Finding maximal flow in a transportation network

1 Formulation of the problem

A transportation network is a directed graph [math]G = (V, E)[/math] in which a nonnegative capacity [math]c(e) \ge 0[/math] is assigned to each edge [math]e \in E[/math]. We assume that, along with each edge [math]e = (v, w) \in E[/math], the graph also contains the reverse edge [math]e^R = (w, v)[/math] (to which, if required, the zero capacity is assigned).

Let two vertices of a graph [math]G[/math] be marked, namely, the source [math]s[/math] and the sink [math]t[/math]. It can be assumed without loss of generality that all the other vertices lie on a path from [math]s[/math] to [math]t[/math]. A flow is a function [math]f: E \to \mathbb{R}[/math] that satisfies the following requirements:

• capacity constraint: [math]f(e) \le c(e)[/math];
• antisymmetry: [math]f(e^R) = -f(e)[/math];
• flow conservation law:

[math] \forall v \ne s, t: \quad \sum_{e = (w, v)} f(e) = \sum_{e = (v, w)} f(e). [/math]

The value of flow is the total amount of flow from the source:

[math] \left \vert f \right \vert = \sum_{e = (s, v)} f(e). [/math]

Maximal flow problem in a transportation network. It is required to find a flow with maximum value:

[math]\left \vert f \right \vert \to \max.[/math]

2 Properties of the problem

The total flow out of the source is equal to the total flow into the sink:

[math] \forall v \ne s, t: \quad \sum_{e = (s, v)} f(e) = \sum_{e = (v, t)} f(e). [/math]

(To prove this statement, it suffices to add up the flow conservation relations for all the vertices except for the source and sink.)

3 Variants of the problem

Depending on constraints on the capacity, the following cases may be distinguished:

• arbitrary positive capacity;
• integral capacity;
• unit capacity (in this case, the maximal flow is equal to the edge connectivity of the graph).

4 Algorithms for solving the problem

• The Ford-Fulkerson algorithm and its versions with the complexity [math]O(n^2m)[/math] (for Dinic's algorithm). In the case of integral capacity bounded by [math]K[/math], the complexity is [math]O(Km)[/math].
• The preflow push algorithm The complexity is [math]O(mn \ln n)[/math] (provided that Sleator-Tarjan dynamic trees are used ).

Notation: [math]m[/math] is the number of edges, [math]n[/math] is the number of vertices.