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 black number next to an arc is its capacity 1 3
1 2
2 4
3 3
2 2 3 1 t
1 1 1 s
2 2
1 4 2
1 4 4
the black number next to an arc is its capacity
the green number next to an arc is the flow on it
There are no s-t paths left on which we can send extra
flow. So is this the maximal flow?

8. 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 capacity of a green arc
the black number is the original capacity of the arc

9. 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

10. 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

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

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