Age of information using Markov-renewal methods

Abstract

When designing a message transmission system, from the point of view of making sure that the information transmitted is as fresh as possible, two rules of thumb seem reasonable: use small buffers and adopt a last-in-first-out policy. In this paper, the freshness of information is interpreted as the recently studied “age of information” performance measure. Considering it as a stochastic process operating in a stationary regime, we compute the whole marginal distribution of the age of information for some well-performing systems. We assume that the arrival process is Poisson and that the messages have independent service times with common distribution, i.e., the M/GI model. We demonstrate the usefulness of Palm and Markov-renewal theory to derive results for Laplace transforms. Our numerical studies address some aspects of open questions regarding the optimality of previously proposed scheduling policies, and a policy newly considered herein, for AoI management.

Appendices

Appendix A: Proof of Theorem 2 by means of Markov embedding

We make use of (11) of Proposition 1 which needs computation of the quantities involving $\alpha (0)$ in its right-hand side. The analog of Lemma 2 is Lemma 10 below which looks conspicuously the same. In fact, the first two formulas are identical. The last two differ.

Lemma 10

For ${\mathcal {P}}_2$,

$$\begin{aligned} {\mathbb {E}}^0[e^{-s \alpha (S_0)} ;{K_{-1}=0,K_0=0}]&= {\hat{G}}(\lambda ){\hat{G}}(s+\lambda ), \end{aligned}$$

(42)

$$\begin{aligned} {\mathbb {E}}^0[e^{-s \alpha (S_0)} ;{K_{-1}=0,K_0=1}]&= {\hat{G}}(\lambda ) \left( {\hat{G}}(s) - {\hat{G}}(s+\lambda )\right) , \end{aligned}$$

(43)

$$\begin{aligned} {\mathbb {E}}^0[e^{-s \alpha (S_0)} ;{K_{-1}=1,K_0=0}]&= \frac{\lambda }{\lambda +s} \left( 1- {\hat{G}}(s+\lambda ) \right) \,{\hat{G}} (s+\lambda ), \end{aligned}$$

(44)

$$\begin{aligned} {\mathbb {E}}^0[e^{-s \alpha (S_0)} ;{K_{-1}=1,K_0=1}]&= \frac{\lambda }{\lambda +s}\left( 1-{\hat{G}}(\lambda +s)\right) \, \left( {\hat{G}}(s)-{\hat{G}}(s+\lambda ) \right) \end{aligned}$$

(45)

Proof

• 1) When $K_{-1}=0, K_0=0$ or when $K_{-1}=0,K_0=1$ the AoI is the same as in the ${\mathcal {B}}_2$ system, the reason being that the number of messages in the system is always at most 1, see Fig. 3.

• 2) Suppose next that $K_{-1}=1,K_0=0$. In Fig. 19, we depict the two scenarios corresponding to the possible values of $K_{-2}$, namely $(K_{-2},K_{-1},K_0)=(0,1,0)$ or (1, 1, 0). In both cases, (3) and the system dynamics imply that

  $$\begin{aligned} \alpha (0) = S_0-S_{-1}+V, \end{aligned}$$

  V is the time elapsed between the last arrival in the interval $(S_{-2},S_{-1})$ and $S_{-1}$. If there is only one arrival in this interval, then $V:=S_{-1}-T_{0}$. In any case,

  $$\begin{aligned}&\text {conditionally on }\{K_{-1}=1,K_0=0\}, \text { the random variables} V \text { and }S_0-S_{-1}\\&\quad \text { are independent.} \end{aligned}$$

  Therefore,

  $$\begin{aligned} {\mathbb {E}}^0[e^{-s (S_0-S_{-1}+V)} \mid K_{-1}= & {} 0,K_0=1] = {\mathbb {E}}^0[e^{-s (S_0-S_{-1})} \mid K_{-1}=0,K_0=1]\,\\ {\mathbb {E}}^0[e^{-s V} \mid K_{-1}= & {} 0,K_0=1] \end{aligned}$$

  and the first factor on the right is easy:

  $$\begin{aligned} {\mathbb {E}}^0[e^{-s (S_0-S_{-1})} \mid K_{-1}=0,K_0=1] \;=\; \int _0^\infty e^{-s x} \frac{e^{-\lambda x} dG(x)}{{\hat{G}}(\lambda )} \;=\; \frac{{\hat{G}}(s+\lambda )}{{\hat{G}}(\lambda )}. \end{aligned}$$

  To evaluate ${\mathbb {E}}^0[e^{-s V} \mid K_{-1}=0,K_0=1]$, we note that V is the distance of the last Poisson point inside the interval $(T_{-1},S_{-1})$ (in the left scenario of Fig. 19) or the interval $(T_{-1},S_{-1})$ in the right scenario. (In both cases, the length of the interval is that of a message size conditioned on containing at least one Poisson point.) To obtain the Laplace transform of V look backward in time starting from $S_{-1}$ until the first Poisson point appears and condition on the event that this occurs between $S_{-1}$ and $S_{-2}$. Thus, the density of V at $v>0$ is

  $$\begin{aligned} \frac{(1-G(v))\lambda e^{-\lambda v}}{1-{\hat{G}}(\lambda )}, \end{aligned}$$

  which gives

  $$\begin{aligned} {\mathbb {E}}^0[e^{-s V} \mid K_{-1}=0,K_0=1]= & {} \int _0^\infty e^{-sv} \frac{\lambda e^{-\lambda v} (1-G(v))}{1-{\hat{G}}(\lambda )} dv \nonumber \\= & {} \frac{\lambda }{\lambda +s} \, \frac{1-{\hat{G}}(s+\lambda )}{1-{\hat{G}}(\lambda )} . \end{aligned}$$

  (46)

  Putting these together, we obtain (44).

• 3) Finally, assume that $K_{-1}=1,K_0=1$. This situation is similar to the previous one and thus, will be treated succinctly. We are guided by Fig. 20. Firstly, we have

  $$\begin{aligned} {\mathbb {E}}^0[e^{-s (S_0-S_{-1})} \mid K_{-1}=1,K_0=1]= & {} \int _0^\infty e^{-s x} \frac{1-e^{-\lambda x} dG(x)}{1-{\hat{G}}(\lambda )}\\= & {} \frac{{\hat{G}}(s)-{\hat{G}}(s+\lambda )}{1-{\hat{G}}(\lambda )}. \end{aligned}$$

Fig. 19

Under ${\mathcal {P}}_2$, the segments $[S_{-2},S_{-1})$, $[S_{-1},S_0)$ when $K_{-1}=1,K_0=0$ in the case where $K_{-2}=0$ (left) and $K_{-2}=1$ (right)

Fig. 20

Under ${\mathcal {P}}_2$, the segments $[S_{-2},S_{-1})$, $[S_{-1},S_0)$ when $K_{-1}=1,K_0=1$ in the case where $K_{-2}=0$ (left) and $K_{-2}=1$ (right)

Secondly, the argument used to derive (46) can be used here again with no changes to obtain

$$\begin{aligned} {\mathbb {E}}^0[e^{-s V} \mid K_{-1}=1,K_0=1]= & {} \int _0^\infty e^{-sv} \frac{\lambda e^{-\lambda v} (1-G(v))}{1-{\hat{G}}(\lambda )} dv\\= & {} \frac{\lambda }{\lambda +s} \, \frac{1-{\hat{G}}(s+\lambda )}{1-{\hat{G}}(\lambda )} . \end{aligned}$$

Putting these together, we obtain (44) as well. $\square$

The formula for ${\mathbb {E}}e^{-s \alpha (0)}$ in Theorem 2 is now clear:

Proof

Adding up (42) and (43) of Lemma 10 and similarly (44) and (45), we obtain

$$\begin{aligned} {\mathbb {E}}[e^{-s \alpha (S_0)} ;{K_0=0}]&= {\hat{G}}(s+\lambda ) \left[ {\hat{G}}(\lambda ) + \frac{\lambda }{\lambda +s} \left( 1- {\hat{G}}(s+\lambda ) \right) \right] \\ {\mathbb {E}}[e^{-s \alpha (S_0)} ;{K_0=1} ]&= \left( {\hat{G}}(s)-{\hat{G}}(s+\lambda ) \right) \, \left[ {\hat{G}}(\lambda ) + \frac{\lambda }{\lambda +s}\left( 1-{\hat{G}}(\lambda +s)\right) \right] . \end{aligned}$$

Substituting these expressions in the numerator of (11), and recalling the definition (19) of $G_I$, we obtain (25). $\square$

An alternative expression for (25) is:

$$\begin{aligned} {\mathbb {E}}[e^{-s\alpha (0)}]= & {} \frac{{\hat{G}}(s) \left( {\hat{G}}(\lambda ) + \frac{\lambda }{\mu } {\hat{G}}_I(s+\lambda ) \right) \left( \frac{\lambda }{\lambda +s}{\hat{G}}(s+\lambda ) + \frac{\lambda }{\mu } {\hat{G}}_I(s)\right) }{\frac{\lambda }{\mu }+{\hat{G}}(\lambda )} . \end{aligned}$$

(47)

Corollary 4

For ${\mathcal {P}}_2$ with exponential message sizes,

$$\begin{aligned} {\mathbb {E}}[e^{-s\alpha (0)}]= & {} \frac{\mu }{\mu +s} \, \left( \frac{\mu }{\mu +\lambda } + \frac{\lambda }{\lambda +\mu +s} \right) \,\nonumber \\&\left( \frac{\mu ^2}{\lambda ^2+\lambda \mu + \mu ^2} \frac{\lambda }{\lambda +s} \frac{\lambda +\mu }{\lambda +\mu +s} + \frac{\lambda ^2+\lambda \mu }{\lambda ^2+\lambda \mu + \mu ^2} \frac{\mu }{\mu +s} \right) , \end{aligned}$$

(48)

$$\begin{aligned} {\mathbb {E}}[\alpha (0)]= & {} \frac{2\lambda ^{5}+7\lambda ^{4}\mu +8\lambda ^{3}\mu ^{2}+7\lambda ^{2}\mu ^{3}+4\lambda \mu ^{4}+\mu ^{5}}{\lambda \mu \left( \lambda +\mu \right) ^{2}\left( \lambda ^{2}+\lambda \mu +\mu ^{2}\right) } , \end{aligned}$$

(49)

and, with $\rho =\lambda /\mu$, the standard deviation of $\alpha (0)$ under ${\mathbb {P}}$ is

$$\begin{aligned}&{\text {sd}}_{\mathbb {P}}(\alpha (0)) \\&\quad = \frac{1}{\mu }\frac{\sqrt{2\rho ^{10}+12\rho ^{9} +35\rho ^{8} +60\rho ^{7} +66\rho ^{6} +56\rho ^{5} +45\rho ^{4} +34\rho ^{3} +18\rho ^{2} +6\rho + 1}}{\rho \left( \rho +1 \right) ^2 \left( \rho ^{2}+\rho +1\right) }. \end{aligned}$$

The expectation and variance have been computed by summing up the expectations and variance of the three independent random variables comprising $\alpha (0)$. Inverting ${\mathbb {E}}e^{-s \alpha (0)}$ shows that $\alpha (0)$ has density (5), i.e., when $\mu =1$:

$$\begin{aligned} f_{{\mathcal {P}}_2}(t)&= q_1(t) e^{-t} + q_2(t) e^{-(\lambda +1)t} - \frac{\lambda }{\lambda -1} e^{-\lambda t}, \end{aligned}$$

if $\lambda \ne 1$ where

$$\begin{aligned} q_1(t) = \frac{( {\lambda }^{3}+{\lambda }^{2}-2\,\lambda ) t-{\lambda }^{2}+\lambda +3}{(\lambda ^2+\lambda +1)(\lambda -1)},&\quad q_2(t) = \frac{( {\lambda }^{2}+\lambda ) t+{\lambda }^{2}+3\,\lambda +3}{\lambda ^2+\lambda +1}, \end{aligned}$$

else if $\lambda = 1$,

$$\begin{aligned} f_{{\mathcal {P}}_2}(t) = \tfrac{1}{3} ( 7+2t) e^{-2t} + \tfrac{1}{3} (6t-7) e^{-t}. \end{aligned}$$

It is easy to see from (48) that $\lim _{\lambda \rightarrow \infty } {\mathbb {E}}[e^{-s\alpha (0)}] = (\mu /(s+\mu ))^2$, the sum of 2 i.i.d. exponentials.

Appendix B: Analysis of the ${\mathcal {P}}_{2,\theta }$ system

We now analyze the system described in Sect. 7.

Given $K_{-1}=0$, consider Fig. 21 at right.

Fig. 21

Number of messages in the system (black line) when $K_{-1}=0,K_0=0$ (left) and $K_{-1}=0,K_0=1$ (right). A black dot indicates a (successful) message departure. A green dot indicates an arrival of a message that will successfully depart. A red dot indicates an arrival of an unsuccessful message. If the fate of an arrival is not indicated in the figure, then it has no indicating dot

A message service period is successful with probability

$$\begin{aligned} 1-q:={\mathbb {P}}(\tau >\theta \wedge \sigma ) = (1-G(\theta ))e^{-\lambda \theta } +\int _0^\theta e^{-\lambda s} dG(s). \end{aligned}$$

So, considering the successful service period which concludes at $S_0$:

$$\begin{aligned} {\mathbb {P}}(K_0=0|K_{-1}=0)= & {} {\mathbb {P}}(\tau>\sigma ~|~\tau >\theta \wedge \sigma )\\= & {} (1-q)^{-1}{\hat{G}}(\lambda ) =1-{\mathbb {P}}(K_1=1|K_{-1}=0). \end{aligned}$$

Given $K_{-1}=1$, consider Fig. 22 below.

Fig. 22

$K_{-1}=1,K_0=1$ cases

Use the memoryless property of interarrivals to similarly obtain

$$\begin{aligned} {\mathbb {P}}(K_0=0|K_{-1}=1)&= (1-q)^{-1}{\hat{G}}(\lambda ) =1-{\mathbb {P}}(K_0=1|K_{-1}=1) \end{aligned}$$

So, $K_n$ is an i.i.d. Bernoulli sequence with

$$\begin{aligned} p_0 := {\mathbb {P}}(K_n=0) = (1-q)^{-1}{\hat{G}}(\lambda ) = 1-{\mathbb {P}}(K_n=1) =: 1-p_1. \end{aligned}$$

The queueing process over consecutive intervals $[S_{i-1},S_i)$ and $[S_i,S_{i+1})$ is conditionally independent given $K_i$. Thus, $\{S_i,K_i\}_{i\in {\mathbb {Z}}}$ is Markov-renewal with renewal times $S_i$ and the queueing process is semi-Markov [4]. In particular, $\alpha (S_i)$ and $S_{i+1}-S_i$ are conditionally independent given $K_i$.

As Proposition 1, we have:

Proposition 3

The Laplace transform of the stationary AoI distribution is

$$\begin{aligned} {\mathbb {E}}e^{-s \alpha (0)}&= \frac{ \sum _{i=0}^1 {\mathbb {E}}^0(e^{-s\alpha (0)}|K_0=i) \Big (1- {\mathbb {E}}^0(e^{-s(S_0-S_{-1})}|K_{-1}=i) \Big )p_i}{s{\mathbb {E}}^0(S_1-S_0)}, \end{aligned}$$

(50)

where ${\mathbb {P}}^0$ and ${\mathbb {E}}^0$ are, respectively, probability and expectation given $S_0=0$.

Proof

By the Palm inversion formula [2],

$$\begin{aligned} {\mathbb {E}}e^{-s \alpha (0)} = \frac{ {\mathbb {E}}^0 \int _{S_0}^{S_1}e^{-s \alpha (t)}dt}{{\mathbb {E}}^0 (S_1-S_0)}, \end{aligned}$$

where the numerator

$$\begin{aligned} {\mathbb {E}}^0 \int _{S_0}^{S_1}e^{-s \alpha (t)}dt&= {\mathbb {E}}^0 \int _{S_0}^{S_1}e^{-s (\alpha (S_0)+t-S_0)}dt\\&= {\mathbb {E}}^0 \int _{S_0}^{S_1} \sum _{i=0}^1 (e^{-s\alpha (0)}|K_0=i) {\mathbb {E}}^0(e^{-s(t-S_0)}| K_0=i) p_i dt \\&= \sum _{i=0}^1 {\mathbb {E}}^0(e^{-s\alpha (0)}|K_0=i) {\mathbb {E}}^0\Bigg (\frac{1-e^{-s(S_1-S_{0})}}{s}\Bigg | K_0=i\Bigg ) p_i \end{aligned}$$

$\square$

To calculate the terms in (50), we need to follow the steps outlined in the lemmas below. Let ${\hat{F}}_0(s)= {\hat{G}}(s+\lambda )/{\hat{G}}(\lambda )$ and ${\hat{J}}(s) = \int _0^\infty e^{-s y} dJ(y)$ where $J(y)= 1$ for $y\ge \theta$ and, for $0\le y<\theta ,$

$$\begin{aligned} dJ(y)&= q^{-1} \lambda e^{-\lambda y}(1-G(y)) dy \end{aligned}$$

Lemma 11

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s(S_0-S_{-1})} \mid K_{-1}=0,K_0=0) = \frac{\lambda }{\lambda +s}\cdot \frac{1-q}{1-q{\hat{J}}(s)} {\hat{F}}_0(s) \end{aligned}$$

(51)

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s \alpha (S_0)} \mid K_{-1}=0,K_0=0) = {\hat{F}}_0(s) \end{aligned}$$

(52)

Proof

See Fig. 21 at left and consider the interval $[S_{-1},S_0)$. Let $\tau _{-1}$ be first message arrival time in this interval minus $S_{-1}$, so that $\tau _{-1}\sim \exp (\lambda )$ by the memoryless property. Note that there is a geometric number N of interarrival times each of which is smaller than both $\theta$ and the associated service time; $N=2$ in Fig. 21. Again, the probability of such unsuccessful service is

$$\begin{aligned} {\mathbb {P}}(\tau <\theta \wedge \sigma ) = q. \end{aligned}$$

So, ${\mathbb {P}}(N=k)=(1-q) q^k$ for $k=0,1,2, \ldots$. Let $Y {\mathop {=}\limits ^{\text {(d)}}}(\tau | \tau < \theta \wedge \sigma )$ so that ${\mathbb {P}}(Y\le y)=J(y)$. Finally, let $\tau _0\sim \exp (\lambda )$ be the duration between the arrival time (green dot) of the message that departs at $S_0$ and the next arrival time. The service time (from the green dot to $S_0$) $\sigma _0$ is independent of $\tau _0$. Considering the prior N unsuccessful service completions, we are given that $\tau _0 >\sigma _0$ or $\tau _0>\theta$. Given $K_0=0$, $\tau _0>\sigma _0$. Let $X_0 {\mathop {=}\limits ^{\text {(d)}}}(\sigma _0|\tau _0>\sigma _0)$ which has distribution

$$\begin{aligned} dF_0(x) = {\hat{G}}(\lambda )^{-1} e^{-\lambda x} dG(x), \quad x > 0, \end{aligned}$$

with ${\mathbb {E}}e^{-s X_0}={\hat{F}}_0(s)$. So,

$$\begin{aligned} (S_0-S_{-1}\mid K_{-1}=0,K_0=0) {\mathop {=}\limits ^{\text {(d)}}}\tau _{-1} + \sum _{n=1}^N Y_n + X_0, \end{aligned}$$

a sum of independent terms with $Y_n {\mathop {=}\limits ^{\text {(d)}}}Y$. Also, $\alpha (S_0) {\mathop {=}\limits ^{\text {(d)}}}X_0$ in this case. $\square$

Define

$$\begin{aligned} dF_1(x)&= \frac{(e^{-\lambda \theta }-e^{-\lambda x}) dG(x)}{\int _\theta ^\infty (e^{-\lambda \theta }-e^{-\lambda z}) dG(z)} , \quad x>\theta \\ {{\hat{F}}}_1(s)&= \int _\theta ^\infty e^{-sx} dF_1(x). \end{aligned}$$

Lemma 12

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s(S_0-S_{-1})} \mid K_{-1}=0,K_0=1) = \frac{\lambda }{\lambda +s}\cdot \frac{1-q}{1-q{\hat{J}}(s)}{\hat{F}}_1(s) \end{aligned}$$

(53)

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s \alpha (S_0)} \mid K_{-1}=0,K_0=1) = {\hat{F}}_1(s). \end{aligned}$$

(54)

Proof

See Fig. 21 at right. The difference between this and the previous case is that here $\sigma _0>\tau _0>\theta$. So, the distribution of $X_1 {\mathop {=}\limits ^{\text {(d)}}}(\sigma _0\mid \sigma _0>\tau _0>\theta )$ is $dF_1(x)$. $\square$

Define

$$\begin{aligned} dH(v)&= \frac{\lambda e^{-\lambda v}(1-G(v+\theta )) dv }{ \int _\theta ^\infty (1-e^{-\lambda (x-\theta )}) dG(x)}, \quad v \ge 0,\\ {\hat{H}}(s)&= \int _0^\infty e^{-sv} dH(v). \end{aligned}$$

Lemma 13

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s(S_0-S_{-1})} \mid K_{-1}=1,K_0=0) = \frac{1-q}{1-q{\hat{J}}(s)}{\hat{F}}_0(s) \end{aligned}$$

(55)

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s \alpha (S_0)} \mid K_{-1}=1,K_0=0) = q{\hat{F}}_0(s) +(1-q){\hat{H}}(s)\times (55) \end{aligned}$$

(56)

Proof

See Fig. 23 below.

Fig. 23

$K_{-1}=1,K_0=0$ cases

For $\alpha (S_0)$, there are two subcases depending of whether there are initial unsuccessful arrivals in the interval $[S_{-1},S_0)$, i.e., whether $N>0$. When $N>0$ (Fig. 23 right), this case is like when $K_{-1}=0,K_0=0$. Otherwise (Fig. 23 left),

$$\begin{aligned} \alpha (S_0) =V+ S_{0}-S_{-1}, \end{aligned}$$

where V is the duration between $S_{-1}$ the last arrival before $S_{-1}$ (green dot), and V and $S_0-S_1$ are independent given $K_{-1}=1$. Starting from $S_{-1}$, look backward in time until the first Poisson point appears (green dot) and condition on the event that this occurs at least $\theta$ units of time before the service time ends. Thus, $V {\mathop {=}\limits ^{\text {(d)}}}(\tau |\tau <\sigma -\theta ,\sigma >\theta )$ with distribution H. Also, $S_0-S_{-1}$ is distributed as for the case where $K_{-1}=0,K_0=0$ except the first interarrival time is absent. $\square$

Lemma 14

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s(S_0-S_{-1})} \mid K_{-1}=1,K_0=1) = \frac{1-q}{1-q{\hat{J}}(s)}{\hat{F}}_1(s) \end{aligned}$$

(57)

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s \alpha (S_0)} \mid K_{-1}=1,K_0=1) = q{\hat{F}}_1(s) +(1-q){\hat{H}}(s)\times (57) \end{aligned}$$

(58)

Proof

See Fig. 22. For $S_0-S_{-1}$, this case is a combination of the previous two cases, and for $\alpha (S_0)$ follow the previous case except use (54) instead of (52). $\square$

The final stage: The formulas obtained in the lemmas above must now be substituted into (50) as follows:

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s(S_0-S_{-1})}|K_{-1}=0) = (53)\times p_1 + (51)\times p_0 \end{aligned}$$

(59)

$$\begin{aligned}&{\mathbb {E}}^0(e^{-s(S_0-S_{-1})}|K_{-1}=1) = (57)\times p_1 + (55)\times p_0 . \end{aligned}$$

(60)

So,

$$\begin{aligned} {\mathbb {E}}^0 (S_1-S_{0}) =-\frac{d}{ds} {\mathbb {E}}^0 e^{-s (S_0-S_{-1})} \Bigg |_{s=0} =-\frac{d}{ds} (p_1\times (60)+p_0\times (59))\Bigg |_{s=0}. \end{aligned}$$

Moreover,

$$\begin{aligned} {\mathbb {E}}^0 (e^{-s\alpha (0)}|K_0=0)&= p_0\times (52) + p_1\times (56) \\ {\mathbb {E}}^0 (e^{-s\alpha (0)}|K_0=1)&= p_0\times (54) + p_1\times (58). \end{aligned}$$

Appendix C: High traffic asymptotics

“High traffic asymptotics” refers to the regime $\rho =\lambda /\mu \rightarrow \infty$. Even though we have no explicit formulas for ${\mathcal {P}}_n$ or ${\mathcal {B}}_n$ when $n \ge 3$, we can easily obtain asymptotics from the system dynamics.

Proposition 4

Let $\sigma , \sigma _1, \sigma _2, \ldots$ be i.i.d. copies of $\sigma$. Let $\sigma _I$ be distributed as ${{\hat{G}}}_I$ as in (19). Then,

$$\begin{aligned} \alpha _{{\mathcal {P}}_n} \xrightarrow [\rho \rightarrow \infty ]{\mathrm{d}} \sigma + \sigma _I, \quad n \ge 2, \end{aligned}$$

while

$$\begin{aligned} \alpha _{{\mathcal {B}}_n} \xrightarrow [\rho \rightarrow \infty ]{\mathrm{d}} \sigma _1+\cdots + \sigma _n + \sigma _I, \quad n \ge 1. \end{aligned}$$

Sketch of proof

In both systems, the buffer consists of n cells. The message being processed sits in cell 1. In ${\mathcal {P}}_n$, the freshest message is either in cell 1, in which case all other cells are empty, or in cell 2. When $\rho$ is high, there is always a message being processed in cell 1 and the freshest message is in cell 2. Hence, at any time t, the AoI $\alpha _{{\mathcal {P}}_n}(t)$ equals the service time of the message in cell 2 plus the remaining service time of the message in cell 1. These are two independent random variables. The first one is distributed as $\sigma$. The second is distributed as $\sigma _I$ since the system is stationary. For ${\mathcal {B}}_1$, we can obtain the limit from the Laplace transform of (28). It is easy to see that $\lim _{\rho \rightarrow \infty } {\mathbb {E}}[e^{-s \alpha _{{\mathcal {B}}_1}}] = {{\hat{G}}}(s) \, {{\hat{G}}}_I(s)$ and so $\alpha _{{\mathcal {B}}_1} \xrightarrow [\rho \rightarrow \infty ]{\mathrm{d}} \sigma + \sigma _I$. For general n, when $\rho$ is high, the AoI $\alpha _{{\mathcal {B}}_n}(t)$ equals the remaining service time of the message in cell 1 (in distribution equal to $\sigma _I$) plus the time elapsed until the beginning of its service which is, in distribution, equal to the sum of n independent service times. $\square$

Remark 2

When $\sigma =1$ with probability 1, $\sigma _I$ is a uniform random variable in the interval [0, 1]. Hence, $\sigma _1+\cdots +\sigma _n+\sigma _I = n + \sigma _I$ and the variance of this random variable is 1/12. Similarly, for ${\mathcal {P}}_2$, the asymptotic variance is again 1/12.