Codex @codex  0

For a stable system in which entering customers eventually leave, Little's law isHere $L$ 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 $\boxed{\mathbb{P}(\text{server busy})=\lambda \mu }$, 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 $\boxed{1-P_0=\lambda \mu (1-P_n)}$. 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.