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In optimization theory, the max-flow min-cut theorem states that in a flow network, the maximum amount of flow passing from the source to the sink is equal to the minimum capacity that, when removed in a specific way from the network, causes the situation that no flow can pass from the source to the sink.

The max-flow min-cut theorem is a special case of the duality theorem for linear programs and can be used to derive Menger's theorem and the König-Egerváry Theorem.

Definition

Let $N=(V,E)$ be a network (directed graph) with $s$ and $t$ being the source and the sink of $N$ respectively.

The capacity of an edge is a mapping $c\colon E\to \mathbb {R} ^{+}$, denoted by c_uv or c(u,v). It represents the maximum amount of flow that can pass through an edge.

A flow is a mapping $f\colon E\to \mathbb {R} ^{+}$, denoted by f_uv or f(u,v), subject to the following two constraints:

1. $f_{uv}\leq c_{uv}$ for each $(u,v)\in E$ (capacity constraint)
2. $\sum _{u:\,\,(u,v)\in E}f_{uv}=\sum _{u:\,\,(v,u)\in E}f_{vu}$ for each $v\in V\setminus \{s,t\}$ (conservation of flows).

The value of flow is defined by $|f|=\sum _{\,\,v\in V}f_{sv}$, where $s$ is the source of $N$. It represents the amount of flow passing from the source to the sink.

The maximum flow problem is to maximize | f |, that is, to route as much flow as possible from s to t.

An s-t cut C = (S,T) is a partition of V such that s∈S and t∈T. The cut-set of C is the set {(u,v)∈E | u∈S, v∈T}. Note that if the edges in the cut-set of C are removed, | f | = 0.

The capacity of an s-t cut is defined by $c(S,T)=\sum _{(u,v)\in S\times T}c_{uv}$.

The minimum s-t cut problem is minimizing $c(S,T)$, that is, to determine S and T such that the capacity of the S-T cut is minimal.

Statement

The max-flow min-cut theorem states

The maximum value of an s-t flow is equal to the minimum capacity over all s-t cuts.

Linear program formulation

The max-flow problem and min-cut problem can be formulated as two primal-dual linear programs.

Max-flow (Primal)	Min-cut (Dual)
maximize $|f|=\nabla _{s}$	minimize $\sum _{(i,j)\in E}c_{ij}d_{ij}$
subject to ${\begin{array}{rclr}f_{ij}&\leq &c_{ij}&(i,j)\in E\\\sum _{j:(j,i)\in E}f_{ji}-\sum _{j:(i,j)\in E}f_{ij}&\leq &0&i\in V,i\neq s,t\\\nabla _{s}+\sum _{j:(j,s)\in E}f_{js}-\sum _{j:(s,j)\in E}f_{sj}&\leq &0&\\-\nabla _{s}+\sum _{j:(j,t)\in E}f_{jt}-\sum _{j:(t,j)\in E}f_{tj}&\leq &0&\\f_{ij}&\geq &0&(i,j)\in E\\\end{array}}$	subject to ${\begin{array}{rclr}d_{ij}-p_{i}+p_{j}&\geq &0&(i,j)\in E\\p_{s}-p_{t}&\geq &1&\\p_{i}&\geq &0&i\in V\\d_{ij}&\geq &0&(i,j)\in E\end{array}}$

The equality in the max-flow min-cut theorem follows from the strong duality theorem in linear programming, which states that if the primal program has an optimal solution, x*, then the dual program also has an optimal solution, y*, such that the optimal values formed by the two solutions are equal.

Example

A network with the value of flow equal to the capacity of an s-t cut

The figure on the right is a network having a value of flow of 7. The vertex in white and the vertices in grey form the subsets S and T of an s-t cut, whose cut-set contains the dashed edges. Since the capacity of the s-t cut is 7, which equals to the value of flow, the max-flow min-cut theorem tells us that the value of flow and the capacity of the s-t cut are both optimal in this network.

Application

Generalized max-flow min-cut theorem

In addition to edge capacity, consider there is capacity at each vertex, that is, a mapping c: V→R⁺, denoted by c(v), such that the flow f has to satisfy not only the capacity constraint and the conservation of flows, but also the vertex capacity constraint

$\sum _{i\in V}f_{iv}\leq c(v)$ for each $v\in V\setminus \{s,t\}.$

In other words, the amount of flow passing through a vertex cannot exceed its capacity. Define an s-t cut to be the set of vertices and edges such that for any path from s to t, the path contains a member of the cut. In this case, the capacity of the cut is the sum the capacity of each edge and vertex in it.

In this new definition, the generalized max-flow min-cut theorem states that the maximum value of an s-t flow is equal to the minimum capacity of an s-t cut in the new sense.

Menger's theorem

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The Menger's theorem states that the maximum number of edge-disjoint s-t paths in an undirected graph is equal to the minimum number of edges in an s-t cut-set.

Project selection problem

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A network formulation of the project selection problem with the optimal solution

In the project selection problem, there are $n$ projects and $m$ equipments. Each project $p_{i}$ yields revenue $r(p_{i})$ and each equipment $q_{j}$ costs $c(q_{j})$ to purchase. Each project requires a number of equipments and each equipment can be shared by several projects. The problem is to determine which projects and equipments should be selected and purchased respectively, so that the profit is maximized.

Let $P$ be the set of projects not selected and $Q$ be the set of equipments purchased, then the problem can be formulated as,

$\max\{g\}=\sum _{i}r(p_{i})-\sum _{p_{i}\in P}r(p_{i})-\sum _{q_{j}\in Q}c(q_{j}).$

Since the first term does not depend on the choice of $P$ and $Q$, this maximization problem can be formulated as a minimization problem instead, that is,

$\min\{g'\}=\sum _{p_{i}\in P}r(p_{i})+\sum _{q_{j}\in Q}c(q_{j}).$

The above minimization problem can then be formulated as a minimum-cut problem by constructing a network, where the source is connected to the projects with capacity $r(p_{i})$, and the sink is connected by the equipments with capacity $c(q_{j})$. An edge ($p_{i}$, $q_{j}$) with infinite capacity is added if project $p_{i}$ requires equipment $q_{j}$. The s-t cut-set represents the projects and equipments in $P$ and $Q$ respectively. By the max-flow min-cut theorem, one can solve the problem as a maximum flow problem.

The figure on the right gives a network formulation of the following project selection problem:

	Project $r(p_{i})$	Equipment $c(q_{j})$
1	100	200	Project 1 requires equipments 1 and 2.
2	200	100	Project 2 requires equipment 2.
3	150	50	Project 3 requires equipment 3.

The minimum capacity of a s-t cut is 250 and the sum of the revenue of each project is 450; therefore the maximum profit g is 450 − 250 = 200, by selecting projects $p_{2}$ and $p_{3}$.

The idea here is to 'flow' the project profits through the 'pipes' of the equipment. If we cannot fill the pipe, the equipment's return is less than its cost, and the min cut algorithm will find it cheaper to cut the project's profit edge instead of the equipment's cost edge.

History

The max-flow min-cut theorem was proven by P. Elias, A. Feinstein, and C.E. Shannon in 1956I am Wyatt and was born on 10 June 1973. My hobbies are Vintage car and Yo-yoing.

Feel free to visit my website - Hostgator Voucher, and independently also by L.R. Ford, Jr. and D.R. Fulkerson in the same yearI am Wyatt and was born on 10 June 1973. My hobbies are Vintage car and Yo-yoing.

Feel free to visit my website - Hostgator Voucher.

See also

• Linear programming
• Maximum flow
• Minimum cut
• Flow network
• Edmonds-Karp algorithm

References

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2. 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
   
   My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
3. 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
   
   My blog: http://www.primaboinca.com/view_profile.php?userid=5889534