1. Network Flow and Connectivity in Wireless Sensor Networks
Youn-Hee Han
Korea University of Technology and Education Laboratory of Intelligent Networks (LINK)
Edited and Copied From Slides by Kevin Wayne

2. Max Flow [Definition] Flow network: Features:
an edge-capacitated directed graph, with two distinguished vertices called the source and the sink.
Features:
Abstraction for material flowing through the edges.
G = (V, E) = directed graph
Two distinguished nodes: s = source, t = sink.
c(e) = capacity of edge e. s 2 3 4 5 6 7 t 15 30 10 8 9
capacity
source
sink

3. Max Flow [Definition] flow:
A flow in a network X is a function f that assigns to each edge e of the network a real number f(e), in such a way that: 4
capacity
flow s 2 3 5 6 7 t 15 30 10 8 9

4. Max Flow [Definition] flow (or s-t flow):
A flow in a network X is a function f that assigns to each edge e of the network a real number f(e), in such a way that: 4
capacity
flow s 2 3 5 6 7 t 15 30 10 8 9

5. Max Flow [Definition] Value Q :
The value Q of a flow in a network X is: 4
capacity
flow s 2 3 5 6 7 t 15 30 10 8 9
Value Q = 4

6. Max Flow [Definition] Value Q :
The value Q of a flow in a network X is: 10 6 11 1 3 8
capacity
flow s 2 4 5 7 t 15 30 9
Value Q = 24

7. Max Flow [Definition] Max Flow Problem:
find the maximum possible value Q of a flow in X and find a flow route of that value. 10 9 14 4 8 1
capacity
flow s 2 3 5 6 7 t 15 30
Value Q = 28 (Max)

8. Min Cut [Definition] cut (A,B) (or s-t cut) :
By a cut (A,B), we mean a partition (A, B) of V with s  A and t  B.
[Definition] capacity of a cut
The capacity of a cut (A, B) is: s 2 3 4 5 6 7 t 15 30 10 8 9 A
cap(A,B) = = 62

9. Min Cut [Definition] Min Cut Problem:
find a cut of the minimum capacity in X s 2 3 4 5 6 7 t 15 30 10 8 9 A
cap(A,B) = = 28

10. Min Cut [Definition] Min Cut Problem:
find a cut of the minimum capacity in X
cut(U,V) is a minimum cut
: its capacity is the sum of the capacities crossing the cut = 1+2=3=cap(U,V).
: the edge (i,j) is not included in cap(U,V) because it is going in the wrong direction.

11. Flows and Cuts
[Lemma 1] Let f be “any” flow, and let cut(A, B) be “any” cut. Then, the net flow sent across the cut is equal to the value Q. 6 2 9 5 10 6 10 15 15 4 4 10 3 8 8 s 5 3 8 6 10 t A 1 10 4 6 15 10 15 11 11
Net flow across the cut = = 24 (= Value Q) 4 30 7

12. Flows and Cuts
[Lemma 1] Let f be “any” flow, and let cut(A, B) be “any” cut. Then, the net flow sent across the cut is equal to the value Q. 6 2 9 5 10 6 10 15 15 4 4 10 3 8 8 s 5 3 8 6 10 t A 1 10 4 6 15 10 15 11 11
Net flow across the cut = = 24 (= Value Q) 4 30 7

13. Flows and Cuts
[Lemma 1] Let f be “any” flow, and let cut(A, B) be “any” cut. Then, the net flow sent across the cut is equal to the value Q. 2 9 5 10 6 10 15 15 4 4 10 3 8 8 s 5 3 8 6 10 t A 1 10 4 6 15 10 15 11 11
Net flow across the cut = = 24 (= Value Q) 4 30 7

14. Flows and Cuts
[Lemma 1] Let f be “any” flow, and let cut(A, B) be “any” cut. Then, the net flow sent across the cut is equal to the value Q.
Proof] The net flow out of s is Q. The net flow out of any other vertex v in A is 0. So, we obtain:

15. Flows and Cuts
[Lemma 2] Let f be “any” flow, and let cut(A, B) be “any” cut. Then the value Q of the flow is at most the capacity cap(A,B) of the cut. 6 2 9 5 10 6 10 15 15 4 4 10 3 8 8 s 5 3 8 6 10 t A 1 10 4 6 15 10 15 11 11 4 30 7
Value Q = 24  cap(A,B) = 30

16. Flows and Cuts
[Lemma 2] Let f be “any” flow, and let cut(A, B) be “any” cut. Then the value Q of the flow is at most the capacity cap(A,B) of the cut. 6 2 9 5 10 6 10 15 15 4 4 10 3 8 8 s 5 3 8 6 10 t A 1 10 4 6 15 10 15 11 11 4 30 7
Value Q = 24  cap(A,B) = 64(= )

17. Flows and Cuts
[Lemma 2] Let f be “any” flow, and let cut(A, B) be “any” cut. Then the value Q of the flow is at most the capacity cap(A,B) of the cut. 2 9 5 10 6 10 15 15 4 4 10 3 8 8 s 5 3 8 6 10 t A 1 10 4 6 15 10 15 11 11 4 30 7
Value Q = 24  cap(A,B) = 28(= )

18. Flows and Cuts
[Lemma 2] Let f be “any” flow, and let cut(A, B) be “any” cut. Then the value Q of the flow is at most the capacity cap(A,B) of the cut.
Proof]

19. Max Flow = Min Cut
[Theorem] The maximum flow value Q in a network X is equal to the minimum capacity of any cut in X. In other words, Let f be “any” flow, and let (A, B) be “any” cut. If Q= cap(A, B), then the flow f is a max flow and cut(A, B) is a min cut.
Value Q = 28 Cut capacity = 28 9 2 9 5 10 1 9 10 15 15 4 10 4 8 9 s 5 3 8 6 10 t 4 10 A 4 6 15 10 15 14 14 4 30 7

20. Max Flow = Min Cut
[Theorem] The maximum flow value Q in a network X is equal to the minimum capacity of any cut in X. In other words, Let f be “any” flow, and let (A, B) be “any” cut. If Q= cap(A, B), then the flow f is a max flow and cut(A, B) is a min cut.
Value Q = 3 Cut capacity = 3 1 1 1 1 1 2 2 2 2 2

21. Max Flow Algorithm: Greedy
Greedy algorithm.
Start with f(e) = 0 for all edge e  E.
Find an s-t path P where each edge has f(e) < c(e).
Augment flow along path P.
Repeat until you get stuck. 1 20 10 s 30 t 10 20
Flow value = 0 2

22. Max Flow Algorithm: Greedy
Greedy algorithm.
Start with f(e) = 0 for all edge e  E.
Find an s-t path P where each edge has f(e) < c(e).
Augment flow along path P.
Repeat until you get stuck. 1 20 X 20 10 s 30 X 20 t 10 20
Flow value = 20 X 20 2

23. Max Flow Algorithm: Greedy
Greedy algorithm.
Start with f(e) = 0 for all edge e  E.
Find an s-t path P where each edge has f(e) < c(e).
Augment flow along path P.
Repeat until you get stuck.
locally optimality ≠ global optimality 1 1 20 20 10 20 10 20 10 s 30 20 t s 30 10 t 10 20 10 20 20 10 20
greedy = 20 2
opt = 30 2

24. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u).
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}.
capacity u 17 v 6
flow
residual capacity u 11 v 6
residual capacity

25. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u)
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}. 1 2 5 1 1 1 1 1 1 s 4 t 1 1 1 3

26. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u)
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}. 1 2 5 1 1 1 1 1 1 s 4 t 1 1 1 3
Find any augmenting path in G(x)
A augmenting path is a path from s to t along with we can push some additional flow

27. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u)
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}. 1 2 5 1 1 1 1 1 1 s 4 t 1 1 1 1 1 1 3
Determine the bottleneck of the path. : the bottleneck is the flow of the path
Update residual capacities.

28. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u)
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}. 1 2 5 1 1 1 1 1 s 4 t 1 1 1 1 3
Find any s-t path in G(x)

29. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u)
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}. 1 2 5 1 1 1 1 1 s 4 t 1 1 2 1 1 1 1 1 3
Determine the bottleneck of the path. : the bottleneck is the flow of the path
Update residual capacities.

30. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u)
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}. 1 2 5 1 1 1 1 2 1 1 1 s 4 t 1 1 1 1 1 1 1 1 1 1 3
Find any s-t path in G(x)

31. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u)
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}. 1 2 5 1 1 1 1 1 1 1 s 4 t 1 1 1 1 1 1 1 1 1 1 3
Determine the bottleneck of the path. : the bottleneck is the flow of the path
Update residual capacities.

32. Max Flow Algorithm: Ford-Fulkerson
Original edge: e = (u, v)  E.
Flow f(e), capacity c(e).
Residual edge.
"Undo" flow sent.
e = (u, v) and eR = (v, u)
Residual capacity:
Residual graph: Gf = (V, Ef ).
Residual edges with positive residual capacity.
Ef = {e}  {eR : c(e) > 0}. 2 5 1 1 1 1 1 s 4 t 1 1 1 1 1 1 1 1 1 1 3
There is no s-t path in the residual network. This flow is optimal

33. Max Flow Algorithm: Ford-Fulkerson1 s 2 4 5 3 t
Here is the optimal flow

34. Ford-Fulkerson Algorithm
flow 2 4 4
capacity G: 10 8 6 2 10 s 10 3 9 5 10 t
value Q = 0 34

35. Ford-Fulkerson Algorithm
flow 2 4 4
capacity G: 8 X 10 8 6 2 10 s 10 3 9 5 10 t
value Q = 0 2 4 4
residual capacity
Gf: 10 8 6 2 10 s 10 3 9 5 10 t 35

36. Ford-Fulkerson Algorithm2 4 4 G: 10 2 X 8 8 10 8 6 2 10 8 s 10 3 9 5 10 t
value Q = 8 2 4 4
Gf: 8 2 8 6 2 10 s 10 3 9 5 2 t 8 36

37. Ford-Fulkerson Algorithm2 4 4 G: X 6 8 10 8 10 8 6 2 10 2 2 10 s 10 3 9 5 10 t
value Q = 10 2 4 4
Gf: 10 8 6 2 10 s 10 3 7 5 10 t 2 37

38. Ford-Fulkerson AlgorithmX 8 2 2 4 4 G: 10 8 6 10 8 6 6 2 10 2 6 8 10 s 10 3 9 5 10 t
value Q = 16 2 4 4
Gf: 6 10 8 6 2 4 s 4 3 1 5 10 t 6 8 38

39. Ford-Fulkerson AlgorithmX 9 7 3 2 2 4 4 G: 10 8 8 10 8 6 6 2 10 8 8 10 s 10 3 9 5 10 t
value Q = 18 2 2 2 4
Gf: 8 10 8 6 2 2 s 2 3 1 5 10 t 8 8 39

40. Ford-Fulkerson Algorithm3 2 4 4 G: 10 7 9 10 8 6 6 2 10 9 9 10 s 10 3 9 5 10 t
value Q = 19 3 2 1 4
Gf: 9 1 10 7 6 2 1 s 1 3 9 5 10 t 9 40

41. Ford-Fulkerson Algorithm3 2 4 4 G: 10 7 9 10 8 6 6 2 10 9 9 10 s 10 3 9 5 10 t
Cut capacity = 19
value Q = 19 3 2 1 4
Gf: 9 1 10 7 6 2 1 s 1 3 9 5 10 t 9
The partition A becomes the set of vertices reachable from s in residual graph. 41

42. Max Flow Algorithm: Ford-Fulkerson
Ford-Fulkerson(G(V,E), s, t, c) {
Gf(V,Ef) G(V,E)
while (there exists augmenting path P) {
b  bottleneck(P)
foreach e  P {
if (e  E) c(e)  c(e) - b
else c(eR)  c(eR) + b }
update G
return the current flow value Q
//forward edge
update Gf
//reverse edge

43. Max-Flow Min-Cut Theorem
Augmenting path theorem. Flow f is a max flow iff there are no augmenting paths.
Max-flow min-cut theorem. [Ford-Fulkerson 1956] The value of the max flow is equal to the value of the min cut.
Proof strategy. We prove both simultaneously by showing the TFAE:
(i) There exists a cut (A, B) such that Q = cap(A, B).
(ii) Flow f is a max flow.
(iii) There is no augmenting path relative to f.
(i)  (ii) This was the corollary to weak duality lemma.
(ii)  (iii) We show contrapositive.
Let f be a flow. If there exists an augmenting path, then we can improve f by sending flow along path.

44. Proof of Max-Flow Min-Cut Theorem
(iii)  (i)
Let f be a flow with no augmenting paths.
Let A be set of vertices reachable from s in residual graph.
By definition of A, s  A.
By definition of f, t  A.

45. K-Edge-Connectivity [Definition] k-edge-connectivity
The network will remain connected after removing any arbitrary k-1 edges from network.
k-edge-connected:  any pair of nodes are connected by k disjoint paths
disjoint paths: 45

46. Edge Disjoint Paths Disjoint path problem.
Given a digraph G = (V, E) and two nodes s and t, find the max number of edge-disjoint s-t paths.
[Def] Two paths are edge-disjoint if they have no edge in common.
Ex: communication networks. 2 5 s 3 6 t 4 7

47. Edge Disjoint Paths Disjoint path problem.
Given a digraph G = (V, E) and two nodes s and t, find the max number of edge-disjoint s-t paths.
[Def] Two paths are edge-disjoint if they have no edge in common.
Ex: communication networks. 2 5 s 3 6 t 4 7

48. Edge Disjoint Paths
Max flow formulation: assign unit capacity to every edge.
[Theorem] Max number edge-disjoint s-t paths equals max flow value.
Proof 
Suppose there are k edge-disjoint paths P1, , Pk.
Set f(e) = 1 if e participates in some path Pi ; else set f(e) = 0.
Since paths are edge-disjoint, f is a flow of value Q=k. ▪ s t 1 1 1 1 1 1 1 1 1 1 1 1 1 1

49. Edge Disjoint Paths
Max flow formulation: assign unit capacity to every edge.
[Theorem] Max number edge-disjoint s-t paths equals max flow value.
Proof 
Suppose max flow value Q is k.
Then, there exists 0-1 flow f of value k.
Consider edge (s, u) with f(s, u) = 1.
by conservation, there exists an edge (u, v) with f(u, v) = 1
continue until reach t, always choosing a new edge
Produces k (not necessarily simple) edge-disjoint paths. s t 1 1 1 1 1 1 1 1 1 1 1 1 1 1
can eliminate cycles to get simple paths if desired

50. Network Connectivity
[Network connectivity] Given a digraph G = (V, E) and two nodes s and t, find min number of edges whose removal disconnects t from s.
Def. A set of edges F  E disconnects t from s if all s-t paths uses at least one edge in F. 2 5 2 5 s 3 6 t s 3 6 t 4 7 4 7 F

51. Edge Disjoint Paths and Network Connectivity
Theorem. [Menger 1927] The max number of edge-disjoint s-t paths is equal to the min number of edges whose removal disconnects t from s.
[Proof] 
Suppose the removal of F  E disconnects t from s, and |F| = k.
All s-t paths use at least one edge of F. Hence, the number of edge-disjoint paths is at most k. ▪ 2 5 2 5 s 3 6 t s 3 6 t 4 7 4 7 F

52. Edge Disjoint Paths and Network Connectivity
Theorem. [Menger 1927] The max number of edge-disjoint s-t paths is equal to the min number of edges whose removal disconnects t from s.
[Proof] 
Suppose max number of edge-disjoint paths is k.
Then max flow value is k.
Max-flow min-cut  cut (A, B) of capacity k.
Let F be set of edges going from A to B.
|F| = k and disconnects t from s. ▪ A 2 5 2 5 s 3 6 t s 3 6 t 4 7 4 7 F