# M/M/1 queues And finding equilibrium probability that the shop is empty

Customers arrive at a barbers shop at the incidents of a Poisson process of rate λ. Each person is served in order of arrival (by the single barber), and takes an exponential, rate μ service time. Services and arrivals are independent, and λ < μ. (a) In equilibrium, how likely is the shop to be empty? (b) Now suppose arrivals are put off by long queues, so that if there are n customers in the shop then a potential arrival decides not to enter with probability n/(n+ 1). What is the equilibrium probability that the shop is empty now?

A) I get that the invariant distribution is $\pi_k = (1-(\lambda/\mu))(\lambda/\mu)^k$ therefore probability the shop is empty is $\pi_0 = 1-(\lambda/\mu)$

For b I changed the arrivals rate to $\lambda n /(n+1)$ and then I solve $\pi Q = 0$ to get $\pi_k = \lambda^k \pi_0/(k\mu^k)$

I'm not sure if this is right, and if it is I cannot solve for $\pi_0$

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Hey, here is a terrific idea: what about accepting answers? – Did Mar 9 '13 at 17:37

Regarding question A), how did you get the invariant distribution you write down? More specifically, did you solve the corresponding linear system $\pi Q=0$ or did you merely copy a formula taken from a manual? If the latter, let me urge you to perform the former, if only as a warm up for...
...Question B), could you explain further how you got the distribution you write down? Two remarks: the identity in your post is undefined for $\pi_0$ since one divides by $0$; the arrival rates are $\lambda/(n+1)$, not $\lambda n/(n+1)$.