An Algorithm for Asymptotic Mean and Variance for Markov Renewal Process of M/G/1 Type with Finite Level

Abstract

The Markov renewal process (MRP) of M/G/1 type has been used for modeling many complex queueing systems with correlated arrivals and the special types of transitions of the MRP process corresponds to the departures from the queueing system. It can be seen from the central limit theorem for regenerative process that the distribution of the number of transitions of MRP is asymptotically normal. Thus, the asymptotic mean and variance of the number of transitions of MRP can be used to estimate the number of departures in the queueing system modelled by MRP.

The aim of this paper is to present an algorithm for computing the asymptotic mean and variance for the number of level-down-transitions in a Markov renewal process of M/G/1 type with finite level. The results are applied to the queueing system with finite buffer and correlated arrivals.

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Funding

This research was supported by Basic Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (Grant Numbers NRF-2018R1D1A1A09083352).

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Appendix: Uniformization Method for Computing A n, \(A^{\prime }_{n}\) and \(A^{\prime \prime }_{n}\)

Appendix: Uniformization Method for Computing A n, \(A^{\prime }_{n}\) and \(A^{\prime \prime }_{n}\)

Let \(\theta =\max \limits _{1\le j\le m}\left \{ [-D_{0}]_{jj}\right \}\) and

$$P_{0}=I+\frac{1}{\theta}D_{0}, P_{1}=\frac{1}{\theta}D_{1},$$

and for n = 0, 1, 2,⋯, k = 0, 1, 2,

$$\gamma_{n,k}={\int}_{0}^{\infty} e^{-\theta x}\frac{\theta^{n}}{n!}x^{n+k} d \hat{H}(x).$$

By repeating the uniflormization arguments (Lucantoni 1991) shows that An, n = 0, 1, 2,⋯ are given by

$$A_{n}=\sum\limits_{j=0}^{\infty} \gamma_{n,0} K^{(j)}_{n} =\sum\limits_{j=n}^{\infty} \gamma_{n,0} K^{(j)}_{n},$$

where the matrices \(K^{(j)}_{n}\) are defined recursively as follows:

  • \(K^{(0)}_{0}=I\), \(K^{(0)}_{n}=0\), n ≥ 1,

  • For j = 1, 2,⋯,

    $$ \begin{array}{@{}rcl@{}} K^{(j)}_{0}&=&K^{(j-1)}_{0}P_{0},\\ K^{(j)}_{n}&=&\left\{\begin{array}{ll} K^{(j-1)}_{n}P_{0}+K^{(j-1)}_{n-1}P_{1}, &1\le n\le j,\\ 0, & n\ge j+1. \end{array}\right. \end{array} $$

The matrices \(A^{\prime }_{n}\) and \(A^{\prime \prime }_{n}\) are obtained similarly by using γn,1 and γn,2 instead of γn,0 in the formula An, respectively.

We present the formulae of γn,k for some distributions.

  1. (i)

    Deterministic distribution of service time with mean h1:

    $$ \gamma_{n,k}=e^{-\theta h_{1}}\frac{\theta^{n}}{n!}h_{1}^{n+k}, n=0,1,2,\cdots, k=0,1,2. $$
  2. (ii)

    Gamma distribution Gam(a,b) of service time with mean h1 and SCV \({c_{s}^{2}}\): The parameters a and b are determined by

    $$a=\frac{1}{{c_{s}^{2}}}, b={c_{s}^{2}} h_{1}$$

    and the LST of Gam(a,b) is H(s) = (1 + bs)a. Let \(\gamma _{k}(z)={\sum }_{n=0}^{\infty } \gamma _{n,k} z^{n}\), k = 0, 1, 2 and c(0) = 1, c(1) = a, c(2) = a(a + 1). Simple algebra yields

    $$ \begin{array}{@{}rcl@{}} \gamma_{k}(z)&=&\frac{{\Gamma}(a+k)}{{\Gamma}(a)}\beta^{k}{\int}_{0}^{\infty} e^{-\theta(1-z)x}\frac{1}{{\Gamma}(a+k)}\frac{1}{b}e^{-x/b}\left( \frac{x}{b}\right)^{a+k-1} dx\\ &=&c(k)b^{k}\left( 1+b \theta(1-z)\right)^{-(a+k)}\\ &=&c(k)\frac{b^{k}}{(1+b\theta)^{a+k}} \left( 1-\frac{b\theta}{1+b\theta}z\right)^{-(a+k)}. \end{array} $$

Expanding the generating function γk(z), we have that

$$\gamma_{k}(z)=c(k)\frac{b^{k}}{(1+b\theta)^{a+k}}\sum\limits_{n=0}^{\infty} \left( \begin{array}{c} a+n+k-1\\ n \end{array}\right) \left( \frac{b\theta}{1+b\theta}\right)^{n}z^{n},$$

where

$$\left( \begin{array}{c} a+n+k-1\\ n \end{array}\right)=\frac{(a+k)(a+k+1){\cdots} (a+k+n-1)}{n!}.$$

Thus for n = 0, 1, 2,⋯, k = 0, 1, 2,

$$\gamma_{n,k}=\frac{1}{n!}\left( \prod\limits_{i=0}^{n+k-1} (a+i)\right) \theta^{-(a+k)}b^{-a} \left( \frac{b\theta}{1+b \theta}\right)^{a+n+k}. $$

As a special case, the exponential distribution with mean h1 is Gam(1,h1) and hence the γn,k for exponential distribution of service time is

$$\gamma_{n,k}=c(k) \theta^{-(k+1)}\frac{1}{h_{1}} \left( \frac{h_{1}\theta}{1+h_{1} \theta}\right)^{n+k+1}$$

with c(0) = c(1) = 1 and c(2) = 2.

(iv) PH(β,S) distribution of service time : The LST of PH(β,S) is

$$H(\theta)={\pmb\beta}(\theta I-{\pmb S})^{-1}{\pmb S}^{0},$$

where S0 = −Se. Note that

$$ \begin{array}{@{}rcl@{}} {\int}_{0}^{\infty} x^{n} e^{-\theta x} d\hat{H}(x)&=&(-1)^{n}H^{(n)}(\theta)\\ &=&n! {\pmb\beta}(\theta I-{\pmb S})^{-(n+1)}{\pmb S}^{0}. \end{array} $$

Thus for n = 0, 1, 2,⋯, k = 0, 1, 2,

$$ \begin{array}{@{}rcl@{}} \gamma_{n,k}&=&\frac{\theta^{n}}{n!}{\pmb\beta}{\int}_{0}^{\infty} x^{n+k} e^{-\theta x}e^{{\pmb S}x} dx {\pmb S}^{0}\\ &=&\frac{\theta^{n}}{n!}(n+k)! {\pmb\beta}(\theta I-{\pmb S})^{-(n+k+1)}{\pmb S}^{0}\\ &=&\frac{(n+k)!}{n!}\theta^{-(k+1)}{\pmb\beta}\left( I-\frac{1}{\theta}{\pmb S}\right)^{-(n+k+1)}{\pmb S}^{0}. \end{array} $$

(v) The γn,k for lognormal distribution LN(a,b) and Weibul distribution WB(a,b) of service time is obtained by numerical integration. The probability density function of LN(a,b) is

$$f(x)= \frac{1}{\sqrt{2\pi b}x}\exp\left[-\frac{(\log x -a)^{2}}{2b}\right], x> 0,$$

and the parameters a and b are determined by h1 and \({c_{s}^{2}}\) as follows

$$b=\log\left( 1+{c_{s}^{2}}\right), a=\log h_{1} -\frac{1}{2}b.$$

The probability density function of WB(a,b) is

$$f(x)= \frac{a}{b}\left( \frac{x}{b}\right)^{a-1} e^{-(x/b)^{a}}, x> 0,$$

and the parameters a and b are determined by h1 and \({c_{s}^{2}}\) as follows

$${c_{s}^{2}}=2a \frac{{\Gamma}(2/a)}{{\Gamma}(1/a)^{2}}-1, b=\frac{a h_{1}}{{\Gamma}(1/a)}.$$

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Shin, Y.W. An Algorithm for Asymptotic Mean and Variance for Markov Renewal Process of M/G/1 Type with Finite Level. Methodol Comput Appl Probab (2021). https://doi.org/10.1007/s11009-021-09846-w

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Keywords

  • First return time
  • Fundamental period
  • Asymptotic mean
  • Asymptotic variance
  • Markov renewal process of M/G/1 type

Mathematics Subject Classification (2010)

  • 60K20
  • 60K25