1. The Ford-Fulkerson Algorithm
aka The Labeling Algorithm

2. Ford-Fulkerson Algorithm
begin
x := 0; label node t;
while t is labeled do
unlabel all nodes; pred(j) := 0 for all j in N;
label s; LIST := {s};
while LIST is not empty and t is not labeled do
remove a node i from LIST;
for all {j in N: (i,j) in A and rij > 0} do
if j is unlabeled then
pred(j) := i, label j, add j to LIST;
end;
if t is labeled then augment flow on path from s to t

3. Labeling Algorithm Examples t 1 2 4 3
(0,2)
(0,6) 5
(0,5) 6
(0,4)
(0,7)

4. The Residual Network G(x)2 2 4 4 5 1 6 4 t s 6 7 3 5 5

5. Iteration 1: LIST = {1}, Labeled = {1}2 2 4 4 5 1 6 4 t s 6 7 3 5 5

6. Iteration 1: LIST = {1}, Labeled = {1}
Arc (1,2)
pred(2) =1
label 2
LIST = {2}
Arc (1,3)
pred (3) = 1
label 3
LIST = {2, 3}

7. Iteration 1: LIST = {2,3}, Labeled= {1,2,3}4 3 5 6 7
pred(2) = 1
pred(3) = 1

8. Iteration 1: LIST = {2,3}, Labeled = {1,2,3}
Arc (2,4)
pred(4) =2
label 4
LIST = {3,4}
Arc (2,5)
pred (5) = 2
label 5
LIST = {3,4,5}
Arc (2,1)
residual capacity of (2,1) = 0

9. Iteration 1: LIST = {3,4,5}, Labeled= {1,2,3,4,5}
pred(2) = 1
pred(4) = 2 2 2 4 4 5 1 6 4 t s 6 7 3 5 5
pred(3) = 1
pred(5) = 2

10. Iteration 1: LIST = {3,4,5}, Labeled = {1,2,3,4,5}
Arc (3,5)
5 is already labeled
Arc (3,1)
residual capacity of (3,1) = 0

11. Iteration 1: LIST = {4,5}, Labeled = {1,2,3,4,5}
Arc (4,2)
residual capacity of (4,2) = 0
Arc (4,6)
pred(6) =4
label 6
LIST = {5,6}

12. Iteration 1: LIST = {5,6}, Labeled= {1,2,3,4,5,6}
pred(2) = 1
pred(4) = 2 2 2 4 4 5
pred(6) = 4 1 6 4 t s 6 7 3 5 5
pred(3) = 1
pred(5) = 2

13. Iteration 1: The sink is labeled
Use pred labels to trace back from the sink to the source to find path P
P = 1 -> 2 -> 4 -> 6
 = min {rij: (i,j) in P) = 2
Send 2 units of flow from to s to t along path P

14. Flow x After Iteration 1 1 2 4 3 5 6 s t (2,2) (0,6) (0,5) (2,5) (2,4)
(0,4)
(0,7)
v = 2

15. The Residual Network G(x)2 4 2 3 2 1 6 2 2 4 t s 6 7 3 5 5

16. Iteration 2: LIST = {1}, Labeled = {1}
Arc (1,2)
pred(2) =1
label 2
LIST = {2}
Arc (1,3)
pred (3) = 1
label 3
LIST = {2, 3}

17. Iteration 2: LIST = {2,3}, Labeled={1,2,3}
p=1 2 4 2 3 2 1 6 2 2 4 t s 6 7 3 5 5
p=1

18. Iteration 2: LIST = {2,3}, Labeled = {1,2,3}
Arc (2,4)
residual cap (2,4) = 0
Arc (2,5)
pred (5) = 2
label 5
LIST = {3,5}
Arc (2,1)
residual capacity of (2,1) = 0

19. Iteration 2: LIST = {3,5}, Labeled={1,2,3,5}
p=1 2 4 2 3 2 1 6 2 2 4 t s 6 7 5 3 5
p=1
p=2

20. Iteration 2: LIST = {3,5}, Labeled = {1,2,3,5}
Arc (3,5)
5 is already labeled
Arc (3,1)
residual capacity of (3,1) = 0
i = 5
LIST = {}
Arc (5,2)
residual cap = 0
Arc (5,3)
Arc (5,6)
pred(6) = 5
label 6
LIST = {6}

21. Iteration 2: LIST = {6}, Labeled={1,2,3,5,6}
p=1 2 4 2 3 2 1 6 2 2 4 t s 6 7
p=5 3 5 5
p=1
p=2

22. Iteration 2: The sink is labeled
Use pred labels to trace back from the sink to the source to find path P
P = 1 -> 2 -> 5 -> 6
 = min {rij: (i,j) in P) = 3
Send 3 units of flow from to s to t along path P

23. Flow x After Iteration 2 1 2 4 3 5 6 s t (2,2) (0,6) (0,5) (5,5) (2,4)
(3,4)
(3,7)
v = 5

24. The Residual Network G(x)2 4 2 2 1 6 5 2 3 1 t s 6 4 3 5 5

25. Iteration 3: LIST = {1}, Labeled = {1}
Arc (1,2)
residual capacity = 0
Arc (1,3)
pred (3) = 1
label 3
LIST = {3}

26. Iteration 3: List = {3}, Labeled = {1,3}2 4 2 2 1 6 5 2 3 1 t s 6 4 3 5 5
p=1

27. Iteration 3: LIST = {3}, Labeled = {1,3}
Arc (3,1)
residual capacity = 0
Arc (3,5)
pred (5) = 3
label 5
LIST = {5}

28. Iteration 3: List = {5}, Labeled = {1,3,5}2 4 2 2 1 6 5 2 3 1 t s 6 4 3 5 5
p=1
p=2

29. Iteration 3: LIST = {5}, Labeled = {1,3,5}
i = 5, LIST = {}
Arc (5,2)
pred(2) = 5
label 2
LIST = {2}
Arc (5,3): residual capacity = 0
Arc (5,6)
pred (6) = 5
label 6
LIST = {2,6}

30. Iteration 3: List = {2,6}, Labeled = {1,2,3,5,6}
p=5 2 4 2 2 1 6 5 2 3 1 t s 6 4
p=5 3 5 5
p=1
p=2

31. Iteration 3: The sink is labeled
Use pred labels to trace back from the sink to the source to find path P
P = 1 -> 3 -> 5 -> 6
 = min {rij: (i,j) in P) = 4
Send 4 units of flow from to s to t along path P

32. Flow x After Iteration 3 1 2 4 3 5 6 s t (2,2) (4,6) (4,5) (5,5) (2,4)
(3,4)
(7,7)
v = 9

33. The Residual Network G(x)2 4 2 2 1 6 5 2 3 1 t s 2 3 1 7 5 4 4

34. Iteration 4: LIST = {1}, Labeled = {1}
Arc (1,2)
residual capacity = 0
Arc (1,3)
pred (3) = 1
label 3
LIST = {3}

35. Iteration 4: List = {3}, Labeled = {1,3}2 4 2 2 1 6 5 2 3 1 t s 2 3 1 7 5 4 4
p=1

36. Iteration 4: LIST = {3}, Labeled = {1,3}
Arc (3,1)
1 is labeled
Arc (3,5)
pred (5) = 3
label 5
LIST = {5}

37. Iteration 4: List = {5}, Labeled = {1,3,5}2 4 2 2 1 6 5 2 3 1 t s 2 7 3 1 5 4 4
p=1
p=3

38. Iteration 4: LIST = {5}, Labeled = {1,3,5}
Arc (5,2)
pred(2) = 5
label 2
LIST = {2}
Arc (5,6)
residual capacity = 0

39. Iteration 4: List = {2}, Labeled = {1,2,3,5}
p=5 2 4 2 2 1 6 5 2 3 1 t s 2 1 7 3 5 4 4
p=3
p=1

40. Iteration 4: LIST = {2}, Labeled = {1,2,3,5}
i = 2 LIST = {}
Arc (2,1)
1 is already labeled
Arc(2,4)
residual capacity = 0
Arc (2,5)
5 is already labeled

41. Iteration 4: List = {} The sink is not labeled
Algorithm ends with optimal flow x

42. Correctness
At the end of each iteration, the algorithm either augments the flow or terminates because it can’t label the sink.
Let S be the set of labeled nodes when the algorithm terminates. Let T = N \ S.
We need to show that when the algorithm terminates v = u[S,T] which implies x is a maximum flow.

43. Correctness: arcs in (S,T)j
rij = 0
rij = uij - xij + xji
0 = uij – xij + xji
xij = uij + xji
Since xji must be at least zero and 0  xij  uij,
it follows that xij = uij.

44. Correctness: arcs in (T,S)j s
Suppose xij > 0
rji = uji – xji + xij
Implies rji > 0 (uji  xji)
Implies s can reach i in residual network which means i should have been labeled.
Thus, xij = 0

45. Correctness
Thus, the flows on the forward arcs in [S,T] are at the upper bound and there is no flow on any of the backwards arcs.
This means x is a maximum flow by Property 6.1.

46. Complexity Let U = max {(i,j) in A} uij.
If S = {s} and T = N\{s}, then u[S,T] is at most nU.
The maximum flow is at most nU.
Iteration of the inner while loop is O(m):
Each arc is inspected at most once
Finding  is O(n)
Updating the flow on P is O(n)
Complexity is O(nU x m) = O(nmU).