Sufficient stability conditions for multi-class constant retrial rate systems

Abstract

We study multi-class retrial queueing systems with Poisson inputs, general service times, and an arbitrary numbers of servers and waiting places. A class-i blocked customer joins orbit i and waits in the orbit for retrial. Orbit i works like a single-server $\cdot /M/1$ queueing system with exponential retrial time regardless of the orbit size. Such retrial systems are referred to as retrial systems with constant retrial rate. Our model is not only motivated by several telecommunication applications, such as wireless multi-access systems, optical networks, and transmission control protocols, but also represents independent theoretical interest. Using a regenerative approach, we provide sufficient stability conditions which have a clear probabilistic interpretation. We show that the provided sufficient conditions are in fact also necessary, in the case of a single-server system without waiting space and in the case of symmetric classes. We also discuss a very interesting case, when one orbit is unstable, whereas the rest of the system is stable.

Appendix

The proof of Lemma 2

Recall definition (11) and define the remaining time for regeneration at instant t as

$$\begin{aligned} T_i(t)=\min _n\big (T_n^{(i)}-t:T_n^{(i)}-t>0\big ). \end{aligned}$$

(56)

Thus, at each instant $z_j$ (satisfying (10)), the coloured orbit i is empty, and we first show that, with a positive probability, the next arrival can start a new i-type regeneration period. That is, its arrival instant will be the first regeneration point, $T_k^{(i)},$ for some value of k, after instant $z_j$. Denote the remaining service time at server k as $S_k(t)$ ($S_k(t)=0$ if the server is empty), $k=1,\ldots , c$. Then it is known that the processes $\{S_k(t),\,t\ge 0\}$ are tight for any value of k [20]. Because the sequence $\{z_j\}$ is deterministic, the sequences $\{S_k(z_j),\,j\ge 1\}$ are tight for any value of k as well. Also note that the number of customers present in the buffer (without servers) is limited by $K-c$. Denote by $S^{(i)}$ a generic service time of a class-i customer. Then the service time of any customer in the system is (stochastically) upper bounded by the random variable $\phi :=\max _i S^{(i)}$ with finite mean. It then follows that the workload accumulated in the buffer is (stochastically) upper bounded by $\phi _1+\cdots +\phi _{K-c}$, where $\{\phi _i\}$ are i.i.d. random variables distributed as $\phi$. It now follows from the tightness and from (10) that there exists a constant $C<\infty$ such that

$$\begin{aligned} \min _j\mathsf{{P}}\left( C_i(z_j)=0,\, \sum _{r=1}^{K-c} \phi _r+\sum _{k=1}^c S_k(z_j)\le C\right) \ge \frac{\varepsilon }{2}. \end{aligned}$$

(57)

We note that the event within parentheses in (57) means that the i-th coloured orbit and the primary system become empty at instant $t+C$, provided no new customer arrives in the interval $[z_j,\,z_j+C]$. If, in addition, a new class-i customer arrives in the interval $[z_j+C,\,z_j+C+x]$, then a regeneration is initiated in this interval. Consequently, it easily follows that for any $x\ge 0$

$$\begin{aligned}&\inf _j\mathsf{{P}}(T_i(z_j)\le x+C)\nonumber \\&\quad \ge \frac{\varepsilon }{2}\left( e^{-(\lambda _i+\mu _0^{(i)}) C} - e^{-(\lambda _i+\mu _0^{(i)}) (C+x)}\right) e^{-\sum _{l\not =i} (\lambda _l+\mu _0^{(l)}) (C+x)}>0, \end{aligned}$$

(58)

where the product of the two exponential terms on the right-hand side represents the probability that class-i, and only class-i, customers arrive in the interval $[z_j+C,\,z_j+C+x]$. Since the lower bound is uniform in j, we have that $T_i(t)\not \Rightarrow \infty$, (where $\Rightarrow$ stands for convergence in probability), and it follows that $\mathsf{{E}}T_i<\infty$, where $T_i$ is the generic regeneration period of the process $\{C_i(t),\,t\ge 0\}$ [16]. In particular, the process $\{C_i(t)\}$ is tight. This conclusion holds for any value of i, and thus the summary orbit size, $\big \{\sum _iC_i(t):=\mathbf{C} (t),\,t\ge 0\big \}$, is a tight process as well. In particular, for each $\varepsilon _1>0$, there exists a constant ${\mathcal {C}}_0$ such that

$$\begin{aligned} \inf _t\mathsf{{P}}(\mathbf{C} (t)\le {\mathcal {C}}_0)\ge 1- \varepsilon _1. \end{aligned}$$

(59)

As a next step we show that, for each instant t, it is possible to unload both the primary system and all coloured orbits, to obtain a new regeneration point $T_n$ in an interval $[t,\, t+ D]$ with a positive probability, where D is a finite constant. The main idea is to unload coloured orbit 1, then orbit 2, etc., within the interval $[t,\, t+ D]$, provided that no new arrivals occur during this interval. Recall that the remaining service times $S_k(t)$ are tight, so, by (59), we can take a finite constant ${\mathcal {D}}_0$ such that

$$\begin{aligned} \inf _t \mathsf{{P}}\left( \mathbf{C} (t)\le {\mathcal {C}}_0,\, \sum _{r=1}^{K-c} \phi _r+\sum _{k=1}^c S_k(t)\le {\mathcal {D}}_0\right) \ge 1-\frac{\varepsilon _1}{2}. \end{aligned}$$

(60)

Fix an arbitrary instant t and a constant $\Delta >0$ and denote

$$\begin{aligned} \gamma =\min _i \big (1-e^{-\Delta \mu _0^{(i)}}\big ). \end{aligned}$$

It is easy to see that the probability that a coloured customer (if any) makes an attempt to enter the primary system in the interval $[t,\,t+\Delta ]$ is not less than $\gamma$. Note that $\mathsf{{P}}(C_i(t)\le {\mathcal {C}}_0)\ge 1- \varepsilon _1$ for any value of i. Now, for a given constant $\zeta >0$, take constant a such that $\mathsf{{P}}(S^{(i)}\le a)\ge \zeta ,\,\,i=1,\ldots ,N$. Assume that an orbit customer enters an empty primary system at an instant z. Then, provided no new arrivals and new retrial attempts from other non-empty orbits (if any) occur, the next orbit customer can start service in the interval $[z,\,z+a+\Delta ]$ with a probability which is lower bounded by $\zeta \Delta$. Denote by $\sigma =(\zeta \Delta ) ^{{\mathcal {C}}_0}$ and note that $\sigma$ is a lower bound of the probability that at least ${\mathcal {C}}_0$ orbit customers of any class can be served one by one, provided the primary system is empty and no new arrivals enter the system. Consider for a moment the case of two orbits, that is, $N=2$. Recall that, for convenience only, we unload the orbits in increasing order, that is, first orbit 1, followed by orbit 2. Thus, provided no new arrivals of both classes and no retrial attempts of class-2 customers happened since instant t, orbit 1 becomes empty during the interval

$$\begin{aligned}{}[t,\,t+{\mathcal {D}}_0+{\mathcal {C}}_0 (\Delta +a)], \end{aligned}$$

with a probability which is not less than

$$\begin{aligned} \left( 1-\frac{\varepsilon _1}{2}\right) \sigma >0. \end{aligned}$$

In a similar way we can completely unload the second orbit during the interval

$$\begin{aligned}{}[t,\,t+{\mathcal {D}}_0+2{\mathcal {C}}_0 (\Delta +a)], \end{aligned}$$

with a probability which is not less than $( 1-\varepsilon _1/2) \sigma ^2,$ provided no new arrivals/attempts happened since instant t. If we now take into account the probability of the event $\{$ no new arrivals/attempts happened since instant $t\}$, then we obtain that the probability of unloading the primary system and both coloured orbits in this interval is lower bounded by the quantity

$$\begin{aligned} ( 1-\varepsilon _1/2) \sigma ^2 \exp \Bigl \{-\Bigl ({\mathcal {D}}_0+2{\mathcal {C}}_0 (\Delta +a)\Bigr )\sum _{i=1}^2\big (\lambda _i+\mu _0^{(i)}\big ) \Bigr \}>0, \end{aligned}$$

(61)

which is independent of t. Note that this lower bound is definitely not tight but simple and suitable for our purpose. Consider now the general case of N orbits. Continuing in a similar way, we find that both the primary system and all coloured orbits are completely unloaded in the interval

$$\begin{aligned}{}[t,\,t+{\mathcal {D}}_0+N{\mathcal {C}}_0 (\Delta +a)], \end{aligned}$$

(62)

with a probability which is lower bounded by

$$\begin{aligned} ( 1-\varepsilon _1/2) \sigma ^N \exp \Bigl \{-\Bigl ({\mathcal {D}}_0+N{\mathcal {C}}_0 (\Delta +a)\Bigr ) \sum _{i=1}^N\big (\lambda _i+\mu _0^{(i)}\big ) \Bigr \}>0. \end{aligned}$$

(63)

It is then easy to see that in a finite interval, with a positive probability, a new customer arrives observing an empty primary system and all coloured orbits empty. Consequently, (24) follows from this property. $\square$