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A busy period is the interval from an arrival that finds the system empty until the next departure leaving it empty. It excludes the preceding idle waiting time. In a stable M/G/1 queue with arrival rate $\lambda$ and mean service time $m$, its mean is $m/(1-\lambda m)$.

In an M/G/1 queue, arrivals during the first service generate independent subtrees of work. Conditional on service time $S=s$, the number of descendants is Poisson with mean $\lambda s$. Thus $B\stackrel{d}{=}S+{\sum }_{j=1}^{N(S)}{B}_j$, with independent descendant busy periods. Conditioning and the Poisson generating function give the displayed transform equation wherever its moment-generating functions are finite. Taking expectations yields $\mathbb{E}B=\mathbb{E}S+\lambda \mathbb{E}S\mathbb{E}B$ in the stable regime.

Let the independent arrivals satisfy $\mathrm{Pr}(G_n=j)=q{p}^j$, with $p+q=1$. During a busy period, each served customer produces $G_n$ further customers, so the total number served has the total-progeny law of a Galton-Watson process with mean $p/q$. Its extinction probability solves $s=q/(1-ps)$ and is the smaller of $1$ and $q/p$. The queue is positive recurrent for $p<1/2$, null recurrent for $p=1/2$, and transient for $p>1/2$. In the positive recurrent case its stationary distribution is ${\pi }_j=(1-p/q)(p/q{)}^j$.

For the busy-period branching equation, during the first service of a busy period, every arriving customer initiates a subtree of further work. The independence and stationary increments of the Poisson process, together with independent services, give the distributional recursionwhere conditional on $S_k=s$, $N(S_k)$ is Poisson with mean $\lambda s$, and the descendant busy periods are independent copies of $B_k$. One may explore these subtrees in order; the server's total workload is unchanged by that ordering. Consequently, wherever the moment-generating functions are finite,For $\lambda <\mu$, this identity in particular holds in a sufficiently small neighbourhood of zero.

The first arrival time $A$ has an exponential distribution of rate $\lambda$ and is independent of the following busy period. ThusSubstitute ${\varphi }_{B_k}=(\lambda -\theta ){\varphi }_{T_k}/\lambda$ into part (c). Its service-transform argument becomes $\theta +\lambda ({\varphi }_{B_k}-1)=(\lambda -\theta )({\varphi }_{T_k}-1)$. ThereforeAs before, positive arguments must lie in the domain of finite moment-generating functions.