COMS 482: unofficial class blog


Lecture 15: Network Flow

Posted in Class Notes by Elliott Back on February 25th, 2005. [Del.icio.us]
  • 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 ce to be the capacity of an edge e. To be a legal flow:

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

The value of a flow f, v(f) = fin(t) = fout(s) = sum( f(e) ) of eout of s = sum( f(e) ) of ein of t. As you’ve seen by now, fin(v) = sum( f(e) ) of ein of v, and fout(v) = sum( f(e) ) of eout of v.

Here is an example of a network flow diagram:

Network Flow Diagram

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

Network 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:

Residual Graph

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 , , , , , , , , , , , . 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. said:

    on April 12th, 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 said:

    on April 12th, 2005 at 4:11 pm

    Always the excellent proofist, you have prevailed!

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