2. Marginal overlapping times in the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ queue

In this section we determine the marginal distribution and LST of $T_a$ (Section 2.1) and $T_d$ (Section 2.2).

2.1. The marginal distribution of $T_a$

Assume that C arrives to find $X_a$ customers. It is well known that its steady-state distribution is geometric with parameter $\rho$ :

(2) \begin{equation} \mathbb{P}(X_a=k) = (1-\rho) \rho^k, \quad k=0,1,\ldots . \end{equation}

If $X_a=k \geq 1$ , then C overlaps the remaining service time R with the customer in service, $R+S_2$ with the second customer, and so on, and finally $R+\sum_{j=2}^k S_j$ with the kth customer. Because of the memoryless property of the exponential distribution, R has the same distribution as $S_1$ . Hence

(3) \begin{equation} T_a \bigg |_{X_a=k} {\buildrel d \over =} \sum_{j=1}^{k} (k- j+1 ) S_{j} \,{\buildrel d \over =}\, \sum_{j=1}^{k} j \cdot S_{j} . \end{equation}

The last equality (in distribution) follows since all service times are i.i.d. (independent and identically distributed).

Obviously,

(4) \begin{equation} \mathbb{P} (\, j \cdot S_j > t ) = \mathbb{P} \biggl( S_j > \dfrac{t}{j} \biggr) = {\mathrm{e}}^{-({{\mu}/{j}}) t}.\end{equation}

The above equations (3) and (4) allow us to conclude that $T_a$ , given $X_a=k$ , is hypo-exponentially distributed:

\begin{equation*} T_a \bigg |_{X_a=k} \,{\buildrel d \over =}\, \sum_{j=1}^{k} \mathcal{E}_{\mu/j} \sim \operatorname{hypo-exp}\biggl(\mu, \dfrac{\mu}{2}, \ldots,\dfrac{\mu}{k} \biggr).\end{equation*}

We summarize these findings in the following proposition.

Proposition 1. If we condition on $X_a$ to be k, then the overlapping time upon arrival, $T_a$ , is $\operatorname{hypo-exp}\!(\mu, {{\mu}/{2}}, \ldots,{{\mu}/{k}} )$ :

\begin{align*} \mathbb{P} ( T_a > t | X_a=k ) &= \boldsymbol{\alpha}\,{\mathrm{e}}^{t\mathcal{M}}\boldsymbol{1} \\[3pt] &= \sum_{j=1}^k \,{\mathrm{e}}^{- ({{\mu}/{j}}) t } \Biggl(\prod_{i=1, i \ne j}^k \dfrac{{{\mu}/{i}}}{{{\mu}/{i}} - {{\mu}/{j}}} \Biggr) \\[3pt] &= \sum_{j=1}^k \,{\mathrm{e}}^{- ({{\mu}/{j}}) t } \Biggl(\prod_{i=1, i \ne j}^k \dfrac{j}{j-i} \Biggr), \end{align*}

where

\begin{gather*} \alpha = \begin{bmatrix}1& \quad 0& \quad 0& \quad \ldots& \quad 0& \quad 0 \end{bmatrix}\!, \\[3pt] \mathcal{M} = \begin{bmatrix}-\mu& \quad \mu& \quad 0& \quad \ldots& \quad 0& \quad 0\\[2pt] 0& \quad -\dfrac{\mu}{2}& \quad \dfrac{\mu}{2}& \quad \ddots& \quad 0& \quad 0\\[2pt] \vdots& \quad \ddots& \quad \ddots& \quad \ddots& \quad \ddots& \quad \vdots\\[2pt] 0& \quad 0& \quad \ddots& \quad -\dfrac{\mu}{k-2}& \quad \dfrac{\mu}{k-2}& \quad 0\\[9pt] 0& \quad 0& \quad \ldots& \quad 0& \quad -\dfrac{\mu}{k-1}& \quad \dfrac{\mu}{k-1}\\[9pt] 0& \quad 0& \quad \ldots& \quad 0& \quad 0& \quad - \dfrac{\mu}{k}\end{bmatrix}\!. \end{gather*}

The unconditional distribution tail of $T_a$ follows by combining (2) and Proposition 1.

Proposition 2. The tail distribution of $T_a$ is given by

\begin{equation*} \mathbb{P}(T_a>t) = \sum_{k=1}^{\infty} (1-\rho) \rho^k \sum_{j=1}^k \,{\mathrm{e}}^{- ({{\mu}/{j}}) t } \Biggl(\prod_{i=1, i \ne j}^k \dfrac{j}{j-i} \Biggr).\end{equation*}

For several purposes we shall find it useful to also determine the LST of $T_a$ . We have the following result.

Lemma 1. The LST of $T_a$ is given by

\begin{equation*} \mathbb{E} \bigl[{\mathrm{e}}^{-\theta T_a} \bigr] = \sum_{k=0}^{\infty} (1-\rho) \biggl(\dfrac{\lambda}{\theta} \biggr)^k \dfrac{\Gamma(1+\mu/\theta)}{\Gamma(k+1+\mu/\theta)}. \end{equation*}

Proof. We have

\begin{align*} \mathbb{E} \bigl[ {\mathrm{e}}^{-\theta T_a} \bigr] &= \sum^{\infty}_{k=0} (1-\rho) \rho^k \mathbb{E} \bigl[ {\mathrm{e}}^{-\theta \sum_{j=1}^{k} j S_j } \bigr] \nonumber \\[3pt] &= \sum^{\infty}_{k=0} (1-\rho) \rho^k \prod^k_{j=1} \biggl( \dfrac{\mu}{\mu + j \theta } \biggr) \end{align*}

\begin{align*} &= \sum^{\infty}_{k=0} (1-\rho) \rho^k \prod^k_{j=1} \biggl( \dfrac{\mu/\theta}{\mu/\theta + j } \biggr) \nonumber \\[3pt] &= \sum^{\infty}_{k=0} (1-\rho) \biggl( \dfrac{\lambda}{\theta} \biggr)^k \dfrac{\Gamma(1+\mu/\theta)}{\Gamma(k + 1 + \mu/\theta)} . \\[-35pt] \end{align*}

In Lemma 2 we present a different derivation of the LST of $T_a$ . This alternative approach is particularly convenient when, in Section 3, we derive the joint LST of $(T_a,T_d)$ . After the proof of the lemma we will show agreement between the results of Lemmas 1 and 2.

Lemma 2. The LST of $T_a$ is given by

(5) \begin{equation} \mathbb{E} \bigl[{\mathrm{e}}^{-\theta T_a} \bigr] = (1-\rho) \int_{t=0}^{\infty} \mu {\mathrm{e}}^{-\mu t} \,{\mathrm{e}}^{({{\lambda}/{\theta}})(1 - {\mathrm{e}}^{-\theta t})} \,{\mathrm{d}} t= (\mu-\lambda) \biggl(\dfrac{\theta}{\lambda} \biggr)^{\mu/\theta} \dfrac{1}{\theta}\,{\mathrm{e}}^{\lambda/\theta} \gamma \biggl(\dfrac{\mu}{\theta},\dfrac{\lambda}{\theta} \biggr) .\end{equation}

Proof. We exploit the following. (i) The number $X_a$ seen by the tagged customer upon arrival is $\operatorname{geom}(\rho)$ -distributed. (ii) If $X_a=k$ , then the sojourn time $\Sigma$ of the tagged customer is $\operatorname{Erlang}(k+1,\mu)$ -distributed, hence with density

\begin{align*}\mu \frac{(\mu t)^k}{k!} \,{\mathrm{e}}^{-\mu t}. \end{align*}

(iii) A Poisson arrival process has the following well-known property (see Section 2.3 of Ross [Reference Ross28]), which we shall henceforth refer to as property $\mathcal{P}$ . If there are k Poisson events in an interval of length t, then the joint distribution of the event epochs equals the joint distribution of the order statistics of k independent, uniformly on (0, t)-distributed random variables. Hence

(6) \begin{align} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_a}\bigr]& = \sum_{k=0}^\infty (1-\rho) \rho^k \int_{t=0}^\infty \mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_a}\mid X_a=k, \Sigma =t\bigr] \,{\mathrm{d}} \mathbb{P}(\Sigma < t\mid X_a=k) \nonumber \\[3pt] & = \sum_{k=0}^\infty (1-\rho) \rho^k \int_{t=0}^\infty \mu \dfrac{(\mu t)^k}{k!} \,{\mathrm{e}}^{-\mu t} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_a}\mid X_a=k, \Sigma = t\bigr] \,{\mathrm{d}} t . \end{align}

Now observe that the k service completions of the $X_a=k$ customers seen by C at arrival occur in (0, t) according to a Poisson process with rate $\mu$ . Hence, exploiting the above-mentioned property $\mathcal{P}$ ,

(7) \begin{equation} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_a}\mid X_a=k,S=t \bigr] = \biggl(\int_{u=0}^t \,{\mathrm{e}}^{-\theta u} \dfrac{{\mathrm{d}} u}{t} \biggr)^k = \biggl(\dfrac{1-{\mathrm{e}}^{-\theta t}}{\theta t} \biggr)^k . \end{equation}

Substitution of (7) in (6) and summation over k gives the first equality in (5). The second equality follows via the substitution $z = ({{\lambda}/{\theta}}) \,{\mathrm{e}}^{-\theta t}$ ; it transforms the first integral in (5) into

(8) \begin{equation} (\mu-\lambda) \,{\mathrm{e}}^{{{\lambda}/{\theta}}} \biggl(\dfrac{\theta}{\lambda} \biggr)^{\mu/\theta} \, \int_{z=0}^{{{\lambda}/{\theta}} }z^{{{\mu}/{\theta}}} \,{\mathrm{e}}^{-z} \dfrac{{\mathrm{d}} z}{\theta z} = (\mu-\lambda) \biggl(\dfrac{\theta}{\lambda} \biggr)^{\mu/\theta} \dfrac{1}{\theta}\,{\mathrm{e}}^{\lambda/\theta}\int_{z=0}^{{{\lambda}/{\theta}}} z^{{{\mu}/{\theta}}-1} \,{\mathrm{e}}^{-z} \,{\mathrm{d}} z .\end{equation}

Now observe (see (1)) that the last integral in (8) equals $\gamma({{\mu}/{\theta}},{{\lambda}/{\theta}})$ .

It should be observed that many identities and asymptotic results are known for incomplete gamma functions; see Abramowitz and Stegun [Reference Abramowitz and Stegun1].

We now show the equivalence between the results of Lemmas 1 and 2. To this end, rewrite the part of (8) after the equality sign as follows, using the transformation $v = 1 -({{\theta}/{\lambda}})z$ and the Taylor series development of ${\mathrm{e}}^{({{\lambda}/{\theta}})v}$ :

\begin{equation*} (\mu-\lambda) \dfrac{1}{\theta} \int_{v=0}^1 (1-v) ^{{{\mu}/{\theta}}-1} \,{\mathrm{e}}^{({{\lambda}/{\theta}})v} \,{\mathrm{d}} v= (\mu-\lambda) \dfrac{1}{\theta} \sum_{k=0}^\infty \dfrac{(\lambda/\theta)^k}{k!} \int_{v=0}^1 (1-v)^{\mu/\theta-1} v^k \,{\mathrm{d}} v .\end{equation*}

Finally, observe that the last integral equals the beta function

\begin{align*}B\biggl(\frac{\mu}{\theta},k+1\biggr) = \frac{\Gamma({{\mu}/{\theta}}) \Gamma(k+1)}{\Gamma(k+1+{{\mu}/{\theta}})}, \end{align*}

and use the fact that

\begin{align*}\Gamma\biggl(1+\frac{\mu}{\theta}\biggr) = \frac{\mu}{\theta} \Gamma\biggl(\frac{\mu}{\theta}\biggr). \end{align*}

We close this subsection by deriving two expressions for the moments of $T_a$ .

Proposition 3. With $X_a$ denoting a random variable that is $\operatorname{geom}(\rho)$ -distributed, we have

\begin{align*} \mathbb{E} \bigl[ T_a^m \bigr] &= \dfrac{m!}{\mu^m} \sum_{k=0}^{\infty} (1-\rho) \rho^k \sum_{j=1}^k \Biggl(\prod_{i=1, i \ne j}^k \dfrac{j}{j-i} \Biggr) j^m \nonumber\\[4pt] &= \mathbb{E} \bigl[ S^m \bigr] \cdot \mathbb{E} \Biggl[ \sum_{j=1}^{X_a} \Biggl(\prod_{i=1, i \ne j}^{X_a} \dfrac{j}{j-i} \Biggr) j^m \Biggr] .\end{align*}

Proof. Use Proposition 2 and the fact that the mth moment of an $\operatorname{Exp}\!(\mu/j)$ distribution equals $({{m!}/{\mu^m}}) j^m$ .

Using the multinomial theorem, we also have the following alternative representation for the moments of $T_a$ .

Proposition 4. We have

\begin{align*} \mathbb{E} \bigl[ T_a^m \bigr] = \dfrac{m!}{\mu^m} \sum^{\infty}_{k=0} (1-\rho) \rho^k\sum_{\substack{ n_1+\cdots+n_k=m \\[3pt] n_1, \ldots, n_k \geq 0}} \prod^k_{j=1} j^{n_j}.\end{align*}

Proof. This follows directly from first conditioning on the queue length and observing the fact that

\begin{align*}\mathbb{E} \bigl[ S_j ^{n_j} \bigr] = \frac{n_j!}{\mu^{n_j}} \end{align*}

and the multinomial theorem. We provide an outline of the proof below:

This completes the proof.

2.2. The marginal distribution of $\boldsymbol{T}_{\boldsymbol{d}}$

As is well known, the sojourn time $\Sigma$ of tagged customer C in the $\mathrm{M}_\lambda/\mathrm{M}_\mu/1$ queue is $\operatorname{Exp}\!(\mu-\lambda)$ -distributed. To obtain the distribution of the total overlapping time $T_d$ of C with the customers who arrive during her sojourn time, we reason as follows. If $\Sigma =x$ , then $N_\lambda(x)$ customers arrive during C’s sojourn time, where $N_\lambda(x)$ is $\operatorname{Poisson}(\lambda x)$ -distributed. Their arrival epochs are distributed as the order statistics of $N_\lambda(x)$ i.i.d. uniformly distributed random variables on (0, x) (see property $\mathcal{P}$ ); we denote them by $U_{(\,j)}(0, x)$ , for $j=1,\ldots,N_\lambda(x)$ . Hence

\begin{equation*} T_d\bigg |_{\Sigma = x, N_\lambda(x) = n } \stackrel{d}{=} \sum^{n}_{j=1} (x - U_{(\,j)}(0, x) ) .\end{equation*}

We shall use a well-known fact about the sum of i.i.d. uniformly distributed random variables.

Lemma 3. Let $G_n$ be the sum of n i.i.d. uniform (0, x) random variables. Then the CDF of $G_n$ is equal to

\begin{equation*} F_n(t) = \dfrac{1}{n!} \sum^{ \lfloor {{t}/{x}} \rfloor}_{k=0} (-1)^k {\binom{ n}{ k }} \biggl( \dfrac{t}{x} - k \biggr)^n, \quad 0 \leq t \leq n x . \end{equation*}

Proof. This is a well-known fact; the CDF is called the Irwin–Hall distribution [Reference Hall14, Reference Irwin16].

Denoting $\bar{F}_n(x) = 1-F_n(x)$ , we have the following expression for $\mathbb{P}(T_d>t)$ .

Proposition 5. The tail distribution of $T_d$ is given by

\begin{equation*} \mathbb{P}(T_d>t) = \int_{x=0}^\infty (\mu-\lambda) \,{\mathrm{e}}^{-(\mu-\lambda )x} \, \mathbb{E} \biggl[\bar{F}_{N_\lambda (x)} \biggl(\dfrac{t}{x} \biggr) \biggr] \,{\mathrm{d}} x . \end{equation*}

Proof. We have

(9)

Remark 1. A more compact expression for $\mathbb{P}(T_d>t)$ immediately follows from the above proposition:

\begin{equation*} \mathbb{P}(T_d>t) = \mathbb{E} \biggl[\bar{F}_{N_\lambda ( \mathcal{E}_{\mu-\lambda} )} \biggl(\dfrac{t}{\mathcal{E}_{\mu-\lambda}} \biggr) \biggr] .\end{equation*}

We now derive an expression for the LST of $T_d$ .

Lemma 4. The LST of $T_d$ is given by

(10) \begin{equation} \mathbb{E} \bigl[{\mathrm{e}}^{-\theta T_d} \bigr] = (1-\rho) \int_{x=0}^\infty \mu\,{\mathrm{e}}^{-\mu x} \,{\mathrm{e}}^{({{\lambda}/{\theta}}) (1-{\mathrm{e}}^{-\theta x})} \,{\mathrm{d}} x = (\mu-\lambda) \biggl(\dfrac{\theta}{\lambda} \biggr)^{{{\mu}/{\theta}}} \dfrac{1}{\theta} \,{\mathrm{e}}^{{{\lambda}/{\theta}}} \gamma \biggl( \dfrac{\mu}{\theta} , \dfrac{\lambda}{\theta} \biggr) . \end{equation}

Proof. Conditioning on the sojourn time $\Sigma$ of C being equal to x and subsequently on the number of arrivals during x being equal to n we can write (very similar to (9) above)

\begin{align*} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_d}\bigr] &= \int_{x=0}^\infty (\mu-\lambda) \,{\mathrm{e}}^{-(\mu-\lambda)x} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_d}\mid \Sigma = x\bigr] \,{\mathrm{d}} x \nonumber\\[3pt] &= \int_{x=0}^\infty (\mu-\lambda) \,{\mathrm{e}}^{-\mu x} \, \sum_{n=0}^\infty \dfrac{(\lambda x)^n}{n!} \biggl( \dfrac{1 - {\mathrm{e}}^{-\theta x}}{\theta x} \biggr)^n \,{\mathrm{d}} x \quad (\textrm{property $\mathcal{P}$}) \nonumber \\[3pt] &= \int_{x=0}^\infty (\mu-\lambda) \,{\mathrm{e}}^{-\mu x} \,{\mathrm{e}}^{({{\lambda}/{\theta}})(1 - {\mathrm{e}}^{-\theta x})} \,{\mathrm{d}} x . \\[-35pt] \end{align*}

Remark 2. The expressions in (5) and (10) reveal that $T_a$ and $T_d$ are identically distributed in the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ FCFS queue. This is a much stronger result than that of Xu et al. [Reference Xu, Ko, Kong and Pender31] or Oz and Perlman [Reference Oz and Perlman21], who only show this for expectations. Moreover, it highlights an important distinction between the single server queue and the infinite server queue. In the infinite server queue $T_a$ and $T_d$ have the same mean; however, their distributions are not identical (see Boxma et al. [Reference Boxma, van Kreveld and Pender8]).

By differentiating the first expression in either (5) or (10) (which perhaps works best by taking the Taylor series development of $({{(1 - {\mathrm{e}}^{-\theta x})}/{\theta}} )$ , it is readily seen that the first moments of $T_a$ and $T_d$ are given by

(11) $$\begin{align}\mathbb{E}[T_a]& = \mathbb{E}[T_d] = \dfrac{\rho}{\mu (1-\rho)^2},\end{align}$$

(12) $$\begin{align}\mathbb{E}[T_a^2]& = \mathbb{E}[T_d^2] = \dfrac{2\rho+4\rho^2}{\mu^2(1-\rho)^4},\end{align}$$

(13) $$\begin{align}\operatorname{Var}\![T_a]& = \operatorname{Var}\![T_d] = \dfrac{2\rho+3\rho^2}{\mu^2(1-\rho)^4}.\end{align}$$

In view of the fact that the study of overlapping times is still in its infancy, it seems justified to close this section by discussing an alternative way to study overlapping times. Suppose that C is customer zero, and that there was an infinite stream of customers $-1,-2,\ldots$ before C, and an infinite stream of customers $1,2,\ldots$ after C. Define $O_n$ as the overlap time of C with customer n, $n= \ldots,-2,-1,1,2,\ldots.$ Then we can write

(14) \begin{equation} T_a = \sum_{n=1}^\infty O_{-n}, \quad T_d = \sum_{n=1}^\infty O_n . \end{equation}

Successive overlap times depend on each other, and it is not a priori clear how to obtain the distribution of $T_a$ and $T_d$ from (14). However, the fact that $O_n$ contributes in the same way to the overlap time of C with customer n as to the overlap time of customer n with C gives us some insight into the fact that $T_a$ and $T_d$ have the same distribution. Furthermore, $\mathbb{E}[T_a]$ and $\mathbb{E}[T_d]$ easily follow from (14) and the following proposition.

Proposition 6. The mean overlap time of C and customer k in the $\mathrm{M}_\lambda/\mathrm{M}_\mu/1$ FCFS queue is given by

\begin{equation*}\mathbb{E} \bigl[ O_{k} \bigr] = \dfrac{\rho^k}{\mu - \lambda} , \quad k=1,2,\ldots .\end{equation*}

Proof. We have

\begin{equation*}\mathbb{E} \bigl[ O_{k} \bigr] = \int^{\infty}_{0} \mathbb{P} ( O_{k} > t ) \,{\mathrm{d}} t = \int^{\infty}_{0} \,{\mathrm{e}}^{-(\mu -\lambda) t} \rho^k \,{\mathrm{d}} t = \dfrac{\rho^k}{\mu - \lambda} .\end{equation*}

In the second equality we have used the memoryless property of the $\operatorname{Exp}\!(\mu-\lambda)$ -distributed sojourn time of C, which implies that $\mathbb{P}(O_k>t\mid O_k>0) = {\mathrm{e}}^{-(\mu-\lambda)t}$ , and we have also used the fact that

\begin{align*}\mathbb{P}(O_k>0) = \biggl(\frac{\lambda}{\lambda+ (\mu-\lambda)} \biggr)^k = \rho^k \end{align*}

(the probability that there are at least k arrivals during the sojourn time of C).

It follows from (14) and the above proposition that the mean overlapping time with customers that arrive after C is equal to

\begin{equation*}\mathbb{E} \bigl[ T_d \bigr] = \mathbb{E} \Biggl[ \sum^{\infty}_{k=1} O_{k} \Biggr]= \sum^{\infty}_{k=1} \mathbb{E} \bigl[ O_{k} \bigr]= \sum^{\infty}_{k=1} \dfrac{\rho^k}{\mu - \lambda}= \dfrac{\rho}{\mu (1-\rho)^2} ,\end{equation*}

in agreement with (11).

3. The joint LST of $\boldsymbol{T}_{\boldsymbol{a}}$ and $\boldsymbol{T}_{\boldsymbol{d}}$

In order to calculate the joint LST of $T_a$ and $T_d$ , we again exploit the following observations about the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ queue.

1. (i) The number of customers $X_a$ seen by the tagged customer upon arrival is $\operatorname{Geometric} (\rho)$ -distributed.

2. (ii) If $X_a=k$ , then the sojourn time $\Sigma$ of the tagged customer is $\operatorname{Erlang}(k+1,\mu)$ -distributed, hence with density $\mu ({{(\mu t)^k}/{k!}}) \,{\mathrm{e}}^{-\mu t}$ .

3. (iii) Property $\mathcal{P}$ , as stated in the proof of Lemma 2.

Now we are ready to give the joint LST of $T_a$ and $T_d$ .

Theorem 1. The joint LST of $(T_a,T_d)$ is given by

(15) \begin{equation} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta_1T_a-\theta_2T_d} \bigr] = (1-\rho) \int_{t=0}^{\infty} \mu {\mathrm{e}}^{-\mu t} \exp\biggl({\lambda \frac{1-{\mathrm{e}}^{-\theta_1t}}{\theta_1}}\biggr) \,{\mathrm{e}}^{-\lambda t} \exp\biggl({\lambda \frac{1-{\mathrm{e}}^{-\theta_2t}}{\theta_2}}\biggr) \,{\mathrm{d}} t .\end{equation}

Proof. We have

(16) \begin{align} \mathbb{E} \bigl[ {\mathrm{e}}^{-\theta_1T_a-\theta_2T_d} \bigr] &= \sum_{k=0}^{\infty} \mathbb{P}(X_a=k) \mathbb{E}\bigl[{\mathrm{e}}^{-\theta_1T_a-\theta_2T_d}\mid X_a=k \bigr] \notag \\[3pt] &= \sum_{k=0}^{\infty} (1-\rho) \rho^k \mathbb{E} \bigl[ {\mathrm{e}}^{-\theta_1T_a-\theta_2T_d}\mid X_a=k,\Sigma=t \bigr] \,{\mathrm{d}} \mathbb{P}(\Sigma < t\mid X_a=k) \nonumber\\[3pt] &= \sum_{k=0}^{\infty} (1-\rho) \rho^k \int_{t=0}^{\infty} \mu \dfrac{(\mu t)^k}{k!} \,{\mathrm{e}}^{-\mu t} \mathbb{E}\bigl[ {\mathrm{e}}^{-\theta_1T_a-\theta_2T_d}\mid X_a=k,\Sigma=t\bigr] \,{\mathrm{d}} t . \end{align}

To determine

\begin{align*}\mathbb{E} \bigl[{\mathrm{e}}^{-\theta_1T_a-\theta_2T_d}\mid X_a=k,\Sigma=t \bigr], \end{align*}

we first observe that $T_a$ and $T_d$ are conditionally independent, given that $X_a=k$ and $\Sigma=t$ . Let us now twice use (iii), i.e. the Poisson property $\mathcal{P}$ . For $T_d$ we have n arrivals during $\Sigma=t$ with probability ${\mathrm{e}}^{-\lambda t} {{(\lambda t)^n}/{n!}}$ , and $T_d$ is the sum of the times from their arrival until the end of $\Sigma$ . Moreover, $T_d$ does not depend on $X_a$ , given $\Sigma=t$ . Hence, as in (7), we have

(17) \begin{equation} \mathbb{E} \bigl[{\mathrm{e}}^{-\theta_2T_d}\mid X_a=k,\Sigma=t \bigr]= \sum_{n=0}^{\infty} \,{\mathrm{e}}^{-\lambda t} \dfrac{(\lambda t)^n}{n!} \biggl(\dfrac{1 - {\mathrm{e}}^{-\theta_2 t}}{\theta_2 t} \biggr)^n = {\mathrm{e}}^{-\lambda t} \exp\biggl({\lambda \frac{1-{\mathrm{e}}^{-\theta_2t}}{\theta_2}}\biggr) . \end{equation}

For $T_a$ , we observe that the k service completions of the $X_a=k$ customers seen at arrival occur in (0, t) according to a Poisson process with rate $\mu$ . Hence

(18) \begin{equation} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta_1T_a}\mid X_a=k,\Sigma=t \bigr] = \biggl(\dfrac{1-{\mathrm{e}}^{-\theta_1t}}{\theta_1 t} \biggr)^k . \end{equation}

The proof now follows by combining (16), (17), and (18), also using the fact that $\rho \mu = \lambda$ .

By successively taking $\theta_2=0$ , $\theta_1=0$ , and $\theta_1=\theta_2=\theta$ in (15), we get the marginal LST of $T_a$ as already given in (5), the marginal LST of $T_d$ as already given in (10), and the LST of the total overlapping time $T \, :\!= \, T_a+T_d$ .

Corollary 1. We have

(19) \begin{equation} \mathbb{E} \bigl[{\mathrm{e}}^{-\theta T} \bigr] = (1-\rho) \int_{t=0}^{\infty} \mu {\mathrm{e}}^{-\mu t} \,{\mathrm{e}}^{-\lambda t} \exp\biggl({2\lambda \frac{1-{\mathrm{e}}^{-\theta t}}{\theta}}\biggr) \,{\mathrm{d}} t .\end{equation}

Remark 3. Just as in (5) and (10), one can implement the substitution ${\mathrm{e}}^{-\theta t} = y$ , with ${{{\mathrm{d}} y}/{{\mathrm{d}} t}} = -\theta y$ . This yields the following expression:

(20) \begin{equation} {\mathbb{E}} \bigl[{\mathrm{e}}^{-\theta T} \bigr] = (1-\rho) \dfrac{\mu}{\theta} {\int}_{\!\!\! y=0}^1 y^{{{(\mu+\lambda)}/{\theta}}-1} \,{\mathrm{e}}^{{{2\lambda(1-y)}/{\theta}}} \,{\mathrm{d}} y = (1-\rho) {\mathrm{e}}^{{{2\lambda}/{\theta}}} \dfrac{\mu}{\theta} \int_{y=0}^1 y^{{{(\mu+\lambda)}/{\theta}}-1} {\mathrm{e}}^{-{{2\lambda y}/{\theta}}}{\mathrm{d}} y. \end{equation}

The integral in (20) again is an incomplete gamma function. In particular, we can write

\begin{equation*}\mathbb{E} \bigl[{\mathrm{e}}^{-\theta T} \bigr] = (1-\rho) \dfrac{\mu}{\theta} \,{\mathrm{e}}^{{{2\lambda}/{\theta}}} \biggl( \dfrac{\theta}{2 \lambda } \biggr)^{{{(\mu+\lambda)}/{\theta}}} \gamma \biggl(\dfrac{\mu+\lambda}{\theta}, \dfrac{2\lambda}{\theta} \biggr) . \end{equation*}

In order to get moments of $T_a$ , $T_d$ , and T, it seems best to differentiate the expressions in Corollary 1, using the Taylor series development of ${{(1-{\mathrm{e}}^{-\theta t})}/{\theta}}$ . We thus obtain (11), (12), and (13), as well as

(21) \begin{align} \mathbb{E}[T_aT_d] &= \dfrac{7 \rho^2}{\mu^2 (1-\rho)^4} , \end{align}

(22) \begin{align} \operatorname{Cov}\![T_a,T_d] & =\dfrac{6 \rho^2}{\mu^2(1-\rho)^4} , \end{align}

(23) \begin{align} \operatorname{Corr}\![T_a,T_d] & = \dfrac{5\rho}{2+3\rho} , \end{align}

(24) \begin{align} \mathbb{E}[T]& = \dfrac{2\rho}{\mu(1-\rho)^2} , \end{align}

(25) \begin{align}\mathbb{E}[T^2] & = \dfrac{4 \rho + 20 \rho^2}{\mu^2(1-\rho)^4},\end{align}

(26) \begin{align}\operatorname{Var}\![T] & = \dfrac{4\rho + 16 \rho^2}{\mu^2 (1-\rho)^4} .\end{align}

4. Heavy-traffic limits

In this section we study T and $(T_a,T_d)$ in the heavy-traffic limit $\rho \uparrow 1$ . In the heavy-traffic analysis of queue lengths and waiting times, which typically have means that behave as a constant over $1-\rho$ for $\rho \uparrow 1$ , one usually has to scale by $1-\rho$ to obtain a non-trivial heavy-traffic limit; an exception is when service times are heavy-tailed (see Boxma and Cohen [Reference Boxma and Cohen6]). However, the fact that the means of $T_a$ , $T_d$ , and T behave as $(1-\rho)^{-2}$ for $\rho \uparrow 1$ suggests that we now have to scale by $(1-\rho)^2$ . Below we shall see that this indeed gives a valid heavy-traffic limit. Remarkably, the resulting limit random variable is a Weibull random variable.

Theorem 2. Let $T= T_a + T_d$ be the overlapping time in the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ queue. Then, as we let $\rho \to 1$ , we have the following heavy-traffic limit result:

\begin{equation*} \rm{lim}_{\rho \uparrow 1} \mathbb{E} \bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta T} \bigr] = \int_{z=0}^{\infty}\dfrac{1}{2} \sqrt{\dfrac{\mu}{z}} \,{\mathrm{e}}^{-\sqrt{\mu z}} \,{\mathrm{e}}^{-\theta z} \,{\mathrm{d}} z = \mathbb{E} \bigl[ {\mathrm{e}}^{-\theta W} \bigr],\end{equation*}

where W has the $\operatorname{Weibull}({{1}/{2}},{{1}/{\mu}})$ distribution with density $\frac{1}{2} \sqrt{{{\mu}/{z}}} \,{\mathrm{e}}^{-\sqrt{\mu z}}$ .

Proof. It follows from (19) that

\begin{equation*} \mathbb{E} \bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta T} \bigr] = (1-\rho) \mu \int_{t=0}^{\infty} \,{\mathrm{e}}^{-\mu(1-\rho)t} \operatorname{exp} \biggl(2\lambda \biggl[\dfrac{1 - {\mathrm{e}}^{-(1-\rho)^2 \theta t}}{(1-\rho)^2 \theta } - t \biggr] \biggr) \,{\mathrm{d}} t . \end{equation*}

Using the Taylor series development of ${\mathrm{e}}^a$ for $a \downarrow 0$ , we can write

\begin{equation*} \rm{lim}_{\rho \uparrow 1} \mathbb{E} \bigl[{\mathrm{e}}^{-(1-\rho)^2\theta T} \bigr]= \lim_{\rho \uparrow 1} \, (1-\rho) \mu\int_{t=0}^{\infty} \,{\mathrm{e}}^{-\mu(1-\rho)t} \,{\mathrm{e}}^{- \rho \mu (1-\rho)^2 \theta t^2} \,{\mathrm{d}} t= \int_{y=0}^{\infty} \mu {\mathrm{e}}^{-\mu y - \mu \theta y^2} \,{\mathrm{d}} y. \end{equation*}

In the last step we have used the transformation $y = (1-\rho) t$ . One can invert this LST by using the transformation $z = \mu y^2$ : for $\rho \uparrow 1$ ,

\begin{equation*}\mathbb{E}\bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta T}\bigr] \rightarrow \int_{z=0}^{\infty}\dfrac{1}{2} \sqrt{\dfrac{\mu}{z}} \,{\mathrm{e}}^{-\sqrt{\mu z}} \,{\mathrm{e}}^{-\theta z} \,{\mathrm{d}} z .\end{equation*}

Hence Feller’s convergence theorem implies that $(1-\rho)^2 T$ converges, for $\rho \uparrow 1$ , to a random variable with density

\begin{align*}t(z) \, :\!= \, \frac{1}{2} \sqrt{\frac{\mu}{z}} \,{\mathrm{e}}^{-\sqrt{\mu z}}. \end{align*}

It is easily seen that this is the density of distribution $T(z) \, :\!= \, 1 - {\mathrm{e}}^{-\sqrt{\mu z}}$ . This, in turn, is the distribution of the square of an $\operatorname{Exp}\!(\sqrt{\mu})$ -distributed random variable; put differently, it is a $\operatorname{Weibull}({{1}/{2}},{{1}/{\mu}})$ -distributed random variable. Interestingly, the heavy-traffic limit in this $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ model has a subexponential tail.

Proceeding in exactly the same way, it follows from (5) and (10) that

\begin{align*} & \rm{lim}_{\rho \uparrow 1} \mathbb{E}\bigl[{\mathrm{e}}^{-(1-\rho)^2\theta T_a}\bigr]\\[3pt]&\quad = \rm{lim}_{\rho \uparrow 1} \mathbb{E}\bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta T_d}\bigr]\\[3pt] &\quad = \lim_{\rho \uparrow 1} \, (1-\rho) \mu\int_{t=0}^{\infty} \,{\mathrm{e}}^{-\mu(1-\rho)t} \,{\mathrm{e}}^{- \frac{1}{2}\rho \mu (1-\rho)^2 \theta t^2} \,{\mathrm{d}} t \\[3pt] &\quad = \int_{y=0}^{\infty} \mu {\mathrm{e}}^{-\mu y - \frac{1}{2} \mu \theta y^2} \,{\mathrm{d}} y, \end{align*}

which yields the following corollary.

Corollary 2. We have

\begin{equation*}\lim_{\rho \uparrow 1} \mathbb{E}\bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta T_a}\bigr] = \lim_{\rho \uparrow 1} \mathbb{E}\bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta T_d}\bigr] = \int_{z=0}^{\infty} \sqrt{\dfrac{\mu}{2z}} \,{\mathrm{e}}^{-\sqrt{2 \mu z}} \,{\mathrm{e}}^{-\theta z} \,{\mathrm{d}} z ,\end{equation*}

that is, both $T_a$ and $T_d$ , scaled by $(1-\rho)^2$ , in the heavy-traffic limit have a $\operatorname{Weibull} ({{1}/{2}},{{1}/{(2\mu)}} )$ distribution.

As we have seen, $T_a$ and $T_d$ are dependent, and their correlation approaches one in the heavy-traffic case (see (23)). In general, it is hard to obtain the distribution of the sum of two Weibull-distributed random variables; but in this heavy-traffic limit, apparently $(1-\rho)^2 (T_a+T_d)$ also has a Weibull distribution. Incidentally, for overlap numbers in the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ queue, a somewhat similar result is found in Boxma [Reference Boxma5]: the overlap number O is geometrically distributed, just like its two components, namely the number of customers $X_a$ seen at arrival of the tagged customer and the number of customers $X_d$ arriving during the sojourn of the tagged customer.

5. Generalizations

In this section we consider a generalization of the analysis in Section 3: we determine the transform of the joint distribution of $X_a$ , $X_d$ , $T_a$ , and $T_d$ . As a by-product this gives the transform of the joint distribution of the total overlap number $X_O=X_a+X_d$ and total overlapping time $T=T_a+T_d$ in the $\mathrm{M}_\lambda/\mathrm{M}_\mu/1$ queue. Here the overlap number of a tagged customer is the number of customers with whom she shares some time in the system.

Theorem 3. The joint transform of $X_a$ , $X_d$ , $T_a$ , and $T_d$ in the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ queue is given by

(27) \begin{equation} \mathbb{E} \Bigl[z_1^{X_a} z_2^{X_d} \,{\mathrm{e}}^{-\theta_1T_a-\theta_2T_d} \Bigr] =(1-\rho) \mu \int_{t=0}^{\infty}\,{\mathrm{e}}^{-(\mu+\lambda)t} \,{\mathrm{e}}^{\lambda ({{z_1}/{\theta_1}})(1-{\mathrm{e}}^{-\theta_1t})} \,{\mathrm{e}}^{\lambda ({{z_2}/{\theta_2}})(1-{\mathrm{e}}^{-\theta_2t})} \,{\mathrm{d}} t ,\end{equation}

and the joint transform of total overlap number $X_O$ and total overlapping time T in the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ queue is given by

(28) \begin{equation}\mathbb{E} \bigl[z^{X_O} \,{\mathrm{e}}^{-\theta T} \bigr] = (1-\rho) \mu\int_{t=0}^{\infty} \, {\mathrm{e}}^{-(\mu+\lambda)t} \,{\mathrm{e}}^{2 \lambda ({{z}/{\theta}}) (1-{\mathrm{e}}^{-\theta t})} \,{\mathrm{d}} t . \end{equation}

Proof. The proof is very similar to that of Theorem 1:

Equation (27) now follows by performing the two summations and realizing that $\rho \mu =\lambda$ . Equation (28) subsequently follows from (27) by taking $z_1=z_2=z$ and $\theta_1=\theta_2=\theta$ .

Remark 5. The marginal transform (probability generating function) of the overlap number $X_O$ for the $\mathrm{M}_{\lambda}/\mathrm{G}/1$ queue was already obtained in Boxma [Reference Boxma5]. For the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ queue, this gives

(29) \begin{equation} \mathbb{E} \bigl[z^{X_O} \bigr] = \dfrac{1-\rho}{1+\rho-2\rho z} , \end{equation}

corresponding to the geometric distribution

\begin{equation*} \mathbb{P}(X_O=k) = \biggl(1 - \dfrac{2\rho}{1+\rho} \biggr) \biggl(\dfrac{2\rho}{1+\rho} \biggr)^k , \quad k=0,1,\ldots , \end{equation*}

with mean

\begin{align*}\mathbb{E} \bigl[X_O \bigr] = \frac{2\rho}{1-\rho} \end{align*}

and variance

\begin{align*}\operatorname{Var}\bigl[X_O \bigr] = \frac{2\rho+2\rho^2}{(1-\rho)^2}. \end{align*}

Expression (29) also follows from (28) by realizing that

\begin{align*}\lim_{\theta \downarrow 0} \frac{1-{\mathrm{e}}^{-\theta t}}{\theta} = t \end{align*}

and performing the ensuing integration in (28).

It readily follows from (28) by differentiation that

\begin{equation*} \operatorname{Cov}\![X_O,T] = \dfrac{2 \rho + 6 \rho^2}{\mu(1-\rho)^3},\end{equation*}

and finally the correlation between total overlap number $X_O$ and total overlapping time T is found:

\begin{equation*} \operatorname{Corr}\![X_O,T] = \dfrac{1+ 3 \rho}{\sqrt{(2+2\rho)(1+4\rho)}} .\end{equation*}

It is readily checked that this correlation is non-negative and increasing in $\rho$ .

In taking the heavy-traffic limit in (28) with respect to both $X_O$ and T, we need to scale $X_O$ by $1-\rho$ and T by $(1-\rho)^2$ . Replacing z with ${\mathrm{e}}^{-s}$ in (28), we can write

(30) \begin{align} & \lim_{\rho \uparrow 1} \, \mathbb{E} \bigl[{\mathrm{e}}^{-(1-\rho)sX_O -(1-\rho)^2 \theta T} \bigr] \notag \\[3pt] &\quad = \lim_{\rho \uparrow 1} \, \mu(1-\rho) \int_0^\infty \,{\mathrm{e}}^{-(\mu+\lambda)t} \exp\biggr({2\lambda \,{\mathrm{e}}^{-(1-\rho)s} \biggl(\frac{1-{\mathrm{e}}^{-(1-\rho)^2 \theta t}}{(1-\rho)^2 \theta}\biggr)}\biggr) \,{\mathrm{d}} t \nonumber\\[3pt] &\quad = \lim_{\rho \uparrow 1} \, \mu(1-\rho) \int_0^\infty \,{\mathrm{e}}^{-\mu(1-\rho)t} \,{\mathrm{e}}^{-2\lambda (1-\rho)st} \,{\mathrm{e}}^{-\lambda (1-\rho)^2 \theta t^2} \,{\mathrm{d}} t \nonumber\\[3pt] &\quad = \int_0^\infty \mu {\mathrm{e}}^{-\mu y} \,{\mathrm{e}}^{-2 \mu sy} \,{\mathrm{e}}^{-\mu \theta y^2} \,{\mathrm{d}} y .\end{align}

In the third line of (30) we have omitted terms of smaller $(1-\rho)$ order in the Taylor series developments of exponents, and in the fourth line we have used the transformation $y=(1-\rho) t$ while also replacing $\lambda$ with $\mu$ .

The expression in the last line of (30), to be called $\psi(s,\theta)= \mathbb{E}\bigl[{\mathrm{e}}^{-sG-\theta H}\bigr]$ , is the double transform of $(G,H) \, :\!= \, \lim_{\rho \uparrow 1} ((1-\rho) X_O,(1-\rho)^2 T)$ . It is readily verified that

\begin{align*}{G \sim \operatorname{Exp} \biggl(\frac{1}{2} \biggr)\quad\text{and}\quad H \sim \operatorname{Weibull} \biggl(\frac{1}{2},\frac{1}{\mu} \biggr),} \end{align*}

and that

\begin{gather*}{\mathbb{E}[G]=2,\quad\operatorname{Var}\![G] = 4,\quad\mathbb{E}[H] = \frac{2}{\mu}, \quad\operatorname{Var} [H] =\frac{20}{\mu^2},}\\[3pt]{\mathbb{E}[GH] = \frac{12}{\mu},\quad\operatorname{Cov}\![G,H] = \frac{8}{\mu}, \quad\operatorname{Corr}\![G,H] = \frac{1}{\sqrt{1.25}}}\end{gather*}

(which is consistent with the above moments of $X_O$ and T).

6. Overlapping time in the $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ queue infection model

In this section we analyze the infection queueing models of Newman and Pender [Reference Newman and Pender19, Reference Newman and Pender20] where each customer has a probability q of being infectious. We are interested in the amount of time that the tagged customer C spends with infectious customers. We now let $Z_a$ be the total time C spends with infectious customers who are already present at her arrival, and $Z_d$ be the total time C spends with infectious customers who arrive during her sojourn. Below, we derive the LST of $(Z_a,Z_d)$ .

Theorem 4. The joint LST of $(Z_a,Z_d)$ is given by

\begin{align*} & \mathbb{E}\bigl[{\mathrm{e}}^{-\theta_1Z_a-\theta_2Z_d} \bigr] \\&\quad = (1-\rho) \int_{t=0}^{\infty} \mu {\mathrm{e}}^{-\mu t} \exp\biggl({\lambda \biggl( q \frac{1-{\mathrm{e}}^{-\theta_1t}}{\theta_1} +(1-q) t \biggr)}\biggr) \,{\mathrm{e}}^{-\lambda q t} \exp\biggl({\lambda q \frac{1-{\mathrm{e}}^{-\theta_2t}}{\theta_2}}\biggr) \,{\mathrm{d}} t . \end{align*}

Proof. We have

(31) \begin{align} \mathbb{E} \bigl[ {\mathrm{e}}^{-\theta_1Z_a-\theta_2Z_d} \bigr] \notag& = \sum_{k=0}^{\infty} \mathbb{P}(X_a=k) \mathbb{E}({\mathrm{e}}^{-\theta_1Z_a-\theta_2Z_d}\mid X_a=k) \\[3pt] & = \sum_{k=0}^{\infty} (1-\rho) \rho^k \mathbb{E} \bigl[{\mathrm{e}}^{-\theta_1Z_a-\theta_2Z_d}\mid X_a=k,\Sigma=t \bigr] \,{\mathrm{d}} \mathbb{P}(\Sigma < t\mid X_a=k) \nonumber\\[3pt] & = \sum_{k=0}^{\infty} (1-\rho) \rho^k \int_{t=0}^{\infty} \mu \dfrac{(\mu t)^k}{k!} \,{\mathrm{e}}^{-\mu t} \mathbb{E} \bigl[{\mathrm{e}}^{-\theta_1Z_a-\theta_2Z_d}\mid X_a=k,\Sigma=t \bigr] \,{\mathrm{d}} t . \end{align}

Just as in the proof of Theorem 1 we observe that $Z_a$ and $Z_d$ are conditionally independent, given that $X_a=k$ and $\Sigma=t$ , and that $Z_d$ does not depend on $X_a$ , given $\Sigma$ . The LST of $Z_d$ given $\Sigma=t$ is equal to the LST of $T_d$ given $\Sigma=t$ , except that $\lambda$ should be replaced by $\lambda q$ , which is the arrival rate of infectious customers (because for $T_d$ it does not make a difference whether or not non-infected customers arrive after C joined the system). Hence, using (17),

(32) \begin{equation} \mathbb{E} \bigl[{\mathrm{e}}^{-\theta_2 Z_d}\mid \Sigma=t \bigr] = {\mathrm{e}}^{-\lambda q t} \exp\biggl({\lambda q \biggl( \frac{1 - {\mathrm{e}}^{-\theta_2 t}}{\theta_2} \biggr)}\biggr) . \end{equation}

Finally, consider $Z_a$ . We can write

(33) \begin{align} \mathbb{E} \bigl[ {\mathrm{e}}^{-\theta_1 Z_a} \mid X_a=k,\Sigma =t \bigr] &= \sum^{k}_{j=0} \binom{k}{j} q^{\,j} \biggl( \dfrac{1}{t} \int^t_0 \,{\mathrm{e}}^{-\theta_1 u} \,{\mathrm{d}} u \biggr)^j (1-q)^{k-j} \nonumber \\[3pt] &= \biggl[ q \biggl( \dfrac{1 - {\mathrm{e}}^{-\theta_1 t}}{\theta_1 t} \biggr) +1-q \biggr]^k . \end{align}

Combining (31), (32), and (33), we get

\begin{align*}& \mathbb{E} \bigl[{\mathrm{e}}^{-\theta_1 Z_a-\theta_2 Z_d} \bigr] \notag\\[3pt] &\quad = \sum_{k=0}^\infty (1-\rho) \rho^k \\[3pt] &\quad\quad \times\int_{t=0}^\infty \mu \dfrac{(\mu t)^k}{k!} \,{\mathrm{e}}^{-\mu t} \biggl[ q \biggl( \dfrac{1 - {\mathrm{e}}^{-\theta_1 t}}{\theta_1 t} \biggr) +1-q \biggr]^k \,{\mathrm{e}}^{-\lambda q t} \exp\biggl({\lambda q \biggl(\frac{1 - {\mathrm{e}}^{-\theta_2 t}}{\theta_2} \biggr)}\biggr) \,{\mathrm{d}} t ,\end{align*}

and summing over k gives the statement of the theorem.

It immediately follows from the above theorem that the total overlapping time Z with infectious customers has LST

(34) \begin{align} \mathbb{E} \bigl[{\mathrm{e}}^{-\theta Z} \bigr] & = \mathbb{E} \bigl[ {\mathrm{e}}^{-\theta (Z_a + Z_d)} \bigr] \nonumber \\[3pt] &= (1-\rho) \mu \int^{\infty}_0 \,{\mathrm{e}}^{-\mu t + \lambda (1-q) t} \exp\biggl({\lambda q \biggl( \frac{1- {\mathrm{e}}^{-\theta t}}{\theta } \biggr)}\biggr) \,{\mathrm{e}}^{ -\lambda q t } \exp\biggl({ \lambda q \biggl( \frac{1- {\mathrm{e}}^{-\theta t}}{\theta } \biggr) }\biggr) \,{\mathrm{d}} t \nonumber \\[3pt] &= (1-\rho) \mu \int^{\infty}_0 \,{\mathrm{e}}^{-(\mu - \lambda) t } \,{\mathrm{e}}^{- 2\lambda q t} \exp\biggl({ 2 \lambda q \biggl( \frac{1- {\mathrm{e}}^{-\theta t}}{\theta } \biggr)}\biggr) \,{\mathrm{d}} t .\end{align}

The LSTs of $Z_a$ and of $Z_d$ immediately follow from Theorem 4 by taking $\theta_2=0$ (resp. $\theta_1=0$ ):

(35) \begin{equation} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta Z_a}\bigr] = \mathbb{E}\bigl[{\mathrm{e}}^{-\theta Z_d}\bigr] = (1-\rho) \mu \int_0^\infty \,{\mathrm{e}}^{-\mu t} \exp\biggl({\lambda q \biggl( \frac{1-{\mathrm{e}}^{-\theta t}}{\theta} \biggr)}\biggr) \,{\mathrm{d}} t .\end{equation}

Moments of Z, $Z_a$ , and $Z_d$ can be derived in exactly the same way in which moments of T, $T_a$ , and $T_d$ were obtained in previous sections. In particular,

\begin{align*} \mathbb{E}[Z_a] & = \mathbb{E}[Z_d] = \dfrac{\rho q}{\mu (1-\rho)^2} ,\\[3pt] \mathbb{E}[Z] & = \dfrac{2\rho q}{\mu (1-\rho)^2}, \\[3pt] \mathbb{E}[Z^2] & = \dfrac{4q\rho(1-\rho) + 24 q^2 \rho^2}{\mu^2 (1-\rho)^4} . \end{align*}

For $q=1$ , we find agreement with results from Section 3.

The heavy-traffic behavior of Z, $Z_a$ , and $Z_d$ can be derived from (34) and (35) in exactly the same way in which the heavy-traffic limits in Section 4 were obtained. We find

\begin{equation*} \lim_{\rho \uparrow 1} \mathbb{E}\bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta Z}\bigr] =\int_0^\infty \mu {\mathrm{e}}^{-\mu y} \,{\mathrm{e}}^{-\mu q \theta y^2} \,{\mathrm{d}} y =\int_0^\infty \dfrac{1}{2} \sqrt{\dfrac{\mu}{qz}} \,{\mathrm{e}}^{-\sqrt{{{\mu z}/{q}}}} \,{\mathrm{e}}^{-\theta z} \,{\mathrm{d}} z ,\end{equation*}

and

\begin{align*} \lim_{\rho \uparrow 1} \mathbb{E}\bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta Z_a}\bigr] & = \lim_{\rho \uparrow 1} \mathbb{E}\bigl[{\mathrm{e}}^{-(1-\rho)^2 \theta Z_d}\bigr] \notag \\[3pt] & =\int_0^\infty \mu {\mathrm{e}}^{-\mu y} \,{\mathrm{e}}^{-\frac{1}{2} \mu q \theta y^2} \,{\mathrm{d}} y \notag \\[3pt] & =\int_0^\infty \sqrt{\dfrac{\mu}{2qz}} \,{\mathrm{e}}^{-\sqrt{{{\mu z}/{q}}}} \,{\mathrm{e}}^{-\theta z} \,{\mathrm{d}} z .\end{align*}

In other words, in the heavy-traffic limit, $(1-\rho)^2 Z$ has a $\operatorname{Weibull}({{1}/{2}},{{q}/{\mu}})$ distribution, and $(1-\rho)^2 Z_a$ and $(1-\rho)^2 Z_d$ have a $\operatorname{Weibull}({{1}/{2}},{{q}/{(2\mu)}})$ distribution.

7. The $\mathrm{M}_{\lambda}/\mathrm{G}/1$ queue

In this section we briefly consider the overlapping time distribution in the $\mathrm{M}_{\lambda}/\mathrm{G}/1$ queue, with a general non-specified service time distribution. We obtain expressions for the LST of $T_a$ and of $T_d$ , and point out why it is difficult to obtain the LST of their joint distribution.

Consider the $\mathrm{M}_{\lambda}/\mathrm{G}/1$ queue with generic service time S. If a tagged customer meets a non-empty system, then we denote the residual part of the ongoing service by R; as before, $X_a$ denotes the number of customers seen upon arrival by the tagged customer. We can write

(36)

Here the expectation term in the first line of the right-hand side follows by using property $\mathcal{P}$ ; that property is also used in the last line for the $\theta_2$ -term, while the $\theta_1$ -term is obtained by realizing that, given $X_a=k$ and $R=v$ , we have $T_a = kv + \sum_{j=1}^{k-1} (k-j) S_j$ .

For $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ , $X_a$ and R are independent, and we have an explicit expression for $\mathbb{P}(X_a=k,R \in(v,v+{\mathrm{d}} v))$ . However, for any other service time distribution, $X_a$ and R are dependent and we only have their transform $\mathbb{E}\bigl[z^{X_a} \,{\mathrm{e}}^{-\theta R}\bigr]$ ; see for example the work by Boxma [Reference Boxma4] and Wishart [Reference Wishart30]. This transform is given by

(37) \begin{equation} \mathbb{E}\bigl[z^{X_a} \,{\mathrm{e}}^{-\theta R}I(X_a>0)\bigr] = (1-\rho)z \dfrac{\lambda(1-z)}{\mathbb{E}\bigl[{\mathrm{e}}^{-\lambda(1-z)S}\bigr] -z} \dfrac{\mathbb{E}\bigl[{\mathrm{e}}^{-\theta S}\bigr] - \mathbb{E}\bigl[{\mathrm{e}}^{-\lambda(1-z) S}\bigr]}{\lambda(1-z)-\theta} ,\end{equation}

with $I(\! \cdot \!)$ denoting an indicator function. It seems very hard to work out the inverse of this double transform, and subsequently (36).

Below we (i) determine the marginal LST of $T_d$ (for which an alternative derivation is possible, resulting in a much simpler expression), (ii) consider the marginal LST of $T_a$ (which is quite complicated), (iii) indicate how (36) simplifies for $\mathrm{M}_{\lambda}/\mathrm{E}_2/1$ and $\mathrm{M}_{\lambda}/\mathrm{D}/1$ , and (iv) obtain expressions for $\mathbb{E}[T_a]$ and $\mathbb{E}[T_d]$ and indicate how higher moments can be derived.

(i) The marginal LST of $T_d$ . To derive the marginal LST of $T_d$ , we do not need to use (36). Instead we can exploit the following. Firstly, the LST of the sojourn time $\Sigma$ in the $\mathrm{M}_{\lambda}/\mathrm{G}/1$ queue is known. Secondly, given that $\Sigma = t$ and that there are j arrivals in (0, t), those arrival epochs are distributed as the order statistics of j random variables which are U(0, t)-distributed. Using property $\mathcal{P}$ , we thus obtain

(38) \begin{equation}\mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_d}\bigr] = \int_{t=0}^\infty \sum_{j=0}^\infty \,{\mathrm{e}}^{-\lambda t} \dfrac{(\lambda t)^{\,j}}{j!} \biggl(\dfrac{1-{\mathrm{e}}^{-\theta t}}{\theta t} \biggr)^j \,{\mathrm{d}} \mathbb{P}(\Sigma < t)= \int_{t=0}^\infty \,{\mathrm{e}}^{-\lambda t + ({{\lambda}/{\theta}})(1-{\mathrm{e}}^{-\theta t})} \,{\mathrm{d}} \mathbb{P}(\Sigma < t) .\end{equation}

For service time distributions for which the distribution of $\Sigma$ is explicitly known, this might allow one to obtain a fairly tractable expression for the LST of $T_d$ . Furthermore, moments of $T_d$ can be derived in a quite straightforward manner (see (iv) below).

(ii) The marginal LST of $T_a$ . From (36),

(39) \begin{align}\mathbb{E} \bigl[{\mathrm{e}}^{-\theta T_a} \bigr] &= \mathbb{P}(X_a=0)+ \sum_{k=1}^\infty \prod_{j=1}^{k-1} \mathbb{E}\bigl[{\mathrm{e}}^{-(k-j) \theta S_j} \bigr] \int_0^\infty \,{\mathrm{e}}^{-k\theta v} \,{\mathrm{d}} _v \mathbb{P}(X_a=k, R < v)\nonumber\\[3pt]&= 1-\rho +\sum_{k=1}^\infty \prod_{j=1}^{k-1} \mathbb{E}\bigl[{\mathrm{e}}^{-j \theta S} \bigr] \mathbb{E}\bigl[{\mathrm{e}}^{-k \theta R} I(X_a=k) \bigr] .\end{align}

Inversion of (37) with respect to the generating function of $X_a$ would give $\mathbb{E}\bigl[{\mathrm{e}}^{-\theta R} I(X_a=k)\bigr]$ , which features in (39). However, even such an inversion in one of the two variables is not straightforward (it could be done formally via contour integration). For a general service time distribution, one could derive moments of $T_a$ from (39); see (iv) below.

(iii) The $\mathrm{M}_{\lambda}/\mathrm{E}_2/1$ and $\mathrm{M}_{\lambda}/\mathrm{D}/1$ cases. In the case of an $\operatorname{Erlang-2}$ -distributed service time with mean $2/\mu$ , one has

\begin{align*} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta S}\bigr] = \biggl(\frac{\mu}{\mu+\theta}\biggr)^2. \end{align*}

Observing that the denominator of (37) contains a factor $1-z$ and the numerator a factor $\theta - \lambda(1-z)$ , (37) simplifies to

\begin{equation*} \mathbb{E}\bigl[z^{X_a} \,{\mathrm{e}}^{-\theta R}I(X_a>0)\bigr] = \biggl(\dfrac{\mu}{\mu+\theta} \biggr)^2\dfrac{\lambda(1-\rho)z}{\lambda^2 z^2 -(\lambda^2+2\lambda \mu )z+ \mu^2} [2\mu+\theta +\lambda -\lambda z] . \end{equation*}

This can be rewritten in the form

\begin{equation*} \mathbb{E}\bigl[z^{X_a} \,{\mathrm{e}}^{-\theta R}I(X_a>0)\bigr] = \biggl(\dfrac{\mu}{\mu+\theta} \biggr)^2 z \biggl[\dfrac{A_1(\theta)}{z-z_1} + \dfrac{A_2(\theta)}{z-z_2}\biggr], \end{equation*}

with $z_1$ and $z_2$ the two zeros of $\lambda^2 z^2 -(\lambda^2 +2\lambda \mu)z+ \mu^2=0$ , while $A_1(\theta), A_2(\theta)$ are also easily determined. It is now trivial to obtain $\mathbb{E}\bigl[{\mathrm{e}}^{-\theta R}I(X_a=k)\bigr]$ , and even $\mathbb{P}(X_a=k,R < v)$ . A similar procedure can be followed for $\mathrm{M}/\mathrm{E}_k/1$ with $k \geq 3$ , but now one has to find the zeros $z_1,\ldots,z_k$ of a kth-order polynomial in z.

If the service time equals a constant, say D, then equation (36) simplifies to

(40) \begin{align} \mathbb{E}\bigl[{\mathrm{e}}^{-\theta_1T_a-\theta_2T_d} \bigr]& = \mathbb{P}(X_a=0) \,{\mathrm{e}}^{-\lambda D} \exp\biggl({\lambda \frac{1-{\mathrm{e}}^{-\theta_2 D}}{\theta_2}}\biggr) \nonumber\\[3pt]&\quad + \sum_{k=1}^{\infty} \int_{v=0}^{\infty}\mathbb{P}(X_a=k,R \in (v,v+{\mathrm{d}} v))\nonumber\\[3pt]&\quad\quad \times{\mathrm{e}}^{-[k v + \frac{1}{2}k(k-1)D]\theta_1} \,{\mathrm{e}}^{-\lambda(v+ kD)}\exp\biggl({\lambda \frac{1 - {\mathrm{e}}^{-(v+kD)\theta_2}}{\theta_2}}\biggr) .\end{align}

For $\mathrm{M}_{\lambda}/\mathrm{D}/1$ , (37) becomes

\begin{equation*}\mathbb{E}\bigl[z^{X_a} \,{\mathrm{e}}^{-\theta R}I(X_a>0) \bigr] = (1-\rho)z\dfrac{\lambda(1-z)}{{\mathrm{e}}^{-\lambda(1-z)}-z} \dfrac{{\mathrm{e}}^{-\theta D} - {\mathrm{e}}^{-\theta \lambda(1-z)}}{\lambda(1-z)-\theta} .\end{equation*}

Even if one is able to invert this transform, for example by using contour integration expressions such as

\begin{align*}\mathbb{P}(X_a=k) = \frac{1}{2 \pi {\mathrm{i}}} \int \frac{\mathbb{E}[z^{X_a}]}{z^{k+1}} \,{\mathrm{d}} z, \end{align*}

then one still has to sum over k, with terms like ${\mathrm{e}}^{-\frac{1}{2} k(k-1) D \theta_1}$ , in order to evaluate (40), and this seems to be very difficult.

(iv) The moments of $T_a$ and $T_d$ for $\mathrm{M}_{\lambda}/\mathrm{G}/1$ . Via differentiation of the expression in (36), or (39) and (38), or by reasonably straightforward probabilistic reasoning, it follows that

(41) $$\begin{align}\mathbb{E}[T_a] & = \sum_{k=1}^{\infty} \int_{v=0}^{\infty} \Biggl(kv+ \sum_{j=1}^{k-1} (k-j) \mathbb{E}[S_j]\Biggr) \,\mathbb{P}(X_a=k,R \in (v,v+{\mathrm{d}} v)) \nonumber\\[3pt] & = \sum_{k=1}^{\infty} \int_{v=0}^{\infty} \biggl(kv+ \dfrac{1}{2} k(k-1) \mathbb{E}[S]\biggr) \mathbb{P}(X_a=k,R \in (v,v+{\mathrm{d}} v)) \nonumber \\[3pt] & = \mathbb{E}[X_aR] + \dfrac{1}{2} \mathbb{E}[X_a(X_a-1)] \mathbb{E}[S],\end{align}$$

(42) $$\begin{align}\mathbb{E}[T_d] & = \mathbb{P}(X_a=0) \lambda \dfrac{\mathbb{E}[S^2]}{2} \nonumber \\[3pt] &\quad +\sum_{k=1}^{\infty} \int_{v=0}^{\infty} \dfrac{\lambda}{2} \mathbb{E}\Biggl[\Biggl(v + \sum_{j=1}^{k-1} S_j + S\Biggr)^2 \Biggr] \,\mathbb{P}(X_a=k,R \in (v,v+{\mathrm{d}} v)) \nonumber\\[3pt] & = (1-\rho) \lambda \dfrac{\mathbb{E}[S^2]}{2} + \dfrac{\lambda}{2} \rho \mathbb{E}[R^2] + \rho \mathbb{E}[X_aR] \nonumber\\[3pt] &\quad + \dfrac{\lambda}{2} \mathbb{E}[X_a(X_a-1)] \mathbb{E}[S]^2 + \dfrac{\lambda}{2} \mathbb{E}[X_a] \mathbb{E}[S^2].\end{align}$$

From formula (2.7) of Boxma [Reference Boxma4] we have

\begin{equation*} \mathbb{E}[X_aR] = \dfrac{\lambda^3 (\mathbb{E}[S^2])^2}{4(1-\rho)} + \dfrac{\lambda^2 \mathbb{E}[S^3]}{6} + \dfrac{\lambda \mathbb{E}[S^2]}{2}.\end{equation*}

An expression for $\mathbb{E}[X_a(X_a-1)]$ rarely occurs in the literature. It readily follows from the well-known expression for the PGF $\mathbb{E}[z^{X_a}]$ (see equation (II.4.17) in Cohen [Reference Cohen11]) that, for $\mathrm{M}_{\lambda}/\mathrm{G}/1$ FCFS,

\begin{equation*} \mathbb{E}[X_a(X_a-1)] = \dfrac{\lambda^4 (\mathbb{E}[S^2])^2}{2(1-\rho)^2} + \dfrac{\lambda^3 \mathbb{E}[S^3]}{3(1-\rho)} + \dfrac{\lambda^2 \mathbb{E}[S^2]}{1-\rho} . \end{equation*}

Now, working out the expressions in (41) and (42) reveals that

(43) \begin{equation} \mathbb{E}[T_a] = \mathbb{E}[T_d] = \dfrac{\lambda^3 (\mathbb{E}[S^2])^2}{4(1-\rho)^2} + \dfrac{\lambda^2 \mathbb{E}[S^3]}{6(1-\rho)} + \dfrac{\lambda \mathbb{E}[S^2]}{2(1-\rho)} ,\end{equation}

and hence the mean total overlapping time in the $\mathrm{M}_{\lambda}/\mathrm{G}/1$ FCFS queue is given by

\begin{equation*} \mathbb{E}[T] = \dfrac{\lambda^3 (\mathbb{E}[S^2])^2}{2(1-\rho)^2} + \dfrac{\lambda^2 \mathbb{E}[S^3]}{3(1-\rho)} + \dfrac{\lambda \mathbb{E}[S^2]}{1-\rho} . \end{equation*}

For $\mathrm{M}_{\lambda}/\mathrm{M}_{\mu}/1$ FCFS, this reduces to the expression obtained in (24).

To determine moments of $T_d$ , it actually seems easier to differentiate the expression in (38) with respect to $\theta$ . In that way it readily follows that

(44) $$\begin{align}\mathbb{E}[T_d] & = \dfrac{\lambda}{2} \mathbb{E}[\Sigma^2],\end{align}$$

(45) $$\begin{align}\mathbb{E}[T_d^2] & = \dfrac{\lambda^2}{4} \mathbb{E}[\Sigma^4] + \dfrac{\lambda}{3} \mathbb{E}[\Sigma^3].\end{align}$$

Equation (44) is in agreement with Corollary 1 of Oz and Perlman [Reference Oz and Perlman21], which states that $\mathbb{E}[T] = \lambda \mathbb{E}[\Sigma^2]$ for overtake-free queues with Poisson( $\lambda$ ) arrivals, where customers depart in the order of their arrival. In combination with an $\mathrm{M}_{\lambda}/\mathrm{G}/1$ result on page 256 of Cohen [Reference Cohen11] we retrieve the second part of (43).

By differentiating (39) twice, and using the identities

\begin{align*}\sum_{j=1}^{k-1} j^2 = \frac{1}{6}(k-1)k(2k-1)\quad\text{and}\quad\sum_{j=1}^{k-1} j^3 = \frac{1}{4}(k-1)^2k^2 , \end{align*}

one can derive the second moment of $T_a$ :

(46) \begin{align} \mathbb{E}\big[T_a^2\big]&= \mathbb{E}\big[X_a^2R^2\big] + 2 \mathbb{E}[S] \mathbb{E}\big[X_a^2(X_a-1) R\big] +\dfrac{1}{3} \mathbb{E}[S^2] \mathbb{E}\big[(X_a-1)X_a(2X_a-1)\big]\nonumber \\[3pt]&\quad + \mathbb{E}[S^2] \mathbb{E} \biggl[\dfrac{1}{4} (X_a-1)^2 X_a^2 - \dfrac{1}{6} (X_a-1)X_a(2X_a-1) \biggr] .\end{align}

Finally, by differentiating (36) once with respect to both z and $\theta$ , we get

(47) \begin{align} \mathbb{E}\big[T_aT_d\big] &= \dfrac{1}{2} \mathbb{E}\bigl[X_aR^2\bigr] + \dfrac{1}{2} \mathbb{E}[S] \mathbb{E}\bigl[X_a^2R\bigr] + \dfrac{1}{4} \mathbb{E}[S] \mathbb{E}\bigl[(X_a^2-X_a)R\bigr] \nonumber \\[3pt] &\quad + \dfrac{1}{4} (\mathbb{E}[S])^2 \mathbb{E}[X_a(X_a-1)] + \dfrac{1}{4} \mathbb{E}[S^2] \mathbb{E}[X_a(X_a-1)] \nonumber \\[3pt] &\quad + \dfrac{1}{4} (\mathbb{E}[S])^2 \mathbb{E}[X_a(X_a-1)(X_a-2)] .\end{align}

All these terms can be evaluated using (37). In particular, using (45)–(47), one can thus obtain the second moment of the total overlapping time for the $\mathrm{M}_{\lambda}/\mathrm{G}/1$ FCFS queue, using $\mathbb{E}[T^2] = \mathbb{E}[T_a^2] + 2 \mathbb{E}[T_aT_d] + \mathbb{E}[T_d^2]$ . We omit the details.

8. The $\mathrm{G}/\mathrm{M}_{\mu}/1$ queue

In this section we describe some extensions of our work to the $\mathrm{G}/\mathrm{M}_{\mu}/1$ queue, with generic interarrival time A and service rate $\mu$ where $\mu \mathbb{E}[A] > 1$ . We show that understanding the overlapping time in this case is also quite non-trivial. We start with the LST of the overlapping time experienced by the tagged customer upon arrival, i.e. $T_a$ . We let $\rho^*$ be the unique solution to $x = \mathbb{E}[{\mathrm{e}}^{-\mu(1-x)A} ]$ in [0,1].

Lemma 5. For the $\mathrm{G}/\mathrm{M}_{\mu}/1$ queue, let $\mu \mathbb{E}[A] > 1$ and $\theta>0$ . Then

\begin{equation*} \mathbb{E}\bigl[ {\mathrm{e}}^{-\theta T_a }\bigr] = (1-\rho^*) \, {}_1F_1\biggl(1;\, 1 + \dfrac{\mu}{\theta};\, \dfrac{\rho^* \mu}{\theta}\biggr),\end{equation*}

where ${}_1F_1(a;\,b;\,z)$ denotes the confluent hypergeometric function of the first kind.

Proof. We have

\begin{align*}\mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_a }\bigr] &= \sum_{k=0}^{\infty} \mathbb{P}( X_a = k ) \mathbb{E}\bigl[{\mathrm{e}}^{-\theta T_a } \mid X_a = k \bigr] \\&= \sum_{k=0}^{\infty} (1-\rho^*) (\rho^*)^k \prod_{j=1}^{k} \dfrac{\mu}{\mu + j\theta} \nonumber \\&= \sum_{k=0}^{\infty} (1-\rho^*) (\rho^*)^k \biggl(\dfrac{\mu}{\theta} \biggr)^k\prod_{j=1}^{k} \dfrac{1}{{{\mu}/{\theta}} + j} \nonumber \\&= (1-\rho^*) \sum_{k=0}^{\infty} \dfrac{ (\rho^*)^k ({{\mu}/{\theta}} )^k }{( {{\mu}/{\theta}} + 1 )_k} \nonumber \\&= (1-\rho^*) \sum_{k=0}^{\infty} \dfrac{ (\rho^*)^k ({{\mu}/{\theta}} )^k ( 1 )_k}{( {{\mu}/{\theta}} + 1 )_k \ k!} \nonumber \\&= (1-\rho^*) \, {}_1F_1\biggl(1;\, 1 + \dfrac{\mu}{\theta};\, \dfrac{\rho^* \mu}{\theta}\biggr). \\[-35pt] \end{align*}

Remark 6. In the case where the arrival process is deterministic with $A \equiv \Delta$ , yielding a $\mathrm{D}/\mathrm{M}_{\mu}/1$ queue, the fixed point for $x = {\mathrm{e}}^{-\mu(1-x) \Delta}$ equals

\begin{align*}\rho^* = -\frac{W_{-1}( - {\mathrm{e}}^{-\Delta \mu} \Delta \mu )}{\Delta \mu}, \end{align*}

where $W_{-1}(x)$ is defined as the principal branch of the Lambert W function when $-1/{\mathrm{e}} \leq x \leq 0$ . Moreover, in the case where the arrival process is an $\operatorname{Erlang}(2,\lambda)$ , yielding an $\mathrm{E}_2/\mathrm{M}_{\mu}/1$ queue, we have the fixed point

\begin{align*}\rho^* = \frac{\lambda}{\mu} + \frac{1}{2} - \sqrt{\frac{\lambda}{\mu} + \frac{1}{4}}. \end{align*}

It is challenging to calculate the LST for $T_d$ for the $\mathrm{G}/\mathrm{M}_{\mu}/1$ queue. However, we show that it is possible to calculate it for the $\mathrm{D}/\mathrm{M}_{\mu}/1$ queue and show a connection to partial theta functions.

Lemma 6. For the $\mathrm{D}/\mathrm{M}_{\mu}/1$ queue where $\Delta$ is the interarrival time and $m = \lfloor {{x}/{\Delta}} \rfloor$ , then we have the following expression for the LST of $T_d$ :

\begin{align*}\mathbb{E}\bigl[ {\mathrm{e}}^{-\theta T_d }\bigr] = \sum^{\infty}_{m=0} \dfrac{\mu(1-\rho^*) \,{\mathrm{e}}^{-\mu(1-\rho^*) \Delta m}}{\mu(1-\rho^*) + \theta m} \bigl( 1 - {\mathrm{e}}^{-\mu(1-\rho^*) \Delta} \,{\mathrm{e}}^{-\theta \Delta m} \bigr)\exp\biggl({-\theta \frac{m(m-1)}{2} \Delta}\biggr) . \end{align*}

Proof. When the sojourn time is equal to x, then it is only possible for $m = \lfloor {{x}/{\Delta}} \rfloor$ customers to enter the queue. This also implies that the overlapping time is equal to

\begin{equation*} T_d = \sum^{m}_{j=1} (x- j \Delta) = mx - \dfrac{m(m+1)}{2}\Delta = \dfrac{m(m-1)}{2} \Delta + m(x-m \Delta).\end{equation*}

Therefore, using the fact that the sojourn time in $\mathrm{G}/\mathrm{M}_{\mu}/1$ is $\operatorname{Exp}\!(\mu(1-\rho^*))$ , we obtain the following LST expression for $T_d$ :

\begin{align*}\mathbb{E}\bigl[ {\mathrm{e}}^{-\theta T_d }\bigr] &= \int^{\infty}_0 \mathbb{E}\bigl[ {\mathrm{e}}^{-\theta T_d } \mid \mathcal{S} = x\bigr] \mu (1-\rho^*) \,{\mathrm{e}}^{-\mu(1 - \rho^*)x} \,{\mathrm{d}} x \\[3pt]&= \sum^{\infty}_{m=0} \int^{\Delta(m+1)}_{\Delta m} \mu (1-\rho^*) \,{\mathrm{e}}^{-(\mu(1-\rho^*) + \theta m) x} \,{\mathrm{e}}^{\theta m^2 \Delta} \exp\biggl({-\theta \frac{m(m-1)}{2} \Delta}\biggr) \,{\mathrm{d}} x \nonumber \\[3pt]&= \sum^{\infty}_{m=0} \dfrac{\mu(1-\rho^*) \,{\mathrm{e}}^{-\mu(1-\rho^*) \Delta m}}{\mu(1-\rho^*) + \theta m} \bigl( 1 - {\mathrm{e}}^{-\mu(1-\rho^*) \Delta} \,{\mathrm{e}}^{-\theta \Delta m} \bigr) \exp\biggl({-\theta \frac{m(m-1)}{2} \Delta}\biggr) . \\[-35pt] \end{align*}

It is important to note that this expression is connected to partial theta functions, which are a class of special functions that arise in number theory and combinatorics. These partial theta functions are given by

\begin{equation*} \psi(a,q) = \sum^{\infty}_{n=0} a^n q^{{{n(n-1)}/{2}}}, \quad |q|<1\end{equation*}

and are the partial sums of the standard Jacobi theta functions, which sum over the positive and negative integers [Reference Andrews and Warnaar2, Reference Warnaar29]. Recently, these partial theta functions have appeared in queueing models with exponential/geometric balking; see for example Pender [Reference Pender26]. Thus this correspondence provides a novel connection between stochastic service systems and classical objects of analytic number theory. Moreover, the quadratic decay of the summand ensures rapid convergence of the infinite series and makes the numerical evaluation via truncation both efficient and highly accurate.