1
$\begingroup$

Consider a queue (1 server only) where arrivals occurs based on a Poisson process with $\lambda > 0$ and the dropout $\mu > 0$. The thing is each arrival can correspond to 1 client with probability $p$ or 2 clients with probability $1-p$.

How to determine equilibrium condition?

$\endgroup$
4
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. $\endgroup$ Jun 25, 2023 at 16:19
  • $\begingroup$ I would consider framing this as having $p\lambda + 2(1-p) \lambda$ arrivals $\endgroup$ Jun 26, 2023 at 4:56
  • 1
    $\begingroup$ @HeiniHøgnason I don't believe that will yield the correct result, as the global balance equations would then imply $\pi_1 = \frac{\lambda(2-p)}\mu \pi_0$ (when it should be $\pi_1 = \frac\lambda\mu \pi_0$). The arrival rate when the system is empty should instead be considered as $\lambda$. $\endgroup$
    – Math1000
    Jun 27, 2023 at 3:43
  • 1
    $\begingroup$ @Math1000 You are correct, of course. Please ignore my comment. $\endgroup$ Jun 28, 2023 at 6:39

1 Answer 1

2
$\begingroup$

The global balance equations $$ \pi_i\sum_{j\in S\setminus\{i\}} q_{ij} = \sum_{j\in S\setminus\{i\}} \pi_jq_{ji} $$ for $i\in S:=\{0,1,2,\ldots\}$ are given by \begin{align} \lambda\pi_0 &=\mu\pi_1\\ (\lambda+\mu)\pi_1 &= \lambda p\pi_0+\mu\pi_2\\ (\lambda+\mu)\pi_n &= \lambda(1-p)\pi_{n-2}+\lambda p\pi_{n-1}+\mu\pi_{n+1},\ n\geqslant 2. \end{align} This recursion yields $$ \pi_n = \frac{\lambda^{n-1}(\lambda+(n-1)(1-p)\mu)}{\mu^n}\pi_0, $$ and from $\sum_{i=0}^\infty \pi_i=1$ it follows that \begin{align} \pi_0 &= \frac{(\lambda -\mu )^2}{\mu (\mu -\lambda p)}\\ \pi_n &= \frac{\lambda^{n-1}(\lambda+(n-1)(1-p)\mu)}{\mu^n}\cdot\frac{(\lambda -\mu )^2}{\mu (\mu -\lambda p)},\ n\geqslant 1. \end{align}

$\endgroup$
3
  • $\begingroup$ Thank you very much!! I understand the balance equations, but If I could just ask you one more thing, can you explain how does this recursion yields to this formula of $\pi_n$? I can't arrive at this conclusion myself, and I also did not understood how you found $\pi_0$. In any case, thank you! Really grateful for your answer, it helped a lot $\endgroup$
    – Davi
    Jun 28, 2023 at 22:24
  • 1
    $\begingroup$ After expressing $\pi_1$, $\pi_2$, and $\pi_3$ in terms of $\pi_0$ I noticed a pattern giving a closed form for $\pi_n$ in terms of $\pi_0$. Then you can write $\sum_{i=0}^ \infty \pi_i = \pi_0 C$, hence $\pi_0 = 1/C$. $\endgroup$
    – Math1000
    Jun 29, 2023 at 6:44
  • 1
    $\begingroup$ I finally got it how this infinite sum leads to $\pi_0$, thank you very much for sharing your knowledge! $\endgroup$
    – Davi
    Jun 29, 2023 at 12:49

You must log in to answer this question.

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