The following is a question from Pinksy and Karlin's $\textit{An Introduction to Stochastic Modelling}$:
Customers arrive at a service facility according to a Poisson process of rate λ. There is a single server, whose service times are exponentially distributed with parameter µ. Suppose that “gridlock” occurs whenever the total number of customers in the system exceeds a capacity C. What is the smallest capacity C that will keep the probability of gridlock, under the limiting distributing of queue length, below 0.001? Express your answer in terms of the traffic intensity ρ = λ/µ.
My approach is to simply the expression:
$$\pi_k = \lim\limits_{t\to \infty} \mathbf{P}(X(t)=k)<0.001$$
which simplifies to
$$(1-\frac{\lambda}{\mu})(\frac{\lambda}{\mu})^k <0.001$$
where $\pi_k$ is the limiting distribution of queue length.
From which, I will then obtain an inequality involving $\rho$. Am I on the right track, or do I have a conceptual misunderstanding? I am currently lost as well, so some insight and clarification will be appreciated.
What you have written is essentially the answer once you solve for $k$, which will be the value of $C$ that you are looking for once you make it an integer.
Really it seems to me that what your after is developing intuition; for this I always find it helpful to try plugging in values to Markov chains and seeing if they make sense. Say $\lambda=1$ and $\mu=2$, so $\rho=\frac{1}{2}$ i.e. customers leave twice as fast as they arrive. Now, we want to look at the limiting distribution of how many people are in the queue. For $k$ people, we have the limiting distribution $$\pi_k = (1-\rho)(\rho)^k = (\frac{1}{2})^{k+1}$$
Now, as we look at higher values of $k$, we are looking at longer queues. This should intuitively make sense, as when $k$ gets large our limiting distribution $\pi_k$ will get smaller. This question is essentially about finding the 'threshold' queue for $k$ where the limiting distribution is small enough.