Referring to Tim Roughgarden's Lecture Note - 2 (page 11) on CS261 at Stanford following statement and the para seem quite confusing.
Claim: d(f) never decreases during the execution of the Edmonds-Karp algorithm.
Can anybody shed some lights on it?
Suppose the Edmonds-Karp algorithm augments the current flow f by routing flow on the path P . Because P is a shortest s-t path in Gf , it is also a path in the layered graph Lf . The only new edges created by augmenting on P are edges that go in the reverse direction of P . These are all backward edges, so any s-t of Gf that uses such an edge has at least d(f ) + 2 hops. Thus, no new shorter paths are formed in Gf after the augmentation.