \ell \in \{L,\dots ,1\},\\ &{} \quad R((j,\ell ),(j-1,\ell ))=1 &{}\quad \text {if} ~ (j,\ell )\in \{1,\ldots , r\}\times \{L,\ldots ,1\},\\ &{} \quad R((0,\ell ),(0,\ell ))=1 &{}\quad \text {if} ~ \ell \in \{L,\ldots ,1\}, \quad \text {and} \quad R(k,j)=0 \\ &{} &{}\qquad \text {if}~ k,j\in K,~\text {otherwise}, \end{array}} \end{aligned}$$

$$\begin{aligned} \text {and} \qquad v((j,\ell ),(j,\ell -1))&=\beta \qquad \text {if } j\in \{0,1,\ldots ,r\},\ell \in \{L,\dots ,2\},\\ \qquad v((j,1),S)&=\beta \qquad \text {if } j\in \{0,1,\ldots , r\},\quad \text {and} \quad v{(k,j)}=0\\&\quad \qquad \qquad ~ \text {if } ~ k,j\in K,~\text {otherwise}. \end{aligned}$$

Theorem 1 yields the product form steady state distribution

$$\begin{aligned} \pi (n,k)=C^{-1}\frac{\lambda ^{n}}{\prod _{i=0}^{n-1}\mu (i+1)}\theta (k),\quad (n,k)\in E\,. \end{aligned}$$

(34)

We have to solve (5). By definition this is, with $R(k,k)=0,\,\forall k\in K\backslash \{0\},\,R(0,0)=1$,

$$\begin{aligned} \theta (k)\left( 1_{[k\in K_\mathrm{W}]}\lambda +\sum _{m\in K\backslash \{k\}}v(k,m)\right)&= \sum _{m\in K_\mathrm{W}\backslash \{k\}}\theta (m)\left( \lambda R(m,k)+v(m,k)\right) \nonumber \\&+\sum _{m\in K_\mathrm{B}\backslash \{k\}}\theta (m)v(m,k). \end{aligned}$$

(35)

(I) For $r>0$, (35) translates into

$$\begin{aligned} \theta (S){\cdot }\lambda&= \sum _{j=0}^{r}\theta (j,1){\cdot }\beta ,\end{aligned}$$

(36)

$$\begin{aligned} \theta (k){\cdot }\lambda&= \theta (k+1){\cdot }\lambda , \,\,\,\quad \qquad \qquad \qquad \qquad \qquad k=r+1,\dots ,S-1,\end{aligned}$$

(37)

$$\begin{aligned} \theta (r,\ell ){\cdot }(\lambda +\beta )&= \theta (r+1){\cdot }\lambda b(\ell )+\theta (r,\ell +1){\cdot }\beta , \,\quad 1\le \ell <L,\end{aligned}$$

(38)

$$\begin{aligned} \theta (r,L){\cdot }(\lambda +\beta )&= \theta (r+1){\cdot }\lambda b(L),\end{aligned}$$

(39)

$$\begin{aligned} \theta (j,L){\cdot }(\lambda +\beta )&= \theta (j+1,L){\cdot }\lambda , \,\,\,\,\,\qquad \qquad \qquad \qquad \quad 1\le j<r,\end{aligned}$$

(40)

$$\begin{aligned} \theta (j,\ell ){\cdot }(\lambda +\beta )&= \theta (j+1,\ell ){\cdot }\lambda +\theta (j,\ell +1){\cdot }\beta , \qquad 1\le j<r,1\le \ell <L,\end{aligned}$$

(41)

$$\begin{aligned} \theta (0,\ell ){\cdot }\beta&= \theta (1,\ell ){\cdot }\lambda +\theta (0,\ell +1){\cdot }\beta , \,\quad \qquad \quad 1\le \ell <L,\end{aligned}$$

(42)

$$\begin{aligned} \theta (0,L){\cdot }\beta&= \theta (1,L){\cdot }\lambda . \end{aligned}$$

(43)

From (37) it follows that

$$\begin{aligned} \theta (S)=\theta (S-1)=\cdots =\theta (r+1), \end{aligned}$$

(44)

and from (39) and (40) it follows that for $1\le j \le r$

$$\begin{aligned} \theta (j,L)=\theta (r+1)b(L)\left( \frac{\lambda }{\lambda +\beta }\right) ^{r+1-j}. \end{aligned}$$

(45)

From (45) (for $j=r$) and it (38) follows directly that

$$\begin{aligned} \theta (r,\ell )=\theta (r+1)\frac{\lambda }{\lambda +\beta }\sum _{i=\ell }^{L}b(i)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell },\quad 1\le \ell <L\,. \end{aligned}$$

(46)

Up to now we have obtained the expressions for the north and west border line of the array $(\theta (j,\ell ):1\le j\le r,1\le \ell \le L)$ which can be filled step by step via (41). The proposed solution is

$$\begin{aligned} \theta (r-h,\ell )=\theta (r+1)\left( \frac{\lambda }{\lambda +\beta }\right) ^{h+1}\sum _{i=\ell }^{L}b(i)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +h\\ h\end{array}}\right) , \end{aligned}$$

(47)

for $h=0,1,\dots ,r-1,\ell =1,\dots ,L$ which fits with (46) ($h=0$ with $\left( {\begin{array}{c}i-l\\ 0\end{array}}\right) =1$) and (45). Inserting (47) into (41) verifies (47) by a two-step induction with help from the elementary formula $\left( {\begin{array}{c}a\\ n\end{array}}\right) +\left( {\begin{array}{c}a\\ n-1\end{array}}\right) =\left( {\begin{array}{c}a+1\\ n\end{array}}\right)$. For computing the residual boundary probabilities $\theta (0,\ell )$, we need some more effort. The proposed solution is, for $\ell =1,\ldots ,L,$

$$\begin{aligned} \theta (0,\ell )=\theta (r+1)\left( \frac{\lambda }{\lambda +\beta }\right) ^{r}\frac{\lambda }{\beta }\left[ \sum _{i=\ell }^{L}\sum _{g=i}^{L}b(g)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +r-1\\ r-1\end{array}}\right) \right] .\nonumber \\ \end{aligned}$$

(48)

From (43) and (45) we obtain

$$\begin{aligned} \theta (0,L)=\theta (r+1)\left( \frac{\lambda }{\lambda +\beta }\right) ^{r}\frac{\lambda }{\beta }b(L), \end{aligned}$$

(49)

which fits into (48), and it remains to check the recursion (42). This amounts to computing

$$\begin{aligned}&\theta (1,\ell ){\cdot }\frac{\lambda }{\beta }+\theta (0,\ell +1)\\&\quad = \theta (r+1)\left( \frac{\lambda }{\lambda +\beta }\right) ^{r}\sum _{i=\ell }^{L}b(i)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +r-1\\ r-1\end{array}}\right) {\cdot }\frac{\lambda }{\beta }\\&\qquad +\,\theta (r+1)\left( \frac{\lambda }{\lambda +\beta }\right) ^{r}\frac{\lambda }{\beta }\left[ \sum _{i=\ell +1}^{L}\sum _{g=i}^{L}b(g)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-(\ell +1)}\left( {\begin{array}{c}i-(\ell +1)+r-1\\ r-1\end{array}}\right) \right] \\&\quad = \theta (r+1)\left( \frac{\lambda }{\lambda +\beta }\right) ^{r}\frac{\lambda }{\beta }\left[ \sum _{i=\ell +1}^{L}\left\{ \sum _{g=i}^{L}b(g)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-(\ell +1)}\left( {\begin{array}{c}\overbrace{i-(\ell +1)}^{=(i-1)-\ell }+r-1\\ r-1\end{array}}\right) \right. \right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. +\,b(i-1)\left( \frac{\beta }{\lambda +\beta }\right) ^{(i-1)-\ell }\left( {\begin{array}{c}(i-1)-\ell +r-1\\ r-1\end{array}}\right) \right\} \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. +\,b(L)\left( \frac{\beta }{\lambda +\beta }\right) ^{L-\ell }\left( {\begin{array}{c}L-\ell +r-1\\ r-1\end{array}}\right) \right] \\&\quad = \theta (r+1)\left( \frac{\lambda }{\lambda +\beta }\right) ^{r}\frac{\lambda }{\beta }\left[ \sum _{i=\ell +1}^{L}\left\{ \sum _{g=i-1}^{L}b(g)\left( \frac{\beta }{\lambda +\beta }\right) ^{(i-1)-\ell }\left( {\begin{array}{c}(i-1)-\ell +r-1\\ r-1\end{array}}\right) \right\} \right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. +\,b(L)\left( \frac{\beta }{\lambda +\beta }\right) ^{L-\ell }\left( {\begin{array}{c}L-\ell +r-1\\ r-1\end{array}}\right) \right] \\&\quad = \theta (r+1)\left( \frac{\lambda }{\lambda +\beta }\right) ^{r}\frac{\lambda }{\beta }\left[ \sum _{i=\ell }^{L-1}\left\{ \sum _{g=i}^{L}b(g)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +r-1\\ r-1\end{array}}\right) \right\} \right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. +\,b(L)\left( \frac{\beta }{\lambda +\beta }\right) ^{L-\ell }\left( {\begin{array}{c}L-\ell +r-1\\ r-1\end{array}}\right) \right] = \theta (0,\ell ).\\&\text {Setting}\ \theta (r+1)=G^{-1}\left( \frac{\lambda +\nu }{\lambda }\right) ^{r} \text { completes the proof in the case}~r=0. \end{aligned}$$

(II) For $r>0$, (35) translates into

$$\begin{aligned} \theta (S){\cdot }\lambda&= \theta (0,1){\cdot }\beta ,\end{aligned}$$

(50)

$$\begin{aligned} \theta (k){\cdot }\lambda&= \theta (k+1){\cdot }\lambda ,\,\quad \qquad \qquad \qquad \qquad \quad k=1,\dots ,S-1,\end{aligned}$$

(51)

$$\begin{aligned} \theta (0,\ell ){\cdot }\beta&= \theta (1){\cdot }\lambda {\cdot } b(L)+\theta (0,\ell +1){\cdot }\beta ,\quad 1\le \ell <L,\end{aligned}$$

(52)

$$\begin{aligned} \theta (0,L){\cdot }\beta&= \theta (1){\cdot }\lambda {\cdot } b(L). \end{aligned}$$

(53)

From (51) it follows that

$$\begin{aligned} \theta (S)=\theta (S-1)=\dots =\theta (1), \end{aligned}$$

(54)

and we will show that

$$\begin{aligned} \theta (0,\ell )=\theta (1){\cdot }\left( \frac{\lambda }{\beta }\right) \left[ \sum _{i=\ell }^{L}b(i)\right] ,\quad \ell =1,\dots ,L, \end{aligned}$$

(55)

holds. For $\ell =L$ this is immediate from (53), and for $\ell <L$ it follows by induction from (52). Setting $\theta (1)=G^{-1}$ completes the proof in the case $r=0$. $\square$

Remark 4

For $r>0$, we can write (26) as

$$\begin{aligned} \theta (0,\ell ) \!=\! G^{-1}\frac{\lambda }{\beta }\left[ \sum _{i=\ell }^{L}\left( \sum _{g=i}^{L}b(g)\right) \left( \frac{\beta }{\lambda \!+\!\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +r-1\\ i-\ell \end{array}}\right) \right] ,\,\, \ell \!=\!1,\dots ,L, \end{aligned}$$

and can extend this formula to the case $r=0$. This yields, with $\left( {\begin{array}{c}-1\\ 0\end{array}}\right) =1$, explicitly

$$\begin{aligned} \theta (0,\ell )&= G^{-1}\frac{\lambda }{\beta }\left[ \sum _{i=\ell }^{L}b(i)\right] ,\quad \ell =1,\dots ,L. \end{aligned}$$

Proof of Corollary 2

For $j=1,\dots ,r$, we have

$$\begin{aligned} P(I=j)&= G^{-1}\left( \frac{\lambda +\beta }{\lambda }\right) ^{j-1}\sum _{\ell =1}^{L}\sum _{i=\ell }^{L}b(i)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +r-j\\ r-j\end{array}}\right) \\&= G^{-1}\left( \frac{\lambda +\beta }{\lambda }\right) ^{j-1}\sum _{i=1}^{L}b(i)\sum _{\ell =1}^{i}\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +r-j\\ r-j\end{array}}\right) \\&= G^{-1}\left( \frac{\lambda +\beta }{\lambda }\right) ^{j-1}\sum _{i=1}^{L}b(i)\sum _{g=0}^{i-1}\left( \frac{\beta }{\lambda +\beta }\right) ^{g}\left( {\begin{array}{c}g+r-j\\ r-j\end{array}}\right) \\&= G^{-1}\left( \frac{\lambda +\beta }{\lambda }\right) ^{j-1}\left( \frac{\lambda +\beta }{\lambda }\right) ^{r+1-j} \sum _{i=1}^{L}b(i)\sum _{g=0}^{i-1}\left( {\begin{array}{c}g+(r+1-j)-1\\ (r+1-j)-1\end{array}}\right) \\&\qquad \left( \frac{\lambda }{\lambda +\beta }\right) ^{r+1-j}\left( \frac{\beta }{\lambda +\beta }\right) ^{g}\\&= G^{-1}\left( \frac{\lambda +\beta }{\lambda }\right) ^{r}\sum _{i=1}^{L}b(i) {\cdot } P\left( W\left( r+1-j,\frac{\lambda }{\lambda +\beta }\right) <i\right) , \end{aligned}$$

and for $j=0$ we have

$$\begin{aligned} P(I=0)&= G^{-1}\left( \frac{\lambda }{\beta }\right) \sum _{\ell =1}^{L}\sum _{i=\ell }^{L}\sum _{g=i}^{L}b(g)\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +r-1\\ r-1\end{array}}\right) \\&= G^{-1}\left( \frac{\lambda }{\beta }\right) \sum _{i=1}^{L}\sum _{\ell =1}^{i}\left( \frac{\beta }{\lambda +\beta }\right) ^{i-\ell }\left( {\begin{array}{c}i-\ell +r-1\\ r-1\end{array}}\right) \sum _{g=i}^{L}b(g)\\&= G^{-1}\underbrace{\left( \frac{\lambda }{\beta }{\cdot } E(V)\right) }_{=\lambda /\nu }\left( \frac{\lambda +\beta }{\lambda }\right) ^{r}\sum _{i=1}^{L}\underbrace{\left( \frac{1}{E(U)}\sum _{g=i}^{L}b(g)\right) }_{=:P(V_{e}=i)}\sum _{f=0}^{i-1}\left( {\begin{array}{c}f+r-1\\ r-1\end{array}}\right) \\&\qquad \left( \frac{\lambda }{\lambda +\beta }\right) ^{r}\left( \frac{\beta }{\lambda +\beta }\right) ^{f}\\&= G^{-1}\left( \frac{\lambda }{\nu }\right) \left( \frac{\lambda +\beta }{\lambda }\right) ^{r} P\left( W\left( r+1-1,\frac{\lambda }{\lambda +\beta }\right) <U_{e}\right) . \end{aligned}$$

$\square$