1
$\begingroup$

I'm posting the following question from my notes as an image because it has a diagram within it. It's from my lecture notes.

Poisson Process, jackson network.

I start to get confused when $\delta $ is choses to be much smaller than $1/\lambda$. I also don't understand when he computes the probability of there being at least one customer in the queue, and I'm pretty sure that's the crux of the problem.

Any help would really be appreciated! Thanks.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

These notes are using the infinitesimal definition of the Poisson process, which requires the following properties of the arrival process $A:=\{A(t):t\geqslant 0\}$:

  • $A$ has independent increments; if $0\leqslant t_1<t_2<\ldots<t_n<\infty$, then $A(t_2)-A(t_1),\ldots,A(t_n)-A(t_{n-1})$ are independent.
  • $A$ has stationary increments; the distribution of $A(t+s)-A(s)$ is independent of $s$.
  • $\mathbb P(A(\delta)=1)=\lambda\delta + o(\delta)$
  • $\mathbb P(A(\delta)\geqslant 2) = o(\delta).$ Here $o(\delta)$ denotes a function $g$ that satisfies $$\lim_{\delta\to 0}\frac{g(\delta)}\delta=0. $$ In other words, $\lim_{\delta\to 0}g(\delta)=0$, and for small $\delta$, $g(\delta)$ is so small compared to $\delta$ that it is negligible.

Recall that a queueing system must have $\lambda<\mu$ in order to be stable. Equivalently, the mean service time $\frac1\mu$ is less than the mean interarrival time $\frac1\lambda$. Under the assumptions that $p\to1$ and $\mu\to\infty$ as time tends to infinity, $\frac1\mu$ becomes negligibly small compared to $\frac1\lambda$ and the probability that a customer leaves the system after completing service becomes negligibly small as well. The choice of $\delta$ such that $$\frac1{\lambda'}\gg \delta\gg \frac1\mu $$ means that $\delta$ is negligible compared to $\frac1{\lambda'}$, and also $\frac1\mu$ is negligibly small compared to $\delta$. The probability of having at least one arrival in $(t,t+\delta]$, conditioned on there being at least one customer in the system at time $t$, is essentially $1$, because the amount of time it takes a customer to complete service and return to the queue is very small in comparison to $\delta$.

However, conditioned on there being no customers in the system at time $t$, from $\delta\ll\frac1{\lambda'}$ we have that $\lambda'$ and $\delta$ are of similar magnitude. The probability of having at least one arrival in $(t,t+\delta]$ is approximately $\lambda'\delta$, since it is proportional to both the arrival rate and the length of the time interval.

$\endgroup$
4
  • $\begingroup$ Ok, I think I'm understanding a little more. Still some questions: From the diagram we see that $\lambda = \lambda'/ 1-p$ Then we say, let p -> 1. But then wouldn't this mean that in the limiting case $\lambda$ would go to 0, and thus there would be no arrivals entering the queue? $\endgroup$
    – Joe
    Apr 30, 2016 at 21:10
  • $\begingroup$ Unless there is a finite buffer on the queue, the value of $p$ should not affect the exogenous arrival process. $\endgroup$
    – Math1000
    May 1, 2016 at 0:06
  • $\begingroup$ But the equation says that $\lambda$ is a function of p. I understand that because it is exogenous it shouldn't be affected by p, but the math seems to disagree with that. Do you have any intuition as to why this is? $\endgroup$
    – Joe
    May 1, 2016 at 0:36
  • $\begingroup$ The exogeneous arrival process has rate $\lambda'$, not $\lambda$. $\endgroup$
    – Math1000
    May 1, 2016 at 1:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .