Ford-Fulkerson max flow algorithm analysis

Question

I am reading Ford-Fulkerson maxflow algorithm in book Algorithms written by Robert Sedgewick. Here author mentioned as below

The number of augmenting paths needed in the shortest-augmenting-path implementation of the Ford-Fulkerson maxflow algorithm for a flow network with V vertices and E edges is at most EV /2.

Proof sketch: Every augmenting path has a critical edge—an edge that is deleted from the residual network because it corresponds either to a forward edge that becomes filled to capacity or a backward edge that is emptied. Each time an edge is a critical edge, the length of the augmenting path through it must increase by 2. Since an augmenting path is of length at most V each edge can be on at most V/2 augmenting paths, and the total number of augmenting paths is at most EV/2.

My questions on above text are

1. How we got by saying if augumenting path length is atmost V, each edge can be in atmost V/2 augumenting paths?

Request to explain above with simple example if possible.

Answer 1

You first need to prove the previous statement

Each time an edge is a critical edge, the length of the augmenting path through it must increase by 2.

A path length is at most V because it doesn't make sense to go through a vertex twice(remove all the edges between the two occurrences of a vertex x on such a path and you will still have a good path, with capacity at least that of the original path).

So, if the path length is at most V and every time an edge is critical the length of the path increases by 2, then an edge can be critical at most V/2 times.