The flow f(u, v) through an edge cannot be negative and cannot exceed the capacity of the edge c(u, v), 0 ≤ f(u, v) ≤ c(u, v). If an edge (u, v) doesn’t exist in the network, then c(u, v) = 0.

Aside from the source vertex s and sink vertex t, each vertex u belongs to V must satisfy the property that the sum of f(v, u) for all edges (v, u) in E (the flow into u) must equal the sum of f(u, w) for all edges (u, w) belongs to E (the flow out of u).

This property ensures that flow is neither produced nor consumed in the network, except at s and t.

For consistency, the quantity f(v, u) represents the net flow from vertex u to v. This means that it must be the case that f(u, v) = −f(v, u); this holds even if both edges (u, v) and (v, u) exist in a directed graph.

Intuitively, given a flow network and a flow, the residual network consists of edges that can admit more flow. More formally, let a flow network with source s and sink t. Let f be a flow in G, and consider a pair of vertices The amount of additional flow which can be pushed from u to v before exceeding the capacity c(u, v) is the residual capacity of (u, v), given by c_f(u, v) = c(u, v) − f(u, v).

Consider the water supply system in a Campus. The pipeline between the supply points of any two areas has a stated capacity in gallons per hour, given as a maximum flow at which water can flow through the pipe between those two areas. To supply water from the source area, A (say) to the sink area, I (say) and water passes through 7 other areas before reaching from source to sink. Suppose these 7 areas are B, C, D, E, F, G, H and pipeline between any two areas has a defined capacity. Table 1 shows the defined capacities of each pipeline between any two areas which can flow between corresponding two areas.

Calculate the maximum amount of water which can flow from A to I.

Now the problem discussed in the Example 1 is converted into directed graph by representing areas as vertices of the graph and pipelines between any two areas as edges of the graph. The capacity of the pipeline in gallons per hour is represented as capacity of an edge in units between vertices.

Now, following correspondence between areas and vertices is used to create the graph.

So, the initial graph corresponding to Table 1 and Table 2 is as follows.

Now we use the modified Edmonds-Karp Method to compute a maximum flow in G.

According to the Figure 1, the maximum capacity is 48. So, the value of the variable C in the above algorithm will be 48.

So, , this will be the value of the variable I in the 1st iteration of the algorithm.

Iteration-1: I = 27. So, the augmenting path with capacity at least 27 will be searched by the Breadth First Search (BFS) procedure. But, there is no augmenting path with capacity at least 27. So, no flow will be added to the initial flow of the graph which is 0.

Iteration-2: I = I/3 = 27/3 = 9. So, now the augmenting path with capacity at least 9 will be searched by the same BFS procedure in the residual graph which is given in Figure 2 corresponding to the initial graph. The augmenting path will be searched till path with capacity at least 9 is found in the graph. Now, in the consecutive figures, Figure 2 shows the residual graph and Figure 3 shows the corresponding flow in the graph.

Whenever the augmenting path is to be found in the graph, if there are more than 1 path satisfying the capacity criteria, then the path is determined by BFS procedure on the basis of sequential ordering of the vertices and corresponding edges which has been given as input.

1^st Augmentation: So, the augmenting path found in 2nd iteration is s - v1 - v3 - v6 - t with capacity 20. So, the initial flow is augmented by 20 units and the flow in the graph is shown in Figure 3 giving maximum flow value f = 20. The residual graph after 1^st augmentation is shown in Figure 4.

2^nd Augmentation: Now, again there is a path with capacity at least 18 and the path found in the same 2^nd iteration is s - v4 - v3 - v7 - t with capacity 18. So, the maximum flow is augmented by 18 units and the flow in the graph is shown in Figure 5 giving maximum flow value f = 20 + 18 = 38. Now, there is no path with capacity at least 18. Same procedure will be followed until variable I becomes < 1. Residual graph after 2^nd augmentation is shown in Figure 6.

3^rd Augmentation: Now, again there is a path with capacity at least 12 and the path found in the same 2^nd iteration is s - v2 - v5 - v7 - t with capacity 12. So, the maximum flow is augmented by 12 units and the flow in the graph is shown in Figure 7 giving maximum flow value f = 38 + 12 = 50. Now, there is no path with capacity at least 12. Same procedure will be followed until variable I becomes < 1. Residual graph after 3rd augmentation is shown in Figure 8.

4^th Augmentation: Now, again there is a path with capacity at least 15 and the path found in the same 2^nd iteration is s - v4 - v5 - v7 - v6 - t with capacity 15. So, the maximum flow is augmented by 15 units and the flow in the graph is shown in Figure 9 giving maximum flow value f = 50 + 15 = 65. Now, there is no path with capacity at least 15. Same procedure will be followed until variable I becomes < 1.

3^rd Iteration: I = I/3 = 9/3 = 3. So, now the augmenting path with capacity at least 3 will be searched in the residual graph which is given in Figure 10 corresponding to the initial graph.

5^th Augmentation: So, the augmenting path found in 3rd iteration is s - v1 - v4 - v3 - v5 - v7 - v6 - t with capacity 5. So, the initial flow is augmented by 5 units and the flow in the graph is shown in Figure 11 gives maximum flow value f = 65 + 5 = 70. The residual graph after 4^th augmentation is shown in Figure 10.

4^th Iteration: I = I/3= 3/3 = 1. So, the augmenting path with capacity at least 1 will be searched in the residual graph which is given in Figure 12 corresponding to the initial graph.

6^th Augmentation: So, the augmenting path found in 4th iteration is s - v2 - v4 - v3 - v6 - t with capacity 2. So, the initial flow is augmented by 2 units and the flow in the graph is shown in Figure 13 giving maximum flow value f = 70 + 2 = 72. The residual graph after 6^th augmentation is shown in Figure 14.

The residual graph after 6^th augmentation is shown below:

Suppose a pipeline system in a Campus to supply gas in different areas of a Campus. The pipeline between any two areas has a stated capacity in per unit per hour, given as a maximum flow at which gas can flow through the pipe between those two areas. Now, suppose we want to supply gas from the source area, suppose A to the sink area, say F and gas passes through 4 other areas before reaching from source to sink. Suppose these 4 areas are B, C, D, E and pipeline between any two areas has defined capacity.

The following table shows the defined capacities of each pipeline between any two areas which can flow between corresponding two areas.

Calculate the maximum amount of gas which can flow from A to F.

Now the problem discussed in the Example 2 is converted into directed graph by representing areas as vertices of the graph and pipelines between any two areas as edges of the graph. The capacity of the pipeline in unit per hour is represented as capacity of an edge in units between vertices.

Now, following correspondence between areas and vertices is used to create the graph.

So, the initial graph corresponding to Table 3 and Table 4 is shown in Figure 15.

Solving the above example using our proposed algorithm, the maximum flow value f = 19.