doc/min_cost_flow.dox
author Alpar Juttner <alpar@cs.elte.hu>
Tue, 26 May 2015 16:54:15 +0200
branch1.2
changeset 1000 d94bb1e50557
parent 788 c92296660262
permissions -rw-r--r--
Merge bugfix #598 to branch 1.2
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/* -*- mode: C++; indent-tabs-mode: nil; -*-
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 *
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 * This file is a part of LEMON, a generic C++ optimization library.
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 *
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 * Copyright (C) 2003-2010
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 * Egervary Jeno Kombinatorikus Optimalizalasi Kutatocsoport
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 * (Egervary Research Group on Combinatorial Optimization, EGRES).
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 *
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 * Permission to use, modify and distribute this software is granted
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 * provided that this copyright notice appears in all copies. For
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 * precise terms see the accompanying LICENSE file.
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 *
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 * This software is provided "AS IS" with no warranty of any kind,
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 * express or implied, and with no claim as to its suitability for any
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 * purpose.
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 *
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 */
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namespace lemon {
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/**
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\page min_cost_flow Minimum Cost Flow Problem
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\section mcf_def Definition (GEQ form)
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The \e minimum \e cost \e flow \e problem is to find a feasible flow of
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minimum total cost from a set of supply nodes to a set of demand nodes
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in a network with capacity constraints (lower and upper bounds)
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and arc costs \ref amo93networkflows.
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Formally, let \f$G=(V,A)\f$ be a digraph, \f$lower: A\rightarrow\mathbf{R}\f$,
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\f$upper: A\rightarrow\mathbf{R}\cup\{+\infty\}\f$ denote the lower and
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upper bounds for the flow values on the arcs, for which
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\f$lower(uv) \leq upper(uv)\f$ must hold for all \f$uv\in A\f$,
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\f$cost: A\rightarrow\mathbf{R}\f$ denotes the cost per unit flow
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on the arcs and \f$sup: V\rightarrow\mathbf{R}\f$ denotes the
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signed supply values of the nodes.
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If \f$sup(u)>0\f$, then \f$u\f$ is a supply node with \f$sup(u)\f$
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supply, if \f$sup(u)<0\f$, then \f$u\f$ is a demand node with
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\f$-sup(u)\f$ demand.
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A minimum cost flow is an \f$f: A\rightarrow\mathbf{R}\f$ solution
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of the following optimization problem.
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\f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]
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\f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \geq
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    sup(u) \quad \forall u\in V \f]
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\f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]
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The sum of the supply values, i.e. \f$\sum_{u\in V} sup(u)\f$ must be
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zero or negative in order to have a feasible solution (since the sum
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of the expressions on the left-hand side of the inequalities is zero).
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It means that the total demand must be greater or equal to the total
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supply and all the supplies have to be carried out from the supply nodes,
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but there could be demands that are not satisfied.
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If \f$\sum_{u\in V} sup(u)\f$ is zero, then all the supply/demand
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constraints have to be satisfied with equality, i.e. all demands
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have to be satisfied and all supplies have to be used.
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\section mcf_algs Algorithms
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LEMON contains several algorithms for solving this problem, for more
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information see \ref min_cost_flow_algs "Minimum Cost Flow Algorithms".
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A feasible solution for this problem can be found using \ref Circulation.
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\section mcf_dual Dual Solution
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The dual solution of the minimum cost flow problem is represented by
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node potentials \f$\pi: V\rightarrow\mathbf{R}\f$.
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An \f$f: A\rightarrow\mathbf{R}\f$ primal feasible solution is optimal
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if and only if for some \f$\pi: V\rightarrow\mathbf{R}\f$ node potentials
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the following \e complementary \e slackness optimality conditions hold.
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 - For all \f$uv\in A\f$ arcs:
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   - if \f$cost^\pi(uv)>0\f$, then \f$f(uv)=lower(uv)\f$;
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   - if \f$lower(uv)<f(uv)<upper(uv)\f$, then \f$cost^\pi(uv)=0\f$;
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   - if \f$cost^\pi(uv)<0\f$, then \f$f(uv)=upper(uv)\f$.
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 - For all \f$u\in V\f$ nodes:
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   - \f$\pi(u)\leq 0\f$;
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   - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,
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     then \f$\pi(u)=0\f$.
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Here \f$cost^\pi(uv)\f$ denotes the \e reduced \e cost of the arc
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\f$uv\in A\f$ with respect to the potential function \f$\pi\f$, i.e.
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\f[ cost^\pi(uv) = cost(uv) + \pi(u) - \pi(v).\f]
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All algorithms provide dual solution (node potentials), as well,
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if an optimal flow is found.
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\section mcf_eq Equality Form
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The above \ref mcf_def "definition" is actually more general than the
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usual formulation of the minimum cost flow problem, in which strict
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equalities are required in the supply/demand contraints.
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\f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]
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\f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) =
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    sup(u) \quad \forall u\in V \f]
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\f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]
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However if the sum of the supply values is zero, then these two problems
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are equivalent.
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The \ref min_cost_flow_algs "algorithms" in LEMON support the general
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form, so if you need the equality form, you have to ensure this additional
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contraint manually.
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\section mcf_leq Opposite Inequalites (LEQ Form)
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Another possible definition of the minimum cost flow problem is
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when there are <em>"less or equal"</em> (LEQ) supply/demand constraints,
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instead of the <em>"greater or equal"</em> (GEQ) constraints.
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\f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]
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\f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \leq
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    sup(u) \quad \forall u\in V \f]
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\f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]
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It means that the total demand must be less or equal to the
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total supply (i.e. \f$\sum_{u\in V} sup(u)\f$ must be zero or
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positive) and all the demands have to be satisfied, but there
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could be supplies that are not carried out from the supply
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nodes.
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The equality form is also a special case of this form, of course.
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You could easily transform this case to the \ref mcf_def "GEQ form"
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of the problem by reversing the direction of the arcs and taking the
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negative of the supply values (e.g. using \ref ReverseDigraph and
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\ref NegMap adaptors).
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However \ref NetworkSimplex algorithm also supports this form directly
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for the sake of convenience.
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Note that the optimality conditions for this supply constraint type are
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slightly differ from the conditions that are discussed for the GEQ form,
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namely the potentials have to be non-negative instead of non-positive.
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An \f$f: A\rightarrow\mathbf{R}\f$ feasible solution of this problem
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is optimal if and only if for some \f$\pi: V\rightarrow\mathbf{R}\f$
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node potentials the following conditions hold.
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 - For all \f$uv\in A\f$ arcs:
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   - if \f$cost^\pi(uv)>0\f$, then \f$f(uv)=lower(uv)\f$;
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   - if \f$lower(uv)<f(uv)<upper(uv)\f$, then \f$cost^\pi(uv)=0\f$;
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   - if \f$cost^\pi(uv)<0\f$, then \f$f(uv)=upper(uv)\f$.
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 - For all \f$u\in V\f$ nodes:
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   - \f$\pi(u)\geq 0\f$;
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   - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,
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     then \f$\pi(u)=0\f$.
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*/
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}