= Solution
For a stable system in which entering customers eventually leave, <Little's law> is
$$
\boxed{L=\lambda_{\mathrm{eff}}W.}
$$
Here $L$ is the long-run mean number of customers in the chosen system, $\lambda_{\mathrm{eff}}$ is its actual entry rate, and $W$ is the mean <customer sojourn time>. The arrival rate counts admitted customers, not rejected attempts. One way to see the identity is to integrate the customer count over time: each customer's time in the system contributes exactly that much area. Dividing by elapsed time gives entry rate times mean <customer sojourn time>, with negligible endpoint terms under the usual stability and integrability assumptions.
For the server alone in a stationary <M M 1 queue>, the customer count is the busy indicator. Customers enter this subsystem when their service begins; in a stable, loss-free queue its entry rate equals $\lambda$. Their mean time in this subsystem is the mean <service time> $\mu$. Thus \b[$\boxed{\mathbb P(\text{server busy})=\lambda\mu}$], with stability requiring $\lambda\mu<1$. In this paper $\mu$ denotes a mean duration, so the exponential service rate is $1/\mu$.
For <finite-buffer server utilization>, assume the server is work conserving and admitted customers are eventually served, with finite mean <service time>. <Poisson arrivals see time averages>, so an offered arrival is rejected with probability $P_n$. Hence the admitted arrival rate, and therefore the service-entry rate, or <queue throughput>, is $\lambda(1-P_n)$. Apply <Little's law> to the server alone again. Its mean population, or <server utilization>, is $1-P_0$, and its mean <customer sojourn time> is $\mu$, giving \b[$\boxed{1-P_0=\lambda\mu(1-P_n)}$.] Neither argument requires exponential service times in the finite-buffer system: Poisson arrivals are used for the blocking probability, while the service calculation needs only the mean and the stated independence. Applying the offered rate $\lambda$ directly to this lossy system would miss the admission factor.
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