 Solution

The work passed downstream in slot $(t-1,t)$ isBecause this work is eligible for downstream service in the same slot, the downstream Lindley recursion isPut $T_t=Q_t+R_t$. If $R_{t-1}+D_t\ge d$, thenIf $R_{t-1}+D_t<d$, 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 yieldThis 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.