Re: [Help-glpk] formulation of a problem

[Andrew Makhorin]

> I'm in trouble with set in a problem in the glpk datastructure and I
> can't find answers to my questions (perhaps I looked at the wrong  
> places).
> The two main problems:
> 1. How can I set a variable to the right hand side?
> If I have a constraint like this: x1+x2 <= y1 I can't set:
> 
> lpx_set_row_name(lp, 3, "constraint");
> lpx_set_row_bnds(lp, 3, LPX_UP, 0, y1);

You should write all your constraint in a standard form:

   a[i,1]*x[1] + ... + a[i,n]*x[n] <rho> b[i]

where <rho> is '=', '<=', or '>='.

In your case:

   x1 + x2 - y1 <= 0

i.e. the right-hand size must be a constant, not a variable.

> 
> 2. How can I set 'holes' in the matrix (if a variable is not needed  
> in one constraint)?
> 
> For instance:
> x1+x2 = 3
> x1 <= 1
> 
> then I would think the array assignments looks like that:
> 
> ia[1] = 1, ja[1] = 1, ar[1] = 1.0;
> ia[2] = 1, ja[2] = 2, ar[2] = 1.0;
> ia[3] = 2, ja[3] = 1, ar[3] = 1.0;
> ia[4] = 2, ja[4] = 2, ar[4] = 0; // <- it doesn't work
> 
> (a similar problem: if I don't need a basic variable in the minimize  
> or maximize funktion, can I set the the coefficient to zero? like:  
> lpx_set_obj_coef(lp,1,0.0);  )

You should omit all zero constraint coefficients.

If it is inconvenient, fill the matrix as is and then use the routine
lpx_remove_tiny to remove zero entries (it is not doc'ed, so see its
description in file glplpx7a.c).