Math 482 (Lecture 16): Max flow and min cut
This lecture discusses Sections 6.1 and 6.2 of the textbook.
* We have studied the primaldual method and shown how it applies to the shortest path
problem.
* We will now study the maxflow problem at greater depth. It is efficiently solved by the
FordFulkerson algorithm.
* Later we will comment on how the FordFulkerson algorithm can be seen from the
point of view of primaldual methods.
We have aleady introduced the max flow problem:
max v
s.t. Af+dv=0, f≤ b, g≥0
where d is defined by d
s
=1, d
t
=1 and d
i
=0 if i is not s,t.
In class exercise:
(a) Write this LP out for the
wiki problem
.
(b) In eq (5.21) of the textbook, it is mentioned that one can replace Af+dv=0 with
Af+dv≤ 0. Why?
Solution:
(a) is straightforward and discussed in class. (b): the textbook says "since a
deficit in the flow balance at any node implies a surplus at some other". Indeed, convince
yourself that it must be an equality in the rows for s and t. Then a strict inequality means
some of the goods being flowed are lost at a vertex. But some must be gained elsewhere
if at the end "t" receives everything "s" sent.
The min cut problem
Definition:
An
st cut
is a partition (A,B) of the nodes V into A and B such that s
∈
A and
t
∈
B.
The
capacity
of an st cut is
C(A,B)=Σ b(i,j)
where the sum is over all arcs (i,j) with i
∈
A and j&isin B.
In class exercise:
What is the minimum cut in the
wiki problem
?
Answer:
15 = the max flow
In class exercise:
"Clearly" the max flow is at most the capacity of any cut. Why?
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 Spring '08
 Staff
 Math, Shortest path problem, Flow network, Maximum flow problem, MAX Flow

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