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Max Flow – Min Cut Problem

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Max Flow – Min Cut Problem

- 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

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

- Capacity constraint :
f(u, v) ≤ c(u, v), for all u, v Î V

- Skew symmetry constraint :
f(u, v) = - f(v, u)

j1

j2

j3

i

jn

k4

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

∑jf(i, j) - ∑kf(k, i) = 0 for all i Î V – {s, t}

flow in equals flow out

k1

k2

k3

1

(8/10)

(7/8)

(1/1)

s

t

(5/6)

(6/10)

2

- We refer a flow f as maximum if it is feasible and maximize ∑kf(s, k).
Where f(s, k) is flow out of source s.

- Problem:
- Objective: To find a maximum flow

(flow/capacity)

Factories

s1

10

Warehouses

3

12

t1

s2

15

5

6

8

t2

s3

20

14

13

7

t3

s4

11

18

2

s5

Convert this problem to single source - single sink.

Factories

s1

10

3

Warehouses

12

t1

s2

15

5

Super-sink

6

Super-source

8

t2

t

s3

20

14

s

13

7

t3

s4

11

18

2

s5

- 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

persons

tasks

1

5

2

6

3

7

4

8

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

persons

tasks

1

5

1

1

1

1

2

6

1

1

s

t

1

3

7

1

4

8

- Key ingredients:-
- Residual Networks
- Augmenting Paths
- Cut

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

1

(7/8)

(8/10)

f (u, v) / c (u, v)

(1/1)

u

u

v

v

s

t

(5/6)

(6/10)

2

1

1

2

c (u, f) – f (u, v)

8

7

1

s

t

f (u, v)

4

1

2

6

5

Flow Network

(flow / capacity)

Residual Network

Residual capacity r (u ,v)

1

1

2

8

7

1

s

t

4

1

1

2

2

6

5

8

8

1

s

t

4

2

6

6

- An augmenting path is a path from s to t in the residual network.
- The Residual capacity of augmented path P is
d(P) = min {r(i, j):(i, j) P}

- Augmentation along P
- Add d(P) in each arc along P inflow network
- Modify residual capacities in residual network
r(u, v) = r(u, v) - d(P)andr(v, u) = r(v, u) + d(P)for u, v P

1

(8/8)

(9/10)

(1/1)

s

t

(6/6)

(7/10)

2

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)

Begin

x := 0;

create the residual network G(x);

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

begin

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

:= d(P);

send units of flow along P;

update the residual capacities;

end

end {the flow x is now maximum}.

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

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

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.

1

(8/8)

(9/10)

(1/1)

s

t

(6/6)

(7/10)

2

1

1

9

8

1

s

t

3

2

7

6

There is no augmenting path in the residual network.

1

(8/8)

(9/10)

(1/1)

s

t

(6/6)

(7/10)

2

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

(Max Flow Min Cut Theorem)

S

T

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)

1

(8/8)

(9/10)

(1/1)

s

t

(6/6)

(7/10)

2

1

(8/8)

(9/10)

S

(1/1)

s

t

(6/6)

(7/10)

T

2

S/T cut

1

(8/8)

(9/10)

(1/1)

s

t

(6/6)

(7/10)

2

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

S

T

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 Î S - {s} to the constraint that the flow leaving s is |f|. The resulting equality is f(S,T) = |f|.

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

Claim:

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

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 Þ 2. (not 2 Þ not 1)

Suppose that there is an augmenting path in G(x). Then x is not maximum.

3 Þ 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 Þ 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.

Saturated

.t

.s

0

Reachable from s

Not Reachable from s

S

T

Max Flow Min Cut Theorem Contd.

Thus iÎ S and j Î T Þf(i, j) = c(i, j)

iÎ T and jÎ S Þ f(i, j) = 0.

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

It follows that

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

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

1

M

M

s

t

1

M

M

2

- Termination even with integral flow and few nodes can take
- large no. of steps.
- Depends on path selection and capacity of flow network.

Î

Flow

M >> 1

Flow Network

27

1

M-1

M

1

s

t

1

M-1

M

1

2

Residual Network

28

1

M-1

M-1

1

1

s

t

1

M-1

M-1

1

1

2

29

1

M-2

M-1

2

1

s

t

1

M-2

M-1

1

2

2

30

1

M

M

s

t

1

M

M

2

31

Limitation of Algorithm (Non termination for irrational flow)

Flow Network

1

m

m

r

s

2

3

t

1

m

m

1

m

m

4

m >> 1

1/2 < r < 1

32

After 1st augmentation

1

S

2

3

t

r

1

1

1

2

3

1

1

4

Flow in Flow Network

Residual Network

33

33

After 2nd augmentation

1

1

r

r

r

2

3

2

3

1-r

1-r

r

4

4

Residual Network

Flow in Flow Network

34

34

34

After 3rd augmentation

1

1

r

r

2

3

2

3

1

r

1-r

r

4

Residual Network

Flow in Flow Network

35

35

35

35

1

2r -1

1-r

1 - r

2

3

r

1

4

After 4th augmentation

1

1 - r

2

3

1 - r

1-r

4

Flow in Flow Network

Residual Network

36

36

36

36

36

1

1

2r -1

1-r

1-r

1-r

2

3

2

3

1

1 - r

r

1-r

4

4

After 5th augmentation

Flow in Flow Network

Residual Network

37

37

37

37

37

37

1

1

ai+1

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

3

2

3

2

ai

ai - ai+1 = ai+2

4

4

And this goes on …………….

38

1. Augmenting path theorem

2. Ford-Fulkerson Algorithm

3. Duality Theory.

4. Computational Speedups.

39