doc/min_cost_flow.dox
 author Peter Kovacs Fri, 10 Jul 2009 09:15:22 +0200 changeset 751 7124b2581f72 child 802 134852d7fb0a child 833 e20173729589 child 1081 f1398882a928 permissions -rw-r--r--
Make K a template parameter in KaryHeap (#301)
1 /* -*- mode: C++; indent-tabs-mode: nil; -*-
2  *
3  * This file is a part of LEMON, a generic C++ optimization library.
4  *
6  * Egervary Jeno Kombinatorikus Optimalizalasi Kutatocsoport
7  * (Egervary Research Group on Combinatorial Optimization, EGRES).
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9  * Permission to use, modify and distribute this software is granted
10  * provided that this copyright notice appears in all copies. For
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13  * This software is provided "AS IS" with no warranty of any kind,
14  * express or implied, and with no claim as to its suitability for any
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16  *
17  */
19 namespace lemon {
21 /**
22 \page min_cost_flow Minimum Cost Flow Problem
24 \section mcf_def Definition (GEQ form)
26 The \e minimum \e cost \e flow \e problem is to find a feasible flow of
27 minimum total cost from a set of supply nodes to a set of demand nodes
28 in a network with capacity constraints (lower and upper bounds)
29 and arc costs.
31 Formally, let \f$G=(V,A)\f$ be a digraph, \f$lower: A\rightarrow\mathbf{R}\f$,
32 \f$upper: A\rightarrow\mathbf{R}\cup\{+\infty\}\f$ denote the lower and
33 upper bounds for the flow values on the arcs, for which
34 \f$lower(uv) \leq upper(uv)\f$ must hold for all \f$uv\in A\f$,
35 \f$cost: A\rightarrow\mathbf{R}\f$ denotes the cost per unit flow
36 on the arcs and \f$sup: V\rightarrow\mathbf{R}\f$ denotes the
37 signed supply values of the nodes.
38 If \f$sup(u)>0\f$, then \f$u\f$ is a supply node with \f$sup(u)\f$
39 supply, if \f$sup(u)<0\f$, then \f$u\f$ is a demand node with
40 \f$-sup(u)\f$ demand.
41 A minimum cost flow is an \f$f: A\rightarrow\mathbf{R}\f$ solution
42 of the following optimization problem.
44 \f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]
45 \f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \geq
46     sup(u) \quad \forall u\in V \f]
47 \f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]
49 The sum of the supply values, i.e. \f$\sum_{u\in V} sup(u)\f$ must be
50 zero or negative in order to have a feasible solution (since the sum
51 of the expressions on the left-hand side of the inequalities is zero).
52 It means that the total demand must be greater or equal to the total
53 supply and all the supplies have to be carried out from the supply nodes,
54 but there could be demands that are not satisfied.
55 If \f$\sum_{u\in V} sup(u)\f$ is zero, then all the supply/demand
56 constraints have to be satisfied with equality, i.e. all demands
57 have to be satisfied and all supplies have to be used.
60 \section mcf_algs Algorithms
62 LEMON contains several algorithms for solving this problem, for more
63 information see \ref min_cost_flow_algs "Minimum Cost Flow Algorithms".
65 A feasible solution for this problem can be found using \ref Circulation.
68 \section mcf_dual Dual Solution
70 The dual solution of the minimum cost flow problem is represented by
71 node potentials \f$\pi: V\rightarrow\mathbf{R}\f$.
72 An \f$f: A\rightarrow\mathbf{R}\f$ primal feasible solution is optimal
73 if and only if for some \f$\pi: V\rightarrow\mathbf{R}\f$ node potentials
74 the following \e complementary \e slackness optimality conditions hold.
76  - For all \f$uv\in A\f$ arcs:
77    - if \f$cost^\pi(uv)>0\f$, then \f$f(uv)=lower(uv)\f$;
78    - if \f$lower(uv)<f(uv)<upper(uv)\f$, then \f$cost^\pi(uv)=0\f$;
79    - if \f$cost^\pi(uv)<0\f$, then \f$f(uv)=upper(uv)\f$.
80  - For all \f$u\in V\f$ nodes:
81    - \f$\pi(u)<=0\f$;
82    - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,
83      then \f$\pi(u)=0\f$.
85 Here \f$cost^\pi(uv)\f$ denotes the \e reduced \e cost of the arc
86 \f$uv\in A\f$ with respect to the potential function \f$\pi\f$, i.e.
87 \f[ cost^\pi(uv) = cost(uv) + \pi(u) - \pi(v).\f]
89 All algorithms provide dual solution (node potentials), as well,
90 if an optimal flow is found.
93 \section mcf_eq Equality Form
95 The above \ref mcf_def "definition" is actually more general than the
96 usual formulation of the minimum cost flow problem, in which strict
97 equalities are required in the supply/demand contraints.
99 \f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]
100 \f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) =
101     sup(u) \quad \forall u\in V \f]
102 \f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]
104 However if the sum of the supply values is zero, then these two problems
105 are equivalent.
106 The \ref min_cost_flow_algs "algorithms" in LEMON support the general
107 form, so if you need the equality form, you have to ensure this additional
108 contraint manually.
111 \section mcf_leq Opposite Inequalites (LEQ Form)
113 Another possible definition of the minimum cost flow problem is
114 when there are <em>"less or equal"</em> (LEQ) supply/demand constraints,
115 instead of the <em>"greater or equal"</em> (GEQ) constraints.
117 \f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]
118 \f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \leq
119     sup(u) \quad \forall u\in V \f]
120 \f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]
122 It means that the total demand must be less or equal to the
123 total supply (i.e. \f$\sum_{u\in V} sup(u)\f$ must be zero or
124 positive) and all the demands have to be satisfied, but there
125 could be supplies that are not carried out from the supply
126 nodes.
127 The equality form is also a special case of this form, of course.
129 You could easily transform this case to the \ref mcf_def "GEQ form"
130 of the problem by reversing the direction of the arcs and taking the
131 negative of the supply values (e.g. using \ref ReverseDigraph and
133 However \ref NetworkSimplex algorithm also supports this form directly
134 for the sake of convenience.
136 Note that the optimality conditions for this supply constraint type are
137 slightly differ from the conditions that are discussed for the GEQ form,
138 namely the potentials have to be non-negative instead of non-positive.
139 An \f$f: A\rightarrow\mathbf{R}\f$ feasible solution of this problem
140 is optimal if and only if for some \f$\pi: V\rightarrow\mathbf{R}\f$
141 node potentials the following conditions hold.
143  - For all \f$uv\in A\f$ arcs:
144    - if \f$cost^\pi(uv)>0\f$, then \f$f(uv)=lower(uv)\f$;
145    - if \f$lower(uv)<f(uv)<upper(uv)\f$, then \f$cost^\pi(uv)=0\f$;
146    - if \f$cost^\pi(uv)<0\f$, then \f$f(uv)=upper(uv)\f$.
147  - For all \f$u\in V\f$ nodes:
148    - \f$\pi(u)>=0\f$;
149    - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,
150      then \f$\pi(u)=0\f$.
152 */
153 }