Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 27I / e / Solution

Passing to the deterministic-service limit in part (c), near zero, givesDifferentiation at zero gives $\mathbb{E}B=(1+\lambda \mathbb{E}B)/\mu$. Alternatively, the same branching expectation equation gives $\mathbb{E}{B}_k=1/(\mu -\lambda )$ for every $k$, because $\mathbb{E}{S}_k=1/\mu$. The assumed convergence of finite moment-generating functions at a positive point near zero bounds exponential moments and makes these means uniformly integrable, so the means pass to the limit. Since $T=A+B$, the answers areThey correspond to an M/D/1 queue with deterministic service time $1/\mu$. The calculation only needs finite transforms near zero; the literal assumption of finite transforms for every $\theta <\lambda$ is stronger than necessary and is not generally true for positive busy-period arguments. For instance, with $k=1$ and $\lambda =\mu /2$, part (c) gives $\lambda b^2-(\mu +\lambda -\theta )b+\mu =0$. Its discriminant is negative for $(\sqrt{\mu }-\sqrt{\lambda }{)}^2<\theta <\lambda$, so no finite busy-period transform can exist there.