1. Network Flows – Minimum capacities
This problem involves a network in which the arcs have both minimum and maximum capacities.
The definition of the value of a cut has to be modified when there are minimum capacities involved.
1, 4 A C
0, 7
2, 8
0, 3
2, 5 T S
4, 8
1, 9
4, 10 B D
1, 8

2. Network Flows – Minimum capacities
The value of a cut is the sum of the maximum capacities for arcs that cross the cut line from S to T minus the sum of the minimum capacities for arcs that cross the cut line from T to S.
The cut shown in the diagram below includes the arcs SA, BC and DT which cross the cut line from S to T,
The cut shown in the diagram below includes the arcs SA, BC and DT which cross the cut line from S to T, and the arcs AB and CD which cross the cut line from T to S.
The value of the cut is therefore – 0 – 4
The value of the cut is therefore
The value of the cut is therefore – 0 – 4 = 19.
1, 4 A C
0, 7
2, 8
2, 8
0, 3
0, 3
2, 5
2, 5 T S
4, 8
4, 8
1, 9
4, 10
4, 10 B D
1, 8

3. Network Flows – Minimum capacities
The minimum cut for this network is shown below.
This cut includes the arcs AC, BC and DT which cross the cut line from S to T, and the arc CD which crosses the cut line from T to S.
This cut includes the arcs AC, BC and DT which cross the cut line from S to T,
The value of the cut is therefore
The value of the cut is therefore – 4 = 15.
The value of the cut is therefore – 4
1, 4
1, 4 A C
0, 7
2, 8
0, 3
2, 5
2, 5 T S
4, 8
4, 8
1, 9
4, 10
4, 10 B D
1, 8

4. Network Flows – Minimum capacities
Since the value of the minimum cut is 15, the maximum flow for this network must be 15 (using the maximum flow – minimum cut theorem).
We can find the maximum flow through the pipes using a modified version of the labelling procedure.
1, 4 S A B C D T
0, 7
2, 8
0, 3
2, 5
4, 8
1, 9
4, 10
1, 8

5. Network Flows – Minimum capacities
The first step is to find an initial flow which satisfies the minimum capacities.
You can start by allocating the minimum possible flows to each arc.
1, 4 S A B C D T 1
0, 7
2, 8 2
0, 3
2, 5 4 2
4, 8 1 4
1, 9
4, 10 1
1, 8

6. Network Flows – Minimum capacities
Now you need to check each vertex and increase the flow where necessary so that the flow into a vertex is the same as the flow out of the vertex.
Vertex D has a flow of = 5 going into it, so the flow out of it needs to be increased to 5.
Vertex C has a flow of 4 coming out of it, but only 3 going into it, so the flow in either AC or BC needs to be increased by 1.
Vertex B has a flow of 3 coming out of it, but only 1 going into it, so the flow in SB needs to be increased by 2.
1, 4 S A B C D T 2 1
0, 7
2, 8 2
0, 3
2, 5 4 2
4, 8 3 1 4 5
1, 9
4, 10 1
1, 8

7. Network Flows – Minimum capacities
The next step is to show the excess capacities and potential backflows for each arc.
The flow in arc SB could increase by up to 6 units, and could decrease by up to 2 units.
The flow in arc SA could increase by up to 6 units, but cannot decrease.
The flow in arc AC could increase by up to 2 units, and could decrease by 1 unit.
The flow in arc BD could increase by up to 7 units, but cannot decrease.
The flow in arc DT could increase by up to 5 units, and could decrease by 1 unit.
The flow in arc CT could increase by up to 7 units, but cannot decrease.
The flow in arc CD could increase by up to 4 units, but cannot decrease.
The flow in arc BC could increase by up to 3 units, but cannot decrease.
The flow in arc AB could increase by up to 3 units, but cannot decrease. 2 1 2
1, 4 S A B C D T
0, 7
2, 8 2 6 7 3 3 4
0, 3
2, 5 2
4, 8 4 6 2 3
1, 9 5
4, 10 5 1 7
1, 8 1

8. Network Flows – Minimum capacities
We now use the labelling procedure in the same way as before.
Find a flow augmenting path. One example is SACT.
The minimum arrow along this path is 2, and so this is the increase in the flow.
We decrease all the arrows in the direction of this path by 2,
and increase all the arrows against the direction of this path by 2. 2 1 S A B C D T 3 4 6 2 7 5 3 3 4 2 6 2 5 1 7

9. Network Flows – Minimum capacities
We now use the labelling procedure in the same way as before.
Find a flow augmenting path. One example is SACT.
Arc AC is now saturated. 2 1 S A B C D T 3 4 6 2 7 5 3 3 4 2 6 2 5 1 7

10. Network Flows – Minimum capacities
Now we choose another flow augmenting path. One example is SABCT.
The minimum arrow along this path is 3, and so this is the increase in the flow.
We decrease all the arrows in the direction of this path by 3,
and increase all the arrows against the direction of this path by 3. 2 1 S A B C D T 3 5 1 4 6 2 7 5 2 3 3 4 2 5 3 6 2 3 5 1 7

11. Network Flows – Minimum capacities
Now we choose another flow augmenting path. One example is SABCT.
Arcs AB and BC are now saturated. 2 1 S A B C D T 3 5 1 4 6 2 7 5 2 3 3 4 2 5 3 6 2 3 5 1 7

12. Network Flows – Minimum capacities
For our next flow augmenting path let’s choose SBDT.
The minimum arrow along this path is 5, and so this is the increase in the flow.
We decrease all the arrows in the direction of this path by 5,
and increase all the arrows against the direction of this path by 5. 2 1 S A B C D T 3 5 1 4 6 2 7 5 2 3 3 4 2 5 1 3 6 2 3 6 5 1 2 7 7 5

13. Network Flows – Minimum capacities
For our next flow augmenting path let’s choose SBDT.
Arc DT is now saturated. 2 1 S A B C D T 3 5 1 4 6 2 7 5 2 3 3 4 2 5 1 3 6 2 3 6 5 1 2 7 7 5

14. Network Flows – Minimum capacities
S and T are now disconnected, so there are no more flow augmenting paths.
However, this diagram does not tell you the actual flows. The back flows (the numbers on the backward arrows) tell you by how much the flow could be reduced – i.e. how much greater the flow is than the minimum flow.
Write the minimum capacity on each arc.
Now add the back flow to the minimum capacity to give the flow on each arc. 4 2 1 S A B C D T 5 7 2 3 1 5 1 4 6 2 7 5 2 3 3 4 2 5 5 1 3 4 6 2 3 4 3 2 4 6 1 5 1 2 7 6 8 7 10 1 5

15. Network Flows – Minimum capacities
The flow out of the source and into the sink is 15 units.
This is the maximum possible flow.
Notice that this maximum flow is the same as the value of the minimum cut, as expected from the maximum flow – minimum cut theorem. 4 2 1 S A B C D T 5 7 2 3 1 5 1 4 6 2 7 5 2 3 3 4 2 5 5 1 3 4 6 2 3 4 3 2 4 6 1 5 1 2 7 6 8 7 10 1 5

16. Network Flows – Minimum capacities
The flow out of the source and into the sink is 15 units.
This is the maximum possible flow.
Notice that this maximum flow is the same as the value of the minimum cut, as expected from the maximum flow – minimum cut theorem.
Notice that all the arcs in the minimum cut which flow from S to T are saturated, and the arc in the minimum cut which flows from T to S has as small a flow as possible. 4
1, 4
1, 4 A C 5 7
0, 7
2, 8
0, 3
2, 5
2, 5 T S 5
4, 8
4, 8 3 4
1, 9
4, 10
4, 10 6 8 B D 10
1, 8