|
|
/* 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;
|
