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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.2011-09-28
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    @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.2011-09-29
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    @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.2011-10-28

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