Large deviations of the waiting time in the GI/G/1 queue with random order service

Abstract

We consider the GI/G/1 queue where customers are served in random order and the service time distribution has a finite exponential moment. We derive the large deviations result for the waiting time distribution by showing that the asymptotic decay rate of the waiting time distribution is the same as that of the busy period distribution.

Appendix: Proof of Theorem 1

For the proof of Theorem 1, we need a series of lemmas. Before presenting the lemmas, we need the following result which is used in the proof of one of the lemmas.

Lemma 2

Let $X$ be a nonnegative random variable with finite mean and let $X^{(e)}$ be the equilibrium random variable of $X$, i.e., $\mathbb{P }(X^{(e)}>t)=\frac{1}{\mathbb{E } X}\int _t^\infty \mathbb{P }(X>s) \mathrm{d}s$. If $\underline{r}^X =\overline{r}^X$, then so is $r^{X^{(e)}}$ and $r^{X^{(e)}}= r^X$.

Proof

Since $\underline{r}^X =\overline{r}^X,\mathbb{P }(X>t)=e^{-t(r^X+o(1))}$ as $t \rightarrow \infty$. Then

$$\begin{aligned} \mathbb{P }(X^{(e)}>t)=\frac{1}{\mathbb{E } X}\int \nolimits _t^\infty e^{-s\left(r^X+o(1)\right)} \mathrm{d}s= e^{-t\left(r^X+o(1)\right)} \quad \text{ as}\; t \rightarrow \infty . \end{aligned}$$

This implies that

$$\begin{aligned} \lim _{t \rightarrow \infty }-\frac{1}{t}\log \mathbb{P }(X^{(e)}>t)= r^X, \end{aligned}$$

which completes the proof. $\square$

Now, we present a series of lemmas which will be used in the proof of Theorem 1. The first lemma can be found in Theorem 2.1 of Palmowski and Rolski [8].

Lemma 3

(Palmowski and Rolski [8]) Let $\tau _{x,z}$ be the time interval from the time when the initial workload is $x$ and the remaining arrival time is $z$ to the time when the system is empty. If $\theta _*<\theta _\gamma$, then we have for $x >z \ge 0$,

$$\begin{aligned} \mathbb{P }(\tau _{x,z}>t)\;\sim\;h(x,z) t^{-\frac{3}{2}}e^{\kappa (\theta _*)t}\quad \text{ as} \;t \rightarrow \infty , \end{aligned}$$

where $h(x,z)>0$ is given explicitly in [8].

The following corollary is immediate from Lemma 3.

Corollary 1

If $\theta _*<\theta _\gamma$, then $r^{\tau _{x,z}} = -\kappa (\theta _*)$ for $x>z \ge 0$.

We note that Theorem 1 is alternatively written as $r^G=r^{G_a}=-\kappa (\theta _*)$. In the following two lemmas, we show that $r^{G} = -\kappa (\theta _*)$.

Lemma 4

If $\theta _*<\theta _\gamma$, then $r^{G} = -\kappa (\theta _*)$.

Proof

For $x >z \ge 0,\,G$ is stochastically smaller than $\tau _{x,z}$, i.e.,

$$\begin{aligned} \mathbb{P }(G>t) \le \mathbb{P }(\tau _{x,z}>t). \end{aligned}$$

(11)

By (1), there exist $x$ and $z$ such that

$$\begin{aligned} \mathbb{P }(B>x)>0,\quad \mathbb{P }(T \le z)>0, \quad x >z \ge 0. \end{aligned}$$

Then

$$\begin{aligned} \mathbb{P }(G>t) \ge \mathbb{P }(B >x) \mathbb{P }(T \le z) \mathbb{P }(\tau _{x,z}>t). \end{aligned}$$

(12)

By Corollary 1, (11) and (12), we have $r^{G} = -\kappa (\theta _*)$. $\square$

Recall that we use $\kappa ^{T,B}$ and $\theta _*^{T,B}$ instead of $\kappa$ and $\theta _*$ when we want to represent the interarrival time $T$ and the service time $B$ explicitly. For the same purpose, we use $\theta _\gamma ^{T,B},\,G^{T,B}$ and $G_a^{T,B}$ instead of $\theta _\gamma , G$ and $G_a$.

Lemma 5

If $\theta _*=\theta _\gamma$, then $r^{G} = -\kappa (\theta _*)$.

Proof

If we let $B^M=B \wedge M$ for $M>0$, then $\mathbb{P }(G>t) \ge \mathbb{P }(G^{T, B^M} >t)$, which implies

$$\begin{aligned} \overline{r}^G \le r^{G^{T, B^M}}. \end{aligned}$$

(13)

Since $B^M$ is bounded, $\theta _\gamma ^{T, B^M} =\infty$ and $\hat{m}_{\!B^M}(\theta _\gamma ^{T, B^M})=\infty$. By Remark 1, we have $\theta _*^{T,B^M}<\theta _\gamma ^{T,B^M}$. Therefore, by Lemma 4,

$$\begin{aligned} r^{G^{T, B^M}}=-\kappa ^{T,B^M}\left(\theta _*^{T, B^M}\right). \end{aligned}$$

(14)

Since $-\kappa ^{T,B^M}(\theta _*^{T, B^M})$ converges to $-\kappa (\theta _*)$ as $M \rightarrow \infty$, combining (13) with (14) yields $\overline{r}^G \le -\kappa (\theta _*)$.

It remains to show that $\underline{r}^G \ge -\kappa (\theta _*)$. For $\epsilon \in (0, (1-\rho )\frac{\theta _*}{\lambda })$, let ${\widetilde{X}}_\epsilon$ be a nonnegative random variable with the complementary distribution function $\mathbb{P }({\widetilde{X}}_\epsilon >t) = \epsilon e^{-\theta _* t},\,t \ge 0$, and assume that it is independent of $B$. Let $\widetilde{B}_\epsilon =B+ \widetilde{X}_\epsilon$. Then $\mathbb{P }(G>t) \le \mathbb{P }(G^{T, \widetilde{B}_\epsilon } >t)$, which implies

$$\begin{aligned} \underline{r}^G \ge r^{G^{T, \widetilde{B}_\epsilon }}. \end{aligned}$$

(15)

Since $\hat{m}_{\widetilde{B}_\epsilon }(\theta )=\hat{m}_B(\theta ) \left(\frac{\epsilon \theta _*}{\theta _*-\theta }+1-\epsilon \right),\,\theta <\theta _*$, we have $\hat{m}_{\widetilde{B}_\epsilon }(\theta _*)=\infty$. By Remark 1, we have $\theta _*^{T, \widetilde{B}_\epsilon }< \theta _\gamma ^{T, \widetilde{B}_\epsilon }$. From (15) and Lemma 4, it follows that

$$\begin{aligned} \underline{r}^G \ge r^{G^{T, \widetilde{B}_\epsilon }}= -\kappa ^{T,\widetilde{B}_\epsilon }\left(\theta _*^{T, \widetilde{B}_\epsilon }\right). \end{aligned}$$

Since $-\kappa ^{T,\widetilde{B}_\epsilon }(\theta _*^{T, \widetilde{B}_\epsilon })$ converges to $-\kappa (\theta _*)$ as $\epsilon \rightarrow 0+$, we get $\underline{r}^G \ge -\kappa (\theta _*)$, which completes the proof. $\square$

In the remainder, we will show that $r^{G_a}=r^G$. Let $t_k$ be the arrival epoch of customer $k,\,k=1,2, \ldots$. For the following two lemmas, we assume that a customer arrives at the empty system at time $0$, i.e., the busy period starts when $t_1=0$. By the discrete-time renewal reward theorem,

$$\begin{aligned} \mathbb{P }(G_a>t)=\frac{1}{\mathbb{E }[N]} \mathbb{E }\left[\sum _{k=1}^N {{\small 1}\!\!1}_{\{G-t_k>t\}}\right]= \frac{1}{\mathbb{E }[N]} \sum _{k=1}^\infty \mathbb{P }(k \le N, G-t_k>t),\qquad \end{aligned}$$

(16)

where $N$ is the number of customers arriving during the busy period.

Lemma 6

We have $\overline{r}^{G_a} \le r^G$.

Proof

This is trivial because $\mathbb{P }(G_a>t) \ge \frac{1}{\mathbb{E }[N]}\mathbb{P }(G>t)$ by (16). $\square$

Lemma 7

Suppose that there exists $\epsilon >0$ such that $\mathbb{P }(T >\epsilon )=1$. Then $\underline{r}^{G_a} \ge r^G$.

Proof

According to (16), we have

$$\begin{aligned} \mathbb{P }(G_a>t)&= \frac{1}{\mathbb{E }[N]}\mathbb{P }(G>t)+ \frac{1}{\mathbb{E }[N]} \sum _{k=2}^\infty \mathbb{P }(G-t_k >t, k \le N). \end{aligned}$$

(17)

The second term of the right-hand side of the above can be written as

$$\begin{aligned} \frac{1}{\mathbb{E }[N]} \sum _{k=2}^\infty \mathbb{E }\left[{{\small 1}\!\!1}_{\{G-t_k >t, k \le N\}}\right]&\le \frac{1}{\epsilon \mathbb{E }[N]} \sum _{k=2}^\infty \mathbb{E }\left[\int \nolimits ^{t_k}_{t_{k-1}}{{\small 1}\!\!1}_{\{G-s >t, k \le N\}} \mathrm{d}s\right] \\&= \frac{1}{\epsilon \mathbb{E }[N]}\mathbb{E }\left[\int \nolimits _0^{t_N}{{\small 1}\!\!1}_{\{G-s >t\}} \mathrm{d}s \right]\\&\le \frac{1}{\epsilon \mathbb{E }[N]}\mathbb{E }\left[\int \nolimits _0^{G}{{\small 1}\!\!1}_{\{G-s >t\}} \mathrm{d}s \right]\\&= \frac{1}{\epsilon \mathbb{E }[N]}\mathbb{E }\left[\int \nolimits _0^G {{\small 1}\!\!1}_{\{s >t\}} \mathrm{d}s \right]\\&= \frac{1}{\epsilon \mathbb{E }[N]}\int \nolimits _t^\infty \mathbb{P }(G>s) \mathrm{d}s. \end{aligned}$$

Therefore, (17) becomes

$$\begin{aligned} \mathbb{P }(G_a>t)&\le \frac{1}{\mathbb{E }[N]}\mathbb{P }(G>t)+ \frac{\mathbb{E }[G]}{\epsilon \mathbb{E }[N]} \mathbb{P }\left(G^{(e)}>t\right), \end{aligned}$$

where $G^{(e)}$ is the equilibrium random variable of $G$. Finally, by using Lemma 2, we complete the proof. $\square$

To prove $\underline{r}^{G_a} \ge r^G$, without the assumption of Lemma 7, let $T_\epsilon =T {{\small 1}\!\!1}_{\{T>\epsilon \}}$ and $\widehat{T}_\epsilon$ be a random variable with the complementary distribution function

$$\begin{aligned} \mathbb{P }(\widehat{T}_\epsilon >t)= \left\{ \begin{array}{ll} \frac{\mathbb{P }(T>t)}{\mathbb{P }(T>\epsilon )}&\quad \text{ if} \; t \ge \epsilon ,\\ 0&\quad \text{ if} \;t<\epsilon .\\ \end{array}\right. \end{aligned}$$

We consider the GI/G/1 ROS queue with interarrival time $T_\epsilon$ and service time $B$. This queue can be thought of as a queue with batch arrivals where the batch interarrival time is $\widehat{T}_\epsilon$, the batch size is geometrically distributed with parameter $\mathbb{P }(T>\epsilon )$, and the service time is $B$. If we consider one batch to be one customer, then this queue is the same as the GI/G/1 ROS queue with interarrival time $\widehat{T}_\epsilon$ and service time $\widehat{B}_\epsilon =B_1+B_2+\cdots +B_{\widehat{L}_\epsilon }$, where $\widehat{L}_\epsilon$ denotes the batch size. It is not difficult to see that $G_a^{T_\epsilon , B}$ has the same distribution as $G_a^{\widehat{T}_\epsilon , \widehat{B}_\epsilon }$. Therefore,

$$\begin{aligned} \mathbb{P }(G_a>t) \le \mathbb{P }\left(G_a^{T_\epsilon , B}>t\right)= \mathbb{P }\left(G_a^{\widehat{T}_\epsilon , \widehat{B}_\epsilon }>t\right), \end{aligned}$$

from which and Lemma 7, it follows that

$$\begin{aligned} \underline{r}^{G_a} \ge r^{G_a^{\widehat{T}_\epsilon , \widehat{B}_\epsilon }}=-\kappa ^{\widehat{T}_\epsilon , \widehat{B}_\epsilon }(\theta _*^{\widehat{T}_\epsilon , \widehat{B}_\epsilon }). \end{aligned}$$

Since $-\kappa ^{\widehat{T}_\epsilon , \widehat{B}_\epsilon }\left(\theta _*^{\widehat{T}_\epsilon , \widehat{B}_\epsilon }\right)$ converges to $-\kappa (\theta _*)$ as $\epsilon \rightarrow 0+$, we have the following lemma.

Lemma 8

We have $\underline{r}^{G_a} \ge r^G$.

Finally, we are ready to prove Theorem 1.

Proof of Theorem 1

By Lemmas 4 and 5, we have $r^G=-\kappa (\theta _*)$. From Lemmas 6 and 8, we obtain $r^{G_a}=r^G$. Therefore, $r^G=r^{G_a}=-\kappa (\theta _*)$, which is the assertion of the theorem. $\square$