Analysis of dynamic service system between regular and retrial queues with impatient customers

Balasubramanian Krishna Kumar; Ramachandran Navaneetha Krishnan; Rathinam Sankar; Ramasamy Rukmani

doi:10.3934/jimo.2020153

Article Contents

2022, Volume 18, Issue 1: 267-295. Doi: 10.3934/jimo.2020153

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Analysis of dynamic service system between regular and retrial queues with impatient customers

1.
Department of Mathematics, College of Engineering, Anna University, Chennai 600 025, India
2.
Department of Mathematics, Pachaiyappa's College, Chennai 600 030, India

^* Corresponding author: B. Krishna Kumar
^* Corresponding author: B. Krishna Kumar

Received: December 31, 2019

Revised: June 30, 2020

Early access: October 2020

Published: January 2022

This research work is supported in part by Department of Science and Technology, New Delhi, India, under research grant INT/RUS/RFBR/377

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Abstract

In this article, we propose a dynamic operating of a single server service system between conventional and retrial queues with impatient customers. Necessary and sufficient conditions for the stability, and an explicit expression for the joint steady-state probability distribution are obtained. We have derived some interesting and important performance measures for the service system under consideration. The first-passage time problems are also investigated. Finally, we have presented extensive numerical examples to demonstrate the effects of the system parameters on the performance measures.

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Mathematics Subject Classification: Primary: 60K25, 90B05; Secondary: 68M20, 90B22.

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Figure 1. Dynamic oscillating queue with threshold and impatience

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Figure 2(a). $\pi(0, 0) ~\text { versus }~ \xi ~\text { for }~ \alpha = 3, \mu = 4, \mathrm{N} = 5$

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Figure 2(b). $\text { E(L) versus }~ \xi~ \text { for } ~\alpha = 3, \mu = 4, N = 5$

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Figure 2(c). $\mathrm{E}\left(\mathrm{W}_{\mathrm{S}}\right) ~~\text { versus } ~~\xi ~~\text { for }~~ \alpha = 3, \mu = 4, \mathrm{N} = 5$

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Figure 2(d). $\mathrm{R}_{\mathrm{A}} ~~\text { versus } ~~\xi ~~\text { for }~~ \alpha = 3, \mu = 4, \mathrm{N} = 5$

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Figure 2(e). $\mathrm{P}_{\mathrm{S}} ~~\text { versus } ~~\xi ~~\text { for }~~ \alpha = 3, \mu = 4, \mathrm{N} = 5$

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Figure 3(a). $\pi(0, 0) \text { versus } \mu \text { for } \alpha = 3, \xi = 5, \mathrm{N} = 5$

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Figure 3(b). $\mathrm{E}(\mathrm{L}) \text { versus } \mu \text { for } \xi = 5, \alpha = 3, \mathrm{N} = 5$

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Figure 3(c). $\mathrm{E}\left(\mathrm{W}_{\mathrm{S}}\right) \text { versus } \mu \text { for } \xi = 5, \alpha = 3, \mathrm{N} = 5$

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Figure 3(d). $\mathrm{R}_{\mathrm{A}} \text { versus } \mu \text { for } \xi = 5, \alpha = 3, \mathrm{N} = 5$

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Figure 3(e). $\mathrm{P}_{\mathrm{S}} \text { versus } \mu \text { for } \xi = 5, \alpha = 3, \mathrm{N} = 5$

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Figure 4(a). $\mathrm{E}\left(\tau_{\mathrm{j}}\right) ~~\text { versus } ~~\xi ~~\text { for }~~ \alpha = 3, \mu = 4, \mathrm{N} = 5, \mathrm{j} = 10$

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Figure 4(b). $\mathrm{E}\left(\tau_{\mathrm{j}}\right) \text { versus } \mu \text { for } \xi = 5, \alpha = 3, \mathrm{N} = 5, \mathrm{j} = 10$

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Figure 4(c). $\mathrm{E}\left(\tau_{\mathrm{j}}\right) ~~\text { versus } ~~\xi ~~\text { for }~~ \lambda = 18, \nu = 20, \alpha = 3, \mu = 4, \mathrm{N} = 5$

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Figure 4(d). $\mathrm{E}\left(\tau_{\mathrm{i}}\right) \text { versus } \mu \text { for } \lambda = 18, \nu = 20, \xi = 5, \alpha = 3, \mathrm{N} = 5$

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Figure 4(e). $\mathrm{E}\left(\tau_{\mathrm{j}}\right) \text { versus } \lambda \text { for } \xi = 5, \nu = 20, \mu = 4, \alpha = 3, \mathrm{N} = 5$

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Figure 4(f). $\mathrm{E}\left(\tau_{\mathrm{j}}\right) \text { versus } \nu \text { for } \lambda = 18, \xi = 5, \alpha = 3, \mu = 4, \mathrm{N} = 5$

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