diff -r d59bea55db9b -r c445c931472f examples/min01ks.mod
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/examples/min01ks.mod	Mon Dec 06 13:09:21 2010 +0100
@@ -0,0 +1,111 @@
+/* min01ks.mod - finding minimal equivalent 0-1 knapsack inequality */
+
+/* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */
+
+/* It is obvious that for a given 0-1 knapsack inequality
+
+      a[1] x[1] + ... + a[n] x[n] <= b,  x[j] in {0, 1}              (1)
+
+   there exist infinitely many equivalent inequalities with exactly the
+   same feasible solutions.
+
+   Given a[j]'s and b this model allows to find an inequality
+
+      alfa[1] x[1] + ... + alfa[n] x[n] <= beta,  x[j] in {0, 1},    (2)
+
+   which is equivalent to (1) and where alfa[j]'s and beta are smallest
+   non-negative integers.
+
+   This model has the following formulation:
+
+      minimize
+
+         z = |alfa[1]| + ... + |alfa[n]| + |beta| =                  (3)
+
+           = alfa[1] + ... + alfa[n] + beta
+
+      subject to
+
+         alfa[1] x[1] + ... + alfa[n] x[n] <= beta                   (4)
+
+                              for all x satisfying to (1)
+
+         alfa[1] x[1] + ... + alfa[n] x[n] >= beta + 1               (5)
+
+                              for all x not satisfying to (1)
+
+         alfa[1], ..., alfa[n], beta are non-negative integers.
+
+   Note that this model has n+1 variables and 2^n constraints.
+
+   It is interesting, as noticed in [1] and explained in [2], that
+   in most cases LP relaxation of the MIP formulation above has integer
+   optimal solution.
+
+   References
+
+   1. G.H.Bradley, P.L.Hammer, L.Wolsey, "Coefficient Reduction for
+      Inequalities in 0-1 Variables", Math.Prog.7 (1974), 263-282.
+
+   2. G.J.Koehler, "A Study on Coefficient Reduction of Binary Knapsack
+      Inequalities", University of Florida, 2001. */
+
+param n, integer, > 0;
+/* number of variables in the knapsack inequality */
+
+set N := 1..n;
+/* set of knapsack items */
+
+/* all binary n-vectors are numbered by 0, 1, ..., 2^n-1, where vector
+   0 is 00...00, vector 1 is 00...01, etc. */
+
+set U := 0..2^n-1;
+/* set of numbers of all binary n-vectors */
+
+param x{i in U, j in N}, binary, := (i div 2^(j-1)) mod 2;
+/* x[i,j] is j-th component of i-th binary n-vector */
+
+param a{j in N}, >= 0;
+/* original coefficients */
+
+param b, >= 0;
+/* original right-hand side */
+
+set D := setof{i in U: sum{j in N} a[j] * x[i,j] <= b} i;
+/* set of numbers of binary n-vectors, which (vectors) are feasible,
+   i.e. satisfy to the original knapsack inequality (1) */
+
+var alfa{j in N}, integer, >= 0;
+/* coefficients to be found */
+
+var beta, integer, >= 0;
+/* right-hand side to be found */
+
+minimize z: sum{j in N} alfa[j] + beta; /* (3) */
+
+phi{i in D}: sum{j in N} alfa[j] * x[i,j] <= beta; /* (4) */
+
+psi{i in U diff D}: sum{j in N} alfa[j] * x[i,j] >= beta + 1; /* (5) */
+
+solve;
+
+printf "\nOriginal 0-1 knapsack inequality:\n";
+for {j in 1..n} printf (if j = 1 then "" else " + ") & "%g x%d",
+   a[j], j;
+printf " <= %g\n", b;
+printf "\nMinimized equivalent inequality:\n";
+for {j in 1..n} printf (if j = 1 then "" else " + ") & "%g x%d",
+   alfa[j], j;
+printf " <= %g\n\n", beta;
+
+data;
+
+/* These data correspond to the very first example from [1]. */
+
+param n := 8;
+
+param a := [1]65, [2]64, [3]41, [4]22, [5]13, [6]12, [7]8, [8]2;
+
+param b := 80;
+
+end;