Re: Ax=b subject to lb<=x<=ub

[Thomas Shores]

Depends.  If by "solve" you only mean "find some solution"  the function glpk will work nicely -- use anything nonzero for the objective function vector C.  (Do "help glpk"  for details.)  However, if you mean "find the general solution" you will need a CAS to do the job, since octave isn't designed for symbolic calculations.  For example,

octave:1> A = [1 2 7;3 0 2;1 2 7];
octave:2> mysoln = [1 -2 3]';
octave:3> lb = [0 -4 2]';
octave:4> ub = [2 2,4]';
octave:5> C = ones(size(mysoln));
octave:6> b = A*mysoln;
octave:7> gsoln = glpk(C,A,b,lb,ub)
xsoln =

   0.57895
  -4.00000
   3.63158

Here gsoln  and mysoln are not multiples of each other, but both are solutions to the problem.



On 06/17/2010 11:28 AM, Jaana Tommiska wrote:

Hi,
how can I solve
Ax=b subject to lb<=x<=ub
A is rectangular, lb,ub and b are vectors?
Thanks
Jaana

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