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

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

• Please use MathJax to typeset your math, instead of linking to an external image. Even so, it is not clear what your question is. Can you be more specific about what you have tried and where you are stuck? May 23, 2019 at 15:06
• @LarrySnyder610 I have edited the problem. I have tried brute force solution but i really don't know how to find the correct optimal value? Can you help me solve these kind of problem , thank you May 23, 2019 at 17:22

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

$$p_2 = p_M - p_1$$

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

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.

• thank you so much! But in my homework, they said : " solve it with brute force solution " . How can this complex function can be solved with brute force ? May 27, 2019 at 12:10
• @HàNguyễn Added a note with the short way to obtain the solution. Remember that it is always necessary to use the shortcuts indicated by theory... May 27, 2019 at 13:05
• Can this be solved with CVX ( matlab tool ) ? Because solving the above set of equations is really hard ? May 28, 2019 at 6:22