• Source s with outflow only
• Source t with inflow only
• Directed graph with non-negative edge capacities

The goal is the minimize the flow from s -> t. A flow is a function f:E -> R+, from the edge set to amount of flow on that edge. We also define c_e to be the capacity of an edge e. To be a legal flow:

1. For each edge, 0 <= f(e) <= c_e
2. For each node v other than s, t, fⁱⁿ(v) = f^out(v)

The value of a flow f, v(f) = fⁱⁿ(t) = f^out(s) = sum( f(e) ) of e_out of s = sum( f(e) ) of e_in of t. As you’ve seen by now, fⁱⁿ(v) = sum( f(e) ) of e_in of v, and f^out(v) = sum( f(e) ) of e_out of v.

Here is an example of a network flow diagram:

Now, imagine pushing 18 units of flow across nodes {s, 1, 3, 4, t} to get the following flow diagram:

This is bad–it uses up too much of our capacity. The most we can push again from s -> t is only 2! Let’s allow the option of “undoing” an edge:

This is our redidual graph. Everywhere we have flow, we potentially have:

1. Some capacity left
2. Some “backflow” we can use to change our mind

Ford-Fulkerson Algorithm (Overview):

Find an s -> t path in graph
Update all flows
Make a new residual graph
Repeat

This entry was posted on Friday, February 25th, 2005 at 3:49 pm and is tagged with ford fulkerson, network flow, negative edge, flow diagram, backflow, inflow, fout, outflow, graph, algorithm, lt, ford. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback.

2 Responses to “Lecture 15: Network Flow”

1. Ivan G. says:

   4/12/2005 at 3:33 pm

   “Now, imagine pushing 18 units of flow across nodes {s, 1, 2, 4, t} to get the following flow diagram:”

   Should be { s, 1, 3, 4, t }
2. Elliott Bäck says:

   4/12/2005 at 4:11 pm

   Always the excellent proofist, you have prevailed!