Department of Management Engineering / Operations Research
Department of Management Engineering / Operations Research
The Max Flow Problem
The Max Flow Problem
Jesper Larsen & Jens Clausen
jla,[email protected]
Department of Management Engineering
Technical University of Denmark
Jesper Larsen & Jens Clausen
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Department of Management Engineering / Operations Research
Example
We consider a digraph G = (V, E) which for
each e has a capacity ue ∈ R+ .
Furthermore two “special” vertices r and s are
given; these are called resp. the source and the
sink.
Aim is to find the maximum flow from the source
to the sink in the graph while respecting the
capacities of the edges.
Jesper Larsen & Jens Clausen
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Department of Management Engineering / Operations Research
Flow
A flow in G is a function x : E → R+ satisfying:
X
X
(1) ∀i ∈ V \ {r, s} :
xji −
xij = 0
(j,i)∈E
(2)
(i,j)∈E
∀(i, j) ∈ E : 0 ≤ xij ≤ uij
For a flow
P we define: P
fx (i) = (j,i)∈E xji − (i,j)∈E xij
fx (i) is the net flow into i or the excess for x in
i, and is called the value of x.
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Jesper Larsen & Jens Clausen
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Department of Management Engineering / Operations Research
Example with a flow
Math. Programming Model of Max Flow
Here we have added a flow of value 4 to our graph.
Jesper Larsen & Jens Clausen
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Department of Management Engineering / Operations Research
Max Flow and Cuts I
max fx (s)
s.t. fx (i) = 0
i ∈ V \ {r, s}
0 ≤ xe ≤ ue e ∈ E
xe ∈ Z+
e∈E
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Department of Management Engineering / Operations Research
Max Flow and Cuts II
We now consider R ⊂ V . δ(R) is the set of
edges incident from a vertex in R to a vertex in
R. δ(R) is also called the cut generated by R:
δ(R) = {(i, j) ∈ E | i ∈ R, j ∈ R}
Note the difference to the cut in an undirected
graph. In a directed graph the cut only consists
of arcs going from R to R and not the opposite
direction.
In an undirected graph we have δ(R) = δ(R),
this does not hold here.
Because the cut is uniquely defined by the
vertex-set R we might simply call R the cut.
R is an r,s-cut if r ∈ R and s ∈
/ R.
The capacity of the cut R is defined as:
X
u(δ(R)) =
uij
(i,j)∈E,i∈R,j∈R
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Jesper Larsen & Jens Clausen
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Department of Management Engineering / Operations Research
Relationship between flows and cuts
Proposition: For any r, s-cut δ(R) and any flow
x we have: x(δ(R)) − x(δ(R̄)) = fx (s)
Corollary: For any feasible flow x and any
r, s-cut δ(R), we have: fx (s) ≤ u(δ(R)).
Department of Management Engineering / Operations Research
The Residual graph
Suppose that x is a flow in G. The residual
graph shows how flow excess can be moved in
G given that the flow x is already present.
The residual graph Gx for G wrt. x is defined by:
V (Gx ) = V
E(Gx ) = {(i, j) : (i, j) ∈ E ∧ xij < uij } ∪
{(j, i) : (i, j) ∈ E ∧ xij > 0}
Short we write Vx = V (Gx ) and Ex = E(Gx )
Some textbooks use the word “auxiliary graph”
instead of residual graph.
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Department of Management Engineering / Operations Research
Incrementing and augmenting path
A path is called x-incrementing if for every
forward arc e xe < ue and for every backward
arc xe > 0.
A x-incrementing path from r to s is called
x-augmenting.
Note: this relates to the original graph. In the
Residual graph all arcs in a path are forward
arcs.
Jesper Larsen & Jens Clausen
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Jesper Larsen & Jens Clausen
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Department of Management Engineering / Operations Research
Basic Approach
1. Build residual graph
2. Build augmenting path
3. If not possible stop
4. Increase flow along the augmenting path.
5. Go to step 1
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Department of Management Engineering / Operations Research
The Augmenting Path Algorithm
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xij ← 0 for all (i, j) ∈ E
repeat
construct Gx
find an augmenting path P
if s is reached then
determine max amount K of augmentation
augment x along the path P by K
8
until s is not reachable from r in Gx
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Department of Management Engineering / Operations Research
Finding an augmenting path II
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while Q 6= ∅ and s is not labeled yet do
select a i ∈ Q and scan (i, j) ∈ Ex
if (i, j) ∈ E and j unlabeled then
label j; Q ← Q ∪ {j}; p[j] ← i
umax [j] ← min{umax [i], uij − xij }
if (j, i) ∈ E and j unlabeled then
label j; Q ← Q ∪ {j}; p[j] ← i
umax [j] ← min{umax [i], xji }
Q ← Q \ {i}; S ← S ∪ {i}
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Department of Management Engineering / Operations Research
Finding an augmenting path I
Input: The network G = (V, E, u) and the
current feasible flow x.
Output: An augmenting path from r to s and
the capacity for the flow augmentation.
Initialize: all vertices are unlabeled;
Q ← {r}, S ← ∅, p[.] ← 0;
umax [i] ← +∞ for i ∈ G
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Department of Management Engineering / Operations Research
Augmenting a feasible flow x
Input: The network G = (V, E, u), the current
feasible flow x, an augmenting path P from r to
s in Gx and the x-width of P .
Output: An “updated” feasible flow x′ .
An updated flow is achieved by:
for (i, j) ∈ P (i, j) ∈ E : x′ij ← xij + umax [s]
for (i, j) ∈ P (j, i) ∈ E : x′ji ← xji − umax [s]
for all other (i, j) ∈ E : x′ij ← xij
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Department of Management Engineering / Operations Research
Department of Management Engineering / Operations Research
Example
Max Flow - Min Cut Theorem
1
Given a network G = (V, E, u) and a current feasible
flow x. The following 3 statements are equivalent:
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3
6
6
2
1
4
3
r
3
2
5
4
s
An r, s-cut R exists with capacity equal to the
value of x, ie. u(R) = fx (s).
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x is a maximum flow in G.
A flow augmenting path does not exist (an
r-s-dipath in Gx ).
1
8
1
3
2
3
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Department of Management Engineering / Operations Research
Concluding Remarks
We call an augmenting path from r to s shortest
if it has the minimum possible number of arcs.
The augmenting path algorithm with a
breadth-first search solves the maximum flow
problem in O(nm2 ).
Breadth-first can easily be established using a
queue.
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Jesper Larsen & Jens Clausen
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