## 3 Flows over Time
### Storing at Vertices
Regarding the maximal number of edges with positive flow value, this means two edges, at most, per non-terminal vertex, apart from the terminal-terminal edges that do not affect stability--a flow with this property has positive value on at most \(2|V(D)\setminus S|\) edges in the time-expanded network. These maximum \(2|V(D)\setminus S|\) edges have to ensure that there is no unsaturated walk between terminal vertices. If the length (w.r.t. transit times) of the shortest walk is denoted by \(\tau(W_{min})\), then \(T-\tau(W_{min})+1\) is a lower bound for the number of disjoint walks between terminal vertices. In order to saturate at least one edge along these walks, the inequality, \(2|V(D)\setminus S|\geq T-\tau(W_{min})+1\), must hold; otherwise, at least one copy of the shortest walk is unsaturated--hence, it blocks the flow. **Remark 3.11**. : _Note that holdover edges can be simulated by introducing loops of unit transit time and sufficiently large capacity at every vertex of the underlying static network. By performing the modification explained in Remark 3.5 on these loops, we obtain an equivalent time-expanded network that fulfills the requirements of Lemma 3.6. We will use this observation in the following theorem._  
**Theorem 3.12**. : _If \(T\in\mathbb{Z}_{>0}\) and the holdover edges stand on the last place on preference lists, then there is a stable flow that has a value of zero on all holdover edges. It can be computed by applying Algorithm 3 on the time-expanded network, with all holdover edges being split according to Remark 3.11._  
Proof. : By contradiction, assume there is a holdover edge with positive flow value at some vertex, \(v\). W.l.o.g., assume \(i\) to be minimal with \(f(v_{i}v_{i+1})>0\). Since the time horizon is finite, there also is a latest point in time \(j\), such that \(f(v_{j}v_{j+1})>0\). Since \(f(v_{i}v_{i+1})>0\), the pointer, \(p[v_{i}]\), must already have passed all outgoing non-holdover edges of \(v_{i}\), offering as much flow as the corresponding vertices are willing to accept. Therefore, by Lemma 3.6, the flow on each of these edges must be at least the flow on the corresponding edges of \(v_{j+1}\) (if they exist). Thus, the total outflow of \(v_{i}\) exceeds the total outflow of \(v_{j+1}\) by at least \(f(v_{i}v_{i+1})\). On the other hand, \(r[v_{j+1}]\) points either to zero or the incoming holdover edge, and thus, \(v_{j+1}\) does not refuse any flow on the regular incoming edges. Again, by Lemma 3.6, the flow on those edges is at least the flow on the corresponding edges of \(v_{i}\). Thus, the total inflow of \(v_{j+1}\) exceeds the total inflow of \(v_{i}\) by at least \(f(v_{j}v_{j+1})\). Putting this together yields \(\operatorname{inflow}(v_{j+1})-\operatorname{outflow}(v_{j+1})\geq \operatorname{inflow}(v_{i})+f(v_{j}v_{j+1})-\operatorname{outflow}(v_{i})+f( v_{i}v_{i+1})>0\), contradicting flow conservation. In the other two cases, when holdover edges are the best for one of the end vertices and the worst for the other one, storing can be used or avoided, depending on the network. There are even networks where all stable flows have positive value on some holdover edges.