Suppose we have a graph with nodes labelled
The Ford-Fulkerson Algorithm
First assign some initial flow with value
s.t. for every
Repeat this until there are no more such augmented paths.
This is optimal by the The max-flow min-cut Theorem.
To prove there are no more augmented paths,
we just need to find a cut with
Note that the algorithm always terminates for integer (hence also rational) values of
Dual problem
Calculate the Lagrangian
Now for
Now pick an optimal solution , and pick: (using notation from The max-flow min-cut Theorem) One can check that Complimentary slackness holds, hence the dual of max-flow problem is indeed the min-cut problem.