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For a work-conserving single-server queue with independent Poisson arrivals, finite buffer and independent identically distributed admitted service times of mean $\mu$, Poisson arrivals see time averages give admitted rate $\lambda (1-P_n)$. Little law for the server alone then gives the displayed busy fraction. The time parameter is the mean service duration, and need not be exponentially distributed.

Immediately after departure $n$, either $Q_n>0$ and one of those customers starts the next service, or $Q_n=0$ and the next arrival both initiates service and is not counted among those waiting after it. During that service $A_n$ further customers arrive. Thus the embedded departure chain of an M-G-1 queue obeys

Conditional on the next service time $S_{n+1}=s$, the Poisson process property givesThe law of total expectation and law of total variance therefore giveand

In equilibrium let $Q$ have the common distribution of $Q_n,Q_{n+1}$ and put $H=1_{\{Q>0\}}=h(Q)$. The fresh pair $(A_n,S_{n+1})$ is independent of $Q_n$. Taking expectations in the recursion givesBecause $QH=Q$ and $H^2=H$, stationarity of the second moment givesSolving,

In stationarity, level crossing together with Poisson arrivals see time averages identifies this as the mean number $L$ in the shop. The Little law gives $L=\lambda (W_q+\mathbb{E}S)$, where $W_q$ is waiting time before service. Since $\rho =\lambda \mathbb{E}S$,This is the Pollaczek-Khinchine mean formula.

Let regeneration cycles have lengths $C_i$, contain $A_i$ arrivals, and accrue queue-length rewardFubini's geometric identity writes this area as the sum, over customers in the cycle, of their time in the system, up to boundary terms whose contribution vanishes over many cycles. The renewal-reward theorem therefore givesHence Little law is