Stability conditions for a multiserver queueing system with a regenerative input flow and simultaneous service of a customer by a random number of servers

Abstract

We investigate the stability condition of a multiserver queueing system. Each customer needs simultaneously a random number of servers to complete the service. The times taken by each server are independent. The input flow is assumed to be a regenerative one. The service time has an exponential, phase-type or hyper-exponential distribution. The stability criteria for the models are established. It turns out that the stability conditions do not depend on the structure of the input flow, but only on the rate of the process. However, the distribution of the service times is a very important factor. We give examples which show that the stability condition cannot be expressed only in terms of the mean of the service time.

Appendices

Appendix

Proof of Lemma 1

We begin with the first statement. Let ${\mathcal {J}}=\bigcup \limits _{l=0}^\infty [\theta _l, \theta _l+v_{l})$ so that $T_1=\min \{s_k>0:s_k\in {\mathcal {J}}\}$. Define the sequence $\{e_n\}_{n=0}^\infty$ ($e_0=0$) letting $e_n=0$ if $s_n\in {\mathcal {J}}$ and $e_n=1$ otherwise. Taking into account the memoryless property of the exponential distribution, the strongly regenerative structure of X and the assumption that $\{\gamma _n\}_{n=1}^\infty$ is a sequence of iid random variables, one may easily notice that $\{e_n\}_{n=1}^\infty$ ($e_0=0$) is a discrete-time regenerative process (see Ch.10, Sec. 3.5 in [34]). Points of the regeneration of this process are determined by the recursion

$$\begin{aligned} n_k=\min \{n>n_{k-1}:s_n\in {\mathcal {J}}\},\quad n_0=0. \end{aligned}$$

Define the events

$$\begin{aligned} B_n=\{e_n=0\},\quad A_n=\{e_n=0,e_k\ne 0, k=1,2,\ldots ,n-1\} \end{aligned}$$

and let $P_n={\mathsf {P}}(B_n)$, $f_n={\mathsf {P}}(A_n)$, $n=1,2,\ldots$. Then

$$\begin{aligned} {\mathsf {P}}(T_1<\infty )=\sum \limits _{n=1}^\infty {\mathsf {P}}(A_n)=\sum \limits _{n=1}^\infty f_n. \end{aligned}$$

(28)

Since $A_j\bigcap A_k=\emptyset$ if $j\ne k$ and $B_n=\bigcup \limits _{j=1}^n A_j\bigcap B_n$ we have

$$\begin{aligned} P_n=\sum \limits _{j=1}^n {\mathsf {P}}\left( A_j\bigcap B_n\right) =\sum \limits _{j=1}^n {\mathsf {P}}(A_j){\mathsf {P}}(B_n|A_j). \end{aligned}$$

Taking into account that the moment of arrival to the state zero is a point of regeneration of $\{e_n\}_{n=1}^\infty$, one may easily get

$$\begin{aligned} {\mathsf {P}}(B_n|A_j)= & {} {\mathsf {P}}(e_n=0|e_j=0,e_k\ne 0, k=1,2,\ldots ,j-1)\\= & {} {\mathsf {P}}(e_n=0|e_j=0)=P_{n-j},\quad (j<n). \end{aligned}$$

Therefore, the recursion relations

$$\begin{aligned}&\displaystyle P_1={\mathsf {P}}(s_1\in {\mathcal {J}}),\\&\displaystyle P_n=f_1 P_{n-1}+f_2 P_{n-2}+\ldots +f_{n-1}P_1+f_n, \quad n=1,2,\ldots \text { and } P_0=1 \end{aligned}$$

hold. Then, for the generating functions $F(z)=\sum \nolimits _{n=1}^\infty z^n f_n$ and $P(z)=\sum \nolimits _{n=1}^\infty z^n P_n$, the following equality is valid:

$$\begin{aligned} P(z)=F(z)(1-P(z)), \end{aligned}$$

and hence

$$\begin{aligned} F(z)=\frac{P(z)}{1+P(z)}. \end{aligned}$$

Since

$$\begin{aligned} {\mathsf {P}}(T_1<\infty )=\lim \limits _{z\uparrow 1}F(z), \end{aligned}$$

then ${\mathsf {P}}(T_1<\infty )=1$ if and only if

$$\begin{aligned} \lim \limits _{z\uparrow 1}P(z)=\sum \limits _{n=1}^\infty {\mathsf {P}}(s_n\in {\mathcal {J}})=\infty . \end{aligned}$$

(29)

Now we estimate $P_n={\mathsf {P}}(s_n\in {\mathcal {J}})$ as $n\rightarrow \infty$. To do this, we define for any (non-random) $t\geqslant 0$ the function

$$\begin{aligned}&h(t)={\mathsf {P}}(t\in {\mathcal {J}})={\mathsf {P}}\left( \bigcup \limits _{l=0}^\infty \{t\in [\theta _l,\theta _l+v_{l})\}\right) =\sum \limits _{l=0}^\infty {\mathsf {P}}(\theta _l\leqslant t,v_{l}>t-\theta _l)=\\&\quad =\sum \limits _{l=0}^\infty \int _{0}^{t}e^{-\alpha (t-y)}\mathrm{d}{\mathsf {P}}(\theta _l<y)=\int _{0}^{t}e^{-\alpha (t-y)}\mathrm{d}H(y), \end{aligned}$$

where $H(y)=\sum \limits _{l=0}^\infty {\mathsf {P}}(\theta _l\leqslant y)$ is a renewal function for the renewal process $\{\theta _l\}_{l=0}^\infty$. From the key renewal Theorem (Ch. 3, Par. 6, Th. 10 in [3] and Ch. 10 in [34]), it follows that there exists

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }h(t)=\frac{1}{{\mathsf {E}}\theta _1}\int _{0}^{\infty }e^{-\alpha y}\mathrm{d}y=\frac{1}{\alpha {\mathsf {E}}\theta _1}=\beta >0. \end{aligned}$$

(30)

Since $s_n\uparrow \infty$ w.p.1 as $n\rightarrow \infty$ and $\{s_k\}_{k=1}^\infty$ and $\{\theta _n,v_n\}_{n=1}^\infty$ are independent sequences, one may easily prove from (30) the convergence

$$\begin{aligned} {\mathsf {P}}(s_n\in {\mathcal {J}})\xrightarrow [n\rightarrow \infty ]{}\beta >0. \end{aligned}$$

(31)

This means that (29) is fulfilled, and therefore, the first statement is valid. The second statement follows from the fact that $\nu _1=\min \{k:s_k\in {\mathcal {J}}\}$ is a Markov moment with respect to the filtration $\{\sigma (\gamma _n,\gamma _{n+1},\ldots )\}_{n=1}^\infty$ (see [3, 9]).

Now we move to the third statement. Because of Remark 1, the strongly regenerative flow X is defined by the sequence $X_j=(x_j(t),t\in (0,v_j+u_j),v_j,u_j)\, (j=0,1,\ldots )$ of independent random elements with the help of (1). Consider the stochastic flow $X_{T_1}(t)=X(T_1+t)-X(T_1),t\geqslant 0$. First, we define the random elements $\{{\tilde{X}}_j\}_{j=0}^\infty$ which determine this process by formula (1). Let $T_1=s_\nu$ and $\zeta$ be the number of the regeneration period of X such that $s_\nu \in [\theta _\zeta ,\theta _\zeta +v_\zeta )$, so that

$$\begin{aligned} \zeta =\min \left\{ j\geqslant 0:\bigcup \limits _{k=1}^\infty \left\{ s_k\in (\theta _{j},\theta _{j}+v_j)\right\} \right\} \end{aligned}$$

and

$$\begin{aligned} {\tilde{v}}_0=\theta _{\zeta }+v_{\zeta }-s_{\nu }. \end{aligned}$$

Then

$$\begin{aligned} {\tilde{X}}_0=({\tilde{x}}_0(t),t\in (0,{\tilde{v}}_0+u_{\zeta }),{\tilde{v}}_0,u_{\zeta }). \end{aligned}$$

Here $\{{\tilde{x}}_0(t),t\in (0,{\tilde{v}}_0)\}$ is a Poisson process with rate $\delta$ not depending on the random element $X_\zeta$ and ${\tilde{x}}_0(t)=x_\zeta (t)-x_\zeta (v_\zeta )+{\tilde{x}}_0({\tilde{v}}_0)$ for $t\in ({\tilde{v}}_0,{\tilde{v}}_0+u_\zeta )$. Because of the properties of the exponential distribution (memoryless property), the Poisson process (independence of the increments) and the strongly regenerative process (properties 1 and 2), we see that ${\tilde{X}}_0$ and $X_0$ have the same distribution. The random element ${\tilde{X}}_j$ for $j>0$ is given by the formula

$$\begin{aligned} {\tilde{X}}_j=(x_{\zeta +j}(t),t\in (0,v_{\zeta +j}+u_{\zeta +j}],v_{\zeta +j}, u_{\zeta +j}). \end{aligned}$$

Since $\zeta$ is a Markov moment with respect to the filtration $\{\sigma (\theta _n,\theta _{n+1},\ldots )\}_{n=1}^\infty$ we get the equality in distribution

$$\begin{aligned} \{X_j\}_{j=0}^\infty {\mathop {=}\limits ^{(d)}}\{{\tilde{X}}_j\}_{j=0}^\infty . \end{aligned}$$

This means that the distributions of the processes $\{X(t),t\geqslant 0\}$ and $\{X_{T_1}(t),t\geqslant 0\}$ are the same.

Now we will prove the independence of $\{X(t),t\leqslant T_1\}$ and $\{X_{T_1}(t),t \geqslant 0\}$. Let us note that the first process is defined by the collection of random elements

$$\begin{aligned} C_1=\{X_j,j=1,\ldots ,\zeta , \{x_\zeta (y),y\leqslant T_1-\theta _{\zeta }\}\} \end{aligned}$$

and the second one by the collection

$$\begin{aligned} C_2=\{\{X_j\}_{j=\zeta +1}^\infty ,{\tilde{X}}_0\}. \end{aligned}$$

Therefore the independence of these processes follows from the independence of the collections $C_1$ and $C_2$, which is almost evident.

Now assume that ${\mathsf {E}}\gamma _1<\infty$. Since $T_1=s_\zeta =\gamma _1+\ldots +\gamma _{\nu _1}$ according to Wald’s identity, we have to prove the finiteness of ${\mathsf {E}}\nu _1$. To do this, define the sequence $\{\nu _n\}_{n=1}^\infty$ by the recursion formula

$$\begin{aligned} \nu _n=\min \{k>\nu _{n-1}:s_k\in {\mathcal {J}}\}, \nu _0=0. \end{aligned}$$

Because of the statements 2 and 3 of the Lemma, the sequence $\{\nu _n-\nu _{n-1}\}_{n=1}^\infty$ consists of iid integer-valued random variables. Therefore, it follows from Blackwell’s Theorem (see Part 1, Par.6, Th.5 in [3], Ch.10, sec.3 in [34]) that there exists the limit

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\sum \limits _{l=0}^\infty {\mathsf {P}}(\nu _l=n)=a \end{aligned}$$

and $a=({\mathsf {E}}\nu _1)^{-1}$ if ${\mathsf {E}}\nu _1<\infty ,a=0$ if ${\mathsf {E}}\nu _1=\infty$ (Ch.10, par 3.6 in [34]). Since

$$\begin{aligned} \sum \limits _{l=0}^\infty {\mathsf {P}}(\nu _l=n)={\mathsf {P}}(s_n\in {\mathcal {J}}), \end{aligned}$$

according to (31) we have ${\mathsf {E}}\nu _1=\alpha {\mathsf {E}}\theta _1$ and therefore ${\mathsf {E}}T_1=\alpha {\mathsf {E}}\gamma _1{\mathsf {E}}\theta _1<\infty$. $\square$

Proof of Lemma 3

Here we give only the proof of (14) for the system $S_1^0$ since the proofs of (13) for $S_1^0$ and (13) and (14) for the systems $S_2^0$ and $S_3^0$ are the same.

Let q(t) be the number of occupied servers in $S_1^0$ at time t. Denote by $S^{(j_0)}$ the system $S_1^0$ under the assumption that the initial state $q(0)=j_0$ and $S^{(k)}$ the system $S_1^0$ if $q(0)=k,\,k>j_0$. Define the processes $Z^{(j_0)}(t)$ and $Z^{(k)}(t)$ as the number of the service completions up to time t in the systems $S^{(j_0)}$ and $S^{(k)}$, respectively. We organize the performance of $S^{(j_0)}$ and $S^{(k)}$ on one probability space in such a way that

$$\begin{aligned} Z^{(j_0)}(t)\leqslant Z^{(k)}(t) \end{aligned}$$

(32)

w.p.1 for any $t\geqslant 0$. Then (14) is a simple corollary of (32).

Let $\eta _1^{(j_0)}$ and $\eta ^{(k)}$ be the moments of the first service completion in the systems $S^{(j_0)}$ and $S^{(k)}$, respectively.

These random variables have an exponential distribution with rates $j_0\mu$ and $k\mu$, respectively. Let $\{\delta _n\}_{n=1}^\infty$ be a sequence of iid random variables with uniform distribution on [0, 1]. We take

$$\begin{aligned} \eta _1^{j_0}=-\frac{1}{j_0\mu }ln(1-\delta _1),\quad \eta ^{(k)}=-\frac{1}{k\mu }ln(1-\delta _1). \end{aligned}$$

(33)

Then $\eta _1^{j_0}<\eta ^{k}$ w.p.1 and the process $Z^{(k)}$ has the first jump at time $\eta ^{k}$ but $Z^{(j_0)}(t)=0$ for $0\leqslant t\leqslant \eta ^{(k)}$. Let ${\tilde{\zeta }}_k$ be the number of servers which requires the first customer in the queue in the system $S^{(k)}$ (at time $t=0$). Then ${\tilde{\zeta }}_k$ has values $\{m-s+1, \text { where } j_0\leqslant s\leqslant k\}$. Firstly, assume that ${\tilde{\zeta }}_k=m-k+1$, i.e., $s=k$. Then in the system $S^{(k)}$ at time $\eta ^{(k)}$ the number of busy servers $q^{(k)}(\eta ^{(k)})=m$.

For the system $S^{(j_0)}$, we consider the system ${\tilde{S}}^{(j_0)}$ with the dominating process ${\tilde{Z}}^{(j_0)}$ which has a unit jump at time $\eta ^{(k)}$.

Then ${\tilde{q}}^{(j_0)}(\eta ^{(k)})=m,$ where ${\tilde{q}}^{(j_0)}$ is the number of occupied servers in the dominating system ${\tilde{S}}^{(j_0)}$. Later we assume that after the moment $\eta ^{(k)}$ the systems ${\tilde{S}}^{(j_0)}$ and $S^{(k)}$ have common sequences $\{\zeta _n\}_{n=1}^\infty$ (the number of required servers) and $\{\overrightarrow{\eta }_n\}_{n=1}^\infty$ (service times on various servers). Therefore ${\tilde{Z}}^{(j_0)}(t)=Z^{(k)}(t)$ w.p.1 for all $t\geqslant 0$. Since $Z^{(j_0)}(t)\leqslant {\tilde{Z}}^{(j_0)}(t)$ w.p.1, inequality (32) is proved.

Now assume that ${\tilde{\zeta }}_k>m-k+1$, i.e., $s<k$. Based on the memoryless property of the exponential distribution, one may easily see that at time $\eta ^{(k)}$ we have two systems $S^{(j_0)}$ and $S^{(k-1)}$ which were described earlier. We define $\eta _2^{(j_0)}$ and $\eta ^{(k-1)}$ with the help of formula (33), taking $\eta _2^{(j_0)}$, $\eta ^{(k-1)}$, $\delta _2$ instead of $\eta _1^{(j_0)}$, $\eta ^{(k)}$, $\delta _1$. If $s=k-1$ we repeat the steps described for the case $s=k$. If $s<k-1$ the process $Z^{(j_0)}$ equals zero on the time interval $(0, \eta ^{(k)}+\eta ^{(k-1)})$, while $Z^{(k)}$ has two jumps at times $\eta ^{(k)}$ and $\eta ^{(k)}+\eta ^{(k-1)}$. In this case, we also define $\eta _3^{(j_0)}$ and $\eta ^{(k-2)}$ by means of formula (33) with $\delta _3$ instead of $\delta _2$ and so on. Finally, at some time $t_0$ we obtain one of two cases:

• 1. $q^{(k)}(t_0)=m$ when ${\tilde{\zeta }}_k<m-j_0+1$;

• 2. $q^{(k)}(t_0)=j_0$ when ${\tilde{\zeta }}_k=m-j_0+1$.

Here $q^{(k)}$ is the number of occupied servers in the system $S^{(k)}$.

In the first case, we proceed in the same way as for the case ${\tilde{\zeta }}_k=m-k+1$$(s=k)$. In the second case, the states of $S^{(j_0)}$ and $S^{(k)}$ are identical but $Z^{(j_0)}(t_0)=0$, $Z^{(k)}(t_0)=k-j_0$. Therefore, inequality (32) takes place w.p.1. This completes the proof of the lemma. $\square$

Conclusion

In this paper, a multiserver queueing system is considered. In the system, each customer requires a random number of servers simultaneously and service times at the occupied servers are independent random variables. The input flow is assumed to be a regenerative one and the service time has an exponential, phase-type or hyper-exponential distribution. By means of the synchronization method, we establish the stability criterion of such systems. The main contribution of this paper is an extension of the stability criterion to the model with a regenerative input flow. Note that the class of regenerative flows is broad and includes Markov arrival process, doubly stochastic Poisson process with a regenerative process as intensity, Markov modulated process, and others. It turns out that the stability condition depends only on the intensity of the input flow and the structure of this flow does not play any role.

The distribution of the service time plays an important role in the stability condition. We give which show that the stability condition cannot be expressed in terms of the mean of the service time.