I have a simple question. How do I use the command qpsolve from Scilab if I only want to use the lower bounds and upper bounds limit?
ci <= x <= cs
The command can be used as this:
[x [,iact [,iter [,f]]]] = qpsolve(Q,p,C,b,ci,cs,me)
But I want to use it like this:
x = qpsolve(Q,p,[],[],ci,cs,[])
Only ci and cs should explain the limits for vector x. Unfortunately, the command cannot take empty []. Should I take [] as a row vector of ones or zeros?
In Scilab 5.5.1 , []
works for C and b but not for me
. so C = [];b = [];me = 0;
should work.
qpsolve is an interface for qp_solve :
function [x, iact, iter, f]=qpsolve(Q,p,C,b,ci,cs,me)
rhs = argn(2);
if rhs <> 7
error(msprintf(gettext("%s: Wrong number of input argument(s): %d expected.\n"), "qpsolve", 7));
end
C(me+1:$, :) = -C(me+1:$, :);
b(me+1:$) = -b(me+1:$);
// replace boundary contraints by linear constraints
Cb = []; bb = [];
if ci <> [] then
Cb = [Cb; speye(Q)]
bb = [bb; ci]
end
if cs <> [] then
Cb = [Cb; -speye(Q)]
bb = [bb; -cs]
end
C = [C; Cb]; b = [b; bb]
[x, iact, iter, f] = qp_solve(Q, -p, C', b, me)
endfunction
It transform every bound constraints into linear constraints. To begin, it swap the sign of the inequality constraints. To do that, it must know me
, ie it must be an integer. Since C
and b
are empty matrices, is value doesn't matter.
if Q
is inversible, you could skip the qpsolve
macro and write
x = -Q\p
x(x<ci) = ci(x<ci)
x(x>cs) = cs(x>cs)