When we use merit function in optimization & why uses this function?

if we use merit function the space must be convex or not?


1 Answer 1


Merit functions can be applied if you have a constrained optimization problem which simultaneously decreases the cost function and violates one of the constraints.

It allows to keep the iteration going while preserving a measure for the speed of convergence and does not require the cost function or the admissible set to be convex.

A simple example for a merit function associated with an equality-constraints problem is the penalty function

$$ p = f + \sigma \sum\limits_{j=1}^p |g_j|, $$

where $f$ is the cost function, $\sigma$ is some real positive value and $g_j = 0$ denote the equality constraints.

The penalty function then serves as a basis for evaluating the optimal step length $\alpha$, given a search direction $\mathbf{s}$:

$$ \alpha = \arg\min_{\alpha} p(\mathbf{x} + \alpha\mathbf{s}) $$

A merit function is a modified cost function to circumvent poor convergence behaviour in iteration regimes where constraints should be weakend.

  • $\begingroup$ can you give me an simple example for merit function? $\endgroup$
    – user157745
    Mar 5, 2015 at 18:39
  • $\begingroup$ sorry,between $P$ and parenthesis is multiply? and If i want to know extra things about merit function which book or site i must read? $\endgroup$
    – user157745
    Mar 6, 2015 at 5:56
  • $\begingroup$ There's no multiplication operator b/w $p$ and $(\mathbf{x} + \alpha\mathbf{s})$. It's an argument like in $f(x) = x + 1$. If you need to know more, here you'll get a free and comprehensive version of lecture notes on that topic. A standard read is Nocedal/Wright's Numerical Optimization which the lecture notes are mainly based on. $\endgroup$ Mar 6, 2015 at 9:17
  • $\begingroup$ thank you for the address,i download the lecture note but didn't say any things about merit function . this lecture has penalty function & ... . Is the relation b/w $P$ from formula 1 and 2? $\endgroup$
    – user157745
    Mar 6, 2015 at 14:45
  • $\begingroup$ See page 79 in section The minimum of $p$ over $\alpha$ defines $\alpha$. $\endgroup$ Mar 6, 2015 at 15:03

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