# Optimization: Constraining through parametrization

I have a non-linear optimization problem formulation in which my objective is convex with a convex equality constraint. However, I parametrized my equality constraint and plugged this parametric form into the objective.

Now, in the new formulation, the objective is constrained in the sense that my parametrization limits the objective to a certain set (the convex constraint set) but there is no explicitly defined constraint.

Strictly speaking, is this a constrained or unconstrained optimization problem? I would appreciate it if you could answer with a reference of citation to a mathematic/scientific paper. This fundamental point will go into my thesis and a reference would be nice to have.

Thank you

• It looks like you used some kind of langrangean. Could you show us how you parametrized your equality constraint? Also, constrained problems are often (not always, not with barrier algorithms for instance) turned into unconstrained problems for resolution Nov 28, 2016 at 10:44
• I didn't use Lagrange multipliers because I wanted to solve it iteratively with least computational effort at each iteration. My constraint is an ellipsoid so I parametrized it using two spatial angles and trigonometric functions. My problem formulation now looks like as if it were an unconstrained problem but the solution set is restricted/constrained by my parameterization. Hope the explanation helps. Nov 28, 2016 at 10:53

Your optimization problem is now $$\min_{x \in X} f(x)$$ where $X \neq \mathbb{R}^n$. There are two ways to look at this. The first way is to say it's constrained optimization by writing it as: $$\min_{x \in \mathbb{R}^n} \{ f(x) : x \in X \}$$ This problem has advantages in terms of solvability when you can explicitly write $X$ in terms of convex functions.
The second way is to define a new function $g$, where $g(x)=f(x)$ when $x\in X$, $g(x)=\infty$ elsewhere. Note that $g$ is convex. Now it is clearly an unconstrained problem: $$\min_{x \in \mathbb{R}^n} g(x)$$ A disadvantage is that you don't have a differentiable objective function.
• I guess that right now I've converted the problem $\min_{\vec{x} \in \mathbb{R}^{3 \times 3}} f( \vec{x} )$ into another problem $\min_{\alpha, \beta \in \mathbb{R}} \ g( \alpha, \beta )$ and $\forall \ \alpha, \beta \in \mathbb{R}, g( \alpha, \beta ) = f( \vec{x} )$. Here, $\vec{x}$ has been parameterized in $\alpha, \beta$. Hope that makes sense. I'm guessing it would still be unconstrained optimization, am I right? I've then solved the problem of $g$ and translated its solution into the solution of $f$, which is what I required in the first place. Thanks for the reply. Nov 28, 2016 at 16:19
• Right, just take $x_1=\alpha$ and $x_2=\beta$ for the first formula in my answer. Nov 28, 2016 at 16:38
• @MotiveHunter $g(\alpha,\beta)=f(x)$ when $x$ contains both $\alpha$ and $\beta$. Nov 28, 2016 at 18:33