Flow Networks: Maximizing Material Flow in Graphs, Study notes of Mathematics

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Mat 3770

4.2 Network Flows

Spring 2009

4.2 Network Flows

I (^) A directed graph can model a ow network where some material (e.g., widgets, current,... ) is produced or enters the network at a source and is consumed at a sink.

I (^) Production and consumption are at a steady rate , which is the same for both.

I (^) The ow of the material at any point in the system is the rate at which the material moves through it.

Flow Conservation

I (^) Vertices are conduit junctions. Other than the source and sink, material ows through the vertices without collecting in them.

I (^) Hence, the rate at which material enters a vertex must equal the rate at which it leaves the vertex.

I (^) This property is called ow conservation , and is similar in concept to Kirchhoff's Current Law concerning electrical current.

Maximumñow

I (^) The Maximumñow problem is the simplest problem concerning ow networks:

What is the greatest rate at which material can be shipped from source to sink without violating any capacity constraints?

I (^) A ow network , G = (V, E), is a directed graph in which each edge (u, v) ∈ E has a nonñnegative capacity c(u, v)  0. I (^) If (u, v) 6 ∈ E, we assume c(u, v) = 0. I (^) In(x) and Out(x) ∈ E, and are the edges into and out of vertex x. I (^) The integer c(e) associated with edge e is a capacity or upper bound.

Flow Cuts

I (^) A cut(S, T) of ow network G = (V, E), is a partition of V into S and T = V - S, such that source ∈ S and sink ∈ T.

I (^) If f is a ow, then the net ow across the cut (S ; T ) is dened to be f (S ; T ).

I (^) The capacity of the cut (S ; T ) is c(S ; T ).

S T

10 1/4 4/9^ 7/

a

v 1 v 3

v 2

z

v 4

cut(fa, v 1 , v 2 g, fv 3 , v 4 , zg) has net ow : f (v 1 ; v 3 ) + f (v 2 ; v 3 ) + F (v 2 ; v 4 ) = 12 4 + 11 = 19.

And, its capacity is: c(v 1 ; v 3 ) + c(v 2 ; v 4 ) = 12 + 14 = 26.

I (^) The value of a ow f is dened as: jf j =

v∈V f^ (s;^ v). That is, the total net ow out of the source.

I (^) In the maximumñow problem , we are given a ow network G with source s and sink t , and we wish to nd a ow of maximum value from s to t.

I (^) Observations:

1. By skew symmetry, the net ow from a vertex to itself is 0, since for all u ∈ V , we have f (u; u) = f (u; u)
2. By skew symmetry, we can rewrite the owñconservation property as: ∀ v ∈ V fs; t g; ∑ u∈V f^ (u;^ v) =^0 That is, the total net ow into a vertex is 0.
3. The∑ positive net ow entering a vertex v is dened by u∈V ;f (u;v)> 0 f^ (u;^ v)

4. The positive net ow is dened symmetrically: the positive net ow entering a vertex (other than the source or sink) must equal the positive net ow leaving the vertex.
5. There can be no net ow between u and v if there is no edge between them.

I (^) If there is no edge between u and v, i.e., (u; v) 6 ∈ E and (v; u) 6 ∈ E , then:

c(u, v) = c(v, u) = 0 ⇒ //no edge, no flow

f(u, v)  0, f(v, u)  0 //capacity constraint

f(u, v) = f(v, u) ⇒ //skew symmetry

f(u, v) = f(v, u) = 0 //flow must be 0

I (^) The Lucky Duck Company has a factory (source s) in Vancouver that manufactures hockey pucks.

I (^) They have a warehouse (sink t ) in Winnipeg that stores them.

I (^) They lease space on trucks from another rm to ship the pucks from the factory to the warehouse ó with capacity c(u; v) crates per day between each pair of cities u and v.

I (^) Goal: determine the largest number p of crates per day that can be shipped, and then produce this amount ó there's no sense in producing more pucks than they can ship to their warehouse.

I (^) The rate at which pucks are shipped along any truck route is a ow. Maximum ow determines the maximum number p of crates per day that can be shipped.

I (^) The pucks leave the factory at the rate of p crates per day, and p crates must arrive at the warehouse each day: p p

Vancouver Winnipeg

s (^) t

I (^) Shipping time is not a concern, only the ow of p crates/day.

I (^) Capacity constraints are given by the restriction that the ow f (u; v) from city u to city v be at most c(u; v) crates per day.

I (^) In a steady state, the number of crates entering and the number leaving an intermediate city must be equal.

Cancellation

I (^) Cancellation allows us to represent the shipments between two cities by a positive net ow along at most one of the two edges between the corresponding vertices.

I (^) If there is zero or negative ow from one vertex to another, no shipments need be made in that direction.

I (^) Any situation in which pucks are shipped in both directions between two cities can be transformed using cancellation into an equivalent situation in which pucks are shipped only in the direction of positive ow.

I (^) No constraints are violated since the net ow between the two vertices is the same.

10 2/

(e)

8/ 3/

(c)

4

5/

(d)

4

8/

(b)

4

10

(a)

v 2

v 1

v 2

v 1

v 2

v 1

v 2

v 1

v 2

v 1

(a) Vertices v 1 and v 2 , with c(v 1 ; v 2 ) = 10 and c(v 2 ; v 1 ) = 4

(b) Net ow when 8 crates per day are shipped from v 1 to v 2.

(c) An additional shipment of 3 crates per day from v 2 to v 1. (d) By cancelling ow going in opposite directions, we can represent the situation in (c) with positive ow in one direction only. (e) Another 7 crates per day shipped from v 2 to v 1 results in a net of 2 crates per day from v 2 to v 1.

Ford-Fulkerson Method (G, s, t)

initialize flow f to 0

while there exists an augmenting path p

augment flow f along p

return f

Residual Networks

I (^) Given a ow network and a ow, the residual network consists of edges that can accommodate more net ow.

I (^) Suppose we have a ow network G = (V, E), with source s and sink t.

Let f be a ow in G, and consider a pair of vertices u, v, ∈ V.

The amount of additional net ow we can push from u to v before exceeding the capacity c(u; v) is the residual capacity of (u, v) given by: cf (u; v) = c(u; v) f (u; v).

I (^) For example, if c(u; v) = 16 and f (u; v) = 11, we can ship cf (u; v) = 5 more units before we exceed capacity.