# Analytically Explicit Results for the Distribution of the Number of Customers Served during a Busy Period for Special Cases of theM/G/1 Queue

Authors: M. L. Chaudhry; Veena Goswami
Journal: Journal of Probability and Statistics
 (2019)

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## Abstract

This paper presents analytically explicit results for the distribution of the number of customers served during a busy period for special cases of theM/G/1 queues when initiated with m customers. The functional equation for the Laplace transform of the number of customers served during a busy period is widely known, but several researchers state that, in general, it is not easy to invert it except for some simple cases such as M/M/1 and M/D/1 queues. Using the Lagrange inversion theorem, we give an elegant solution to this equation. We obtain the distribution of the number of customers served during a busy period for various service-time distributions such as exponential, deterministic, Erlang-k, gamma, chi-square, inverse Gaussian, generalized Erlang, matrix exponential, hyperexponential, uniform, Coxian, phase-type, Markov-modulated Poisson process, and interrupted Poisson process. Further, we also provide computational results using our method. The derivations are very fast and robust due to the lucidity of the expressions.


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## Body

## 1. Introduction

Emphasis on numerical transform inversion has become an important method for computing probability distributions in stochastic models [1–3]. In queueing theory, as transforms can often represent probability distributions, it is necessary to find the numerical inversion technique. Generally, the analytical inverse transform is difficult to evaluate or obtain in closed explicit form. For example, in an M/G/1 queueing system, the probability generating function (PGF) of the distribution of the number of customers served during a busy period and the Laplace transform (LT) of the busy period distribution are represented as functional equations.Queueing systems are directly applicable to many fields such as telecommunication, production, and computer systems. Most performance measures analyze the system to reduce the delays experienced by a customer. The analysis of the busy period is a significant component of any queueing system from the server’s viewpoint, which helps in the efficient planning of the system. Understanding the characteristics of the busy period could manage to prevent server overload. When an arriving customer gets the system empty, a busy period begins and ends when the system next becomes empty.In the case ofM/D/1 queue, Borel [4] first found the distribution of the number of customers served during the busy period when m=1 by a geometric argument, and Tanner [5] showed that Borel’s argument is valid for general m, where m is the number that initiates the busy period. Kendall [6] also obtained a continuous analogue of this distribution. Haight [7] studied the distribution in the case of M/M/1 queue, analogous to the method used by Borel–Tanner. Using different approaches, the distribution of the number served during a busy period has been considered by Prabhu [8, 9], Haight [7, 10, 11], Takács [12, 13], Kemp and Kemp [14], and Feller [15]. Takács studied using a generalization of the classical ballot theorem [16]. Heyman [17] obtained an approximation for the busy period of the M/G/1 queue using a diffusion model. The joint distribution of the length of the busy period and the number of customers served during that busy period for the M/G/1 queues has been studied by Prabhu [9]. Consul and Shenton [18] showed that Lagrange distributions give a general form for the probability distribution of the number of customers served at a single counter in a busy period with the inputs following a discrete distribution and the service-time being constant.For the queueing systemM/G/1, one of the important distributions is the number of customers served during a busy period. This distribution has been discussed by many researchers. In particular, see [19, 20, 21] and references therein. Most of the researchers mentioned there either give only the functional equation or using that they give a few moments or solve that equation numerically (see, e.g., [21]). In general, as stated by them that getting an explicit form of this distribution being difficult, they give results only in some simple cases such as M/M/1 and M/D/1 (see, e.g., [20, 21]). In this paper, for the queueing system M/G/1, using the well-known Lagrange inversion theorem, we give explicit results for the distribution of the number of customers served during a busy period by considering many service-time distributions such as exponential, deterministic, Erlang-k, gamma, chi-square, inverse Gaussian, generalized Erlang, matrix exponential, hyperexponential, uniform, Coxian, phase-type, Markov-modulated Poisson process, and interrupted Poisson process. In addition, we also give numerical results. The benefit of these closed-form expressions is that they are comparatively simple, elegant, and can be easily computed numerically. The problem of finding the number of customers served during a busy period in various systems such as computer communication systems is useful in developing congestion control strategies.The paper is organized as follows. In Section2, we describe the model in detail, derive an equation for the Laplace transform of the number of customers served during a busy period, and find the moments of the number served during a busy period using functional equation. We also present an analytical derivation of the distribution of the number of customers served during a busy period for the M/G/1 system using Lagrange inversion theorem. Section 3 gives analytically explicit results for the distribution of the number of customers served during a busy period by considering many service-time distributions. Section 4 provides computational results. Section 5 concludes the paper.


## 2.M/G/1 Queue

In this section, we consider the distribution of the number of customers served during a busy period for the continuous-timeM/G/1 queueing system when initiated by m customers. We assume that the interarrival times of customers are independent and follow an exponential distribution with common mean interarrival time 1/λ. The service-times are independent and identically distributed as a random variable S with a general distribution function Btt≥0, probability density function (pdf) btt≥0, Laplace transform (LT) b¯s=∫0∞e−stbtdt,s≥0, and mean service-time 1/μ=−b¯10, where b¯10 denotes the first derivative of b¯s evaluated at s=0. We assume that ρ=λ/μ<1. For any integer j≥0, let αj be the probability that j customers arrive during the service period of a single customer. Let αz be the generating function of αj and it is given by αz=∑j=0∞αjzj; that is,(1)αz=∑j=0∞αjzj=∑j=0∞∫0∞λtje−λtj!btdt=∫0∞e−λt∑j=0∞λtzjj!btdt=b¯λ−λz.In this paper, we assume that customers initiate a busy period whenm≥1. The case m>1 can be considered as the initial busy period where a queue is permitted to build before the server starts working. The case m=1 presents subsequent busy periods where the arrival of a single customer terminates the idle period. We also assume that ρ<1, so that all busy periods will terminate with probability one. For the queueing system M/G/1, one of the important distributions is the number of customers served during a busy period. Most of the researchers either give only the functional equation which we derive below for completeness sake or using that equation, they give a few moments or solve that equation numerically [19, 20, 21].


### 2.1. Functional Equation for the Number of Customers Served during a Busy Period

We present the distribution of the number of customers served during a busy period for the continuous-timeM/G/1 queueing system. A busy period is initiated by a single customer. It begins at the instant of the arrival of a customer to an empty system and ends at the instant the server becomes idle for the first time. When the busy period ends, the idle period starts and ends when the next customer arrives. Let NT≡N= be the number of customers served during a busy period T. Since the arrivals are Poisson distributed, it implies that the lengths of subbusy periods Ti are independent and identically distributed random variables and equal to the distribution of the length of the total busy period T. The number of customers served during each busy period is independent and identically distributed and equal to the distribution of number of customers served during a complete busy period. Let Ni,i=1,2,3,…,A, be the number of customers served during the busy period generated by the ith arrival, A a random variable representing the number of arrivals that arrive during S, N the number served during the entire busy period, and its probability generating function Rz=EzN. Then, since each arrival during the initial service-time of the busy period will generate its own distribution of the number served, we may write N=1+∑i=0ANi with N0=0. Now, since the arrivals follow a Poisson process, Ni are independent of A and S and are also independent random variables identically distributed as N, we may write the following by conditioning N by A and S:(2)Rz=∑n=1∞PN=nzn=EzN=EEzNA,S=∫0∞∑n=0∞Ez1+∑j=1ANjA=n,S=tPA=nS=tbtdt=z∑n=0∞Rzn∫0∞λtnn!e−λtbtdt=z∫0∞e−λ−λRztbtdt=zb¯λ−λRz,which shows that the probability generating function for the number of customers served during a busy period satisfies the above functional equation.


### 2.2. Distribution of the Number of Customers Served during a Busy Period for the M/G/1 System

In this subsection, we present the derivation of the distribution of a number of customers served during a busy period for theM/G/1 system using Lagrange inversion theorem. In mathematical analysis, the Lagrange inversion theorem provides the Taylor series expansion of the inverse of an analytic function. Closed-form solutions of (2) can be found using Lagrange’s inversion theorem. Applying (1) in (2), we may rewrite as(3)Rz=zαRz,z≤1.By Lagrange inversion theorem, we get(4)Rz=∑n=1∞znn!Dzn−1αnzz=0,where Dzxyz denotes the xth derivative of yz with respect to z evaluated at z=0 and Dz0yz=yz. For a particular service-time distribution b¯s, we can easily get an expression for αz and thus an expression for PN=n. For more details, one may refer studies of Consul and Famoye [22] and Jain et al. [23]. Suppose that there are m≥1 customers at the beginning of the busy period T. Then, the PGF of the number NmT served during the busy period is given by Rmz, where Rz is the PGF of the number served during the busy period that begins by a single customer. The probability generating function of Nm is ∑n=m∞PNm=nzn=Rmz, and its Lagrange’s series expansion simplifies to(5)Rmz=∑n=m∞mznnn−m!Dzn−mαnzz=0.This provides(6)PNm=n=mnn−m!Dzn−mαnzz=0,n=m,m+1,m+2,….Equation (6) provides the distribution of a number of customers served during a busy period for the M/G/1 system when initiated with m customers. Taking into consideration various service-time distributions, we present the distribution of a number of customers served during a busy period in Section 3. In the derivation of analytical results, we also use Leibniz’s rule. The general Leibniz’s rule states that if fx and gx are n-times differentiable functions, then the product fxgx is also n-times differentiable and its nth derivative is given by(7)Dxnfxgx=∑i=0nniDxn−ifxDxigx.The formula can be generalized to the product ofk differentiable functions f1,f2,…,fk:(8)Dxnf1xf2x…fkx=∑i1+i2+…+ik=nni1,i2,…,ik∏1≤j≤kDxijfj,where the sum extends over all k-tuples i1,i2,…,ik of nonnegative integers with ∑j=1kij=n and ni1,i2,…,ik=n!/i1!i2!…ik! are the multinomial coefficients.


### 2.3. Moments of the Number Served during a Busy Period Using Functional Equation

Here, we obtain mean and variance of the number of customers served during a busy period for the queueing systemM/G/1. Let vk be the kth moment of the service-time; that is, νk=ESk=−1kdkb¯s/dsks=0 and variance σ2=ν2−ν12. Here, ν1=1/μ. Let ENr,r=1,2,… be the rth moment of the number served during a busy period when initiated by single customer. We obtain from (2) the following(9)R11=b¯0−λb¯10R11=b¯0+λν1R11,implying(10)R11≡EN=11−λν1=11−ρ,R21=−2λb¯10R11+b¯20−λR112−λb¯10R21=2λμR11+ν2−λR112+λμR21,giving(11)R21=2ρ1−ρ+λ2ν21−ρ3.SinceR21=EN2−EN, we have(12)EN2=2ρ1−ρ+λ2ν21−ρ3+11−ρ.The variance of the number of customers served during a busy period when initiated by single customer is(13)VarN=ρ1−ρ+λ2ν21−ρ3.When a busy period begins with an initial number ofm customers in the system, the transition of the underlying Markov process from m to 0 can be made up of m intervals with the same distribution corresponding the transitions from m⟶m−1,m−1⟶m−2,…,1⟶0 because of the Markovian property of the arrival process. These m-independent busy periods start with single customer in the system so that Nm=∑j=1mN1j. Hence, if Nm is the random variable representing a busy period initiated by m customers, we have(14)ENm=∑j=1mEN1j=mEN1=m1−ρ,(15)VarNm=mρ1−ρ+λ2ν21−ρ3.Note that higher order moments can be obtained similarly.


## 2.1. Functional Equation for the Number of Customers Served during a Busy Period

We present the distribution of the number of customers served during a busy period for the continuous-timeM/G/1 queueing system. A busy period is initiated by a single customer. It begins at the instant of the arrival of a customer to an empty system and ends at the instant the server becomes idle for the first time. When the busy period ends, the idle period starts and ends when the next customer arrives. Let NT≡N= be the number of customers served during a busy period T. Since the arrivals are Poisson distributed, it implies that the lengths of subbusy periods Ti are independent and identically distributed random variables and equal to the distribution of the length of the total busy period T. The number of customers served during each busy period is independent and identically distributed and equal to the distribution of number of customers served during a complete busy period. Let Ni,i=1,2,3,…,A, be the number of customers served during the busy period generated by the ith arrival, A a random variable representing the number of arrivals that arrive during S, N the number served during the entire busy period, and its probability generating function Rz=EzN. Then, since each arrival during the initial service-time of the busy period will generate its own distribution of the number served, we may write N=1+∑i=0ANi with N0=0. Now, since the arrivals follow a Poisson process, Ni are independent of A and S and are also independent random variables identically distributed as N, we may write the following by conditioning N by A and S:(2)Rz=∑n=1∞PN=nzn=EzN=EEzNA,S=∫0∞∑n=0∞Ez1+∑j=1ANjA=n,S=tPA=nS=tbtdt=z∑n=0∞Rzn∫0∞λtnn!e−λtbtdt=z∫0∞e−λ−λRztbtdt=zb¯λ−λRz,which shows that the probability generating function for the number of customers served during a busy period satisfies the above functional equation.


## 2.2. Distribution of the Number of Customers Served during a Busy Period for the M/G/1 System

In this subsection, we present the derivation of the distribution of a number of customers served during a busy period for theM/G/1 system using Lagrange inversion theorem. In mathematical analysis, the Lagrange inversion theorem provides the Taylor series expansion of the inverse of an analytic function. Closed-form solutions of (2) can be found using Lagrange’s inversion theorem. Applying (1) in (2), we may rewrite as(3)Rz=zαRz,z≤1.By Lagrange inversion theorem, we get(4)Rz=∑n=1∞znn!Dzn−1αnzz=0,where Dzxyz denotes the xth derivative of yz with respect to z evaluated at z=0 and Dz0yz=yz. For a particular service-time distribution b¯s, we can easily get an expression for αz and thus an expression for PN=n. For more details, one may refer studies of Consul and Famoye [22] and Jain et al. [23]. Suppose that there are m≥1 customers at the beginning of the busy period T. Then, the PGF of the number NmT served during the busy period is given by Rmz, where Rz is the PGF of the number served during the busy period that begins by a single customer. The probability generating function of Nm is ∑n=m∞PNm=nzn=Rmz, and its Lagrange’s series expansion simplifies to(5)Rmz=∑n=m∞mznnn−m!Dzn−mαnzz=0.This provides(6)PNm=n=mnn−m!Dzn−mαnzz=0,n=m,m+1,m+2,….Equation (6) provides the distribution of a number of customers served during a busy period for the M/G/1 system when initiated with m customers. Taking into consideration various service-time distributions, we present the distribution of a number of customers served during a busy period in Section 3. In the derivation of analytical results, we also use Leibniz’s rule. The general Leibniz’s rule states that if fx and gx are n-times differentiable functions, then the product fxgx is also n-times differentiable and its nth derivative is given by(7)Dxnfxgx=∑i=0nniDxn−ifxDxigx.The formula can be generalized to the product ofk differentiable functions f1,f2,…,fk:(8)Dxnf1xf2x…fkx=∑i1+i2+…+ik=nni1,i2,…,ik∏1≤j≤kDxijfj,where the sum extends over all k-tuples i1,i2,…,ik of nonnegative integers with ∑j=1kij=n and ni1,i2,…,ik=n!/i1!i2!…ik! are the multinomial coefficients.


## 2.3. Moments of the Number Served during a Busy Period Using Functional Equation

Here, we obtain mean and variance of the number of customers served during a busy period for the queueing systemM/G/1. Let vk be the kth moment of the service-time; that is, νk=ESk=−1kdkb¯s/dsks=0 and variance σ2=ν2−ν12. Here, ν1=1/μ. Let ENr,r=1,2,… be the rth moment of the number served during a busy period when initiated by single customer. We obtain from (2) the following(9)R11=b¯0−λb¯10R11=b¯0+λν1R11,implying(10)R11≡EN=11−λν1=11−ρ,R21=−2λb¯10R11+b¯20−λR112−λb¯10R21=2λμR11+ν2−λR112+λμR21,giving(11)R21=2ρ1−ρ+λ2ν21−ρ3.SinceR21=EN2−EN, we have(12)EN2=2ρ1−ρ+λ2ν21−ρ3+11−ρ.The variance of the number of customers served during a busy period when initiated by single customer is(13)VarN=ρ1−ρ+λ2ν21−ρ3.When a busy period begins with an initial number ofm customers in the system, the transition of the underlying Markov process from m to 0 can be made up of m intervals with the same distribution corresponding the transitions from m⟶m−1,m−1⟶m−2,…,1⟶0 because of the Markovian property of the arrival process. These m-independent busy periods start with single customer in the system so that Nm=∑j=1mN1j. Hence, if Nm is the random variable representing a busy period initiated by m customers, we have(14)ENm=∑j=1mEN1j=mEN1=m1−ρ,(15)VarNm=mρ1−ρ+λ2ν21−ρ3.Note that higher order moments can be obtained similarly.


## 3. Distributions

In this section, we derive the distribution of the number of customers served during a busy period for different service-time distributions. Using Lagrange’s expansion, we find the number of customers served during a busy period when initiated withm customers for the exponential, deterministic, gamma, chi-square, mixed generalized Erlang, matrix exponential, inverse Gaussian, hyperexponential, uniform, Coxian, phase-type, Markov-modulated Poisson process, and interrupted Poisson process service-time distributions. In the literature, explicit results are available only for exponential and deterministic service-time distributions.


### 3.1. Exponential Distribution

The exponential distribution has probability density function (pdf)bt=μe−μt,t≥0, where the parameter µ is a real, positive constant. The Laplace transform (LT) is b¯s=μ/μ+s. The mean, second moment, and variance of the exponential distribution are ν1=1/μ, ν2=2/μ2, and σ2=1/μ2, respectively. Substituting the LT of exponential service-time distribution b¯s in functional equation (2), we have(16)Rz=zb¯λ−λRz=μzμ+λ−λRz.Using (16) in (5) and (6), we get(17)PNm=n=mnn−m!Dzn−mμμ+λ−λzz=0n=mnn−m!μnDzn−mμ+λ−λz−nz=0=mnn−m!μnλn−mnn+1…2n−m−1μ+λ−λz−2n+mz=0=mn2n−m−1n−1ρn−m1+ρ2n−m,n=m,m+1,…,which matches with the results of Medhi [20]. From (14) and (15), the mean and variance of the number of customers served during a busy period when initiated by m customers can be obtained as follows:(18)ENm=m1−ρ,VarNm=mρ1+ρ1−ρ3.Whenm=1, the distribution of the number of customers served during a busy period initiated by a single customer is(19)PN=n=1n2n−2n−1ρn−11+ρ2n−1.Remark.
We could have obtained mean and variance from the distribution presented above in equation (17), but we have used (14) and (15) to find the mean and variance, respectively. This is because, in simple cases like the one presented in equation (17), we can find the mean and variance easily, but in more complex situations that arise later, it is not that simple to get the mean and variance from the distributions derived there. However, the accuracy of the analytic results was checked numerically.


### 3.2. Deterministic Distribution

We consider the deterministic distribution with pdfbt=δt−k, where δ⋅ is the Dirac delta function. The LT is given by b¯s=e−s⋅k. The mean, second moment, and variance of the deterministic distribution are ν1=k, ν2=k2, and σ2=0, respectively. Substituting the LT of deterministic service-time distribution in functional equation (2), we have(20)Rz=ze−λ−λRzk.Using (20) in (5) and (6), we get(21)PNm=n=mnn−m!Dzn−me−λ−λzknz=0=mnn−m!e−λnkDzn−meλnkzz=0=mnn−m!e−λnkλnkn−m=mn−m!nn−m−1e−nρρn−mn=m,m+1,…,which is the Borel–Tanner distribution with ρ=λk. This corresponds to the distribution derived by Tanner [5] as the distribution of the number of customers served during a busy period. The mean and variance of the distribution of the number of customers served during a busy period can be obtained as(22)ENm=m1−ρ,VarNm=mρ1−ρ3,which match with the results reported in [20].


### 3.3. Erlang-k Distribution

A random variableX has an Erlang-k k=1,2,… distribution if X is the sum of k independent random variables X1,…,Xk having a common exponential distribution with mean 1/μ. The probability density function of the Erlang-k distribution is bt=μμtk−1e−μt/k−1!,μ>0andt≥0,k=1,2,… The Laplace transform is given by b¯s=μ/μ+sk. The Erlang distribution is a special case of the gamma distribution. It is defined by two parameters, µ and k, where µ is the scale parameter and k is the shape parameter which must be a positive integer. In the gamma distribution, k can be any real number, including fractions. The mean, second moment, and variance of the Erlang-k distribution are ν1=k/μ, ν2=k/μ21+k, and σ2=k/μ2, respectively. Substituting the LT of Erlang-k service-time distribution in functional equation (2), we have(23)Rz=zμμ+λ−λRzk.Using (23) in (5) and (6), we get(24)PNm=n=mnn−m!Dzn−mμμ+λ−λznkz=0=mnn−m!μnkDzn−mμ+λ−λz−nkz=0=mnnknk+1…nk+n−m−1n−m!λμn−mμμ+λnk+n−m,n=m,m+1,…=mnnk+n−m−1n−mλμn−mμμ+λnk+n−m,n=m,m+1,….The mean and variance of the number of customers served during a busy period can be obtained as(25)ENm=m1−ρ,VarNm=mρk+ρk1−ρ3,where ρ=λk/μ.


### 3.4. Gamma Distribution

Let us consider the gamma distribution with probability density functionbt=μμtγ−1e−μt/Γγ,μ>0,γ>0,t≥0, and Γγ=∫0∞tγ−1e−tdt for γ>0. The parameters are μ>0 and γ>0. The parameters γ and µ are called the shape and scale parameters, respectively. The mean, second moment, and variance of the gamma distribution are ν1=γ/μ, ν2=γ/μ21+γ, and σ2=γ/μ2, respectively. The Laplace transform is b¯s=μ/μ+sγ. It resembles the function involved in defining the Erlang-k distribution. Substituting the LT of gamma service-time distribution in functional equation (2) and using Lagrange’s inversion theorem, the distribution of the number of customers served during a busy period initiated by m customers is(26)PNm=n=mnn−m!Dzn−mμzμ+λ−λzγnz=0=mnnγ+n−m−1n−mμμ+λγnλμ+λn−m.The mean and variance of the number of customers served during a busy period can be obtained as(27)ENm=m1−ρ,VarNm=mργ+ργ1−ρ3.


### 3.5. Chi-Square Distribution

LetY1,Y2,…,Yk be normal independent random variables with mean 0 and variance 1. Then, the random variable χ2=Y12+Y22+…Yk2 has the probability density function bt=tk/2−1e−t/2/2k/2Γk/2,t>0, and is said to follow the chi-square distribution with k degrees of freedom, abbreviated χk2. The mean, second moment, and variance of the χk2 distribution are ν1=k, ν2=k2−k, and σ2=2k, respectively. The special case of the gamma distribution is the chi-square distribution by setting γ=k/2 and μ=1/2. The LT is b¯s=1/1+2sk/2. Substituting the LT of chi-square service-time distribution in functional equation (2) and using Lagrange’s inversion theorem, the distribution of the number of customers served during a busy period initiated by m customers is(28)PNm=n=mnn−m!Dzn−m11+2λ−λzk/2nz=0=mnn−m!11+2λnk/2Dzn−m1−2λz1+2λnk/2z=0=mnnk2+n−m−1n−m11+2λnk/22λ1+2λn−m.The mean and variance of the number of customers served during a busy period can be obtained as(29)ENm=m1−λk,VarNm=mkλ1+2λ1−kλ3.


### 3.6. Inverse Gaussian Distribution

We consider the inverse Gaussian distribution with pdfbt=α/2πt3e−αt−β2/2β2t and LT b¯s=eϕ1−1+2s/γ, where ϕ=α/β and γ=β2/α. The mean, second moment, and variance of the inverse Gaussian distribution are ν1=ϕ/γ, ν2=ϕ1+ϕ/γ2, and σ2=ϕ/γ2, respectively. Substituting the LT of inverse Gaussian service-time distribution in functional equation (2) and applying Lagrange’s inversion theorem, we get(30)PNm=n=mnn−m!Dzn−meϕ1−1+2λ−λz/γ1/2z=0n=mnn−m!Dzn−menϕe−nϕ1+2λ−λz/γ1/2z=0=mnn−m!enϕDzn−me−nϕ1+2λ/γ1/21−2λz/2λ+γ1/2z=0=mnn−m!enϕ∑i=0∞−nϕi1+2λ/γi/2i!Dzn−m1−2λz2λ+γi/2z=0=mnenϕ∑i=0∞−1n−m+inϕi1+2λ/γi/2i!i2n−m2λ2λ+γn−m.The mean and variance of the number of customers served during a busy period can be obtained as(31)ENm=mγγ−λϕ,VarNm=mλϕγλ+γγ−λϕ3.


### 3.7. Generalized Erlang Distribution

We assume the generalized Erlang distribution or hypoexponential distribution of a series of 2 exponential distributions each with their own ratesμ1andμ2 with probability density function bt=μ1μ2/μ1−μ2e−μ2t−e−μ1t,μ1,μ2>0,t>0, and Laplace transform b¯s=μ1μ2/s+μ1s+μ2. The mean, second moment, and variance of the Generalized Erlang distribution are ν1=1/μ1+1/μ2, ν2=2/μ12+2/μ1μ2+2/μ22, and σ2=1/μ12+1/μ22, respectively.Substituting the LT of generalized Erlang service-time distributionb¯s in functional equation (2) and applying Lagrange’s inversion theorem, we get(32)PNm=n=mnn−m!∏i=12μiμi+λnDzn−m1−λzμ1+λ−n1−λzμ2+λ−nz=0=mnn−m!∏i=12μiμi+λn∑k=0n−mn−mkDzn−m−k1−λzμ1+λ−nz=0×Dzk1−λzμ2+λ−nz=0=mn∏i=12μiμi+λn∑k=0n−mn+k−1k2n−m−k−1n−m−kλμ1+λn−m−k×λμ2+λk.The mean and variance of the number of customers served during a busy period can be obtained as(33)ENm=mμ1μ2μ1μ2−λμ1+μ2,VarNm=mλμ1μ2μ1μ2μ1+μ2+λμ12+μ22μ1μ2−λμ1+μ23.


### 3.8. Matrix Exponential Distribution

We consider the matrix exponential distribution with pdfbt=1+1/4π21−cos2πte−t and Laplace transform b¯s=a+1/s+1s+12+a, where a=4π2. The mean, second moment, and variance of the matrix exponential distribution are ν1=a+3/a+1, ν2=2a2+3a+6/a+12, and σ2=a2+3/a+12, respectively. Substituting the LT of matrix exponential service-time distribution b¯s in functional equation (2) and using Lagrange’s expansion, we have(34)PNm=n=mnn−m!Dzn−ma+1λ−λz+1λ−λz+12+anz=0=mnn−m!a+1n1+λ−n1+λ2+a−n×Dzn−m1−λz1+λ+−a−n1−λz1+λ−−a−n1−λz1+λ−nz=0=mnn−m!a+1n1+λ−n1+λ2+a−n∑i=0n−m∑j=0n−m−in−m!i!j!n−m−i−j!×Dzi1−λz1+λ+−a−nz=0Dzj1−λz1+λ−−a−nz=0×Dzn−m−i−j1−λz1+λ−nz=0=mna+11+λn1+λ2+a−n∑i=0n−m∑j=0n−m−in+i−1in+j−1j×2n−m−i−j−1n−m−i−jλ1+λ+−aiλ1+λ−−ajλ1+λn−m−i−j.The mean and variance of the number of customers served during a busy period is(35)ENm=ma+1a+1−λa+3,VarNm=ma+1λa+1a+3+λ2a2+3a+1−λa+33.


### 3.9. Hyperexponential Distribution

We assume the hyperexponential distribution with probability density functionbt=∑i=1kpiμie−μit,t>0, such that ∑i=0kpi=1 and Laplace transform b¯s=∑i=1kpiμi/μi+s. The mean, second moment, and variance of the hyperexponential distribution are ν1=∑i=1kpi/μi, ν2=∑i=1k2pi/μi2, and σ2=∑i=1kpi/μi, respectively. Taking k=2 and substituting the LT of hyperexponential service-time distribution b¯s=∑i=12piμi/μi+s in functional equation (2), we have(36)Rz=zp1μ1μ1+λ−λRz+p2μ2μ2+λ−λRz.Applying Lagrange’s expansion and Leibniz rule, we get(37)PNm=n=mnn−m!Dzn−mp1μ1μ1+λ−λz+p2μ2μ2+λ−λznz=0=mnn−m!Dzn−mp1μ1μ2+λ−λz+p2μ2μ1+λ−λzμ1+λ−λzμ2+λ−λzz=0n=mnn−m!∏i=12μi+λ−n∑i=0n−m∑j=0n−m−in−m!i!j!n−m−i−j!Dzn−m−i−j1−λzμ1+λz=0−nDzi1−λzμ2+λz=0−nDzjp1μ1μ2+λ1−λzμ2+λ+p2μ2μ1+λ1−λzμ1+λz=0n=mn∏i=12μi+λ−n∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1iλμ1+λn−m−i−jλμ2+λi∑k=0nnkp1μ1μ2+λn−k−1jp2μ2μ1+λk∑l=0jkj−lλμ1+λj−ln−klλμ2+λl.The mean and variance of the number of customers served during a busy period can be obtained from as(38)ENm=mμ1μ2μ1μ2−λp1μ2+p2μ1,VarNm=mλμ1μ2p1μ2+p2μ1μ1μ2−λp1μ2−λp2μ1+2λp1μ22+p2μ12μ1μ2−λp1μ2−λp2μ13.


### 3.10. Uniform Distribution

We assume the uniform distribution with probability density functionbt=1/b−a,a≤t≤b, where a and b are real constants with a<b. The Laplace transform is b¯s=e−as−e−bs/b−as. The mean, second moment, and variance of the uniform distribution are ν1=a+b/2, ν2=b2+ab+a2/3, and σ2=b−a2/12, respectively. Substituting the LT of uniform service-time distribution b¯s in functional equation (2) and applying Lagrange’s inversion theorem, we get(39)PNm=n=mnn−m!Dzn−me−aλ−λz−e−bλ−λzb−aλ−λzz=0n=mnn−m!b−a−nλ−nDzn−me−aλ−λz−e−bλ−λzn1−z−nz=0=mnn−m!b−a−nλ−ne−aλnDzn−meaλnz1−e−b−aλ1−zn1−z−nz=0=mnn−m!b−a−nλ−ne−aλn∑i=0n−m∑j=0n−m−in−m!i!j!n−m−i−j!×Dzieaλnzz=0Dzj1−e−b−aλ1−zz=0nDzn−m−i−j1−zz=0−n=mb−a−nλ−ne−aλnn∑i=0n−maλnii!∑j=0n−m−i2n−m−i−j−1n−m−i−j×∑k=0nnk−1ke−b−aλkb−aλkjj!.The mean and variance of the number of customers served during a busy period can be obtained as(40)ENm=2m2−λa+b,VarNm=2mλ6a+b+λb−a232−λa+b3.


### 3.11. Two-Phase Coxian Distribution: Cox-C2

The Coxian distribution is a generalization of the hypoexponential distribution. Instead of only being able to get into the absorbing state from statek, it can be arrived from any phase. Here, we assume that bt follows the two-phase Coxian distribution and its Laplace transform is b¯s=1−αμ1/μ1+s+αμ1μ2/μ1+sμ2+s. The mean, second moment, and variance of the C2 distribution are ν1=1/μ1+α/μ2, ν2=2/μ12+2α/μ22+2α/μ1μ2, and σ2=μ22+αμ122−α/μ12μ22, respectively.Substituting the LT of CoxC2 service-time distribution b¯s in functional equation (2) and using Lagrange’s inversion theorem and Leibniz’s rule, we have(41)PNm=n=mnn−m!Dzn−m1−αμ1μ1+λ−λz+αμ1μ2μ1+λ−λzμ2+λ−λznz=0=mnn−m!Dzn−m1−αμ1μ2+λ−λz+αμ1μ2μ1+λ−λzμ2+λ−λzz=0n=mnn−m!∏i=12μi+λ−nDzn−m1−λzμ1+λ−n1−λzμ2+λ−n×μ1μ2+1−αμ1λ−1−αμ1λznz=0=mnn−m!∏i=12μi+λ−nμ1nμ2+1−αλn∑i=0n−m∑j=0n−m−in−m!i!j!n−m−i−j!×Dzn−m−i−j1−λzμ1+λz=0−nDzi1−λzμ2+λz=0−nDzj1−1−αλzμ2+1−αλz=0n=mn∏i=12μi+λ−nμ1nμ2+1−αλn∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1i×nj−1jλμ1+λn−m−i−jλμ2+λi1−αλμ2+1−αλj.The mean and variance of the number of customers served during a busy period can be given as(42)ENm=mμ1μ2μ1μ2−λμ2+αμ1,VarNm=mλμ1μ2μ22λ+μ1+αμ12μ2+2λ−αλμ1μ2−λμ2+αμ13.


### 3.12. Phase-Type Distribution

A phase-type (PH) distribution is constructed by a mixture of exponentially distributed phases. It results from a system of one or more inter-related Poisson processes occurring in sequence, or phases. Letω denote the number of phases. The process begins in phase i with probability αi,i=1,2,…,ω. Let μi,i=1,2,…,ω be the transition rate out of phase i. After completion of a phase, the process may join the absorbing state or continue with a nonabsorbing phase. Let riω be the probability of transition from phase i to phase j, i,j=1,2,…,ω, and let ri0 be the probability of transition from phase i to the absorbing state, i=1,2,…,ω. The parameters must satisfy the following conditions: ∑i=1ωαi=1andri0=1−∑j=1ωrij,i=1,2,…,ω. Let the row vector α denote the ω-dimensional vector of the initial state distribution of the underlying Markov process, the square matrix T denote the sub-generator of the underlying Markov process held to the transient states 1,…,ω. Let r=T1 denote the ω-dimensional column vector of transition rates to the absorbing state. Hence,(43)α=α1α2⋯αω,T=−μ1μ1r12…μ1r1ωμ2r21−μ2…μ2r2ω⋮⋮⋱⋮μωrω1μωrω2…−μω,r=−μ1r10−μ2r20⋮−μωrω0.The LT of phase-type distribution for the caseω=2, that is, PH2 is given by(44)b¯s=α1r10μ1+α2r20μ2s+1−r12r21μ1μ2s+μ1s+μ2−r12r21μ1μ2,s≥0.Substituting the LT ofPH2 service-time distribution b¯s in functional equation (2) and using Lagrange’s inversion theorem, we get(45)PNm=n=mnn−m!Dzn−mα1r10μ1+α2r20μ2λ−λz+1−r12r21μ1μ2λ−λz+μ1λ−λz+μ2−r12r21μ1μ2z=0n=mnn−m!Dzn−mα1r10μ1+α2r20μ2λ−λz+1−r12r21μ1μ2λ2ψ1−zψ2−zz=0n=mnn−m!K1nλ2ψ1ψ2−nDzn−m1−zψ1−n1−zψ2−n1−K2znz=0=mnK1nλ2ψ1ψ2−n∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1i×nj−1jψ1m+i+j−nψ2−iK2j,where(46)ψ1,2=μ1+μ2+2λ±μ1−μ22+4μ1μ2r12r212λ,K1=λα1r10μ1+α2r20μ2+μ1μ21−r12r21,K2=λα1r10μ1+α2r20μ2K1.The mean and second moment of thePH2 distribution are ν1=α1μ2+μ1r12+α2μ1+μ2r21/μ1μ21−r12r21 and ν2=2μ1α1r12+μ2α2r21−μ1α1−μ2α2/μ1μ21−r12r21, respectively. Here, ρ=λν1. Thus, the mean and variance of the number of customers served during a busy period is(47)ENm=mμ1μ21−r12r21μ1μ21−r12r21−λα1μ2+μ1r12+α2μ1+μ2r21,VarNm=mρ1−ρ+λ2ν21−ρ3.


### 3.13. Markov-Modulated Poisson Process

A Markov-modulated Poisson process (MMPP) is a doubly stochastic Poisson process whose intensities vary according to the state of a Markov chain. An MMPP can be categorized by the number of states the modulating Markov chain contains, for example, a Markov chain with two states and two intensities is denoted MMPP-2, or sometimes switched Poisson process (SPP). If the two intensities are equal, the model becomes an ordinary Poisson process. A two-state MMPP has only two states in its underlying modulating Markov chain, described by(48)Q=−q1q1q2−q2,M=μ100μ2.The Laplace transform is given by(49)b¯s=sq1μ22+q2μ12+q1μ2+q2μ1q1μ2+μ1μ2+q2μ1q1μ2+q2μ1s2+q1+q2+μ1+μ2s+q1μ2+μ1μ2+q2μ1.Substituting the LT of MMPP-2 service-time distributionb¯s in functional equation (2) and applying Lagrange’s inversion theorem, we get(50)PNm=n=mnn−m!K1nλ2ψ1ψ2−nDzn−m1−zψ1−n1−zψ2−n1−K2znz=0=mnK1nλ2ψ1ψ2−n∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1i×nj−1jψ1m+i+j−nψ2−iK2j,where(51)ψ1,2=q1+q2+μ1+μ2+2λ±q1+q2+μ1+μ22−4μ1μ2+q1μ2+q2μ12λ,K1=λq1μ22+q2μ12+q1μ2+q2μ1q1μ2+μ1μ2+q2μ1,K2=λq1μ22+q2μ12K1.The mean and second moment of the MMPP-2 distribution areν1=q12μ2+q22μ1+q1q2μ1+μ2+μ1μ2q1+q2/q1μ2+μ1μ2+q2μ1 and ν2=2q1+q2+μ1+μ22/q1μ2+μ1μ2+q2μ12−q12μ2+q22μ1q1+q2+μ1+μ2/q1μ2+q2μ1q1μ2+μ1μ2+q2μ12−1/q1μ2+μ1μ2+q2μ1, respectively. The mean and variance of the number of customers served during a busy period when initiated by m customers can be obtained from (14) and (23).


### 3.14. Interrupted Poisson Distribution

The interrupted Poisson distribution (IPP) is a special case of the two-state MMPP, described by the following two matrices:(52)Q=−q1q1q2−q2,M=μ00μ.The Laplace transform is given byb¯s=s+q2μ/s2+q1+q2+μs+q2μ. An IPP is stochastically equivalent to a hyperexponential distribution [24]. The following parameter transformations relate the IPP to the hyperexponential distribution:(53)IPP⟶H2,μ1=12μ+q1+q2+μ+q1+q22−4μq2,μ2=12μ+q1+q2−μ+q1+q22−4μq2,p1=μ−μ2μ1−μ2,p2=1−p1.Similarly, the following parameter transformations relate the hyperexponential to the IPP distribution:(54)H2⟶IPP,μ=p1μ1+p2μ2,q1=p1p2μ1−μ22μ,q2=μ1μ2μ.Substituting the LT of IPP service-time distributionb¯s in functional equation (2) and applying Lagrange’s inversion theorem, we get the distribution of the number of customers served during a busy period when initiated by m customers as follows:(55)PNm=n=mnK1nλ2ψ1ψ2−n∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1i×nj−1jψ1m+i+j−nψ2−iK2j,where(56)ψ1,2=q1+q2+μ+2λ±q1+q2+μ2−4q2μ2λ,K1=q2μ2λ+q2,K2=λλ+q2.The mean, second moment, and variance of the IPP distribution areν1=q1+q2/μq2, ν2=2μq1+q1+q22/μq22, and σ2=2μq1+q1+q22/μq22, respectively. The mean and variance of the distribution of the number of customers served during a busy period when initiated by m customers can be obtained from (14) and (15). The mean and variance of the number of customers served during a busy period can be obtained as(57)ENm=mμq2μq2−λq1+q2,VarNm=mλμq2q1+q2μq2+λq1+q2+2λμq1μq2−λq1+q23.


## 3.1. Exponential Distribution

The exponential distribution has probability density function (pdf)bt=μe−μt,t≥0, where the parameter µ is a real, positive constant. The Laplace transform (LT) is b¯s=μ/μ+s. The mean, second moment, and variance of the exponential distribution are ν1=1/μ, ν2=2/μ2, and σ2=1/μ2, respectively. Substituting the LT of exponential service-time distribution b¯s in functional equation (2), we have(16)Rz=zb¯λ−λRz=μzμ+λ−λRz.Using (16) in (5) and (6), we get(17)PNm=n=mnn−m!Dzn−mμμ+λ−λzz=0n=mnn−m!μnDzn−mμ+λ−λz−nz=0=mnn−m!μnλn−mnn+1…2n−m−1μ+λ−λz−2n+mz=0=mn2n−m−1n−1ρn−m1+ρ2n−m,n=m,m+1,…,which matches with the results of Medhi [20]. From (14) and (15), the mean and variance of the number of customers served during a busy period when initiated by m customers can be obtained as follows:(18)ENm=m1−ρ,VarNm=mρ1+ρ1−ρ3.Whenm=1, the distribution of the number of customers served during a busy period initiated by a single customer is(19)PN=n=1n2n−2n−1ρn−11+ρ2n−1.Remark.
We could have obtained mean and variance from the distribution presented above in equation (17), but we have used (14) and (15) to find the mean and variance, respectively. This is because, in simple cases like the one presented in equation (17), we can find the mean and variance easily, but in more complex situations that arise later, it is not that simple to get the mean and variance from the distributions derived there. However, the accuracy of the analytic results was checked numerically.


## 3.2. Deterministic Distribution

We consider the deterministic distribution with pdfbt=δt−k, where δ⋅ is the Dirac delta function. The LT is given by b¯s=e−s⋅k. The mean, second moment, and variance of the deterministic distribution are ν1=k, ν2=k2, and σ2=0, respectively. Substituting the LT of deterministic service-time distribution in functional equation (2), we have(20)Rz=ze−λ−λRzk.Using (20) in (5) and (6), we get(21)PNm=n=mnn−m!Dzn−me−λ−λzknz=0=mnn−m!e−λnkDzn−meλnkzz=0=mnn−m!e−λnkλnkn−m=mn−m!nn−m−1e−nρρn−mn=m,m+1,…,which is the Borel–Tanner distribution with ρ=λk. This corresponds to the distribution derived by Tanner [5] as the distribution of the number of customers served during a busy period. The mean and variance of the distribution of the number of customers served during a busy period can be obtained as(22)ENm=m1−ρ,VarNm=mρ1−ρ3,which match with the results reported in [20].


## 3.3. Erlang-k Distribution

A random variableX has an Erlang-k k=1,2,… distribution if X is the sum of k independent random variables X1,…,Xk having a common exponential distribution with mean 1/μ. The probability density function of the Erlang-k distribution is bt=μμtk−1e−μt/k−1!,μ>0andt≥0,k=1,2,… The Laplace transform is given by b¯s=μ/μ+sk. The Erlang distribution is a special case of the gamma distribution. It is defined by two parameters, µ and k, where µ is the scale parameter and k is the shape parameter which must be a positive integer. In the gamma distribution, k can be any real number, including fractions. The mean, second moment, and variance of the Erlang-k distribution are ν1=k/μ, ν2=k/μ21+k, and σ2=k/μ2, respectively. Substituting the LT of Erlang-k service-time distribution in functional equation (2), we have(23)Rz=zμμ+λ−λRzk.Using (23) in (5) and (6), we get(24)PNm=n=mnn−m!Dzn−mμμ+λ−λznkz=0=mnn−m!μnkDzn−mμ+λ−λz−nkz=0=mnnknk+1…nk+n−m−1n−m!λμn−mμμ+λnk+n−m,n=m,m+1,…=mnnk+n−m−1n−mλμn−mμμ+λnk+n−m,n=m,m+1,….The mean and variance of the number of customers served during a busy period can be obtained as(25)ENm=m1−ρ,VarNm=mρk+ρk1−ρ3,where ρ=λk/μ.


## 3.4. Gamma Distribution

Let us consider the gamma distribution with probability density functionbt=μμtγ−1e−μt/Γγ,μ>0,γ>0,t≥0, and Γγ=∫0∞tγ−1e−tdt for γ>0. The parameters are μ>0 and γ>0. The parameters γ and µ are called the shape and scale parameters, respectively. The mean, second moment, and variance of the gamma distribution are ν1=γ/μ, ν2=γ/μ21+γ, and σ2=γ/μ2, respectively. The Laplace transform is b¯s=μ/μ+sγ. It resembles the function involved in defining the Erlang-k distribution. Substituting the LT of gamma service-time distribution in functional equation (2) and using Lagrange’s inversion theorem, the distribution of the number of customers served during a busy period initiated by m customers is(26)PNm=n=mnn−m!Dzn−mμzμ+λ−λzγnz=0=mnnγ+n−m−1n−mμμ+λγnλμ+λn−m.The mean and variance of the number of customers served during a busy period can be obtained as(27)ENm=m1−ρ,VarNm=mργ+ργ1−ρ3.


## 3.5. Chi-Square Distribution

LetY1,Y2,…,Yk be normal independent random variables with mean 0 and variance 1. Then, the random variable χ2=Y12+Y22+…Yk2 has the probability density function bt=tk/2−1e−t/2/2k/2Γk/2,t>0, and is said to follow the chi-square distribution with k degrees of freedom, abbreviated χk2. The mean, second moment, and variance of the χk2 distribution are ν1=k, ν2=k2−k, and σ2=2k, respectively. The special case of the gamma distribution is the chi-square distribution by setting γ=k/2 and μ=1/2. The LT is b¯s=1/1+2sk/2. Substituting the LT of chi-square service-time distribution in functional equation (2) and using Lagrange’s inversion theorem, the distribution of the number of customers served during a busy period initiated by m customers is(28)PNm=n=mnn−m!Dzn−m11+2λ−λzk/2nz=0=mnn−m!11+2λnk/2Dzn−m1−2λz1+2λnk/2z=0=mnnk2+n−m−1n−m11+2λnk/22λ1+2λn−m.The mean and variance of the number of customers served during a busy period can be obtained as(29)ENm=m1−λk,VarNm=mkλ1+2λ1−kλ3.


## 3.6. Inverse Gaussian Distribution

We consider the inverse Gaussian distribution with pdfbt=α/2πt3e−αt−β2/2β2t and LT b¯s=eϕ1−1+2s/γ, where ϕ=α/β and γ=β2/α. The mean, second moment, and variance of the inverse Gaussian distribution are ν1=ϕ/γ, ν2=ϕ1+ϕ/γ2, and σ2=ϕ/γ2, respectively. Substituting the LT of inverse Gaussian service-time distribution in functional equation (2) and applying Lagrange’s inversion theorem, we get(30)PNm=n=mnn−m!Dzn−meϕ1−1+2λ−λz/γ1/2z=0n=mnn−m!Dzn−menϕe−nϕ1+2λ−λz/γ1/2z=0=mnn−m!enϕDzn−me−nϕ1+2λ/γ1/21−2λz/2λ+γ1/2z=0=mnn−m!enϕ∑i=0∞−nϕi1+2λ/γi/2i!Dzn−m1−2λz2λ+γi/2z=0=mnenϕ∑i=0∞−1n−m+inϕi1+2λ/γi/2i!i2n−m2λ2λ+γn−m.The mean and variance of the number of customers served during a busy period can be obtained as(31)ENm=mγγ−λϕ,VarNm=mλϕγλ+γγ−λϕ3.


## 3.7. Generalized Erlang Distribution

We assume the generalized Erlang distribution or hypoexponential distribution of a series of 2 exponential distributions each with their own ratesμ1andμ2 with probability density function bt=μ1μ2/μ1−μ2e−μ2t−e−μ1t,μ1,μ2>0,t>0, and Laplace transform b¯s=μ1μ2/s+μ1s+μ2. The mean, second moment, and variance of the Generalized Erlang distribution are ν1=1/μ1+1/μ2, ν2=2/μ12+2/μ1μ2+2/μ22, and σ2=1/μ12+1/μ22, respectively.Substituting the LT of generalized Erlang service-time distributionb¯s in functional equation (2) and applying Lagrange’s inversion theorem, we get(32)PNm=n=mnn−m!∏i=12μiμi+λnDzn−m1−λzμ1+λ−n1−λzμ2+λ−nz=0=mnn−m!∏i=12μiμi+λn∑k=0n−mn−mkDzn−m−k1−λzμ1+λ−nz=0×Dzk1−λzμ2+λ−nz=0=mn∏i=12μiμi+λn∑k=0n−mn+k−1k2n−m−k−1n−m−kλμ1+λn−m−k×λμ2+λk.The mean and variance of the number of customers served during a busy period can be obtained as(33)ENm=mμ1μ2μ1μ2−λμ1+μ2,VarNm=mλμ1μ2μ1μ2μ1+μ2+λμ12+μ22μ1μ2−λμ1+μ23.


## 3.8. Matrix Exponential Distribution

We consider the matrix exponential distribution with pdfbt=1+1/4π21−cos2πte−t and Laplace transform b¯s=a+1/s+1s+12+a, where a=4π2. The mean, second moment, and variance of the matrix exponential distribution are ν1=a+3/a+1, ν2=2a2+3a+6/a+12, and σ2=a2+3/a+12, respectively. Substituting the LT of matrix exponential service-time distribution b¯s in functional equation (2) and using Lagrange’s expansion, we have(34)PNm=n=mnn−m!Dzn−ma+1λ−λz+1λ−λz+12+anz=0=mnn−m!a+1n1+λ−n1+λ2+a−n×Dzn−m1−λz1+λ+−a−n1−λz1+λ−−a−n1−λz1+λ−nz=0=mnn−m!a+1n1+λ−n1+λ2+a−n∑i=0n−m∑j=0n−m−in−m!i!j!n−m−i−j!×Dzi1−λz1+λ+−a−nz=0Dzj1−λz1+λ−−a−nz=0×Dzn−m−i−j1−λz1+λ−nz=0=mna+11+λn1+λ2+a−n∑i=0n−m∑j=0n−m−in+i−1in+j−1j×2n−m−i−j−1n−m−i−jλ1+λ+−aiλ1+λ−−ajλ1+λn−m−i−j.The mean and variance of the number of customers served during a busy period is(35)ENm=ma+1a+1−λa+3,VarNm=ma+1λa+1a+3+λ2a2+3a+1−λa+33.


## 3.9. Hyperexponential Distribution

We assume the hyperexponential distribution with probability density functionbt=∑i=1kpiμie−μit,t>0, such that ∑i=0kpi=1 and Laplace transform b¯s=∑i=1kpiμi/μi+s. The mean, second moment, and variance of the hyperexponential distribution are ν1=∑i=1kpi/μi, ν2=∑i=1k2pi/μi2, and σ2=∑i=1kpi/μi, respectively. Taking k=2 and substituting the LT of hyperexponential service-time distribution b¯s=∑i=12piμi/μi+s in functional equation (2), we have(36)Rz=zp1μ1μ1+λ−λRz+p2μ2μ2+λ−λRz.Applying Lagrange’s expansion and Leibniz rule, we get(37)PNm=n=mnn−m!Dzn−mp1μ1μ1+λ−λz+p2μ2μ2+λ−λznz=0=mnn−m!Dzn−mp1μ1μ2+λ−λz+p2μ2μ1+λ−λzμ1+λ−λzμ2+λ−λzz=0n=mnn−m!∏i=12μi+λ−n∑i=0n−m∑j=0n−m−in−m!i!j!n−m−i−j!Dzn−m−i−j1−λzμ1+λz=0−nDzi1−λzμ2+λz=0−nDzjp1μ1μ2+λ1−λzμ2+λ+p2μ2μ1+λ1−λzμ1+λz=0n=mn∏i=12μi+λ−n∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1iλμ1+λn−m−i−jλμ2+λi∑k=0nnkp1μ1μ2+λn−k−1jp2μ2μ1+λk∑l=0jkj−lλμ1+λj−ln−klλμ2+λl.The mean and variance of the number of customers served during a busy period can be obtained from as(38)ENm=mμ1μ2μ1μ2−λp1μ2+p2μ1,VarNm=mλμ1μ2p1μ2+p2μ1μ1μ2−λp1μ2−λp2μ1+2λp1μ22+p2μ12μ1μ2−λp1μ2−λp2μ13.


## 3.10. Uniform Distribution

We assume the uniform distribution with probability density functionbt=1/b−a,a≤t≤b, where a and b are real constants with a<b. The Laplace transform is b¯s=e−as−e−bs/b−as. The mean, second moment, and variance of the uniform distribution are ν1=a+b/2, ν2=b2+ab+a2/3, and σ2=b−a2/12, respectively. Substituting the LT of uniform service-time distribution b¯s in functional equation (2) and applying Lagrange’s inversion theorem, we get(39)PNm=n=mnn−m!Dzn−me−aλ−λz−e−bλ−λzb−aλ−λzz=0n=mnn−m!b−a−nλ−nDzn−me−aλ−λz−e−bλ−λzn1−z−nz=0=mnn−m!b−a−nλ−ne−aλnDzn−meaλnz1−e−b−aλ1−zn1−z−nz=0=mnn−m!b−a−nλ−ne−aλn∑i=0n−m∑j=0n−m−in−m!i!j!n−m−i−j!×Dzieaλnzz=0Dzj1−e−b−aλ1−zz=0nDzn−m−i−j1−zz=0−n=mb−a−nλ−ne−aλnn∑i=0n−maλnii!∑j=0n−m−i2n−m−i−j−1n−m−i−j×∑k=0nnk−1ke−b−aλkb−aλkjj!.The mean and variance of the number of customers served during a busy period can be obtained as(40)ENm=2m2−λa+b,VarNm=2mλ6a+b+λb−a232−λa+b3.


## 3.11. Two-Phase Coxian Distribution: Cox-C2

The Coxian distribution is a generalization of the hypoexponential distribution. Instead of only being able to get into the absorbing state from statek, it can be arrived from any phase. Here, we assume that bt follows the two-phase Coxian distribution and its Laplace transform is b¯s=1−αμ1/μ1+s+αμ1μ2/μ1+sμ2+s. The mean, second moment, and variance of the C2 distribution are ν1=1/μ1+α/μ2, ν2=2/μ12+2α/μ22+2α/μ1μ2, and σ2=μ22+αμ122−α/μ12μ22, respectively.Substituting the LT of CoxC2 service-time distribution b¯s in functional equation (2) and using Lagrange’s inversion theorem and Leibniz’s rule, we have(41)PNm=n=mnn−m!Dzn−m1−αμ1μ1+λ−λz+αμ1μ2μ1+λ−λzμ2+λ−λznz=0=mnn−m!Dzn−m1−αμ1μ2+λ−λz+αμ1μ2μ1+λ−λzμ2+λ−λzz=0n=mnn−m!∏i=12μi+λ−nDzn−m1−λzμ1+λ−n1−λzμ2+λ−n×μ1μ2+1−αμ1λ−1−αμ1λznz=0=mnn−m!∏i=12μi+λ−nμ1nμ2+1−αλn∑i=0n−m∑j=0n−m−in−m!i!j!n−m−i−j!×Dzn−m−i−j1−λzμ1+λz=0−nDzi1−λzμ2+λz=0−nDzj1−1−αλzμ2+1−αλz=0n=mn∏i=12μi+λ−nμ1nμ2+1−αλn∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1i×nj−1jλμ1+λn−m−i−jλμ2+λi1−αλμ2+1−αλj.The mean and variance of the number of customers served during a busy period can be given as(42)ENm=mμ1μ2μ1μ2−λμ2+αμ1,VarNm=mλμ1μ2μ22λ+μ1+αμ12μ2+2λ−αλμ1μ2−λμ2+αμ13.


## 3.12. Phase-Type Distribution

A phase-type (PH) distribution is constructed by a mixture of exponentially distributed phases. It results from a system of one or more inter-related Poisson processes occurring in sequence, or phases. Letω denote the number of phases. The process begins in phase i with probability αi,i=1,2,…,ω. Let μi,i=1,2,…,ω be the transition rate out of phase i. After completion of a phase, the process may join the absorbing state or continue with a nonabsorbing phase. Let riω be the probability of transition from phase i to phase j, i,j=1,2,…,ω, and let ri0 be the probability of transition from phase i to the absorbing state, i=1,2,…,ω. The parameters must satisfy the following conditions: ∑i=1ωαi=1andri0=1−∑j=1ωrij,i=1,2,…,ω. Let the row vector α denote the ω-dimensional vector of the initial state distribution of the underlying Markov process, the square matrix T denote the sub-generator of the underlying Markov process held to the transient states 1,…,ω. Let r=T1 denote the ω-dimensional column vector of transition rates to the absorbing state. Hence,(43)α=α1α2⋯αω,T=−μ1μ1r12…μ1r1ωμ2r21−μ2…μ2r2ω⋮⋮⋱⋮μωrω1μωrω2…−μω,r=−μ1r10−μ2r20⋮−μωrω0.The LT of phase-type distribution for the caseω=2, that is, PH2 is given by(44)b¯s=α1r10μ1+α2r20μ2s+1−r12r21μ1μ2s+μ1s+μ2−r12r21μ1μ2,s≥0.Substituting the LT ofPH2 service-time distribution b¯s in functional equation (2) and using Lagrange’s inversion theorem, we get(45)PNm=n=mnn−m!Dzn−mα1r10μ1+α2r20μ2λ−λz+1−r12r21μ1μ2λ−λz+μ1λ−λz+μ2−r12r21μ1μ2z=0n=mnn−m!Dzn−mα1r10μ1+α2r20μ2λ−λz+1−r12r21μ1μ2λ2ψ1−zψ2−zz=0n=mnn−m!K1nλ2ψ1ψ2−nDzn−m1−zψ1−n1−zψ2−n1−K2znz=0=mnK1nλ2ψ1ψ2−n∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1i×nj−1jψ1m+i+j−nψ2−iK2j,where(46)ψ1,2=μ1+μ2+2λ±μ1−μ22+4μ1μ2r12r212λ,K1=λα1r10μ1+α2r20μ2+μ1μ21−r12r21,K2=λα1r10μ1+α2r20μ2K1.The mean and second moment of thePH2 distribution are ν1=α1μ2+μ1r12+α2μ1+μ2r21/μ1μ21−r12r21 and ν2=2μ1α1r12+μ2α2r21−μ1α1−μ2α2/μ1μ21−r12r21, respectively. Here, ρ=λν1. Thus, the mean and variance of the number of customers served during a busy period is(47)ENm=mμ1μ21−r12r21μ1μ21−r12r21−λα1μ2+μ1r12+α2μ1+μ2r21,VarNm=mρ1−ρ+λ2ν21−ρ3.


## 3.13. Markov-Modulated Poisson Process

A Markov-modulated Poisson process (MMPP) is a doubly stochastic Poisson process whose intensities vary according to the state of a Markov chain. An MMPP can be categorized by the number of states the modulating Markov chain contains, for example, a Markov chain with two states and two intensities is denoted MMPP-2, or sometimes switched Poisson process (SPP). If the two intensities are equal, the model becomes an ordinary Poisson process. A two-state MMPP has only two states in its underlying modulating Markov chain, described by(48)Q=−q1q1q2−q2,M=μ100μ2.The Laplace transform is given by(49)b¯s=sq1μ22+q2μ12+q1μ2+q2μ1q1μ2+μ1μ2+q2μ1q1μ2+q2μ1s2+q1+q2+μ1+μ2s+q1μ2+μ1μ2+q2μ1.Substituting the LT of MMPP-2 service-time distributionb¯s in functional equation (2) and applying Lagrange’s inversion theorem, we get(50)PNm=n=mnn−m!K1nλ2ψ1ψ2−nDzn−m1−zψ1−n1−zψ2−n1−K2znz=0=mnK1nλ2ψ1ψ2−n∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1i×nj−1jψ1m+i+j−nψ2−iK2j,where(51)ψ1,2=q1+q2+μ1+μ2+2λ±q1+q2+μ1+μ22−4μ1μ2+q1μ2+q2μ12λ,K1=λq1μ22+q2μ12+q1μ2+q2μ1q1μ2+μ1μ2+q2μ1,K2=λq1μ22+q2μ12K1.The mean and second moment of the MMPP-2 distribution areν1=q12μ2+q22μ1+q1q2μ1+μ2+μ1μ2q1+q2/q1μ2+μ1μ2+q2μ1 and ν2=2q1+q2+μ1+μ22/q1μ2+μ1μ2+q2μ12−q12μ2+q22μ1q1+q2+μ1+μ2/q1μ2+q2μ1q1μ2+μ1μ2+q2μ12−1/q1μ2+μ1μ2+q2μ1, respectively. The mean and variance of the number of customers served during a busy period when initiated by m customers can be obtained from (14) and (23).


## 3.14. Interrupted Poisson Distribution

The interrupted Poisson distribution (IPP) is a special case of the two-state MMPP, described by the following two matrices:(52)Q=−q1q1q2−q2,M=μ00μ.The Laplace transform is given byb¯s=s+q2μ/s2+q1+q2+μs+q2μ. An IPP is stochastically equivalent to a hyperexponential distribution [24]. The following parameter transformations relate the IPP to the hyperexponential distribution:(53)IPP⟶H2,μ1=12μ+q1+q2+μ+q1+q22−4μq2,μ2=12μ+q1+q2−μ+q1+q22−4μq2,p1=μ−μ2μ1−μ2,p2=1−p1.Similarly, the following parameter transformations relate the hyperexponential to the IPP distribution:(54)H2⟶IPP,μ=p1μ1+p2μ2,q1=p1p2μ1−μ22μ,q2=μ1μ2μ.Substituting the LT of IPP service-time distributionb¯s in functional equation (2) and applying Lagrange’s inversion theorem, we get the distribution of the number of customers served during a busy period when initiated by m customers as follows:(55)PNm=n=mnK1nλ2ψ1ψ2−n∑i=0n−m∑j=0n−m−i2n−m−i−j−1n−m−i−jn+i−1i×nj−1jψ1m+i+j−nψ2−iK2j,where(56)ψ1,2=q1+q2+μ+2λ±q1+q2+μ2−4q2μ2λ,K1=q2μ2λ+q2,K2=λλ+q2.The mean, second moment, and variance of the IPP distribution areν1=q1+q2/μq2, ν2=2μq1+q1+q22/μq22, and σ2=2μq1+q1+q22/μq22, respectively. The mean and variance of the distribution of the number of customers served during a busy period when initiated by m customers can be obtained from (14) and (15). The mean and variance of the number of customers served during a busy period can be obtained as(57)ENm=mμq2μq2−λq1+q2,VarNm=mλμq2q1+q2μq2+λq1+q2+2λμq1μq2−λq1+q23.


## 4. Numerical Results

In this section, we present numerical results, for a few cases, for the distribution of the number served during a busy period when initiated bym customers in Tables 1–5. We consider that service-times are distributed according to an exponential, deterministic, generalized Erlang, hyperexponential, matrix exponential, chi-square, phase-type, Coxian, Markov-modulated Poisson process, and interrupted Poisson process. The mean and variance of the number of customers served during a busy period were calculated using both the distributions as well as the corresponding formulas. The matching of the results was excellent. The behavior of the probability distribution is basically the same for various service-time distributions. The most frequent number of customers served during the busy period is when initiated with less number of customers. The mean and variance of the number of customers served during a busy period increase with the increase of m, where m is the number that initiates the busy period.Table 1
Distribution of the number of customers served during a busy period.
n
Exponential distribution
Deterministic distribution
ρ
=
0.75, λ=0.2, μ=0.8
ρ
=
0.4, λ=0.2, k=2
m
=
3
m
=
7
m
=
9
m
=
3
m
=
7
m
=
9
P
n
P
n
P
n
P
n
P
n
P
n
3
0.512000
0.301194
4
0.245760
0.242276
5
0.117965
0.162402
6
0.058720
0.104507
7
0.030199
0.209715
0.066745
0.060810
8
0.015945
0.234881
0.042742
0.114134
9
0.008599
0.187905
0.134218
0.027536
0.137712
0.027324
10
0.004717
0.132285
0.193274
0.017862
0.136757
0.065936
20
0.000019
0.001148
0.005804
0.000319
0.010366
0.032485
30
0.000000
0.000009
0.000052
0.000008
0.000367
0.001660
⋮
⋮
⋮
⋮
⋮
⋮
⋮
E
N
4.000
9.3333
12.000
5.000
11.6667
15.000
E
N
2
18.222
92.296
150.67
30.556
149.074
241.667
Var
N
2.2222
5.1852
6.6667
5.5556
12.963
16.6667Table 2
Distribution of the number of customers served during a busy period.
n
Generalized Erlang distribution
Hyperexponential distribution
ρ
=
0.75, λ=0.5, μ1=1, μ2=2
ρ
=
0.4, λ=0.5, μ1=1, μ2=2, p1=0.6, p2=0.4
m
=
3
m
=
7
m
=
9
m
=
3
m
=
7
m
=
9
P
n
P
n
P
n
P
n
P
n
P
n
3
0.151704
0.373248
4
0.129454
0.220963
5
0.101837
0.133358
6
0.080573
0.084303
7
0.064895
0.012274
0.055373
0.100306
8
0.053219
0.024439
0.037470
0.138556
9
0.044348
0.033129
0.003491
0.025955
0.138307
0.051999
10
0.037468
0.038404
0.008938
0.018318
0.122466
0.092350
20
0.010875
0.029860
0.030692
0.000971
0.012428
0.029950
50
0.001310
0.005343
0.008297
0.000001
0.000014
0.000044
100
0.000129
0.000588
0.001014
0.000000
0.000000
0.000000
200
0.000003
0.000017
0.000030
0.000000
0.000000
0.000000
⋮
⋮
⋮
⋮
⋮
⋮
⋮
E
N
12.000
28.000
36.000
5.000
11.667
15.000
E
N
2
348.00
1260.00
1908.00
33.194
155.231
249.583
Var
N
204.00
476.000
612.00
8.1944
19.1204
24.5833Table 3
Distribution of the number of customers served during a busy period.
n
Matrix exponential distribution
Chi-square distribution
ρ
=
0.42, λ=0.4, μ=1.049
ρ
=
0.6, λ=0.2, k=3
m
=
3
m
=
7
m
=
9
m
=
3
m
=
7
m
=
9
P
n
P
n
P
n
P
n
P
n
P
n
3
0.339685
0.220001
4
0.222371
0.170756
5
0.139825
0.125171
6
0.090430
0.093132
7
0.060262
0.080509
0.070848
0.029218
8
0.041179
0.122977
0.055009
0.052916
9
0.028722
0.130997
0.039195
0.043465
0.066170
0.010648
10
0.020371
0.121409
0.076975
0.034852
0.071365
0.024794
20
0.001091
0.014350
0.033872
0.006110
0.031203
0.044522
50
0.000001
0.000016
0.000054
0.000167
0.001229
0.002577
100
0.000000
0.000000
0.000000
0.000001
0.000011
0.000026
⋮
⋮
⋮
⋮
⋮
⋮
⋮
E
N
5.1703
12.0640
15.5109
7.5000
17.5000
22.5000
E
N
2
35.5201
166.047
266.953
95.625
398.125
118.125
Var
N
8.7880
20.5054
26.3641
39.375
91.875
624.375Table 4
Distribution of the number of customers served during a busy period.
n
Phase-type distribution
Two-phase Coxian distribution
λ
=
0.5, μ1=3, μ2=1, r10=1/3, r12=2/3, r20=1, r21=0
λ
=
0.5, μ1=1.5, μ2=2, p1=0.75, p2=0.25
m
=
3
m
=
7
m
=
9
m
=
3
m
=
7
m
=
9
P
n
P
n
P
n
P
n
P
n
P
n
3
0.296296
0.361705
4
0.197531
0.225840
5
0.131687
0.138872
6
0.091043
0.087950
7
0.065031
0.058528
0.057338
0.093217
8
0.047689
0.091043
0.038294
0.135805
9
0.035718
0.101159
0.026012
0.026084
0.140038
0.047322
10
0.027214
0.098911
0.052025
0.018057
0.126330
0.088640
20
0.003003
0.022933
0.041731
0.000749
0.011701
0.029999
50
0.000022
0.000226
0.000565
0.000000
0.000006
0.000022
⋮
⋮
⋮
⋮
⋮
⋮
⋮
E
N
6.000
14.000
18.000
4.9660
11.586
14.897
E
N
2
54.000
238.00
378.00
31.925
151.20
243.71
Var
N
18.000
42.000
54.000
7.2680
16.959
21.805Table 5
Distribution of the number of customers served during a busy period.
n
Markov-modulated Poisson process
Interrupted Poisson process
ρ
=
0.214, λ=0.5, μ1=5
ρ
=
0.35, λ=0.5, μ1=2
μ
2
=
1, r1=1, r2=0.5
μ
2
=
1, r1=0.15, r2=1.25
m
=
3
m
=
7
m
=
9
m
=
3
m
=
7
m
=
9
P
n
P
n
P
n
P
n
P
n
P
n
3
0.580690
0.416275
4
0.225374
0.230894
5
0.097992
0.130911
6
0.046357
0.077951
7
0.023164
0.281322
0.048328
0.129386
8
0.012020
0.254765
0.030916
0.167454
9
0.006413
0.176690
0.195809
0.020268
0.156863
0.072134
10
0.003494
0.112568
0.227989
0.013549
0.130565
0.120031
20
0.000015
0.000724
0.003611
0.000426
0.007541
0.021595
50
0.000000
0.000000
0.000000
0.000000
0.000002
0.000007
⋮
⋮
⋮
⋮
⋮
⋮
⋮
E
N
3.81818
8.9091
11.4545
4.6154
10.7692
13.8462
E
N
2
16.4254
83.681
136.747
26.7911
128.785
208.184
Var
N
1.8469
4.3095
5.5407
5.4893
12.8084
16.4679


## 5. Conclusion

The analysis of the distribution of the number of customers served during a busy period is important as it helps us to study the traffic intensity and the cycle time of the queueing system. In this paper, we have developed the analytically explicit results for the distribution of the number of customers served during a busy period for special cases of theM/G/1 queues. The functional equation for the Laplace transform of the number of customers served during a busy period for the M/G/1 queue is widely known, but several researchers state that, in general, it is not easy to invert it except for some simple cases such M/M/1 and M/D/1. We present an elegant solution to determine the distribution of the number of customers served during a busy period and the computation of distributions as well as their moments using the Lagrange inversion theorem and MAPLE. We obtain the distribution of the number of customers served during a busy period for various service-time distributions, such as exponential, deterministic, Erlang-k, gamma, chi-square, inverse Gaussian, generalized Erlang, matrix exponential, hyperexponential, uniform, Coxian, phase-type, Markov-modulated Poisson process, and interrupted Poisson process. We have also presented computational results using our method. The associated derivations are very fast and robust due to the lucidity of the expressions. Using this approach, the method can be applied to other service-time distributions such as Weibull and Gaussian.



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*Source: 7398658-2019-08-27.xml*