Network Flow

1
Network Flow

• The Maximum Flow Problem
• and
• The Ford-Fulkerson Algorithm

2
Types of Networks

• Internet
• Telephone
• Cell
• Highways
• Rail
• Electrical Power
• Water
• Sewer
• Gas

3
Maximum Flow Problem

• How can we maximize the flow in a network from a
  source or set of sources to a destination of set
  of destinations?
• The problem reportedly rose to prominence in
  relation to the rail networks of the Soviet
  Union, during the 1950's. The US wanted to know
  how quickly the Soviet Union could get supplies
  through its rail network to its satellite states
  in Eastern Europe.
• In addition, the US wanted to know which rails it
  could destroy most easily to cut off the
  satellite states from the rest of the Soviet
  Union.
• It turned out that these two problems were
  closely related, and that solving the max flow
  problem also solves the min cut problem of
  figuring out the cheapest way to cut off the
  Soviet Union from its satellites.
• The first efficient algorithm for finding the
  maximum flow was conceived by two Computer
  Scientists, named Ford and Fulkerson. The
  algorithm was subsequently named the
  Ford-Fulkerson algorithm, and is one of the more
  famous algorithms in computer science.

Source lbackstrom, The Importance of
Algorithms, at www.topcoder.com
4
Network Flow

• A Network is a directed graph G
• Edges represent pipes that carry flow
• Each edge ltu,vgt has a maximum capacity cltu,vgt
• A source node s in which flow arrives
• A sink node t out which flow leaves

Goal Max Flow
5
Network Flow

• The network flow problem is as follows
• Given a connected directed graph G
• with non-negative integer weights,
• (where each edge stands for the capacity of that
  edge),
• 2 different vertices, s and t, called the source
  and the sink,
• such that the source only has out-edges and the
  sink only has in-edges,
• Find the maximum amount of some commodity that
  can flow through the network from source to sink.

12
a
b
20
16
9
4
10
7
s
t
4
13
c
d
14
Each edge stands for the capacity of that edge.
6
Network Flow

• One way to imagine the situation is imagining
  each edge as a pipe that allows a certain flow of
  a liquid per second.
• The source is where the liquid is pouring from,
  and the sink is where it ends up.
• Each edge weight specifies the maximal amount of
  liquid that can flow through that pipe per
  second.
• Given that information, what is the most liquid
  that can flow from source to sink per second, in
  the steady state?

12
a
b
20
16
9
4
10
7
s
t
4
13
c
d
14
Each edge stands for the capacity of that edge.
7
Network Flow
12
a
b
12/12
a
b
20
16
19/20
12/16
0/9
9
4
10
7
s
t
0/4
0/10
s
t
7/7
13
4
4/4
11/13
c
d
c
d
14
11/14
This graph contains the capacities of each edge
in the graph.
Here is an example of a flow in the graph.

• The flow of the network is defined as the flow
  from the source, or into the sink.
• For the situation above, the network flow is 23.

8
12
12/12
a
b
a
b
20
19/20
16
12/16
0/9
9
4
0/4
10
7
0/10
s
t
s
t
7/7
13
4
4/4
11/13
c
d
c
d
14
11/14
capacities
flow

• The Conservation Rule
• In order for the assignment of flows to be valid,
  we must have the sum of flow coming into a vertex
  equal to the flow coming out of a vertex, for
  each vertex in the graph except the source and
  the sink.
• The Capacity Rule
• Also, each flow must be less than or equal to the
  capacity of the edge.
• The flow of the network is defined as the flow
  from the source, or into the sink.
• For the situation above, the network flow is 23.

9
Network Flow

• In order to determine the maximum flow of a
  network, we will use the following terms
• Residual capacity is simply an edges unused
  capacity.
• Initially none of the capacities will have been
  used, so all of the residual capacities will be
  just the original capacity.

0/12
a
b
0/20
0/16
Using the notation used / capacity. Residual
Capacity capacity - used.
0/9
0/4
0/10
s
t
0/7
0/4
0/13
c
d
0/14
10
Network Flow

• Residual capacity of a path the minimum of the
  residual capacities of the edges on that path,
  which will end up being the max excess flow we
  can push down that path.
• Augmenting path defined as one where you have a
  path from the source to the sink where every edge
  has a non-zero residual capacity.

0/12
a
b
0/20
0/16
Using the notation used / unused. Residual
Capacity unused - used.
0/9
0/4
0/10
s
t
0/7
0/4
0/13
c
d
0/14
11
Ford-Fulkerson Algorithm

• While there exists an augmenting path
• Add the appropriate flow to that augmenting path
• So were going to arbitrarily choose the
  augmenting path s,c,d,t in the graph below
• And add the flow to that path.

• Residual capacity of a path the minimum of the
  residual capacities of the edges on that path.

0/12
a
b
0/20
0/16
0/9

• 4 in this case, which is the limiting factor for
  this paths flow.

0/4
0/10
s
t
0/7
4/4
0/4
4/13
0/13
c
d
4/14
0/14
12
Ford-Fulkerson Algorithm

• While there exists an augmenting path
• Add the appropriate flow to that augmenting path
• Choose another augmenting path (one where you
  have a path from the source to the sink where
  every edge has a non-zero residual capacity.)
• s,a,b,t

• Residual capacity of a path the minimum of the
  residual capacities of the edges on that path.

12/12
0/12
a
b
12/20
0/20
12/16
0/16
0/9

• 12 in this case, which is the limiting factor for
  this paths flow.

0/4
0/10
s
t
0/7
4/4
4/13
c
d
4/14
13
Ford-Fulkerson Algorithm

• While there exists an augmenting path
• Add the appropriate flow to that augmenting path
• Choose another augmenting path (one where you
  have a path from the source to the sink where
  every edge has a non-zero residual capacity.)
• s,c, d, b, t

• Residual capacity of a path the minimum of the
  residual capacities of the edges on that path.

12/12
a
b
12/20
12/16
19/20
0/9

• 7 in this case, which is the limiting factor for
  this paths flow.

0/4
0/10
s
t
0/7
7/7
4/4
4/13
11/13
c
d
4/14
11/14
14
Ford-Fulkerson Algorithm

• While there exists an augmenting path
• Add the appropriate flow to that augmenting path
• Are there any more augmenting paths?
• No! Were done
• The maximum flow 19 4 23

12/12
a
b
19/20
12/16
0/9
0/4
0/10
s
t
7/7
4/4
11/13
c
d
11/14
15
Ford-Fulkerson Algorithm - Runtime

• While there exists an augmenting path
• Add the appropriate flow to that augmenting path
• We can check the existence of an augmenting path
  by doing a graph traversal on the network (with
  all full capacity edges removed.)
• This graph, a subgraph with all edges of full
  capacity removed is called a residual graph.
• It is difficult to analyze the true running time
  of this algorithm because it is unclear exactly
  how many augmenting paths can be found in an
  arbitrary flow network.
• In the worst case, each augmenting path adds 1 to
  the flow of a network, and each search for an
  augmenting path takes O(E) time, where E is the
  number of edges in the graph.
• Thus, at worst-case, the algorithm takes O(fE)
  time, where f is the maximal flow of the
  network.

16
Edmonds-Karp Algorithm

• This algorithm is a variation on the
  Ford-Fulkerson method which is intended to
  increase the speed of the first algorithm.
• The idea is to try to choose good augmenting
  paths.
• In this algorithm, the augmenting path suggested
  is the augmenting path with the minimal number of
  edges.
• (We can find this using BFS, since this finds all
  paths of a certain length before moving on to
  longer paths.)
• The total number of iterations of the algorithm
  using this strategy is O(VE). Thus, its total
  running time is O(VE2).

17
Network Flow

• Consider a network flow problem where the goal
  was to figure out how to change the capacities to
  increase the maximum flow in the network.
• This might be applicable if you wanted to get a
  greater throughput of data in a network and you
  wanted to figure out where in the network you
  needed to add a router that could handle more
  data per unit of time.

18
Network Flow

• How to increase the maximum flow of a network.
• Lets say you restrict your search in your
  network (that is already at max flow) to paths
  that have at most one edge that is being used to
  capacity.
• For example, s? v1 ? v2 ? ? vn ? t, where each
  edge has extra capacity EXCEPT for the edge vi ?
  vi1.
• What is a simple fix to add extra capacity to one
  of the edges that will add maximum flow to the
  network?
• How much will this simple fix add?

19
Network Flow

• Take a look at the extra capacity available
  through each of the other edges
• s ? v1, v1 ? v2, , vi-1 ? vi, vi1 ? vi2, , vn
  ? t.
• Take the minimum of these values.
• Then, add this much capacity to the edge vi ?
  vi1.
• Adding more than this would do no good, because
  then some other edge would become the bottleneck.
  
• This just adds enough capacity so that the
  original edge is no more of a bottleneck than the
  second worst edge in this particular path.

20
References

• Slides adapted from Arup Guhas Computer Science
  II Lecture notes http//www.cs.ucf.edu/dmarino/
  ucf/cop3503/lectures/
• Additional material from the textbook
• Data Structures and Algorithm Analysis in Java
  (Second Edition) by Mark Allen Weiss
• J Elders Network Flow slides, York University