I have a non-convex optimize question that have to solve it with multiple way?

Question

$$maximize\text{ }: \frac{1}{2}log_2\Biggl(1+\frac{P_1H_1^2P_2H_2^2}{P_1H_1^2+P_2H_2^2+1}\Biggr)$$ $$ s.t\text{ }: P_1 + P_2 \le P_{max}\text{ ; } P_1,P_2\ge0 $$ For $$ H_1^2 = 30 , H_2^2 = 15, P_{max} = 10 $$ Solve the problem by any method possible? Thank you very much.

Answer 1

As $\log$ is a monotonic increasing function and we need a maximization, we will consider instead the simpler problem

$$ \max_{p_1,p_2}\frac{p_1p_2h_1^2h_2^2}{1+p_1h_1^2+p_2h_2^2}\ \ \mbox{s. t. }p_1+p_2\le p_M,\ \ p_1, p_2 \ge 0 $$

Forming the Lagrangian with help of some slack variables $\epsilon_i$

$$ L = \frac{p_1p_2h_1^2h_2^2}{1+p_1h_1^2+p_2h_2^2}+\lambda(p_1+p_2-p_M+\epsilon_0^2)+\mu_1(p_1-\epsilon_1^2)+\mu_2(p_2-\epsilon_2^2) $$

now the stationary conditions dictate

$$ \nabla L = 0 = \left\{ \begin{array}{l} -\frac{h_2^2 p_1 p_2 h_1^4}{\left(p_1 h_1^2+h_2^2 p_2+1\right){}^2}+\frac{h_2^2 p_2 h_1^2}{p_1 h_1^2+h_2^2 p_2+1}+\lambda +\mu _1 \\ -\frac{h_1^2 p_1 p_2 h_2^4}{\left(p_1 h_1^2+h_2^2 p_2+1\right){}^2}+\frac{h_1^2 p_1 h_2^2}{p_1 h_1^2+h_2^2 p_2+1}+\lambda +\mu _2 \\ \epsilon _0^2-p_M+p_1+p_2 \\ p_1-\epsilon _1^2 \\ p_2-\epsilon _2^2 \\ 2 \lambda \epsilon _0 \\ -2 \epsilon _1 \mu _1 \\ -2 \epsilon _2 \mu _2 \\ \end{array} \right. $$

after solving we get at the solution

$$ \left\{ \begin{array}{rcl} p_1 & = & 4.14616 \\ p_2 & = & 5.85384 \\ \lambda & = & -5.14707 \\ \mu_1 & = & 0. \\ \mu_2 & = & 0. \\ \epsilon_0 & = & 0. \\ \epsilon_1 & = & \pm 2.03621 \\ \epsilon_2 & = & \pm 2.41947 \\ \end{array} \right. $$

$\epsilon_0 = 0$ indicates that the restriction $p_1+p_2 = p_M$

The found solution gives

$$ \frac 12\log_2\left(1+\frac{p_1p_2h_1^2h_2^2}{1+p_1h_1^2+p_2h_2^2}\right) = 2.85341 $$

NOTE

Keeping in mind that $\log(\cdot)$ is a monotonic increasing function and we need a maximization, we can assume that the solution is at the restriction border or at $p_1+p_2 = p_M$ so our problem resumes to

$$ \max_{p_1,p_2}\frac{p_1p_2h_1^2h_2^2}{1+p_1h_1^2+p_2h_2^2}\ \ \mbox{s. t. }p_1+p_2 = p_M $$

now making the substitution

$$ p_2 = p_M - p_1 $$

we follow with

$$ \max_{p_1}\frac{p_1(p_M-p_1)}{1+p_1h_1^2+(p_M-p_1)h_2^2} $$

and deriving we get the condition

$$ p_1^2 \left(h_2^2-h_1^2\right)-2 p_1 \left(h_2^2 p_M+1\right)+h_2^2p_M^2+p_M=0 $$

then finally

$$ p_1 = \frac{p_M \left(h_2^2 p_M+1\right)}{\sqrt{\left(h_1^2 p_M+1\right) \left(h_2^2 p_M+1\right)}+h_2^2 p_M+1} = 4.14616 $$

as expected.