So I've set up the following equations for this queueing problem (discrete-time Markov chain) to solve for the stationary distribution. First off, I'm not sure if I'm equations are correct. I'm trying to solve for the stationary distribution $\pi$. So I want to solve for $\pi =\pi P$. The state space is countably infinite: $0,1,2,…$
Here are the transition probabilities:
$P_{00}=(1-\lambda)+\lambda(\mu)$
$P_{01}=\lambda(1-\mu)$
For $i \geq 1$
$P_{ii}=\lambda\mu+(1-\mu)(1-\lambda)$
$P_{i,i-1}=(1-\lambda)\mu$
$P_{i,i+1}=\lambda(1-\mu)$
I want to solve for the stationary distribution $\pi$. I thus set up $\pi=\pi P$.
I get:
$\pi_0=[(1-\lambda)+\lambda \mu]\pi_0+[\lambda(1-\mu)]\pi_1$
$\pi_1=[(1-\lambda)\mu]\pi_0+[\lambda(\mu)+(1-\lambda)(1-\mu)]\pi_1+[\lambda(1-\mu)]\pi_2$
$\pi_2=[(1-\lambda)\mu]\pi_1+[\lambda(\mu)+(1-\lambda)(1-\mu)]\pi_2+[\lambda(1-\mu)]\pi_3$
…
….
And I know $\sum_{i=0}^{\infty} \pi_i=1$
First off, please let me know if these equations are correct. However, my main problem is I have no idea how to solve these equations for a general form for $\pi_i$; I have no idea how to solve these equations to get a general equation for $\pi_{i}$. I tried substituting but its almost impossible me for to get an expression.
Best Answer
This is a birth-death process, so we may instead use the detailed balance equations $$ \lambda(1-\mu)\pi_{n-1} = (1-\lambda)\mu \pi_n,\quad n\geqslant 1. $$ This yields the recurrence $$ \pi_n = \left(\frac{\lambda(1-\mu)}{\mu(1-\lambda)}\right)^n\pi_0,\quad n\geqslant 0. $$ Now, assuming that $\frac{\lambda(1-\mu)}{\mu(1-\lambda)}<1$, we have convergence of the following series which gives us the value of $\pi_0$: $$ 1 = \sum_{n=0}^\infty\left(\frac{\lambda(1-\mu)}{\mu(1-\lambda)}\right)^n\pi_0\implies \pi_0 = \frac{\mu-\lambda}{\mu(1-\lambda)}. $$ Plugging this into the recurrence gives us $$ \pi_n = \left(\frac{\lambda(1-\mu)}{\mu(1-\lambda)}\right)^n\frac{\mu-\lambda}{\mu(1-\lambda)}. $$