= Solution
The work passed downstream in slot $(t-1,t)$ is
$$
D_t=\min\{c,Q_{t-1}+a_t\}=Q_{t-1}+a_t-Q_t.
$$
Because this work is eligible for downstream service in the same slot, the downstream <Lindley recursion> is
$$
\boxed{R_t=(R_{t-1}+D_t-d)^+.}
$$
Put $T_t=Q_t+R_t$. If $R_{t-1}+D_t\geq d$, then
$$
T_t=Q_{t-1}+R_{t-1}+a_t-d\geq0.
$$
If $R_{t-1}+D_t<d$, then $D_t<d<c$, so the upstream server did not use its full capacity. Consequently $D_t=Q_{t-1}+a_t$ and $Q_t=0$. Also $R_t=0$ and $Q_{t-1}+R_{t-1}+a_t-d<0$. Both cases yield
$$
\boxed{Q_t+R_t=(Q_{t-1}+R_{t-1}+a_t-d)^+.}
$$
This <slower-server tandem workload identity> uses both $d<c$ and eligibility for same-slot downstream service. It would not describe a system with an obligatory one-slot transfer delay.
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