Theorem: Cycle velocities with a finite-time offset

Let $q=|V(Q)|$ and $D_{\max }={\max }_e\Vert {\Delta }_e\Vert$. Every dependency path completed within elapsed time $T$ satisfies

The set of limiting velocities along paths with elapsed times tending to infinity is exactly $K_{\mathsf{Sch}}$. These are statements about potential scheduled reachability, not about the channel carried by every path.

Proof. Remove directed cycles from a finite walk until its remaining walk has no repeated vertex. Decompose each removed cycle into simple cycles. The remainder has at most $q-1$ edges.

The cycle contribution is

The sum of those cycle times is at most the elapsed time $T$. Including $0$, the contribution belongs to $T{K}_{\mathsf{Sch}}$. The remaining displacement has norm at most $(q-1)D_{\max }$, proving (eq:finiteoffset).

The offset divided by $T$ tends to zero, so every limiting velocity lies in $K_{\mathsf{Sch}}$. Conversely, choose any finite convex combination of cycle velocities and zero. Repeat each chosen cycle sufficiently many times in the prescribed time proportions, connecting their quotient base vertices by fixed paths. Strong connectivity supplies the connectors. Their total duration and displacement are negligible relative to the repeated cycles. Additional waiting realizes the zero contribution. Integer repetition counts approximate arbitrary real proportions. This realizes every point of $K_{\mathsf{Sch}}$.