1. Max Flow Problem
Given network N=(V,A), two nodes s,t of V, and capacities on the arcs: uij is the capacity on arc (i,j).
Find non-negative flow fij for each arc (i,j) such that for each node, except s and t, the total incoming flow is equal to the total outgoing flow (flow conservation), such that
fij is at most the capacity uij on arc (i,j);
The total amount of flow going out of s (which is equal to the total amount of flow coming into t ) is maximized

2. the black numbers next to an arc is its capacity1 3 2 4 3 2 3 1 t 1 1 s 2 4 2 4 4
the black numbers next to an arc is its capacity

3. Set costs all other arcs at 0 The minimum cost flow circulation (Af=0) 1 3 2 4 3 2 3 1 t 1 1 s 2 4 2 4 4
Cts= -1
Set costs all other arcs at 0
The minimum cost flow circulation (Af=0)
maximises the s-t flow

4. the black numbers next to an arc is its capacity 1 3 2 4 3 2 3 1 t 1 1 s 2 4 2 4 4
the black numbers next to an arc is its capacity
Push flow over the path s-1-4-t
The bottlenecks on this path are the edges {1,4} and
{4,t}. So we can send flow 2 along this path

5. the black number next to an arc is its capacity 1 3 2 4
2 3
2 2 3 1 t 1 1 s
2 2 4 2 4 4
the black number next to an arc is its capacity
the green number next to an arc is the flow on it
We can push another extra flow of 1 on the path
s-1-3-t. The bottleneck is now {s,1} with remaining
capacity 1.

6. the black number next to an arc is its capacity 1 3
1 2
1 4
3 3
2 2 3 1 t 1 1 s
2 2 4 2 4 4
the black number next to an arc is its capacity
the green number next to an arc is the flow on it
Since {4,t} is on its capacity, the only path remaining through
which we can send extra flow is s t. Here {4,3} is the
bottleneck and thus we can send extra flow of 1

7. the residual graph blue arcs (i,j) are forward arcs (fij<uij)1 3
3 3
2 2
3 3
1 1 t
1 1
1 1 s
2 2 2 4
blue arcs (i,j) are forward arcs (fij<uij)
green arcs (j,i) are backward arcs (fij>0)
the blue number is the residual capacity of a blue arc
the green number is the present flow on the opposite
of the green arc (hence, the capacity of the green arc)
the black number is the original capacity of the arc

8. the residual graph red arcs form an s-t-path in the residual graph1 3
1 1
2 2 3 2 3 1 t 1 1 s 2
1 3 2
1 3 4
red arcs form an s-t-path in the residual graph
and therefore a flow-augmenting path in the
original network

9. the residual graph red arcs form an augmenting path1 3
1 1
2 2 3 2 3 1 t 1 1 s 2
1 3 2
1 3 4
red arcs form an augmenting path
Augment the flow by the minimum capacity of a
red arc, i.e, 1

10. the residual graph Augment the flow by the minimum capacity of a1 3
1 1
2 2 3 2 3 1 t 1 1 s 2
1 3 2
1 3 4
Augment the flow by the minimum capacity of a
red arc, i.e, 1:
- increase the flow by 1 on all arcs corresponding to
forward red arcs
- decrease the flow by 1 on all arcs corresponding to
backward red arcs

11. the residual graph Augment the flow by the minimum capacity of a1 3
1 1
2 2 3 2 3 1 t 1 1 s 2
1 3 2
1 3 4
Augment the flow by the minimum capacity of a
red arc, i.e, 1:
- increase the flow by 1 on all arcs corresponding to
forward red arcs
- decrease the flow by 1 on all arcs corresponding to
backward red arcs
Construct the new residual graph

12. the residual graph Augment the flow by the minimum capacity of a1 3 2
3 1 3 3 1
1 1 t 1 1 s 2
2 2 2
2 2 4
Augment the flow by the minimum capacity of a
red arc, i.e, 1:
- increase the flow by 1 on all arcs corresponding to
forward red arcs
- decrease the flow by 1 on all arcs corresponding to
backward red arcs
Construct the new residual graph

13. An s-t cut is defined by a set S of the nodes with 1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4
An s-t cut is defined by a set S of the nodes with
s in S and t not in S.

14. An s-t cut is defined by a set S of the nodes with 1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4
S={s,1,2}
An s-t cut is defined by a set S of the nodes with
s in S and t not in S.

15. An s-t cut is defined by a set S of the nodes with 1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4
S={s,1,2}
An s-t cut is defined by a set S of the nodes with
s in S and t not in S
Size of cut S is the sum of the capacities on the
arcs from S to N\S.

16. An s-t cut is defined by a set S of the nodes with 1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4
C(S)=u13+u14+u24= 2+2+4=8
An s-t cut is defined by a set S of the nodes with
s in S and t not in S
Size of cut S is the sum of the capacities on the
arcs from S to N\S.

17. S1 1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4
C(S1)=us1+u24= 2+4=6

18. S2 S1 C(S1)=us1+u24= 2+4=6 C(S2)=u13+u43+u4t= 2+1+2=5 t s 1 3 2 4 2 2

19. S2 C(S2)=u13+u43+u4t= 2+1+2=5 Max Flow ≤ Min Cut t s 1 3 2 4 2 2 3 1 1

20. S2 C(S2)=u13+u43+u4t= 2+1+2=5 Max Flow ≤ Min Cut fs1+fs2 ≤ Min Cut ≤ 5

21. = the residual graph S2 fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5
3 1 3 3 1
1 1 t 1 1 s 2
2 2 2
2 2 4 =
fs1+fs2=2+3=5
C(S2)=u13+u43+u4t= 2+1+2=5
Max Flow = Min Cut

22. = the residual graph S2 fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5
3 1 3 3 1
1 1 t 1 1 s 2
2 2 2
2 2 4 =
fs1+fs2=2+3=5
C(S2)=u13+u43+u4t= 2+1+2=5
Max Flow = Min Cut
Theorem: Max Flow = Min Cut

23. = Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5
3 1 3 3 1
1 1 t 1 1 s 2
2 2 2
2 2 4 =
fs1+fs2=2+3=5
C(S2)=u13+u43+u4t= 2+1+2=5
There is no s-t-path in the residual graph!!!
S2={s,1,2,4} the set of nodes reachable from S

24. = Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5
2 2
4 3
2 2
3 0
2 1
1 0 t
1 1
1 0 s
2 2
4 3 2
4 2 4 =
fs1+fs2=2+3=5
C(S2)=u13+u43+u4t= 2+1+2=5
from S2={s,1,2,4} to {3,t}
f13 = u13 = 2
f43 = u43 = 1
f4t = u4t = 2

25. = Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5
2 2
4 3
2 2
3 0
2 1
1 0 t
1 1
1 0 s
2 2
4 3 2
4 2 4 =
fs1+fs2=2+3=5
C(S2)=u13+u43+u4t= 2+1+2=5
from S2={s,1,2,4} to {3,t}
f13 = u13 = 2
f43 = u43 = 1
f4t = u4t = 2
Full capacity of cut is used

26. = Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5
2 2
4 3
2 2
3 0
2 1
1 0 t
1 1
1 0 s
2 2
4 3 2
4 2 4 =
fs1+fs2=2+3=5
C(S2)=u13+u43+u4t= 2+1+2=5
from S2={s,1,2,4} to {3,t}
f13 = u13 = 2
f43 = u43 = 1
f4t = u4t = 2
Full capacity of cut is used
from {3,t} to {S2={s,1,2,4}
f32 = 0
f34 = 0

27. = Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5
2 2
4 3
2 2
3 0
2 1
1 0 t
1 1
1 0 s
2 2
4 3 2
4 2 4 =
fs1+fs2=2+3=5
C(S2)=u13+u43+u4t= 2+1+2=5
from S2={s,1,2,4} to {3,t}
f13 = u13 = 2
f43 = u43 = 1
f4t = u4t = 2
Full capacity of cut is used
from {3,t} to {S2={s,1,2,4}
f32 = 0
f34 = 0
nothing flows back across the cut

28. = Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5
2 2
4 3
2 2
3 0
2 1
1 0 t
1 1
1 0 s
2 2
4 3 2
4 2 4 =
fs1+fs2=2+3=5
C(S2)=u13+u43+u4t= 2+1+2=5
from S2={s,1,2,4} to {3,t}
f13 = u13 = 2
f43 = u43 = 1
f4t = u4t = 2
from {3,t} to {S2={s,1,2,4}
f32 = 0
f34 = 0
Capacity of the cut is equal to the flow from s to t