Customers arrive at a single-server station with Poisson rate $\lambda$. A customer enters the bank if the server is available; otherwise, the customer leaves. The service times of successive customers are independent and have a common distribution $G$ and mean $\mu$. What is the rate at which the customer enters the system?
I am unable to figure out the answer. I assume that renewal reward process is to be applied here with regeneration happening at every service completion. Could anyone please help. Thanks in advance!
In your question it is implicitly assumed that the system operates in the steady state. For the steady state to exist the system's load (denote it by $\rho$) must be less than one i.e. $\rho=\lambda \mu<1$.
The system you are asking about, when operating in the steady state, follows the cycle $empty \rightarrow busy \rightarrow empty \rightarrow busy \rightarrow ...$. Thus the proportion of time that the system is empty is $$ {(mean \,\, empty \,\, time) \over (mean \,\, empty \,\, time) + (mean \,\, busy \,\, time)}= {{1 \over \lambda} \over {1 \over \lambda} + \mu}. $$
Thus the arrival rate of customers which find the system empty is $\lambda \times {{1 \over \lambda} \over {1 \over \lambda} + \mu}$.
p.s. The latter conclusion uses the PASTA property of the Poisson process. So in general it is not valid.
p.p.s. in Kendall's notation, you are asking about the $M/G/1/0$ queueing system. In any classic queueing theory book, there are many results on this queue.