Analysis of the waiting time distribution for polling systems with retrials and glue periods

Abstract

We consider a single-server multi-station polling system with retrials and glue periods. Just before the server arrives at a station, there is a deterministic glue period. During a glue period, arriving customers (either newly arriving customers or retrying customers) at the station stick in the queue of that station and will be served during the following service period of that station. Whereas during any other period, arriving customers at the station join the orbit of that station and will retry after an exponentially distributed time. In this paper, we derive the Laplace–Stieltjes transform of the waiting time distribution of an arbitrary customer. This transform allows us to obtain the mean and variance of the waiting time.

Appendix: Stability condition

We define the periodicity of our polling system as follows: Let D be the set of all positive real numbers $\delta$ such that $\sum _{i=1}^K(g_i+S_i)$, and $B_1,\ldots ,B_K$ have their supports in $\delta \mathbb {Z}_+ =\{0,\delta ,2\delta ,\ldots \}$. Our polling system is aperiodic if the set D is empty; otherwise the polling system is periodic. When our polling system is periodic, i.e., when D is nonempty, it can be shown that D has a maximum. This maximum is the period of the polling system.

In the following proposition, we obtain the stability condition of the system.

Proposition 2

Let ${\varvec{N}}(t)=(N_1(t), \ldots , N_K(t))$, where $N_i(t)$ is the number of customers in station i at time t. Suppose that $\rho <1$. If the system is aperiodic, then ${\varvec{N}}(t)$ converges in distribution as $t\rightarrow \infty$. If the system is periodic with period d, then ${\varvec{N}}(nd)$ converges in distribution as $n\rightarrow \infty$.

Proposition 2 implies that if $\rho < 1$, then $\{\mathcal {L}({\varvec{N}}(t)): t\ge 0\}$ is tight, where $\mathcal {L}({\varvec{N}}(t))$ is the distribution of ${\varvec{N}}(t)$.

Proof of Proposition 2

Let

$$\begin{aligned} \tau _0^L=\inf \left\{ t\ge 0: \sum _{i=1}^K N_i(t)\mathbb {E}[B_i]\le L, t \hbox { is the beginning epoch of a glue period of station } 1\right\} , \end{aligned}$$

and for $n \ge 1$,

$$\begin{aligned} \tau _n^L=\inf \left\{ t>\tau _{n-1}^L: \sum _{i=1}^K N_i(t)\mathbb {E}[B_i]\le L, t \hbox { is the beginning epoch of a glue period of station } 1\right\} . \end{aligned}$$

First, we show that there exist positive real numbers L, C, and $\epsilon$ such that for all ${\varvec{l}}=(l_1,\ldots ,l_K)$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]>L$,

$$\begin{aligned}&\mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _1^\infty )\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \le (1-\epsilon )\sum _{i=1}^K l_i\mathbb {E}[B_i], \end{aligned}$$

(9)

$$\begin{aligned}&\mathbb {E}\left[ \tau _1^\infty -\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \le C \sum _{i=1}^K l_i\mathbb {E}[B_i], \end{aligned}$$

(10)

and for all ${\varvec{l}}$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]\le L$,

$$\begin{aligned}&\mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _1^\infty )\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \le C, \end{aligned}$$

(11)

$$\begin{aligned}&\mathbb {E}\left[ \tau _1^\infty -\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \le C. \end{aligned}$$

(12)

Note that $N_i(\tau _1^\infty ) = N_i(\tau _0^\infty )-\mathcal {D}_i+\mathcal {A}_i$, where $\mathcal {D}_i$ is the number of service completions at station i during $(\tau _0^\infty , \tau _1^\infty )$, and $\mathcal {A}_i$ is the number of arrivals at station i during $(\tau _0^\infty , \tau _1^\infty )$. Thus

$$\begin{aligned}&{\mathbb {E}\left[ {\varvec{N}}(\tau _1^\infty ) \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] }\\&\quad = {\varvec{l}}-\mathbb {E}\left[ (\mathcal {D}_1, \ldots ,\mathcal {D}_K) \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] +\mathbb {E}\left[ (\mathcal {A}_1, \ldots ,\mathcal {A}_K) \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] . \end{aligned}$$

Since

$$\begin{aligned} \mathbb {E}\left[ (\mathcal {A}_1, \ldots ,\mathcal {A}_K) \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] =\sum _{i=1}^K(g_i+\mathbb {E}[S_i]+\mathbb {E} [\mathcal {D}_i|{\varvec{N}}(\tau _0^\infty )={\varvec{l}}]\mathbb {E}[B_i]){\varvec{\lambda }}, \end{aligned}$$

we have

$$\begin{aligned}&\mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _1^\infty )\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \\&\quad = \sum _{i=1}^K l_i\mathbb {E}[B_i]-\mathbb {E}\left[ \sum _{i=1}^K \mathcal {D}_i\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \\&\qquad +\, \sum _{i=1}^K(g_i+\mathbb {E}[S_i] +\,\mathbb {E}[\mathcal {D}_i|{\varvec{N}}(\tau _0^\infty )={\varvec{l}}]\mathbb {E}[B_i]) \sum _{i=1}^K \lambda _i \mathbb {E}[B_i] \\&\quad = \rho \sum _{i=1}^K(g_i+\mathbb {E}[S_i]) + \sum _{i=1}^K l_i\mathbb {E}[B_i] -(1-\rho )\mathbb {E}\left[ \sum _{i=1}^K \mathcal {D}_i\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] . \end{aligned}$$

Note that

$$\begin{aligned} \mathbb {E}[\mathcal {D}_i|{\varvec{N}}(\tau _0^\infty )={\varvec{l}}] \ge l_i(1-e^{-\nu _i g_i}) \ge \min _{1\le j\le K}(1-e^{-\nu _j g_j}) l_i. \end{aligned}$$

Hence

$$\begin{aligned}&{\mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _1^\infty )\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] } \nonumber \\&\quad \le \rho \sum _{i=1}^K (g_i+\mathbb {E}[S_i])+\left( 1-(1-\rho )\min _{1\le j\le K}(1-e^{-\nu _j g_j})\right) \sum _{i=1}^K l_i \mathbb {E}[B_i] \nonumber \\&\quad = \rho \sum _{i=1}^K (g_i+\mathbb {E}[S_i])+(1-2\epsilon ) \sum _{i=1}^K l_i \mathbb {E}[B_i], \end{aligned}$$

(13)

where $\epsilon = \frac{1-\rho }{2} \min _{1\le j\le K}(1-e^{-\nu _j g_j})$. Therefore,

$$\begin{aligned} \limsup _{\sum _{i=1}^K l_i\mathbb {E}[B_i] \rightarrow \infty }\frac{\mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _1^\infty )\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] }{\sum _{i=1}^K l_i\mathbb {E}[B_i]} \le 1-2\epsilon , \end{aligned}$$

and so there exists a positive number L such that

$$\begin{aligned} \frac{\mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _1^\infty )\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] }{\sum _{i=1}^K l_i\mathbb {E}[B_i]} \le 1-\epsilon \end{aligned}$$

for all ${\varvec{l}}$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]> L$. This proves (9). We have from (13) that there exists C satisfying (11). To prove (10) and (12), we note that $\tau _1^\infty -\tau _0^\infty$ is stochastically less than the sum of $\sum _{i=1}^K (g_i+S_i)$, and the busy period with initial workload $W_0$ in the standard M / G / 1 queue (where $W_0$ has the same distribution as the workload at $\tau ^\infty _0$ in our polling system). Thus

$$\begin{aligned} \mathbb {E}\left[ \tau _1^\infty -\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \le \frac{\sum _{i=1}^K l_i\mathbb {E}[B_i]}{1-\rho }+\sum _{i=1}^K (g_i+\mathbb {E}[S_i]). \end{aligned}$$

This proves that there exists C satisfying (10) and (12).

Next, we prove

$$\begin{aligned} \mathbb {E}[\tau _{1}^L-\tau _0^L \vert {\varvec{N}}(\tau _0^L)={\varvec{l}}]<\infty \end{aligned}$$

for all ${\varvec{l}}$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]\le L$. We can see from (9) that for ${\varvec{l}}$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]>L$,

$$\begin{aligned}&\mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _n^\infty )\mathbb {E}[B_i] \mathbb {1}_{\{\tau _n^\infty <\tau _0^L\}} \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \le (1-\epsilon )^n\sum _{i=1}^K l_i\mathbb {E}[B_i]. \end{aligned}$$

(14)

Also, for ${\varvec{l}}$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]>L$, by (10),

$$\begin{aligned}&\mathbb {E}\left[ (\tau _{n+1}^\infty -\tau _n^\infty )\mathbb {1}_{\{\tau _n^\infty<\tau _0^L\}} \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \nonumber \\&\quad = \sum _{{\varvec{l}}': \sum _{i=1}^K l_i'\mathbb {E}[B_i]>L} \mathbb {E}\left[ \tau _{n+1}^\infty -\tau _n^\infty \vert {\varvec{N}}(\tau _n^\infty )={\varvec{l}}',\tau _n^\infty<\tau _0^L\right] \nonumber \\&\qquad \times \mathbb {P}({\varvec{N}}(\tau _n^\infty )={\varvec{l}}',\tau _n^\infty<\tau _0^L\vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}) \nonumber \\&\quad \le \sum _{{\varvec{l}}': \sum _{i=1}^K l_i'\mathbb {E}[B_i]>L} C\sum _{i=1}^K l_i'\mathbb {E}[B_i] \mathbb {P}({\varvec{N}}(\tau _n^\infty )={\varvec{l}}',\tau _n^\infty<\tau _0^L\vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}) \nonumber \\&\quad = C \mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _n^\infty )\mathbb {E}[B_i] \mathbb {1}_{\{\tau _n^\infty <\tau _0^L\}} \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] . \end{aligned}$$

Thus, by (14),

$$\begin{aligned} \mathbb {E}\left[ (\tau _{n+1}^\infty -\tau _n^\infty )\mathbb {1}_{\{\tau _n^\infty <\tau _0^L\}} \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \le C (1-\epsilon )^n\sum _{i=1}^K l_i\mathbb {E}[B_i]. \end{aligned}$$

Hence, for ${\varvec{l}}$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]>L$,

$$\begin{aligned} \mathbb {E}\left[ \tau _0^L-\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right]&= \mathbb {E}\left[ \sum _{n=0}^\infty (\tau _{n+1}^\infty -\tau _n^\infty )\mathbb {1}_{\{\tau _n^\infty <\tau _0^L\}} \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \nonumber \\&\le \frac{C}{\epsilon }\sum _{i=1}^K l_i\mathbb {E}[B_i]. \end{aligned}$$

(15)

Now, for ${\varvec{l}}$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]\le L$,

$$\begin{aligned}&\mathbb {E}\left[ \tau _1^L-\tau _0^L \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \\&\quad = \mathbb {E}\left[ \tau _1^\infty -\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \\&\qquad + \sum _{{\varvec{l}}': \sum _{i=1}^K l_i'\mathbb {E}[B_i]>L} \mathbb {P}({\varvec{N}}(\tau _1^\infty )={\varvec{l}}' \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}) \mathbb {E}\left[ \tau _1^L-\tau _1^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}},{\varvec{N}}(\tau _1^\infty )={\varvec{l}}'\right] \\&\quad = \mathbb {E}\left[ \tau _1^\infty -\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] \\&\qquad + \sum _{{\varvec{l}}': \sum _{i=1}^K l_i'\mathbb {E}[B_i]>L} \mathbb {P}({\varvec{N}}(\tau _1^\infty )={\varvec{l}}' \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}) \mathbb {E}\left[ \tau _0^L-\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}'\right] \\&\quad \le \mathbb {E}\left[ \tau _1^\infty -\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] + \sum _{{\varvec{l}}': \sum _{i=1}^K l_i'\mathbb {E}[B_i]>L} \mathbb {P}({\varvec{N}}(\tau _1^\infty )={\varvec{l}}' \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}) \frac{C}{\epsilon }\sum _{i=1}^K l_i'\mathbb {E}[B_i] \\&\quad \le \mathbb {E}\left[ \tau _1^\infty -\tau _0^\infty \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] + \frac{C}{\epsilon }\mathbb {E}\left[ \sum _{i=1}^K N_i(\tau _1^\infty )\mathbb {E}[B_i] \vert {\varvec{N}}(\tau _0^\infty )={\varvec{l}}\right] , \end{aligned}$$

which is finite by (11) and (12). Here we have used (15) in the second last inequality. Therefore, we have proved $\mathbb {E}[\tau _{1}^L-\tau _0^L \vert {\varvec{N}}(\tau _0^L)={\varvec{l}}]<\infty$ for all ${\varvec{l}}$ with $\sum _{i=1}^K l_i\mathbb {E}[B_i]\le L$.

Note that

• ${\varvec{N}}(t)$ is a Markov regenerative process with Markov renewal sequence $({\varvec{N}}(\tau _n^L), \tau _n^L)$.

• The Markov process ${\varvec{N}}(\tau _n^L)$ has a finite state space and is irreducible.

• The semi-Markov process (SMP) corresponding to this Markov renewal sequence is positive recurrent.

Refer to Chapter 9 of Kulkarni (1995) for further details on the theory of Markov regenerative processes, Markov renewal sequences, and SMPs.

If the polling system is aperiodic, then the SMP is aperiodic. In this case ${\varvec{N}}(t)$ converges in distribution as $t\rightarrow \infty$ (refer to Theorem 9.30 of Kulkarni 1995). If the polling system is periodic with period d, then the SMP is periodic with period d. In this case ${\varvec{N}}(nd)$ converges in distribution as $n\rightarrow \infty$. $\square$

In the following proposition, we obtain the instability condition of the system.

Proposition 3

Let U(t) be the workload at time t in our polling system. If $\rho \ge 1$, then

$$\begin{aligned} U(t) \rightarrow \infty \hbox { in distribution as } t \rightarrow \infty . \end{aligned}$$

Proposition 3 implies that if $\rho \ge 1$, then $\{\mathcal {L}(U(t)): t \ge 0\}$ is not tight, where $\mathcal {L}(U(t))$ is the distribution of U(t). This also ensures that if $\rho \ge 1$, then $\{\mathcal {L}({\varvec{N}}(t)): t\ge 0\}$ is not tight.

Proof of Proposition 3

Let $\tilde{U}(t)$ be the workload at time t in the standard M / G / 1 queue with arrival rate $\lambda$ and service time distribution $\sum _{i=1}^K \frac{\lambda _i}{\lambda }\mathbb {P}(B_i \le x)$. If $U(0)=\tilde{U}(0)$ in distribution, then $U(t) \ge \tilde{U}(t)$ stochastically for all $t \ge 0$. If $\rho \ge 1$, then $\tilde{U}(t) \rightarrow \infty$ in distribution as $t \rightarrow \infty$, and so $U(t) \rightarrow \infty$ in distribution as $t \rightarrow \infty$. $\square$