# M/M/1 with balking

I have the following $M/M/1$ queue where:

$\cdot$ customers arrive at a rate $\lambda$ and they require an exponential service time with rate $\mu$.

$\cdot$ customers leave the queue at a rate $\delta$, independently of their position in the queue.

I am asked to prove that for any $\delta>0$ the queue has an invariant distribution.

My approach: I am trying to find the generator(Q-matrix) in order to apply detail balance equation.

The process of leaving the queue and the service process are independent so $g_{i,i-1}=\mu+\delta$. And as customers arrive as a Poisson process with rate $\lambda$ we have that $g_{i,i+1}=\lambda$.

Applying detail balance we have that:

$\pi_{i}g_{i,i-1}=\pi_{i-1}g_{i-1,i}$

$\pi_{i}=\frac{\lambda}{\mu+\delta}\pi_{i-1}$

So we have that: $\pi_{n}=\Big(\frac{\lambda}{\mu+\delta}\Big)^n\pi_{0}$

However this is a distribution only if $\lambda<\mu+\delta$. And as this has to be true for any $\delta>0$, we must have that $\lambda<\delta$. But this information is not given to me in the problem.

• It seems like if all customers leave the queue at a rate of $\delta$, then a state with $n$ queued customers transitions to a state with $n-1$ queued customers at a rate of $\mu + n\delta$, not $\mu+\delta$. – Misha Lavrov Mar 10 '18 at 21:53
• This is actually known as "reneging." "Balking" refers to the phenomenon when customers decide not to enter the system if the queue is too long. – Math1000 Mar 19 '18 at 23:45

As Mishra Lavrov noted in the comments, the transition rate from state $n$ to state $n-1$ is $\mu+n\delta$. So the detailed balance equations are $$\lambda\pi_n = (\mu+n\delta)\pi_{n+1}.$$ This yields the recursion $\pi_{n+1}=\frac\lambda{\mu+ n\delta}\pi_n$, with solution $$\pi_{n+1}=\prod_{k=0}^{n-1}\left(\frac\lambda{\mu+ k\delta}\right)^k\pi_0.$$ Since $\sum_{n=0}^\infty \pi_n=1$, it follows that $$\pi_0 = \left(1 + \sum_{n=1}^\infty\prod_{k=0}^{n-1}\left(\frac\lambda{\mu+ k\delta}\right)^k\right)^{-1},$$ and hence $$\pi_{n+1} = \prod_{k=0}^{n-1}\left(\frac\lambda{\mu+ k\delta}\right)^k\left(1 + \sum_{n=1}^\infty\prod_{k=0}^{n-1}\left(\frac\lambda{\mu+ k\delta}\right)^k\right)^{-1},\; n\geqslant0.$$