A queueing model with independent arrivals, and its fluid and diffusion limits

Abstract

We study a queueing model with ordered arrivals, which can be called the $\varDelta _{(i)}/GI/1$ queue. Here, customers from a fixed, finite, population independently sample a time to arrive from some given distribution $F$, and enter the queue in order of the sampled arrival times. Thus, the arrival times are order statistics, and the inter-arrival times are differences of consecutive order statistics. They are served by a single server with independent and identically distributed service times, with general service distribution $G$. The discrete event model is analytically intractable. Thus, we develop fluid and diffusion limits for the performance metrics of the queue. The fluid limit of the queue length is observed to be a reflection of a ‘fluid netput’ process, while the diffusion limit is observed to be a function of a Brownian motion and a Brownian bridge process or ‘diffusion netput’ process. The diffusion limit can be seen as being reflected through the directional derivative of the Skorokhod regulator of the fluid netput process in the direction of the diffusion netput process. We also observe what may be interpreted as a sample path Little’s Law. Sample path analysis reveals various operating regimes where the diffusion limit switches between a free diffusion, a reflected diffusion process, and the zero process, with possible discontinuities during regime switches. The weak convergence results are established in the $M_1$ topology.

Appendix

1.1 Proof of Lemma 2

Rewrite $\tilde{y}_n$ as $\tilde{y}_n = (\varPsi (\sqrt{n}x + y_n) - \varPsi (\sqrt{n}x + y)) - (\varPsi (\sqrt{n}x + y) - \sqrt{n}\varPsi (x))$. Now, using the fact that the Skorokhod reflection map is Lipschitz continuous under the uniform metric (see Lemma 13.4.1 and Theorem 13.4.1 of [12]), we have $(\varPsi (\sqrt{n}x + y_n) - \varPsi (\sqrt{n}x + y)) \le \Vert y_n - y \Vert$, where $\Vert \cdot \Vert$ is the uniform metric. It follows that $\tilde{y}_n \le \Vert y_n - y \Vert + (\varPsi (\sqrt{n}x + y) - \sqrt{n}\varPsi (x))$. Now, by Theorem 9.5.1 of [19], we know that as $n \rightarrow \infty$

$$\begin{aligned} (\varPsi (\sqrt{n}x + y) - \sqrt{n}\varPsi (x)) \mathop {\rightarrow }\limits ^{a.s.} \tilde{y}, \text { in } (\mathcal D_{\lim },M_1). \end{aligned}$$

Using this result, and the fact that by hypothesis $y_n$ converges to $y$ in $(\mathcal D_{lim},J_1)$, we have $\tilde{y}_n \mathop {\rightarrow }\limits ^{a.s.} \tilde{y}, \text { in } (\mathcal D_{\lim },M_1)$.$\square$

1.2 Proof of Lemma 3

First, suppose $\bar{Q}(t) > 0$. It follows that $\bar{F}(t) -\mu t > \inf _{-T_0 \le s \le t} (\bar{F}(s) - \mu s) = w$, where the latter equality follows because the queue starts empty at time $0$, and the fluid netput is positive before time $0$ (Note that we ignore the positive part operator in the definition of $\varPsi$, as the systems starts empty at time $-T_0$). Now, let $t^* = \sup \{0 \le s \le t | (\bar{F}(s) - \mu s) = \inf _{0 \le s \le t} (\bar{F}(s) - \mu s)\}$ be the point at which the infimum is achieved, on the right- hand side. It follows that $\bar{F}(t) - \mu t > \bar{F}(t^*) - \mu t^*$, in turn yielding

$$\begin{aligned} \rho (t) = \sup _{0 \le s \le t} \frac{\bar{F}(t) - \bar{F}(s)}{\mu (t-s)} > 1. \end{aligned}$$

Next, suppose $\bar{Q}(t) = 0$, $\bar{X}(t) = \varPsi (\bar{X})(t)$ and there exists an $r < t$ such that $\varPsi (\bar{X})(t) = \varPsi (\bar{X})(s)$ for all $s \in [r,t]$. It follows that $\bar{F}(t) - \mu t = - \sup _{-T_0 \le s \le t} (-(\bar{F}(s) - \mu s))$, implying there exists a point $r^* \in [0,t]$ such that $\bar{F}(t) - \mu t = \bar{F}(r^*) - \mu r^*$. This, in turn, implies that

$$\begin{aligned} \sup _{0 \le s \le t} \frac{\bar{F}(t) - \bar{F}(s)}{\mu (t - r)} \ge \frac{\bar{F}(t) - \bar{F}(r^*)}{\mu (t - r^*)} = 1. \end{aligned}$$

However, a simple contradiction argument shows that

$$\begin{aligned} \sup _{0 \le s \le t} \frac{\bar{F}(t) - \bar{F}(s)}{\mu (t - r)} > 1 \end{aligned}$$

is impossible, implying that

$$\begin{aligned} \sup _{0 \le s \le t} \frac{\bar{F}(t) - \bar{F}(s)}{\mu (t - r)} = 1. \end{aligned}$$

Finally, consider case (iii). We have $\forall r < t$,

$$\begin{aligned} -(\bar{F}(t) - \mu t) = \sup _{-T_0 \le s \le t} (-(\bar{F}(s) - \mu s)) > \sup _{-T_0 \le s \le r} (-(\bar{F}(s) - \mu s)). \end{aligned}$$

It follows that $-(\bar{F}(t) - \mu t) > -(\bar{F}(r) - \mu r)$,

implying

$$\begin{aligned} 1 > \frac{\bar{F}(t) - \bar{F}(r)}{\mu (t - r)} ~ \forall r \in [0,t). \end{aligned}$$

$\square$

1.3 Proof of Theorem 3

• (i) Overloaded regime

Proof

First, note that $\tau$ is the first instant of an end of overloading phase, and the current overloaded phase ends at $\tau$. In the overloaded state $\bar{Q}(t) > 0$, implying that $\varPsi (\bar{X})(t)$ is a constant. Using the definition of $\nabla _t^{\bar{X}}$, it follows that $\varPsi (\bar{X})(t) = -\bar{X}(t^*)$, and $\bar{Q}(t) = \bar{X}(t) - \bar{X}(t^*) = (\bar{F}(t) - \bar{F}(t^*) - \mu (t - t^*))$. Next, from Theorem 2, it is obvious that $\frac{Q^n(t)}{\sqrt{n}} \mathop {\approx }\limits ^{d} \tilde{Z}^n_t$.

Next, from Remark 1 after Lemma 1, $\hat{X}(t) - \hat{X}(t^*) = \int _{t^*}^t \sqrt{g^{'}(s)}\, \mathrm{d}W_s$, which can be seen to be a diffusion process that starts from $0$ at $t^*$. Noting that $\nabla _t^{\bar{X}}$ does not change on the interval $(t^*,\tau )$, it follows that $X^* = \sup _{s \in \nabla _{t^*}^{\bar{X}}} \{-\hat{X}(s))$ is a fixed random variable, and $\tilde{Z}^n_t$ has an initial condition $\tilde{Z}^n_{t^*} = \hat{X}(t^*) - X^*$. It is straightforward to see that $\tilde{Z}_n^{\cdot }$ is the strong solution to the mentioned SDE, since it is adapted to the filtration generated by $W$. $\square$

• (ii) Underloaded regime This result is immediate from the definition of the limit processes.

• (iii) Middle and end of critically loaded state

Proof

For any $t \in (t^*,\tau )$ we have $\bar{Q}(t) = 0$. From the weak convergence result in Theorem 2, we have $Q^n(t) \mathop {\approx }\limits ^{d} n \bar{Q}(t) + \sqrt{n}\hat{Q}(t)$, and expanding the definition of $\hat{Q}$, it follows that $Q^n(t) \mathop {\approx }\limits ^{d} \sqrt{n}(\hat{X}(s) + \sup _{s \in \nabla _t^{\bar{X}}} (-\hat{X}(s)))$. Using the fact that $\varPsi (\bar{X})(t) = w = -\bar{X}(t) ~\forall ~ t \in (t^*,\tau )$ in a critically loaded regime, it follows that $\nabla _t^{\bar{X}} = (t^*,t]$ for $t \in (t^*,\tau )$. Thus, we have $Q^n(t) \mathop {\approx }\limits ^{d} \sqrt{n}(\hat{X}(s) + \sup _{t^* < s \le t} (-\hat{X}(s)))$. Let $u = t - t^*$. Then, after a change of variables, we obtain $Q^n(u+t^*) \mathop {\approx }\limits ^{d} \sqrt{n}(\hat{X}(u+t^*) + \sup _{0 \le s < u} (-\hat{X}(s)))$.

Since $W^0$ is a Brownian Bridge process, the strong Markov property of Brownian motion ([18]) implies that $\hat{X}(u+t^*) -\hat{X}(t^*) = \hat{X}(u)$. Substituting this into the expression above we see that we have $Q^n(u+t^*) = Q^n(u) + \hat{X}(t^*)$, where $\hat{X}(t^*)$ is the starting state of the process in the middle-of-critically loaded state. A simple change of variables will provide the desired result. A similar argument will hold for the end of critical loading state as well. $\square$

• (iv) End of overloading state

Proof

By definition for any $\tau >0$, $t - \frac{\tau }{\sqrt{n}}$ is a point of overloading. Therefore,

$$\begin{aligned} \frac{Q^n\big (t-\frac{\tau }{\sqrt{n}}\big )}{\sqrt{n}}&= \hat{X}^n\Big (t - \frac{\tau }{\sqrt{n}}\Big ) + \sqrt{n}\Big (F(t) - \frac{\tau }{\sqrt{n}}\Big ) - \mu \Big (t - \frac{\tau }{\sqrt{n}} \Big )\\&+\, \varPsi (\hat{X}^n + \sqrt{n}\bar{X})\Big (t - \frac{\tau }{\sqrt{n}}\Big ) - \sqrt{n}\varPsi (\bar{X})\Big (t - \frac{\tau }{\sqrt{n}}\Big ). \end{aligned}$$

Without loss of generality, we assume that service started when the queue was in the overloaded state, so that $\varPsi (\bar{X})\left( t - \frac{\tau }{\sqrt{n}}\right) = 0$. Now, using the fact the derivative $f$ exists, the mean value theorem implies the existence of a point $\tilde{t} \in \left[ t-\frac{\tau }{\sqrt{n}}, t\right]$ such that $F\left( t - \frac{\tau }{\sqrt{n}}\right) = F(t) - f(\tilde{t})\frac{\tau }{\sqrt{n}}$. Adding and subtracting the term $f(t) \tau /\sqrt{n}$ to the expression above, we have

$$\begin{aligned} F\Big (t - \frac{\tau }{\sqrt{n}}\Big ) = F(t) - f(t) \frac{\tau }{\sqrt{n}} + f(t) \frac{\tau }{\sqrt{n}} - f(\tilde{t})\frac{\tau }{\sqrt{n}}. \end{aligned}$$

Substituting this into the expression for $Q^n$ above, and introducing the term $\hat{X}^n(t)$, we obtain

$$\begin{aligned} \frac{Q^n\Big (t-\frac{\tau }{\sqrt{n}}\Big )}{\sqrt{n}}&= \hat{X}^n\Big (t - \frac{\tau }{\sqrt{n}}\Big ) - \hat{X}^n(t) + \hat{X}^n(t)+ \sqrt{n}(F(t) - \mu t) -(f(t) - \mu ) \tau \\&\quad +\, \varPsi (\hat{X}^n + \sqrt{n}\bar{X})\Big (t - \frac{\tau }{\sqrt{n}}\Big ) + (f(t) - f(\tilde{t})) \frac{\tau }{\sqrt{n}}. \end{aligned}$$

Now, using Lemma 1 and the continuity of the limit process we see that $\hat{X}^n(t - \frac{\tau }{\sqrt{n}}) - \hat{X}^n(t) \Rightarrow 0$. Further, since $f$ is bounded by virtue of being defined on a finite interval, we have $\tau (f(t) - f(\tilde{t}))/\sqrt{n}\rightarrow \infty$ as $n \rightarrow \infty$. Next, consider the term $\hat{Z}(t) := \hat{X}^n(t) + \sqrt{n}(F(t) - \mu t) + \varPsi (\hat{X}^n + \sqrt{n}\bar{X})\left( t - \frac{t}{\sqrt{n}}\right)$.

Let $\delta > 0$ be sufficiently small, so that the following decomposition of the expression above holds:

$$\begin{aligned} \hat{Z}^n(t)&= \sup _{-T_0 \le s < t-\delta } (\hat{X}^n(t) + \sqrt{n}(F(t) - \mu t) - \hat{X}^n(s) - \sqrt{n}\bar{X}(s) )\\&\vee \sup _{t-\delta \le s \le t - \frac{\tau }{\sqrt{n}}} (\hat{X}^n(t) + \sqrt{n}(F(t) - \mu t) - \hat{X}^n(s) - \sqrt{n}\bar{X}(s)). \end{aligned}$$

Let $t^* = \sup \{\nabla _t^{\bar{X}} \backslash \{t\} \}$. Consider the first term on the RHS above, and call it $\hat{Z}^n_1(t)$. Since the queue is overloaded before $t$ no points are “added” to the correspondence $\nabla _t^{\bar{X}}$; it follows from the definition of an end of overloading point that $(F(t) - \mu t) = - \varPsi (\bar{X})(t) \equiv - \varPsi (\bar{X})(t^*+\delta )$. This, in turn, provides $\hat{Z}_1^n(t) = \hat{X}^n(t) + \varPsi (\hat{X}+\sqrt{n}\bar{X})(t^*+\delta ) - \sqrt{n}\varPsi (\bar{X})(t^*+\delta )$. Using Lemma 2, it follows that $\hat{Z}^n_1(t) \Rightarrow \hat{X}(t) + \sup _{s \in \nabla _t^{\bar{X}} \backslash \{t\}} ( -\hat{X}(s))$ as $n \rightarrow \infty$, followed by letting $\delta \rightarrow 0$. Next, consider the second term

$$\begin{aligned} \hat{Z}_2^n(t)&= \sup _{t-\delta \le s \le t - \frac{\tau }{\sqrt{n}}} ( \hat{X}^n(t) + \sqrt{n}(F(t) - \mu t) - \hat{X}^n(s) - \sqrt{n}\bar{X}(s) )\\&\le \sup _{t - \delta \le s \le t - \frac{\tau }{\sqrt{n}}} (\hat{X}^n(t) - \hat{X}^n(s)) + \sup _{t - \delta \le s \le t - \frac{\tau }{\sqrt{n}}} \sqrt{n}(\bar{X}(t) - \bar{X}(s))\\&\le \sup _{t - \delta \le s \le t} (\hat{X}^n(t) - \hat{X}^n(s)) + \sup _{t - \delta \le s \le t - \frac{\tau }{\sqrt{n}}} \sqrt{n}(\bar{X}(t) - \bar{X}(s)). \end{aligned}$$

For large $n$, as the queue is overloaded at $t-\frac{\tau }{\sqrt{n}}$, it follows that

$$\begin{aligned} \hat{Z}_2^n(t) \le \sup _{t-\delta \le s \le t} (\hat{X}(t) - \hat{X}(s)) + \sqrt{n}\Big (\bar{X}(t) - \bar{X}\Big (t - \frac{\tau }{\sqrt{n}}\Big )\Big ). \end{aligned}$$

Again, by the mean value theorem,

$$\begin{aligned} \sqrt{n}\Big (\bar{X}(t) - \bar{X}\Big (t - \frac{\tau }{\sqrt{n}}\Big )\Big )&= \sqrt{n}\Big (F(t) - F\Big (t - \frac{\tau }{\sqrt{n}}\Big ) - \mu \frac{\tau }{\sqrt{n}}\Big )\\&= \sqrt{n}(f(t) - \mu ) \frac{\tau }{\sqrt{n}} + (f(t) - f(\tilde{t})) \tau , \end{aligned}$$

where $\tilde{t} \in [t-\frac{\tau }{\sqrt{n}},t]$. Since, $\tilde{t} \rightarrow t$ as $n \rightarrow \infty$, by the (right) continuity of $f$, it follows that $f(t) - f(\tilde{t}) \rightarrow 0$ as $n \rightarrow \infty$. Then it follows by an application of Lemma 1 (and using Skorokhod’s almost sure representation) that $\overline{\lim }_{n \rightarrow \infty } \hat{Z}^n_2(t) \le \hat{X}(t) + \sup _{t - \delta \le s \le t} (-\hat{X}(s)) + (f(t) - \mu ) \tau$. On the other hand, for a lower bound, using the mean value theorem again, we have $\hat{Z}^n_2(t) \ge \hat{X}^n(t) - \hat{X}^n(t - \frac{\tau }{\sqrt{n}}) + (f(t) - \mu ) \tau + (f(t) - f(\tilde{t})) \tau$. Once again, using the continuity of $f$, the almost sure representation theorem and Lemma 1, and noting the continuity of the limit process $\hat{X}$, we have

$$\begin{aligned} \underline{\lim }_{n \rightarrow \infty } \hat{Z}^n_2(t) \ge (f(t) - \mu ) \tau ~ a.s. \end{aligned}$$

Now, using the limits derived for $\hat{Z}^n_1 \text { and } \hat{Z}^n_2$, it follows that

$$\begin{aligned}&\frac{Q^n(t - \frac{\tau }{\sqrt{n}})}{\sqrt{n}} \Longrightarrow -(f(t) - \mu ) \tau + \sup _{s \in \nabla _t^{{\bar{X}}} \backslash \{t\}} (\hat{X}(t) - \hat{X}(s))\vee ( f(t) - \mu ) \tau \\&\quad = \left( \hat{X}(t) + \sup _{s \in \nabla _t^{\bar{X}} \backslash \{t\}} (-\hat{X}(s)) - (f(t) - \mu )\tau \right) _+. \end{aligned}$$

$\square$

1.4 Proof of Proposition 5

The proof is a consequence of the following lemma, which consolidates Lemmas 6.5, 6.6, and 6.7 in [6]. The lemma characterizes the points of discontinuity (and continuity) of the process $\tilde{Y}(t) = \sup _{s \in \nabla _t^{\bar{X}}}(-\hat{X}(s))$ in relation to the correspondence $\nabla _t^{\bar{X}}$. We do not prove these conditions, but direct the interested reader to [6].

Lemma 4

A point $t \in [-T_0,\infty )$ is characterized as follows.

• (i) Continuity Conditions. The following are equivalent:

  • (1) $t$ is a continuity point.

  • (2) $t \in \nabla _t^{\bar{X}} = \{t\}$, or $t \not \in \nabla _t^{\bar{X}}$, or $t \in \nabla _t^{\bar{X}} \not = \{t\}$, and $t$ is not isolated in $\nabla _t^{\bar{X}}$ and $\nabla _t^{\bar{X}} \subseteq \nabla _u^{\bar{X}} ~ for ~ some~ u > t$.

• (ii) Right-discontinuity Conditions. The following are equivalent:

  • (1) $t$ is a point of right-discontinuity.

  • (2) $t \in \nabla _t^{\bar{X}} \not = \{t\}$ and $\nabla _u^{\bar{X}} \subseteq (t,u] ~ \forall ~ u > r$.

  • (3) $\tilde{Y}(t) = \tilde{Y}(t-) > \tilde{Y}(t+) = -\hat{X}(t)$.

• (iii) Left-discontinuity Conditions. The following are equivalent:

  • (1) $t$ is a point of left-discontinuity.

  • (2) $t \in \nabla _t^{\bar{X}} \not = \{t\}$ and $t$ is isolated in $\nabla _t^{\bar{X}}$.

  • (3) $\tilde{Y}(t) = \tilde{Y}(t+) = -\hat{X}(t) > \tilde{Y}(t-)$.

A point of right-discontinuity can be seen to be left-continuous, coupled with an ordering on the right and left limits, such that $\tilde{Y}(t-) > \tilde{Y}(t+)$. Similarly, a point of left-discontinuity is right-continuous, and the limits are ordered such that $\tilde{Y}(t+) > \tilde{Y}(t-)$. Using these definitions, we proceed to prove the upper-semicontinuity of the limit process.

Proof (Proposition 5)

[Proposition 5] By definition, $\hat{X}$ is continuous, and it suffices to check that a sample path of the component $\tilde{Y}(t) = \sup _{s \in \nabla _t^{\bar{X}}} ( -\hat{X}(s) )$ is upper-semicontinuous. To see this, consider the pullback of the level set $\tilde{Y}^{-1}[a,\infty ) = \{t \in [-T_0,\infty ) | \tilde{Y}(t) \ge a \}$. It suffices to check that this is a closed set; see [28]. Let $\{\tau _n\} \subseteq \{t \in [-T_0,\infty ) | \tilde{Y}(t) \ge a \}$ be a sequence of points such that $\tau _n \rightarrow \tau$ as $n \rightarrow \infty$, where $\tau \in [-T_0,\infty )$ is an arbitrary point in the domain of $\tilde{Y}$. Thus, if $\epsilon > 0$, then there exists an $n_0 \in \mathbb {N}$ such that $\forall$ $n \ge n_0$, $\epsilon \ge \tau - \tau _n \ge -\epsilon$. If $\tau$ is a continuity point, then the conclusion is obvious. On the other hand, suppose that $\tau$ is a left-discontinuity point. By part (iii) of Lemma 4, it follows that $\tilde{Y}(\tau -) < \tilde{Y}(\tau +) = \tilde{Y}(\tau )$. By the definition of a left-discontinuity, there exits an interval $[t^*,\tau )$, where $t^* = \sup \nabla _{\tau }^{\bar{X}} \backslash \{\tau \}$, on which $\tilde{Y}$ is (locally) continuous. Fix $\delta > 0$, then there exists an $\eta > 0$ such that if $\ge -\eta$, then $\delta \ge \tilde{Y}(\tau -) - \tilde{Y}(t) \ge -\delta$. If $\epsilon$ is small enough, then there exists $n_0$ such that $\forall ~\, n \ge n_0$, $\tau$ $-\tau _n > -\eta$. It follows that $\delta \ge \tilde{Y}(\tau _n) - \tilde{Y}(\tau -) \ge a - \tilde{Y}(\tau -)$, implying that $\tilde{Y}(\tau -) \ge a - \delta$. Since $\delta$ is arbitrary, it follows that $\tilde{Y}(\tau -) \ge a$, in turn implying that $\tilde{Y}(\tau ) \ge 0$. Thus, $\tau \in \tilde{Y}^{-1}[a,\infty )$. Next, suppose that $\tau$ is a right-discontinuity point. Then, from part (ii) of Lemma 4, we have $\tilde{Y}(\tau ) = \tilde{Y}(\tau -) < \tilde{Y}(\tau +)$. Furthermore, for any $u > \tau$, we have $\nabla _u^{\bar{X}} \subseteq (\tau ,u]$ implying that these are continuity points (by part (i) of Lemma 4). Using an argument similar to that for a left-discontinuity, on points to the right of $\tau$, it follows that $\tilde{Y}(\tau ) \ge a$. This implies that the pullback set $\tilde{Y}^{-1}[a,\infty )$ is closed. As $\{\tau _n\}$ is an arbitrary sequence in $\tilde{Y}^{-1}[a,\infty )$ it is necessarily true that $\tilde{Y}$ is upper-semicontinuous. $\square$

1.5 Proof of Corollary 4

The proof of the corollary depends on Lemma 4 above.

Proof (Corollary 4)

[Corollary 4] Recall that $\hat{Q} = \hat{X} + \tilde{Y}$, where $\tilde{Y}(t) = \sup _{s \in \nabla _t^{\bar{X}}} (-\hat{X}(s))$. The proof of (i) follows directly from part (i) of Lemma 4. Next, recall from the proof of Corollary 3 that $\nabla _{\tau }^{\bar{X}} = \{-T_0,\tau \}$. Thus, $\tau$ is isolated in the set and it follows that part (iii) of Lemma 4 is satisfied. On the other hand, recall that $\nabla _t^{\bar{X}} = \{t\} \subset (\tau ,t], ~ \forall t > \tau$, and $\tau$ can also be a point of right-discontinuity, by part (ii) of Lemma 4. Thus, $\tau$ is one or the other depending on the path of $\hat{X}$. If $\hat{X}(\tau ) < 0$ then $\tilde{Y}(\tau +) = \tilde{Y}(\tau ) > \tilde{Y}(\tau -)$ and $\tau$ is a point of left-discontinuity. Otherwise, if $\hat{X}(\tau ) \ge 0$, then s$\tilde{Y}(\tau ) = \tilde{Y}(\tau -) = 0 > \tilde{Y}(\tau +)$ and $\tau$ is a point of right-discontinuity. $\square$