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.

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 A_n, 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 γ_n,1 and γ_n,2 instead of γ_n,0 in the formula A_n, respectively.

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

• (i) Deterministic distribution of service time with mean h₁:

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

• (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 γ_k(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 γ_n,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 γ_n,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

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