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RETRACTED ARTICLE: Transient analysis of an M/M/1 queue with variant impatient behavior and working vacations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

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Change history

  • 30 October 2019

    The editor has retracted this article [1] because it has significant overlap with a work published by Sudhesh and Azhagappan [2] and is therefore redundant. The authors do not agree to this retraction.

  • 30 October 2019

    The editor has retracted this article [1] because it has significant overlap with a work published by Sudhesh and Azhagappan [2] and is therefore redundant. The authors do not agree to this retraction.

References

  1. 1.

    Altman, E., Yechiali, U.: Analysis of customers’ impatience in queues with server vacation. Queueing Syst. 52, 261–279 (2006)

  2. 2.

    Ammar, S.I.: Transient analysis of an M/M/1 queue with impatient behavior and multiple vacations. Appl. Math. Comput. 260, 97–105 (2015)

  3. 3.

    Choi, B.D., Kim, B., Chung, J.: M/M/1 queue with impatient customers of higher priority. Queueing Syst. 38, 49–66 (2001)

  4. 4.

    Choudhury, A.: Impatience in single server queuing model. Am. J. Math. Manag. Sci. 28, 177–211 (2008)

  5. 5.

    Gradshteyn, I., Ryzhik, I., Jeffrey, A., Zwillinger, D.: Table of Integrals, Series and Products, 7th edn. Academic Press, Elsevier, Cambridge (2007)

  6. 6.

    Kalidass, K., Ramanath, K.: Transient analysis of an M/M/1 queue with multiple vacations. Pak. J. Stat. Oper. Res. 10, 121–130 (2014)

  7. 7.

    Kim, B., Kim, J.: A single server queue with Markov modulated service rates and impatient customers. Perform. Eval. 83–84, 1–15 (2015)

  8. 8.

    Li, J., Tian, N.: The M/M/1 queue with working vacations and vacation interruptions. J. Syst. Sci. Syst. Eng. 16(1), 121–127 (2007)

  9. 9.

    Parthasarathy, P.R.: A transient solution to an M/M/1 queue: a simple approach. Adv. Appl. Probab. 19(4), 997–998 (1987)

  10. 10.

    Perel, N., Yechiali, U.: Queues with slow servers and impatient customers. Eur. J. Oper. Res. 201, 247–258 (2010)

  11. 11.

    Selvaraju, N., Goswami, Cosmika: Impatient customers in an M/M/1 queue with single and multiple working vacations. Comput. Ind. Eng. 65, 207–215 (2013)

  12. 12.

    Servi, L.D., Finn, S.G.: M/M/1 queue with working vacations (M/M/1/WV). Perform. Eval. 50, 41–52 (2002)

  13. 13.

    Sudhesh, R.: Transient analysis of a queue with system disasters and customer impatience. Queueing Syst. 66, 95–105 (2010)

  14. 14.

    Sudhesh, R., Raj, L.F.: Computational analysis of stationary and transient distribution of single server queue with working vacation. Global Trends Comput. Commun. Syst. Commun. Comput. Inf. Sci. 269, 480–489 (2012)

  15. 15.

    Tian, N., Zhao, X., Wang, K.: The M/M/1 queue with single working vacation. Int. J. Inf. Manag. Sci. 19, 621–634 (2008)

  16. 16.

    Vijaya Laxmi, P., Jyothsna, K.: Analysis of finite buffer renewal input queue with balking and multiple working vacations. Opsearch 50(4), 548–565 (2013)

  17. 17.

    Vijaya Laxmi, P., Jyothsna, K.: Impatient customer queue with Bernoulli schedule vacation interruption. Comput. Oper. Res. 56, 1–7 (2015)

  18. 18.

    Xiu, C., Tian, N., Liu, Y.: The M/M/1 queue with single working vacation serving at a slower rate during the start-up period. J. Math. Res. 2, 98–102 (2010)

  19. 19.

    Yechiali, U.: Queues with system disasters and impatient customers when system is down. Queueing Syst. 56, 195–202 (2007)

  20. 20.

    Yue, D., Yue, W., Xu, G.: Analysis of a queueing system with impatient customers and working vacations. In: Proceedings of the 6th international conference on queueing theory and network applications, pp. 208–211 (2011)

Download references

Author information

Correspondence to R. Sudhesh.

Additional information

The editor has retracted this article [1] because it has significant overlap with a work published by Sudhesh and Azhagappan [2] and is therefore redundant. The authors do not agree to this retraction.

[1] Sudhesh, R. and Azhagappan, A. OPSEARCH (2018) 55: 787. https://doi.org/10.1007/s12597-018-0339-8

[2] Sudhesh, R and Azhagappan, A. Asian Journal of Research in Social Sciences and Humanities. (2016) ;6(9):1096-104. https://doi.org/10.5958/2249-7315.2016.00857.1

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). \(\square \)

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Sudhesh, R., Azhagappan, A. RETRACTED ARTICLE: Transient analysis of an M/M/1 queue with variant impatient behavior and working vacations. OPSEARCH 55, 787–806 (2018). https://doi.org/10.1007/s12597-018-0339-8

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Keywords

  • M/M/1 queue
  • Single and multiple working vacations
  • Impatient behavior
  • Continued fraction