# Optimization - Get value of Lagrangian

We know that $f(x) \to \min$ subject to $g(x) = t$ and $h(x) \leq m$ can be written as $f(x) + \lambda g(x)\to\min$ subject to $h(x) \leq m$.

How do we get value of lambda so that the two problems are equivalent.

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first of all, if $t\neq 0$ you have to solve $f(x)+\lambda (t-g(x))\to \min$. Second, the value of the multiplier $\lambda$ you find solving the new optimization problem, you don't know it in advance. – Ilya Sep 28 '11 at 14:53
@Gortaur When you do not have inequality constraint, I understand what you wrote is correct. With the presence of the inequality constraint, we will have to use KKT and I do not know if they will yield the same solution. How can I prove that there exists such a lambda. Also, if all functions are convex, does is mean that inequality can be converted to equality in both problem formulations? Thanks a ton. – Pratik Poddar Sep 29 '11 at 3:37
@PratikPoddar: Don't you mean that you minimize $f(x) + \lambda \|g(x)\|$ or $f(x) + \lambda \|g(x)\|^2$ as in penalty methods. You'll find how to determine the optimal value of $\lambda$ for regular problems in any good optimization book. – Dominique Oct 28 '11 at 20:53

You may formulate the inequality constraint using a log-barrier, as it does not allow $m-h(x)$ to be less than or equal to zero. This is the formulation of the minimization problem :$f(x)+\lambda_{1}(g(x)-t)-\lambda_{2}log(m-h(x))$
Now consider this to be $x^{*}=c(x)$. Now substitute the closed form for $x^{*}$ in the constraint as, $g(x^{*})-t=0$ and solve for $\lambda_{1}$ . Similarly solve for $\lambda_{2}$ by substituting $x^{*}$ in the second constraint.
In a statistical modeling scenario, the $\lambda's$ are estimated by cross-validation. But I am not sure about the domain of your work.