= Solution
Define the downstream <queue workload> by starting both <queues> empty at time $-N$ and taking the increasing remote-past limit. For each finite $N$, part (c) makes their total <queue workload> exactly the single-server workload with service $d$, while the upstream <queue workload> is the single-server workload with service $c$. Therefore
$$
R_0^{(N)}(a,c,d)=Q_0^{(N)}(a,d)-Q_0^{(N)}(a,c).
$$
The downstream limit is increasing: the coupled upstream and downstream recursions are increasing in their initial state, because $D_t=\min(c,Q_{t-1}+a_t)$ is increasing in $Q_{t-1}$. With finite upstream <queue workload>, taking the limit gives
$$
\boxed{R_0(a,c,d)=Q_0(a,d)-Q_0(a,c).}
$$
The result is nonnegative because reducing service cannot reduce <queue workload>. It can be $+\infty$ when the slower server is unstable; under $\lambda<d<c$, both quantities are finite. The finite-upstream condition matters: subtracting two infinite <queue workloads> would not define a downstream queue.
Back to article page
