# Constraint optimisation

Consider the constraint optimization $\text{argmin}_{\beta}(f(\beta)+\lambda g(\beta))$ can someone define $\beta(\lambda)$. That is, what is the relationship between $\lambda$ and $\beta$?.

• is this a Lagrangean optimization? Did you start with $\max f(\beta)$ subject to the constraint that $g(\beta)=0$? Jul 14 '14 at 11:23
• It is a Lagrangian yes. Jul 14 '14 at 11:26

If my hunch is correct this is a Lagrangean optimization. The idea is to maximize function:
$f(x)$
subject to $g_1(x)=0,g_2(x)=0,\dots,g_n(x)=0$.

The trick is to derive the function:
$f(x) - \sum \lambda_i g_i(x)$
And solve for $x$, which involves solving for all the $\lambda$.

The "meaning" of the $\lambda$ depends on the application. In general $\lambda_i$ represents by how much $f(x)$ could have increased if we had relaxed the $g_i(x)$ constraint a little. In economics this translates usually into the concept of "shadow prices". But I tell you now that in many cases $\lambda$ has no meaning besides being there to help find the optimal $x$

• Thanks!. The problem came in this flavor $\text{argmin}_{\beta}(\omega f(\beta)+(1-\omega) g(\beta))$. The objective is to get $\beta$ and secondarily what the relationship between $\beta$ and $\omega$. I tried to transform it into a Lagrangean to see the relationship between $\lambda=\frac{1-\omega}{\omega}$ and $\beta$ from the equivalent minimization problem $\text{argmin}_{\beta}(f(\beta)+\lambda g(\beta))$ Jul 14 '14 at 12:08
• okay, it's not a lagrangean then! It seems like the question is rather straightforward. Find $\beta$ to minimize that weighted average between $f(\beta)$ and $g(\beta$). You don't need to solve or derive by $\omega$, just derive directly for $\beta$ and its solution should be a function of $\omega$ directly. Jul 14 '14 at 12:11
• Thanks a lot for the comment. But $f$ and $g$ are non-linear functions with no explicit solutions. And my interest is in $\beta(\omega)$ because I have derived a working New-Raphson for the minimisation problem for $\beta$. Jul 14 '14 at 12:16
• Oh, it's a numerical optimization! Okay, well, if you can solve it numerically, why don't you just try finding $\beta$ by fixing $\omega$ at various levels? Say $\omega=0,0.1,\dots,0.9,1$? It's a rudimentary grid search but it should give you a glimpse of what $\beta(\omega)$ looks like. Jul 14 '14 at 12:23