RETRACTED ARTICLE: Transient analysis of an M/M/1 queue with variant impatient behavior and working vacations

Abstract

This paper studies an M/M/1 queue with single, multiple working vacations and customers’ variant impatient behavior. During working vacations, the arriving customers are served with slower service rate than the service rate of non-vacation period. An arriving customer, during working vacation, finds the system empty and gets his service immediately, does not become impatient. The only customers who are waiting for service, during working vacation, become impatient. The transient system size probabilities of this model are derived explicitly in the closed form using continued fraction. The time-dependent mean and variance are also computed. Numerical examples are provided to visualize the analytical results.

Appendices

Appendix A

Proof of Theorem 1

Let

$$\begin{aligned} Q_{1}(t,z)= & {} \sum _{n=1}^\infty P_{b,n}(t) z^n, \ Q_{1}(0,z)= 0. \end{aligned}$$

From (2.4), we obtain

$$\begin{aligned} \frac{\partial Q_{1}(t,z)}{\partial t} = \left[ -(\lambda + \mu _{b})+\lambda z + \frac{\mu _{b}}{z}\right] Q_{1}(t,z)+ \mu _{b}\left( 1-\frac{1}{z}\right) P_{b,0}(t)+ \gamma \sum \limits _{n=0}^\infty P_{v,n}(t) z^n. \end{aligned}$$

(6.1)

Solving the partial differential equation (6.1) yields,

$$\begin{aligned} Q_{1}(t,z)&= {} \, \gamma \int \limits _{0}^{t}\left[ \sum \limits _{n=0}^\infty P_{v,m}(t) z^m\right] e^{-(\lambda + \mu _{b})(t-u)}e^{(\lambda z + \frac{\mu _{b}}{z})(t-u)}du \nonumber \\&\quad +\mu _{b}\left( 1-\frac{1}{z}\right) \int \limits _{0}^{t}P_{b,0}(u)e^{-(\lambda + \mu _{b})(t-u)}e^{(\lambda z + \frac{\mu _{b}}{z})(t-u)}du. \end{aligned}$$

(6.2)

It is well known that, if $\alpha = 2 \sqrt{\lambda \mu _{b}}$ and $\beta =\sqrt{\frac{\lambda }{\mu _{b}}}$, then

$$\begin{aligned} e^{\left[ (\lambda z + \frac{\mu _{b}}{z})t\right] } = \sum \limits _{n=-\infty }^{\infty } (\beta z)^{n}I_{n}(\alpha t), \end{aligned}$$

(6.3)

Using (6.3) in (6.2), for $n=1, 2, 3, \ldots$, we get

$$\begin{aligned} P_{b,n}(t)&= {} \, \gamma \int \limits _{0}^{t} \sum \limits _{m=0}^{\infty } P_{v,m}(u) \beta ^{n-m}I_{n-m}(\alpha (t-u))e^{-(\lambda + \mu _{b})(t-u)}du \nonumber \\&\quad +\mu _{b} \int \limits _{0}^{t}P_{b,0}(u)e^{-(\lambda + \mu _{b})(t-u)}\beta ^{n}\left[ I_{n}(\alpha (t-u))-\beta I_{n+1}(\alpha (t-u))\right] du. \end{aligned}$$

(6.4)

The above equation holds for $n = -1, -2, -3, \ldots ,$ with the left hand side replaced by zero. Using $I_{-n}(x)= I_{n}(x)$, for $n = 1, 2, 3,\ldots ,$

$$\begin{aligned} 0&= {} \, \gamma \int \limits _{0}^{t} \sum \limits _{m=0}^{\infty } P_{v,m}(u) \beta ^{-n-m}I_{n+m}(\alpha (t-u))e^{-(\lambda + \mu _{b})(t-u)}du \nonumber \\&\quad +\mu _{b} \int \limits _{0}^{t}P_{b,0}(u)e^{-(\lambda + \mu _{b})(t-u)}\beta ^{-n}\left[ I_{n}(\alpha (t-u))-\beta I_{n-1}(\alpha (t-u))\right] du. \end{aligned}$$

(6.5)

From (6.4) and (6.5), we obtain (2.5), for $n = 1, 2, 3, \ldots .$

Evaluation of $P_{1,0}(t)$

Let ${\hat{f}}(s)$ represents the Laplace transform of f(t). Taking Laplace transforms on (2.3), we get

$$\begin{aligned} {\hat{P}}_{b,0}(s)= \frac{\gamma }{s+\lambda } {\hat{P}}_{v,0}(s). \end{aligned}$$

Laplace inversion of the above equation yields (2.6). Thus, we have expressed the probabilities $P_{b,n}(t)$ in terms of $P_{v,n}(t)$ and $P_{b,0}(t)$, for $n = 1, 2, 3, \ldots$ and $P_{b,0}(t)$ is obtained in terms of $P_{v,0}(t)$. $\square$

Appendix B

Proof of Theorem 2

Taking Laplace transform on (2.1) and (2.2), we get

$$\begin{aligned} \hat{P}_{v,0}(s)= & {} \frac{1}{(s+\lambda +\gamma )-\mu _{b}\frac{\hat{P}_{b,1}(s)}{\hat{P}_{v,0}(s)}-\mu _{v}\frac{\hat{P}_{v,1}(s)}{\hat{P}_{v,0}(s)}}. \end{aligned}$$

(6.6)

$$\begin{aligned} \frac{\hat{P}_{v,n}(s)}{\hat{P}_{v,n-1}(s)}= & {} \frac{\lambda }{(s+\lambda +\mu _{v}+(n-1)\xi +\gamma )-(\mu _{v}+n \xi ) \frac{\hat{P}_{v,n+1}(s)}{\hat{P}_{v,n}(s)}}, \quad n \ge 1. \end{aligned}$$

(6.7)

Solving (6.7), for $n = 1, 2, 3, \ldots$, we obtain

$$\begin{aligned} \hat{P}_{v,n}(s)= & {} \left( \frac{\lambda }{\xi }\right) ^{n} \frac{1}{\prod \limits _{i=0}^{n-1}\left( \frac{s+\gamma +\mu _{v}}{\xi }+i\right) } \frac{_1F_{1}\left( \frac{\mu _{v}}{\xi }+n;\frac{s+\gamma +\mu _{v}}{\xi }+n;-\frac{\lambda }{\xi }\right) }{_1F_{1}\left( \frac{\mu _{v}}{\xi };\frac{s+\gamma + \mu _{v}}{\xi };-\frac{\lambda }{\xi }\right) }\hat{P}_{v,0}(s)= \hat{V}_{n}(s) \hat{P}_{v,0}(s). \nonumber \\&\end{aligned}$$

(6.8)

On Laplace inversion, we get (2.7). Using (2.5) and (6.7) for $n = 1,$ in (6.6), we obtain

$$\begin{aligned} \hat{P}_{v,0}(s)=\frac{\frac{1}{(s+\lambda +\gamma )}}{\left\{ 1-\frac{1}{(s+\lambda +\gamma )}\left[ \mu _{v}\hat{V}_{1}(s)+\gamma \sum \limits _{m=0}^\infty d_{2}^{m} \hat{V}_{m}(s) +\frac{\mu _{b}\beta ^{2}\gamma }{(s+\lambda )}d_{2} \right] \right\} }, \end{aligned}$$

where

$$\begin{aligned} d_{2} = \frac{d_{1}-\sqrt{d_{1}^{2}-\alpha ^{2}}}{\alpha \beta }, \quad d_{1} = s +\lambda + \mu _{b}. \end{aligned}$$

(6.9)

After some manipulations yield,

$$\begin{aligned} \hat{P}_{v,0}(s)= & {} \sum \limits _{k=0}^{\infty } \sum \limits _{r=0}^{k} \sum \limits _{i=0}^{r}\left( {\begin{array}{c}{k} \\ r \\ \end{array}} \right) \left( {\begin{array}{c}{r} \\ i \\ \end{array}} \right) \frac{\mu _{v}^{i}\lambda ^{r-i}\gamma ^{k-i}\hat{V}_{1}^{i}(s)}{(s+\lambda +\gamma )^{k+1}(s+\lambda )^{r-i}} d_{2}^{r-i}\left[ \sum \limits _{m=0}^{\infty }d_{2}^{m}\hat{V}_{m}(s) \right] ^{k-r}. \nonumber \\&\end{aligned}$$

(6.10)

On Laplace inversion, we get (2.8). From (6.8), we obtain

$$\begin{aligned} \hat{V}_{n}(s)= \left( \frac{\lambda }{\xi }\right) ^{n} \frac{1}{\prod \limits _{i=0}^{n-1}\left( \frac{s+\gamma +\mu _{v}}{\xi }+i\right) } \frac{_1F_{1}\left( \frac{\mu _{v}}{\xi }+n;\frac{s+\gamma +\mu _{v}}{\xi }+n;-\frac{\lambda }{\xi }\right) }{_1F_{1}\left( \frac{\mu _{v}}{\xi };\frac{s+\gamma + \mu _{v}}{\xi };-\frac{\lambda }{\xi }\right) }. \end{aligned}$$

(6.11)

Using the definition of confluent hypergeometric functions, we obtain

$$\begin{aligned} \frac{_1F_{1}\left( \frac{\mu _{v}}{\xi }+n;\frac{s+\gamma +\mu _{v}}{\xi } +n;-\frac{\lambda }{\xi }\right) }{\prod \limits _{i=0}^{n-1} \left( \frac{s+\gamma +\mu _{v}}{\xi }+i\right) }= \sum \limits _{k=0}^{\infty }\frac{\prod \limits _{j=0}^{k-1} \left( \mu _{v}+(n+j)\xi \right) }{\prod \limits _{i=0}^{k-1} \left( s+\gamma +\mu _{v}+(n+i)\xi \right) } \frac{(-\lambda )^{k}}{\xi ^{k-n}k!}. \end{aligned}$$

(6.12)

Applying partial fraction in the above equation, we get

$$\begin{aligned} \frac{_1F_{1}\left( \frac{\mu _{v}}{\xi }+n;\frac{s+\gamma +\mu _{v}}{\xi }+n;-\frac{\lambda }{\xi }\right) }{\prod \limits _{i=0}^{n-1}\left( \frac{s+\gamma +\mu _{v}}{\xi }+i\right) }= & {} \sum \limits _{k=0}^{\infty }\frac{(-\lambda )^{k}\xi ^{n-2k+1}}{k!} {\prod \limits _{j=0}^{k-1}\left( \mu _{v}+(n+j)\xi \right) } \nonumber \\&\times \sum \limits _{i=1}^{k}\frac{(-1)^{i-1}}{(k-i)!(i-1)!}\frac{1}{(s+\gamma +\mu _{v}+n \xi +(i-1)\xi )}. \nonumber \\&\end{aligned}$$

(6.13)

Further,

$$\begin{aligned} {_1F_{1}\left( \frac{\mu _{v}}{\xi };\frac{s+\gamma +\mu _{v}}{\xi };-\frac{\lambda }{\xi }\right) }=\sum \limits _{k=0}^{\infty }\frac{\prod \limits _{j=0}^{k-1}\left( \mu _{v}+j\xi \right) }{\prod \limits _{i=0}^{k-1}\left( s+\gamma +\mu _{v}+i\xi \right) } \frac{(-\lambda )^{k}}{{\xi ^{k}}k!} =\sum \limits _{k=0}^{\infty }{(-\lambda )^{k}}\hat{\phi }_{k}(s), \end{aligned}$$

where

$$\begin{aligned} {{\hat{\phi }}}_{k}(s)= \sum \limits _{k=0}^{\infty }\frac{\prod \limits _{j=0}^{k-1}\left( \mu _{v}+j\xi \right) }{\prod \limits _{i=0}^{k-1}\left( s+\gamma +\mu _{v}+i\xi \right) } \frac{1}{{\xi ^{k}}k!}, \ k \ge 1. \end{aligned}$$

Using partial fractions, for $k \ge 1$, we have

$$\begin{aligned} {\hat{\phi }}_{k}(s)= \frac{\xi ^{-k}}{k!}\prod \limits _{j=0}^{k-1}\left( \mu _{v}+j\xi \right) \sum \limits _{i=1}^{k}\frac{(-1)^{i-1}}{\xi ^{k-1}(i-1)! (k-i)!}\frac{1}{s+\gamma +\mu _{v}+(i-1)\xi }, \quad {\hat{\phi }}_{0}(s)=1, \end{aligned}$$

(6.14)

Using the identity given in [5], we obtain

$$\begin{aligned} \left[ _1F_{1}\left( \frac{\mu _{v}}{\xi };\frac{s+\gamma +\mu _{v}}{\xi };-\frac{\lambda }{\xi }\right) \right] ^{-1}= \sum \limits _{k=0}^{\infty }\hat{\chi }_{k}(s)\lambda ^{k}, \end{aligned}$$

(6.15)

where ${\hat{\chi }}_{0}(s)= 1$ and for $k = 1, 2, 3, \ldots$,

$$\begin{aligned} {\hat{\chi }}_{k}(s)= & {} \left| \begin{array}{cccccc} {\hat{\phi }}_{1}(s) &{} 1 &{} 0 &{} \dots &{} 0 &{} 0 \\ {\hat{\phi }}_{2}(s) &{} {\hat{\phi }}_{1}(s) &{} 1 &{} \dots &{} 0 &{} 0 \\ {\hat{\phi }}_{3}(s) &{} {\hat{\phi }}_{2}(s) &{} {\hat{\phi }}_{1}(s) &{} \dots &{} 0 &{} 0\\ \ldots &{}\ldots &{}\ldots &{}\ldots &{}\ldots &{}\ldots \\ {\hat{\phi }}_{k-1}(s) &{} {\hat{\phi }}_{k-2}(s) &{} {\hat{\phi }}_{k-3}(s) &{} \dots &{} {\hat{\phi }}_{1}(s) &{} 1\\ {\hat{\phi }}_{k}(s) &{} {\hat{\phi }}_{k-1}(s) &{} {\hat{\phi }}_{k-2}(s) &{} \dots &{} {\hat{\phi }}_{2}(s) &{} {\hat{\phi }}_{1}(s) \end{array}\right| \nonumber \\= & {} \sum \limits _{i=1}^{k}(-1)^{i-1}\hat{\phi }_{i}(s)\hat{\chi }_{k-i}(s). \end{aligned}$$

(6.16)

Substitute (6.13) and (6.15) in (6.11), we get

$$\begin{aligned} \hat{V}_{n}(s) = {\frac{\lambda }{\xi }}^{n}\sum \limits _{k=0}^{\infty }(-\lambda )^{k} \hat{\phi }_{n+k}(s)\sum \limits _{r=0}^{\infty }\lambda ^{r}\hat{\chi }_{r}(s). \end{aligned}$$

(6.17)

On taking inverse Laplace transforms, we get (2.9). Thus, we have obtained the time-dependent system size probabilities $P_{v,n}(t)$ in terms of $P_{v,0}(t)$, for $n = 1, 2, 3, \ldots$ and $P_{v,0}(t)$ is derived explicitly. $\square$

Appendix C

Proof of Theorem 3

Define

$$\begin{aligned} Q_{2}(t,z)= & {} \sum \limits _{n=1}^\infty P_{b,n}(t) z^n, \quad Q_{2}(0,z)= 0. \end{aligned}$$

From (4.3) and (4.4), we obtain

$$\begin{aligned} \frac{\partial Q_{2}(t,z)}{\partial t} = \left[ -(\lambda + \mu _{b})+\lambda z + \frac{\mu _{b}}{z}\right] Q_{2}(t,z)- \mu _{b}P_{b,1}(t)+ \gamma \sum \limits _{n=1}^\infty P_{v,n}(t) z^n. \end{aligned}$$

(6.18)

Using the methodology given in the proof of Theorem 1, we obtain $P_{b,n}(t)$ which resemble with (4.5), for $n = 1, 2, 3, \ldots$. Thus, we have derived the probabilities $P_{b,n}(t)$ as a function of the probabilities $P_{v,n}(t)$, for $n = 1, 2, 3, \ldots .$ $\square$

Appendix D

Proof of Theorem 4

Taking Laplace transform on (4.1), we get

$$\begin{aligned} \hat{P}_{v,0}(s)=\frac{1}{(s+\lambda )-\mu _{b}\frac{\hat{P}_{b,1}(s)}{\hat{P}_{v,0}(s)}-\mu _{v}\frac{\hat{P}_{v,1}(s)}{\hat{P}_{v,0}(s)}}. \end{aligned}$$

(6.19)

Using (4.5) and (6.8) for $n = 1,$ in (6.19), we obtain

$$\begin{aligned} \hat{P}_{v,0}(s)= & {} \frac{\frac{1}{(s+\lambda )}}{\left\{ 1-\frac{1}{\lambda (s+\lambda )}\left[ \mu _{b}\gamma \sum \limits _{m=1}^\infty \beta ^{2}d_{2}^{m} \hat{V}_{m}(s) + \lambda \mu _{v} \hat{V}_{1}(s) \right] \right\} }. \end{aligned}$$

Algebraic simplification yields,

$$\begin{aligned} \hat{P}_{v,0}(s) = \sum \limits _{k=0}^{\infty } \sum \limits _{i=0}^{k}\gamma ^{k} \left( {\begin{array}{c} {k} \\ i \\ \end{array}} \right) \left( \frac{\mu _{v}}{\gamma }\right) ^{i} \frac{\hat{V}_{1}^{i}(s)}{(s+\lambda )^{k+1}}\left[ \sum \limits _{m=1}^{\infty }d_{2}^{m} \hat{V}_{m}(s)\right] ^{k-i}, \end{aligned}$$

(6.20)

where $d_{2}$ is as in (6.9). On Laplace inversion, we get (4.6). $\square$