NLopt optimization with inequality constraint where objective function is undefined outside of the constraint

optimization
#1

Hello,

I would like to do nonlinear optimization with inequality constraint by NLopt, in which my objective function is not defined when the constraint is not satisfied. I’m using LN_COBYLA because I cannot compute the derivative.
However, the solver sometimes try to evaluate the objective function outside of the constraint and, as a consequence, the solver fails.

The minimal example below shows the problem.
I know I can impose a bounds constraint for this simple example but not in general.

Is there any way to search values only inside of the constraint? (What I can come up with is to let the objective function return huge value when undefined, although I don’t think it’s a beautiful solution.)

Thank you.

=======================
Example: \min x_1+x_2 subject to x_1\geq0 and x_2\geq0 but x_1+x_2 is undefined for x_1<0 or x_2<0

using NLopt
function obj(x, grad)
    if any(x .<0)
        println("negative element")
        println(x)
        return error("undefined")
    else
        return x[1]+x[2]
    end
end
function myconstraint(result, x, grad)
    result .= -x
end
opt = Opt(:LN_COBYLA, 2)
min_objective!(opt, obj)
inequality_constraint!(opt, myconstraint, [0, 0])

optimize(opt, [0.001, 0.001])

return

undefined
[-0.000320083, 0.000507602]
(0.000585786437626905, [0.000292893, 0.000292893], :FORCED_STOP)
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#2

Do a domain transformation, such as using x^2 or exp(x) as the input, so that way it is defined on the full real lined.

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#3

For general nonlinear constraints, NLopt makes no guarantee that it will only evaluate points inside the feasible set (i.e. satisfying the constraints).

However, for a constraint like x₁ ≥ 0, you should instead specify the constraint as a bound constraint via opt.lower_bounds = ... Not only is this much more efficient, but also NLopt guarantees that it will never evaluate a point outside of the bound constraints.

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#4

Thank you both,

It’s good to know there is no way to force the solver to use a value satisfying the constraint.

I’ll think about change of variables so that I can use box constraints, although I’m not sure if I can.

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