/* -*- mode: C++; indent-tabs-mode: nil; -*-

 *

 * This file is a part of LEMON, a generic C++ optimization library.

 *

 * Copyright (C) 2003-2011

 * Egervary Jeno Kombinatorikus Optimalizalasi Kutatocsoport

 * (Egervary Research Group on Combinatorial Optimization, EGRES).

 *

 * Permission to use, modify and distribute this software is granted

 * provided that this copyright notice appears in all copies. For

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 * This software is provided "AS IS" with no warranty of any kind,

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 */

namespace lemon {

/**

\page min_cost_flow Minimum Cost Flow Problem

\section mcf_def Definition (GEQ form)

The \e minimum \e cost \e flow \e problem is to find a feasible flow of

minimum total cost from a set of supply nodes to a set of demand nodes

in a network with capacity constraints (lower and upper bounds)

and arc costs.

Formally, let \f$G=(V,A)\f$ be a digraph, \f$lower: A\rightarrow\mathbf{R}\f$,

\f$upper: A\rightarrow\mathbf{R}\cup\{+\infty\}\f$ denote the lower and

upper bounds for the flow values on the arcs, for which

\f$lower(uv) \leq upper(uv)\f$ must hold for all \f$uv\in A\f$,

\f$cost: A\rightarrow\mathbf{R}\f$ denotes the cost per unit flow

on the arcs and \f$sup: V\rightarrow\mathbf{R}\f$ denotes the

signed supply values of the nodes.

If \f$sup(u)>0\f$, then \f$u\f$ is a supply node with \f$sup(u)\f$

supply, if \f$sup(u)<0\f$, then \f$u\f$ is a demand node with

\f$-sup(u)\f$ demand.

A minimum cost flow is an \f$f: A\rightarrow\mathbf{R}\f$ solution

of the following optimization problem.

\f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]

\f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \geq

    sup(u) \quad \forall u\in V \f]

\f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]

The sum of the supply values, i.e. \f$\sum_{u\in V} sup(u)\f$ must be

zero or negative in order to have a feasible solution (since the sum

of the expressions on the left-hand side of the inequalities is zero).

It means that the total demand must be greater or equal to the total

supply and all the supplies have to be carried out from the supply nodes,

but there could be demands that are not satisfied.

If \f$\sum_{u\in V} sup(u)\f$ is zero, then all the supply/demand

constraints have to be satisfied with equality, i.e. all demands

have to be satisfied and all supplies have to be used.

\section mcf_algs Algorithms

LEMON contains several algorithms for solving this problem, for more

information see \ref min_cost_flow_algs "Minimum Cost Flow Algorithms".

A feasible solution for this problem can be found using \ref Circulation.

\section mcf_dual Dual Solution

The dual solution of the minimum cost flow problem is represented by

node potentials \f$\pi: V\rightarrow\mathbf{R}\f$.

An \f$f: A\rightarrow\mathbf{R}\f$ primal feasible solution is optimal

if and only if for some \f$\pi: V\rightarrow\mathbf{R}\f$ node potentials

the following \e complementary \e slackness optimality conditions hold.

 - For all \f$uv\in A\f$ arcs:

   - if \f$cost^\pi(uv)>0\f$, then \f$f(uv)=lower(uv)\f$;

   - if \f$lower(uv)<f(uv)<upper(uv)\f$, then \f$cost^\pi(uv)=0\f$;

   - if \f$cost^\pi(uv)<0\f$, then \f$f(uv)=upper(uv)\f$.

 - For all \f$u\in V\f$ nodes:

   - \f$\pi(u)<=0\f$;

   - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,

     then \f$\pi(u)=0\f$.

Here \f$cost^\pi(uv)\f$ denotes the \e reduced \e cost of the arc

\f$uv\in A\f$ with respect to the potential function \f$\pi\f$, i.e.

\f[ cost^\pi(uv) = cost(uv) + \pi(u) - \pi(v).\f]

All algorithms provide dual solution (node potentials), as well,

if an optimal flow is found.

\section mcf_eq Equality Form

The above \ref mcf_def "definition" is actually more general than the

usual formulation of the minimum cost flow problem, in which strict

equalities are required in the supply/demand contraints.

\f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]

\f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) =

    sup(u) \quad \forall u\in V \f]

\f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]

However if the sum of the supply values is zero, then these two problems

are equivalent.

The \ref min_cost_flow_algs "algorithms" in LEMON support the general

form, so if you need the equality form, you have to ensure this additional

contraint manually.

\section mcf_leq Opposite Inequalites (LEQ Form)

Another possible definition of the minimum cost flow problem is

when there are <em>"less or equal"</em> (LEQ) supply/demand constraints,

instead of the <em>"greater or equal"</em> (GEQ) constraints.

\f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]

\f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \leq

    sup(u) \quad \forall u\in V \f]

\f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]

It means that the total demand must be less or equal to the

total supply (i.e. \f$\sum_{u\in V} sup(u)\f$ must be zero or

positive) and all the demands have to be satisfied, but there

could be supplies that are not carried out from the supply

nodes.

The equality form is also a special case of this form, of course.

You could easily transform this case to the \ref mcf_def "GEQ form"

of the problem by reversing the direction of the arcs and taking the

negative of the supply values (e.g. using \ref ReverseDigraph and

\ref NegMap adaptors).

However \ref NetworkSimplex algorithm also supports this form directly

for the sake of convenience.

Note that the optimality conditions for this supply constraint type are

slightly differ from the conditions that are discussed for the GEQ form,

namely the potentials have to be non-negative instead of non-positive.

An \f$f: A\rightarrow\mathbf{R}\f$ feasible solution of this problem

is optimal if and only if for some \f$\pi: V\rightarrow\mathbf{R}\f$

node potentials the following conditions hold.

 - For all \f$uv\in A\f$ arcs:

   - if \f$cost^\pi(uv)>0\f$, then \f$f(uv)=lower(uv)\f$;

   - if \f$lower(uv)<f(uv)<upper(uv)\f$, then \f$cost^\pi(uv)=0\f$;

   - if \f$cost^\pi(uv)<0\f$, then \f$f(uv)=upper(uv)\f$.

 - For all \f$u\in V\f$ nodes:

   - \f$\pi(u)>=0\f$;

   - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,

     then \f$\pi(u)=0\f$.

*/

}