linprog (Matlab) vs glpk(Octave) - Why do I get different solutions? - matlab

Question

I need to find an initial feasible solution for a network flow problem which is given by Nx = b, x >= 0. I used to the following OCTAVE code to do the job:
function [x,exitflag] = FIND_EXTREME_POINT(A,b)
%
% Function FIND_EXTREME_POINT
%
% Call: [x,fval] = FIND_EXTREME_POINT(A,b)
%
% MATLAB code for finding a extreme point of the set X = { x | Ax = b,
% x >= 0} by solving the auxiliary problem
%
% min 1'*y
% s.t. A*x + y = b
% x >= 0
% y >= 0
%
% either to receive a feasible basis solution or to return infeasibility
%
% Inputs:
% A - m*n matrix
% b = m*1 rhs
%
% Output:
% x - n*1 vector respresenting an extreme point
% exitflag - 0 if problem is infeasibile, 1 if the problem is feasible
%
% setup parameters for LP solver
[m,n] = size(A);
A_aux = [A eye(m)];
c_aux = [zeros(1,n) ones(1,m)]';
lb = [zeros(1,n+m)]';
ub = [];
ctype = repmat('S',1,m);
vartype = repmat('C',1,n+m);
% solve auxiliary LP
[xmin, fmin] = glpk(c_aux, A_aux, b, lb, ub, ctype, vartype, 1);
...% driving out artifical variables
The solution given by the glpk-solver matches to my handwritten notes.
For some reason I'm using MATLAB and I've got a problem with the following code:
function [x,exitflag] = FIND_EXTREME_POINT(A,b)
%
% Function FIND_EXTREME_POINT
%
% Call: [x,fval] = FIND_EXTREME_POINT(A,b)
%
% ....
%
% setup parameters for LP solver
[m,n] = size(A);
x = []; % output vector
A_aux = [A eye(m)];
c_aux = [zeros(1,n) ones(1,m)]';
lb = [zeros(1,n+m)]';
ub = [];
% solve auxiliary LP
[xmin,fval] = linprog(c_aux,[],[],A_aux,b,lb,ub);
... % driving out artifical variables
On the one hand the optimal solution xmin computed by the linprog-solver is feasible but on the other hand the function value fval is not zero but it should be so... Where is the mistake in my MATLAB code?
A little example:
N = [1 1 0 0 0 0 0;
-1 0 0 -1 0 0 0;
0 -1 1 0 0 -1 0;
0 0 -1 1 1 0 0;
0 0 0 0 0 1 1;
0 0 0 0 -1 0 -1];
b = [5 -5 0 0 0 0]';
OCTAVE gives me:
xmin =
5
0
0
0
0
0
0
0
0
0
0
0
0
fval =
0
MATLAB gives me:
xmin =
0.2184
4.7816
4.7816
4.7816
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
fval =
5.8775e-016