Max Flow – Min Cut Problem

Directed Graph Applications

Shortest Path Problem

(Shortest path from one point to another) Max Flow problems

(Maximum material flow through conduit) Liquid flow through pipes Parts through assembly line Current through electrical network, etc

Flow Networks

Directed graph G = (V, E) Only one Source(s) and one Sink(t) Weight on each edge = Capacity of the edge if (u, v) E then Capacity is non-negative,

i.e. c(u, v) ≥ 0 if (u, v) E then Capacity is assumed zero,

i.e. c(u, v) = 0

Flow Networks and Flows

flow(f) in Flow Network(G) is a real-valued function f: V x V → R

f(u, v) is flow from vertex u to vertex v. flow f(u, v) can be positive, negative or zero

Constraints on flow :

1. Capacity constraint :

f(u, v) ≤ c(u, v), for all u, v V

2. Skew symmetry constraint :

f(u, v) = - f(v, u)

Flow Networks and Flows (Contd.)

3. Flow conservation constraint :Total net flow at vertex must equal 0.

∑j f(i, j) - ∑

k f(k, i) = 0 for all i V – {s, t}

flow in equals flow out

i

j1

j3

j2

k3

k2

k1

jnk4

Maximum Flow

We refer a flow f as maximum if it is feasible and maximize ∑

k f(s, k).

Where f(s, k) is flow out of source s. Problem:

Objective: To find a maximum flow

s t

2

1

(6/10)

(7/8)

(1/1)

(5/6)

(8/10)

(flow/capacity)

Multiple source – Multiple sink Network

Convert this problem to single source - single sink.

s3

s4

s5

s2

s1

t3

t2

t1

103

12

5 15

8

14 20

6

711

13

18

2

Factories

Warehouses

Conversion to flow Network

Add a super – source with infinite weighted edges emanting out to original sources

Add a super – sink with edges of infinite weight from original sinks

s3

s4

s5

s2

s1

t3

t2

t1

103

12

5 15

8

14 20

6

711

13

18

2

Super-source

Factories

Warehouses

Super-sink

st

Another Example : Find a feasible match

Assign persons to tasks such that each task is assigned to a person and each person is assigned to a task in feasible manner.

1

2

3

4

5

6

7

8

persons tasks

Transformation to maximum flow problem

1

2

3

4

5

6

7

8

persons tasks

s

1

1

1

1

t

1

1

1

1

Ford-Fulkerson Algorithm for Max Flow

Key ingredients:- Residual Networks Augmenting Paths Cut

Limitations :- Flow should be integral or rational On each iteration residual capacity should be

integral

The Residual Network

s

1

2

t

(8/10) (7/8)

(1/1)

(6/10)(5/6)

s

1

2

t

2 1

141

8

5 6

7

u vf (u, v) / c (u, v)

u vc (u, f) – f (u, v)

f (u, v)

Flow Network

Residual Network

(flow / capacity)

Residual capacity r (u ,v)

Augmenting Paths

An augmenting path is a path from s to t in the residual network.

The Residual capacity of augmented path P is P= min {r(i, j): (i, j) P}

Augmentation along P Add Pin each arc along P inflow network Modify residual capacities in residual networkr(u, v) = r(u, v) - Pandr(v, u) = r(v, u) + Pfor u,

v P

s

1

2

t

2 1

141

8

5 6

7s

1

2

t

2

14

8

6 6

8

Cut

An (S,T)-cut in a flow network G = (V,E) is a partition of

vertices V into two disjoint subsets S and T such that

s S, t T

e.g., S = { s, 1 } and T = { 2, t }.

The capacity of a cut (S,T) is

CAP(S,T) = uSvT c(u,v)

s

1

2

t

(9/10) (8/8)

(1/1)

(7/10)(6/6)

The Ford Fulkerson Maximum Flow Method

Beginx := 0;

create the residual network G(x);

while there is some directed path from s to t in G(x) do

beginlet P be a path from s to t in G(x);

:= (P);

send units of flow along P;

update the residual capacities;

end

end {the flow x is now maximum}.

Proof of Correctness of the Algorithm

Assume that all data are integral.

Lemma:

At each iteration all residual capacities are integral.

Proof:

By assumption it is true at beginning.

Assume it is true after the first k-1 augmentations, and consider augmentation k along path P.

The residual capacity of P is the smallest residual capacity on P, which is integral.

After updating, we modify residual capacities by 0, or , and thus residual capacities stay integral.

Proof of finiteness of Algorithm

Proof:

The capacity of each augmenting path is at least 1.

The augmentation reduces the residual capacity of some arc (s, j) and does not increase the residual capacity of (s, i) for any i.

So, the sum of the residual capacities of arcs out of s keeps decreasing, and is bounded below by 0.

Number of augmentations is O(nU), where U is the largest capacity in the network.

Indication of Optimum Flow

There is no augmenting path in the residual network.

s

1

2

t

(9/10) (8/8)

(1/1)

(7/10)(6/6)

s

1

2

t

1

13

9

6 7

8

Indication of Optimum Flow (Contd.)

Flow across the network is equal to the capacity of some cut

(Max Flow Min Cut Theorem)

s

1

2

t

(9/10) (8/8)

(1/1)

(7/10)(6/6)

Weak Duality Theorem for the Max Flow Problem

Theorem:

If f is any feasible flow and if (S,T) is an (s,t)-cut, then the flow | f | from source to sink in the network is at most CAP(S,T).

Proof: We define the flow across the cut (S,T) to be

f(S,T) = iSjT f(i, j) - iSjT f(j, i)

S T

Flows across different cuts

s

1

2

t

(9/10) (8/8)

(1/1)

(7/10)(6/6)

s

1

2

t

(9/10) (8/8)

(1/1)

(7/10)(6/6)

s

1

2

t

(9/10) (8/8)

(1/1)

(7/10)(6/6)

S

T

S/T cut

If S = {s}, then the flow across (S, T) is 9 + 6 = 15

If S = {s,1}, then the flow across (S, T) is 8 + 1 + 6 = 15

If S = {s,2}, then the flow across (S, T) is 9 + 7 – 1 = 15

More on Flows Across Cuts

Claim:

Let (S,T) be any s-t cut. Then f(S,T) = | f | = flow into t.

Proof:

Add the conservation of flow constraints for each node i S - {s} to the constraint that the flow leaving s is |f|. The resulting equality is f(S,T) = |f|.

j f(i, j) - k f(k, i) = 0 for each i {S} - s

j f(s, j) = | f | S T

More on Flows Across CutsClaim:

The flow across (S,T) is at most the capacity of a cut.Proof:

If i S, and j T, then f(i, j) c(i, j). If i T, and j S, then f(i, j) 0.

f(S,T) = iSjT f(i, j) - iSjT f(j, i)

CAP(S,T) = iSjT c(i, j) - iSjT 0

Max Flow Min Cut Theorem

Theorem: (Optimality conditions for max flows). The following are equivalent.1. A flow x is maximum.2. There is no augmenting path in G(x). 3. There is an s-t cutset (S, T) whose capacity is the flow value

of x.

Corollary: (Max-flow Min-Cut). The maximum flow value is the minimum value of a cut.

Proof of Theorem: 1 2. (not 2 not 1)Suppose that there is an augmenting path in G(x). Then x is not maximum.

Max Flow Min Cut Theorem Contd.

3 1.

Let v = Fx(S, T) be the flow from s to t. By assumption, v = CAP(S, T). By weak duality, the maximum flow is at most CAP(S, T). Thus the flow is maximum.

2 3. Suppose there is no augmenting path in G(x).

Claim: Let S be the set of nodes reachable from s in G(x). Let T = N\S. Then there is no arc in G(x) from S to T.

Thus i S and j T f(i, j) = c(i, j)

i T and j S f(i, j) = 0.

It follows that

Fx(S,T) = iSjT f(i, j) - iSjT f(j, i)

= iSjT c(i, j) - iSjT 0 = CAP(S,T)

There is no arc from S to T in G(x)

Max Flow Min Cut Theorem Contd.

S T

Saturated

Reachable from s

Not Reachable from s

.s .t0

27

Limitation of Algorithm : example 1

s

1

2

t

M M

MM1

M >> 1

• Termination even with integral flow and few nodes can take

large no. of steps.• Depends on path selection and capacity of flow network.

Flow Network

Flow

28

After 1st augmentation

s

1

2

t

M-1 M

M-1M1

1

1

Residual Network

29

After 2nd augmentation

s

1

2

t

M-1 M-1

M-1M-11

1

1

1

1

30

After 3rd augmentation

s

1

2

t

M-2 M-1

M-2M-11

2

2

1

1

31

After 2M augmentations

s

1

2

tM

M

M

M

1

32

s 3

4

t

m r

mm

Flow Network 1

2

m m

m

1

1

m >> 1

1/2 < r < 1

Limitation of Algorithm (Non termination for irrational flow)

3333

Flow in Flow Network Residual Network

3

4

r

1

2 1

1

32 tS 1 1 1

After 1st augmentation

343434

Flow in Flow Network Residual Network

r

r

3

4

r

1

2 1-r

1-r

3

4

r

1

2

After 2nd augmentation

35353535

Flow in Flow Network Residual Network

1-r

3

4

r

1

2 1

r

r3

r

1

2

After 3rd augmentation

3636363636

Flow in Flow Network Residual Network

3

4

2r -1

1

2 r

1

1-r

1 - r

1-r

3

4

1 - r

1

2 1 - r

After 4th augmentation

373737373737Residual NetworkFlow in Flow Network

3

4

2r -1

1

2 1

r

1-r

1-r

3

4

1

2

1 - r

1-r

1-r

After 5th augmentation

Non termination in irrational flow

38

3

4

1

2

ai

ai+1

3

4

1

2

ai - ai+1 = ai+2

ai+1 - (ai - ai+1) = ai+3

And this goes on …………….

39

Summary and Extensions

1. Augmenting path theorem

2. Ford-Fulkerson Algorithm

3. Duality Theory.

4. Computational Speedups.