# applying KKT to non-convex optimization

I am trying to find a solution for following

\begin{eqnarray} \text{minimize }~~ -\sum_{i=1}^K \frac{1}{a_i+b_i2^{(-2x_i)}} \\ \text{s.t.} ~~~ \sum_{i=1}^Kx_i=C \end{eqnarray}

where $a_i,b_i,x_i\geq 0$.

when I check the convexity it requires $b_i < a_i 2^{(2x_i)}$ or $x_i>log_2(\sqrt{b_i/a_i})$.

applying KKT conditions yield the solution like following:

\begin{eqnarray} x_i^*=\frac 12\cdot log_2\left({\frac{b_i}{\lambda-2a_i-\sqrt{\lambda^2-4\lambda a_i}}}\right) \end{eqnarray} where $\lambda$ is some Lagrange multiplier.

I have two questions:

1- if all $x_i^*$'s satisfy the constraint $x_i^*>log_2(\sqrt{b_i/a_i})$ can I claim that the solution is optimal?

2-what if some of $x_i^*$'s violate the constraint $x_i^*>log_2(\sqrt{b_i/a_i})$? Can I still claim that solution is optimal? if not so what is the optimal solution !

my simulations corroborate that the solution coming from KKT is optimal for all scenarios, but I am confused with the math.

• Did you use Lagrange multipliers for the non-negativity contraints $x_i\ge 0$? Can we rule out that some $x_i=0$? Oct 3 '15 at 0:38
• @SergioParreiras actually I didn't use Lagrange multipliers for $x_i>0$, and rule out $x_i<0$ , does it help? Oct 5 '15 at 15:20