 Solution

Measure $Q_t$ immediately after service in slot $(t-1,t)$, and let $a_t\ge 0$ denote work eligible for that slot's service. Writing $x^{+}=\max (x,0)$, the Lindley recursion isStart with an empty queue at time $-N$. Repeated substitution gives its time-zero queue workloadwhere the empty sum is zero. The increasing limit defines the minimal causal queue workload:If the mean of the remote past is $\lambda <c$, then $A_n(a)/n\to \lambda$, so $A_n(a)-cn\to -\infty$. The supremum is then finite and is attained at a finite window. The same formula with indices shifted gives $Q_t(a,c)$ and satisfies the Lindley recursion directly: split the supremum into the zero-window term and all positive-window terms.