1.1 --- /dev/null Thu Jan 01 00:00:00 1970 +0000
1.2 +++ b/examples/mfasp.mod Mon Dec 06 13:09:21 2010 +0100
1.3 @@ -0,0 +1,62 @@
1.4 +/* MFASP, Minimum Feedback Arc Set Problem */
1.5 +
1.6 +/* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */
1.7 +
1.8 +/* The Minimum Feedback Arc Set Problem for a given directed graph
1.9 + G = (V, E), where V is a set of vertices and E is a set of arcs, is
1.10 + to find a minimal subset of arcs, which being removed from the graph
1.11 + make it acyclic.
1.12 +
1.13 + Reference:
1.14 + Garey, M.R., and Johnson, D.S. (1979), Computers and Intractability:
1.15 + A guide to the theory of NP-completeness [Graph Theory, Covering and
1.16 + Partitioning, Minimum Feedback Arc Set, GT9]. */
1.17 +
1.18 +param n, integer, >= 0;
1.19 +/* number of vertices */
1.20 +
1.21 +set V, default 1..n;
1.22 +/* set of vertices */
1.23 +
1.24 +set E, within V cross V,
1.25 +default setof{i in V, j in V: i <> j and Uniform(0,1) <= 0.15} (i,j);
1.26 +/* set of arcs */
1.27 +
1.28 +printf "Graph has %d vertices and %d arcs\n", card(V), card(E);
1.29 +
1.30 +var x{(i,j) in E}, binary;
1.31 +/* x[i,j] = 1 means that (i->j) is a feedback arc */
1.32 +
1.33 +/* It is known that a digraph G = (V, E) is acyclic if and only if its
1.34 + vertices can be assigned numbers from 1 to |V| in such a way that
1.35 + k[i] + 1 <= k[j] for every arc (i->j) in E, where k[i] is a number
1.36 + assigned to vertex i. We may use this condition to require that the
1.37 + digraph G = (V, E \ E'), where E' is a subset of feedback arcs, is
1.38 + acyclic. */
1.39 +
1.40 +var k{i in V}, >= 1, <= card(V);
1.41 +/* k[i] is a number assigned to vertex i */
1.42 +
1.43 +s.t. r{(i,j) in E}: k[j] - k[i] >= 1 - card(V) * x[i,j];
1.44 +/* note that x[i,j] = 1 leads to a redundant constraint */
1.45 +
1.46 +minimize obj: sum{(i,j) in E} x[i,j];
1.47 +/* the objective is to minimize the cardinality of a subset of feedback
1.48 + arcs */
1.49 +
1.50 +solve;
1.51 +
1.52 +printf "Minimum feedback arc set:\n";
1.53 +printf{(i,j) in E: x[i,j]} "%d %d\n", i, j;
1.54 +
1.55 +data;
1.56 +
1.57 +/* The optimal solution is 3 */
1.58 +
1.59 +param n := 15;
1.60 +
1.61 +set E := 1 2, 2 3, 3 4, 3 8, 4 9, 5 1, 6 5, 7 5, 8 6, 8 7, 8 9, 9 10,
1.62 + 10 11, 10 14, 11 15, 12 7, 12 8, 12 13, 13 8, 13 12, 13 14,
1.63 + 14 9, 15 14;
1.64 +
1.65 +end;