Sojourn-time Distribution for $M/G^a/1$ Queue with Batch Service of Fixed Size - Revisited

Abstract

This paper presents an explicit and straightforward method for finding the sojourn-time distribution of a random customer in an $M/G^a/1$ queue with a fixed-size batch service. The exhibited process is much more straightforward than the approach discussed by Yu and Tang (Methodology and Computing in Applied Probability 20(4):1503–1514, 2018). We obtain two closed-form expressions for probability density functions by using the inside and outside roots of the underlying characteristic equation. Applying partial fractions and residue theorem, we determine an explicit form of sojourn-time distribution and evaluate the distribution function for any specific time. In illustrative examples, we compare the results obtained by both methods and find that the results match excellently.

Appendix

An alternative approach to invert (15) is as follows. If Eq. (15) is put in partial fraction form, it is easy to find inverse Laplace transform. Equation (15) is an improper rational function if the degree of the numerator is greater than the degree of the denominator. In such a case, we divide the numerator by the denominator to make it a proper rational function (Kobayashi et al. (2011)). This leads to

$$\begin{aligned} W^{*}(s)&= R(s) +\frac{N_1(s)}{D(s)}, \end{aligned}$$

(26)

where $R(s)=\sum \limits _{i=1}^{d} k_i s^i$ is a proper polynomial which can be dealt with by using the transform pair (Dirac delta function) and $\frac{N_1(s)}{D(s)}$ is a proper rational function which we deal with by using partial fraction, where

$$\begin{aligned} \delta ^{(n)}(t)&= \frac{d^n}{dt^n} \delta (t) \ \leftrightarrow \ s^n,~~n=1,2,\ldots \end{aligned}$$

Thus,

$$\begin{aligned} W^{*}(s)&= \sum _{i=1}^{d} k_i s^i \ +\ \sum _{i=1}^{m} \frac{T_i}{s- \xi _i}, \end{aligned}$$

(27)

The constants $T_i$ can be found as

$$\begin{aligned} T_i=\frac{C \left( \lambda ^a-(\lambda -\xi _i)^a\right) }{\xi _i }\cdot \frac{F(1-\xi _i/\lambda )}{\prod \limits _{j=1, j\ne i}^{m} (\xi _i-\xi _j)}, ~\forall i. \end{aligned}$$

Let us define the probability density function of the sojourn time to be w(t). Taking the inverse Laplace transforms of Eq. (27), we have

$$\begin{aligned} w(t)&= \sum _{i=0}^{d} k_i \delta ^{(i)}(t) +\ \sum _{i=1}^{m} T_i\ e^{\xi _i t}. \end{aligned}$$

(28)

Example 5

We consider the service time distribution as phase-type (PH) having the representation of $\left( \alpha , \mathbf {T} \right)$ where $\alpha$ and $\mathbf {T}$ are assumed as $\alpha =\begin{pmatrix} 0.25&0.75 \end{pmatrix}, \mathbf {T}=\begin{pmatrix} -1 &{} 0\\ 0 &{} -2 \end{pmatrix}$ with $\mu =1.6$. The LST $B^{*}(s)=\mathbf {\alpha }\left( s\ \mathbf {I}-\mathbf {T}\right) ^{-1} \mathbf {T^{o}},~\mathfrak {Re}(s)\ge 0,$ and the other parameters are $\lambda =6$ and $a=5$. From the denominator of Eq. (6), the roots outside the unit circle are $\gamma _1=1.080640$ and $\gamma _2=1.265010$. The LST of the sojourn time of a random customer is

$$\begin{aligned} W^{*}(s)&=\frac{0.000059(2+1.75s) \left( s^4-30 s^3+360 s^2-2160s +6480\right) }{(0.483838+s)(1.590599+s)}\\&=0.000104s^3-0.003214s^2+0.040436s-0.263117+\frac{0.471068}{0.483838+s} +\frac{0.460496}{1.590598+s} \end{aligned}$$

The probability density function of the sojourn time by taking the inverse Laplace transforms of $W^{*}(s)$ is

$$\begin{aligned} w(t)=& \ 0.000104\ \delta ^{(3)}(t) -0.003214\ \delta ^{(2)}(t)+0.040436\ \delta ^{(1)}(t)-0.263117\ \delta (t)\\&+0.471068\ e^{-0.483838 t}+0.460496\ e^{-1.590598 t} \end{aligned}$$

We obtain $E(W)=2.153832$, which is matching with the result of Yu and Tang (2018). We also check the correctness of the result using the Padé approximation of order [4/5]. Here, we have

$$\begin{aligned} W^{*}(s)&=\frac{ 1.0+0.68702 s-0.1468852 s^2+0.0155368 s^3}{1.0+2.840852 s+1.70168 s^2+0.217345 s^3+0.0140274 s^4}. \end{aligned}$$

From Eq. (17), we obtain

$$\begin{aligned} w(t)=& \ e^{-6.710681 t}\left[ 0.178441 \cos (6.905605t)-3.689684 \sin (6.905605t)\right] \\ {}&+0.458098 e^{-1.589077t}+0.471068 e^{-0.483839t}. \end{aligned}$$

Table 3 presents a cumulative distribution function (CDF) of the sojourn time and compares the results with an alternative method, Padé approximation, and Yu and Tang (2018). We obtain $E(W)=\int _{0}^{\infty } t \cdot w(t)dt =2.153832$, in both the methods which verifies the correctness of the result.

Table 3 CDF of sojourn time of a random customer from Example 5