1. Max Flow – Min Cut Problem

2. 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

3. Sample Networks Network Nodes Arcs Flow communication
telephone exchanges, computers, satellites
cables, fiber optics, microwave relays
voice, video, packets
circuits
gates, registers, processors
wires
current
mechanical
joints
rods, beams, springs
heat, energy
hydraulic
reservoirs, pumping stations, lakes
pipelines
fluid, oil
financial
stocks, companies
transactions
money
transportation
airports, rail yards, street intersections
highways, railbeds, airway routes
freight, vehicles, passengers
chemical
sites
bonds
energy

4. 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) Î 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

5. 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 :
Capacity constraint :
f(u, v) ≤ c(u, v), for all u, v Î V
Skew symmetry constraint :
f(u, v) = - f(v, u)

6. Flow Networks and Flows (Contd.)
Flow conservation constraint :
Total net flow at vertex must equal 0.
∑j f(i, j) - ∑k f(k, i) = 0 for all i Î V – {s, t}
flow in equals flow out

7. Maximum Flow
We refer a flow f(s,t) as maximum if it is feasible and maximizes ∑k f(s, k) OR ∑k f(k, t)
Problem:
Objective: To find a maximum flow f from s to t
(flow/capacity) s t 2 1
(6/10)
(7/8)
(1/1)
(5/6)
(8/10)

8. Max Flow
We want to find a flow of maximum value from the source to the sink
Denoted by |f|
Max Flow, |f| = 19
Or is it?
Best we can do?
Lucky Puck Distribution Network

9. Max Flow Current Flow, |f| = 13 Is it Max Flow?2 1
(6/10)
(7/8)
(1/1)
(5/6)
(8/11)
Current Flow, |f| = 13
Is it Max Flow?
Can it be increased further?
By how much? s t 2 1
(6/10)
(7/8)
(1/5)
(5/6)
(8/11)
Current Flow, |f| = 13
Is it Max Flow?
Can it be increased further?
By how much?

10. Cancellation of flows
We would like to avoid two positive flows in opposite directions between the same pair of vertices
Such flows cancel (maybe partially) each other due to skew symmetry
2/15 a
2/15 a
5/19
5/19 t s t s
2/9
5/5
0/9
3/5
2/3
5/14
2/3
5/14 b b

11. Ford-Fulkerson method
Contains several terms:
Residual networks
Augmenting paths
Find a path p from s to t (augmenting path), such that there is some value (maximum value) x > 0, and
for each edge (u,v) in p we can add x units of flow
f(u,v) + x  c(u,v)
Augmenting Path?
8/13 a b
10/15
13/19 10 t s
2/4 9
5/5
6/14
8/11
3/3 c d

12. Residual Network
To find augmenting path we can find any path in the residual network:
Residual capacities: cf(u,v) = c(u,v) – f(u,v)
i.e. the actual capacity minus the net flow from u to v
Net flow may be negative
Residual network: Gf =(V,Ef), where
Ef = {(u,v) Î V ´ V : cf(u,v) > 0}
5/15 10
Sub-graph
With
c(u,v) and
f(u,v)
Residual
Sub-Graph a b
5/6 a b 1 c c
0/14 19 5

13. Residual Graph
Compute the residual graph of the graph with the following flow:
Residual graph

14. Residual Capacity and Augmenting Path
Finding an Augmenting Path
Find a path from s to t in the residual graph
The residual capacity of a path p in Gf:
cf(p) = min{cf(u,v): (u,v) is in p}
i.e. find the minimum capacity along p
Doing augmentation: for all (u,v) in p, we just add this cf(p) to f(u,v) (and subtract it from f(v,u))
Resulting flow is a valid flow with a larger value.

15. Residual network and augmenting path
Before
After
Flow Graph Residual Graph

16. 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

17. The Residual Network Flow Network Residual Network s 1 2 t (8/10)
(7/8)
(1/1)
(6/10)
(5/6)
(flow / capacity) u v
f (u, v) / c (u, v)
Residual Network
Residual capacity r (u ,v) s 1 2 t 4 8 5 6 7 u v
c (u, f) – f (u, v)
f (u, v)

18. Augmenting Paths
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 in flow network
Modify residual capacities in residual network
r(u, v) = r(u, v) - d(P) and r(v, u) = r(v, u) + d(P) for u, v  P s 1 2 t 4 8 5 6 7 s 1 2 t 4 8 6

19. Cut An (S,T)-cut in a flow network G = (V,E) is a partition of1 2 t
(9/10)
(8/8)
(1/1)
(7/10)
(6/6)
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) = uSvT c(u,v)

20. The Ford-Fulkerson method
Ford-Fulkerson(G,s,t)
1 for each edge (u,v) in G.E do
2 f(u,v) ¬ f(v,u) ¬ 0
3 while there exists a path p from s to t in residual network Gf do
4 cp = min{cf(u,v): (u,v) is in p}
5 for each edge (u,v) in p do
f(u,v) ¬ f(u,v) + cf
f(v,u) ¬ -f(u,v)
8 return f
The algorithms based on this method differ in how they choose p in step 3.
If chosen poorly the algorithm might not terminate.

21. Execution of Ford-Fulkerson (1)
Residual Graph Augmented Flow

22. Execution of Ford-Fulkerson (2)
Residual Graph Augmented Flow

23. Multiple source – Multiple sink Network10 3 12 5 15 8 14 20 6 7 11 13 18 2
Factories
Warehouses
Convert this problem to single source - single sink.

24. 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 10 3 12 5 15 8 14 20 6 7 11 13 18 2
Super-source
Factories
Warehouses
Super-sink s t 

25. 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

26. Transformation to maximum flow problem1 2 3 4 5 6 7 8
persons
tasks s t

27. The Ford Fulkerson Maximum Flow Method
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}.

28. 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 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.

29. Proof of finiteness of Algorithm
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.

30. Indication of Optimum Flows 1 2 t
(9/10)
(8/8)
(1/1)
(7/10)
(6/6)
There is no augmenting path in the residual network. s 1 2 t 3 9 6 7 8

31. Indication of Optimum Flow (Contd.)s 1 2 t
(9/10)
(8/8)
(1/1)
(7/10)
(6/6)
Flow across the network is equal to the capacity of some cut
(Max Flow Min Cut Theorem)

32. Weak Duality Theorem for the Max Flow Problem
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) = iSjT f(i, j) - iSjT f(j, i) S T

33. Flows across different cuts
 S/T cut s 1 2 t
(9/10)
(8/8)
(1/1)
(7/10)
(6/6)
 S
If S = {s}, then the flow across (S, T) is = 15
 T s 1 2 t
(9/10)
(8/8)
(1/1)
(7/10)
(6/6)
If S = {s,1}, then the flow across (S, T) is = 15 s 1 2 t
(9/10)
(8/8)
(1/1)
(7/10)
(6/6)
If S = {s,2}, then the flow across (S, T) is – 1 = 15

34. 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 Î S - {s} to the constraint that the flow leaving s is |f|. The resulting equality is f(S,T) = |f|.
Sj f(i, j) - Sk f(k, i) = for each i  {S} - s
Sj f(s, j) = | f | S T

35. More on Flows Across Cuts
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) = iSjT f(i, j) - iSjT f(j, i)
CAP(S,T) = iSjT c(i, j) - iSjT 0

36. 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 Þ 2. (not 2 Þ not 1)
Suppose that there is an augmenting path in G(x). Then x is not maximum.

37. Max Flow Min Cut Theorem Contd.
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.

38. Max Flow Min Cut Theorem Contd.
Thus i Î S and j Î T Þ f(i, j) = c(i, j)
i Î T and j Î S Þ f(i, j) = 0. S T
Saturated
Reachable from s
Not Reachable from s .s .t
There is no arc from S to T in G(x)
It follows that
Fx(S,T) = iSjT f(i, j) iSjT f(j, i)
= iSjT c(i, j) iSjT = CAP(S,T)

39. Limitation of Algorithm : example 1
Termination even with integral flow and few nodes can take
large no. of steps.
Depends on path selection and capacity of flow network. s 1 2 t M Î
Flow
M >> 1
Flow Network 39

40. After 1st augmentation
Residual Network s 1 2 t
M-1 M 40

41. After 2nd augmentation s 1 2 t
M-1 41

42. After 3rd augmentation s 1 2 t
M-2
M-1 42

43. After 2M augmentations s 1 2 t M 43

44. 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 44

45. After 1st augmentation 1 S 2 3 t 2 3 4 r 1 1 1 1 1
Flow in Flow Network
Residual Network 45 45

46. After 2nd augmentation 1 1 2 3 2 3 4 4 r r r 1-r 1-r r
Residual Network
Flow in Flow Network 46 46 46

47. After 3rd augmentation 1 1 2 3 2 3 4 r r 1 r 1-r r Residual Network
Flow in Flow Network 47 47 47 47

48. After 4th augmentation 3 4 1 2 1 2 3 4 2r -1 r 1-r 1 - r 1 - r 1 - r
Flow in Flow Network
Residual Network 48 48 48 48 48

49. After 5th augmentation 3 4 1 2 2r -1 r 1-r 1 - r Flow in Flow Network
Residual Network 49 49 49 49 49 49

50. Non termination in irrational flow3 4 1 2 ai
ai+1
ai - ai+1 = ai+2
ai+1 - (ai - ai+1) = ai+3
And this goes on ……………. 50

51. Summary and Extensions
1. Augmenting path theorem
2. Ford-Fulkerson Algorithm
3. Duality Theory.
4. Computational Speedups. 51