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Maximum Flow Problem (MFP) discusses the maximum amount of flow that can be sent from the source to sink. Edmonds-Karp algorithm is the modified version of Ford-Fulkerson algorithm to solve the MFP. This paper presents some modifications of Edmonds-Karp algorithm for solving MFP. Solution of MFP has also been illustrated by using the proposed algorithm to justify the usefulness of proposed method.

The maximum amount of commodities that can be shipped through any network from source to sink is called maximum flow and this problem is called MFP, which is a classical network flow problem. Network flows problem has got a vast application in the field of Mathematics, Computer Science, Management and Operations Research.

At first the effective solution procedure to obtain the maximum flow in a flow network was introduced by Lester R. Ford and Delbert R. Fulkerson [

In this paper, a modified Edmonds-Karp algorithm is proposed to compute maximum amount of flow from source to sink for a MFP. Numerical illustration of the proposed algorithm is also done by solving a good number of examples to test the effectiveness and usefulness of the proposed algorithm.

Some of the basic ideas related to maximum flow problems are given below with an aim to accustom the readers with the article.

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 _{f}(u, v) = c(u, v) − f(u, v).

It is to be mentioned that the proposed algorithm is a modified version of Ford-Fulkerson algorithm. The basic steps of this algorithm are explained below.

Two numerical examples have been solved for finding the maximum value of a MFP by using proposed algorithm which is given below.

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.

Source Area | Destination Area | Capacity (Gallons/hour) | Source Area | Destination Area | Capacity (Gallons/hour) |
---|---|---|---|---|---|

A | B | 26 | D | G | 22 |

A | C | 17 | D | H | 18 |

A | E | 37 | E | D | 35 |

B | D | 20 | E | F | 15 |

B | E | 10 | F | H | 34 |

C | E | 20 | G | I | 48 |

C | F | 15 | H | G | 20 |

D | F | 10 | H | I | 30 |

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

Solution of Example-1Now 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.

Area | A | B | C | D | E | F | G | H | I |
---|---|---|---|---|---|---|---|---|---|

Vertex | s | 1 | 2 | 3 | 4 | 5 | 6 | 7 | t |

So, the initial graph corresponding to

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

According to the

So,

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

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 ^{st} augmentation is shown in

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 ^{nd} augmentation is shown in

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

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

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

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 ^{th} augmentation is shown in

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

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 ^{th} augmentation is shown in

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.

Source Area | Destination Area | Capacity (Gallons/hour) | Source Area | Destination Area | Capacity (Gallons/hour) |
---|---|---|---|---|---|

A | B | 10 | C | E | 9 |

A | C | 10 | D | F | 10 |

B | C | 2 | E | D | 6 |

B | D | 4 | E | F | 10 |

B | E | 8 |

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

Solution of Example-2Now 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.

Area | A | B | C | D | E | F |
---|---|---|---|---|---|---|

Vertex | S | 1 | 2 | 3 | 4 | t |

So, the initial graph corresponding to

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

In this paper, we have modified Edmonds-Karp algorithm to compute maximum amount of flow from source to

Name of the Algorithm | Number of Iterations | Number of Augmentation | ||
---|---|---|---|---|

Example-1 | Example-2 | Example-1 | Example-2 | |

Ford-Fulkerson | 9 | 6 | 8 | 5 |

Edmonds-Karp | 7 | 5 | 7 | 5 |

Md. Al-Amin Khan et al.’s | 7 | 5 | 7 | 5 |

Chintan J. & Deepak G.’s | 6 | 4 | 6 | 3 |

Faruque Ahmed et al.’s | 6 | 4 | 6 | 5 |

Modified Edmonds-Karp (Proposed) | 4 | 2 | 6 | 3 |

sink in a flow network. We have also solved several number of maximum flow problems by using our proposed algorithm, Ford-Fulkerson algorithm, Md. Al-Amin Khan et al.’s Algorithm, and Faruque Ahmed et al.’s algorithm and Edmonds-Karp algorithm to test the efficiency of the proposed algorithm. During this process, it is observed that our proposed algorithm is faster and taking less number of iterations and less number of augmentations to determine the maximum flow in maximum flow problems in comparison with the other well known algorithms.

Kalyan KumarMallick,Aminur RahmanKhan,Mollah MesbahuddinAhmed,Md. ShamsulArefin,Md. SharifUddin, (2016) Modified EDMONDS-KARP Algorithm to Solve Maximum Flow Problems. Open Journal of Applied Sciences,06,131-140. doi: 10.4236/ojapps.2016.62014