/* 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;