I am a researcher in Telecommunication and currently I am running into a difficult optimization with a huge square root.
Minimize $A$ such that the following constraints $C_1$ and $C_2$ are satisfied
$\begin{gathered}
{C_1}:{A_{Min}} \leqslant A \leqslant {A_{Max}} \hfill \\
{C_2}:\frac{{a{A^2}\sin (\theta )\left( {\cos (\theta )\sqrt {\frac{{{B^2}\left( {\frac{1}{{{{\sin }^2}(\theta )}}} \right)}}{{{A^2}}} – 1} + \sin (\theta )} \right) – a{B^2}}}{{{A^2} – {B^2}}} + D \leqslant \frac{{a{C^2}\sin (\theta )\left( {\cos (\theta )\sqrt {\frac{{{B^2}\left( {\frac{1}{{{{\sin }^2}(\theta )}}} \right)}}{{{C^2}}} – 1} + \sin (\theta )} \right) – a{B^2}}}{{{C^2} – {B^2}}} \hfill \\
\end{gathered}$
$\theta$ is an angle that satisfied $0 \leqslant \theta \leqslant \frac{\pi }{2}$
$a,A,B,C$ are all positive
$D$ is strictly negative
$A_{min}$ and $A_{max}$ are the upper bound and lower bound of $A$ respectively.
From the first glance, it seems that there are a lot of symmetry in the left and right hand side of constraint $C_2$ of this problem but I do not know how to exploit it.
The only effort so far is to get rid of the trigonometry function by using the Weierstrass substitution by letting $t = \tan \left( {\frac{\theta }{2}} \right)$ where $t \in \left[ {0,1} \right]$ because $0 \leqslant \theta \leqslant \frac{\pi }{2}$. After that, we have the following:
$\begin{gathered}
\sin (\theta ) = \frac{{2t}}{{1 + {t^2}}} \hfill \\
\cos (\theta ) = \frac{{1 – {t^2}}}{{1 + {t^2}}} \hfill \\
\end{gathered}$
With the Weierstrass substitution, I guess the optimization problem is now purely in polynomial form but nonetheless still posed a major challenge for me. I have some guts feeling that this problem might have something to do with all the square term (i.e quadratic optimization flavor).
Constraint $C_2$ then becomes
$\frac{{a{A^2}(2t)\left( {\frac{{{S_1}\left( {1 – {t^2}} \right)}}{{{t^2} + 1}} + \frac{{2t}}{{{t^2} + 1}}} \right)}}{{\left( {{t^2} + 1} \right)\left( {{A^2} – {B^2}} \right)}} + D \leqslant \frac{{a{C^2}(2t)\left( {\frac{{{S_2}\left( {1 – {t^2}} \right)}}{{{t^2} + 1}} + \frac{{2t}}{{{t^2} + 1}}} \right)}}{{\left( {{t^2} + 1} \right)\left( {{C^2} – {B^2}} \right)}}$
Where $\begin{gathered}
{S_1}^2 = \frac{{{B^2}{{\left( {\frac{{1 + {t^2}}}{{2t}}} \right)}^2}}}{{{A^2}}} \hfill \\
{S_2}^2 = \frac{{{B^2}{{\left( {\frac{{1 + {t^2}}}{{2t}}} \right)}^2}}}{{{C^2}}} \hfill \\
\end{gathered} $
Please help me with this, thank you for your enthusiasm !
Clarification:
1/ Regarding the question of fixed and varied parameter. $A$ is the decision variable that needs to be minimized and the rest of the variable $a,B,C$ are fixed (Physical interpretation: they are some measurements value that can be captured very quickly and accurate so I considered them as fixed value). The answer of this problem should be something in the form $A$ equal to some combination of the other fixed parameters.
Also,${A_{\min }}$ and ${A_{\max }}$ are fixed values depend on the transmission standard and not obtain from measure.
However, in practice the angle $\theta$ will also varied as well albeit very very slowly. Therefore, I plan to solve for the case when $\theta$ is fixed first.
2/ It is not always possible to guarantee ${A^2} – {B^2}$ to be positive but if this assumption make the analysis easier I think it is possible to accept it although it limit the generality of the analysis.
Note that, from my domain knowledge $0<B<1$ is always guarantee.
Also, I think it is fine to assume $C^2 – B^2$ to be positive too.
Best Answer
It seems from your question that all parameters are free to vary. Well, take $\theta = \pi/2$. Then the second constraint becomes $$ a + D \leqslant a $$ which is true since D is strictly negative. So, picking that value for $\theta$, $A$ can be made equal to $A_{min}$.
If, on the other hand, the value of $\theta$ is fixed, then the problem can't be solved without knowing which other parameters are fixed.
It's been clarified that all parameters other that $A$ are fixed. This is a straightforward case to handle numerically. Note that the left and right hand sides of $C_2$ have terms of the same functional form, with $C$ substituted for $A$ on the right: $$ f_{\theta,B}(A) \leq f_{\theta,B}(C) + \frac{\left| D \right|}{a} $$ (the negative value $D$ has been moved to the r.h.s.), where the deomain $f_{\theta,B}(x)$ is restricted to values $$ x^2 \leq B^2 \csc^2 \theta \leq B, x \neq B $$ due to the radical term. Now there are a couple promising strategies:
One edge case is if $\csc \theta$ has its maximum value $1$, in which case the inequality becomes $x^2 < B^2$ due to the denominator. That doesn't end up mattering, though, because this is just the simple case considered above where $C_2$ introduces no constraint, and $so $A is free to take on the value $A_{min}$. (It doesn't even matter if $A_{min} = B$, because this restriction is just a removable singularity in the constraint).
A final note is that pen-and-paper inspection of $f_{\theta,B}$ indicates that the $C_2$ constraint may often be quite slack, unless $\left| D \right|/a$ can become very small. Note again the special case tackled above, where it imposes no constraint on $A$ if $\theta = \pi/2$. Now instead let $\theta$ be very close to $\pi/2$, say $\pi/2-\delta$. Then $\sin \theta = 1 + \mathcal{O}(\delta^2)$ and $\cos \theta = \delta + \mathcal{O}(\delta^2)$, and so to second order in $\delta$, the constraint $C_2$ becomes $$ -\frac{A}{\sqrt{B^2-A^2}} \leq - \frac{C}{\sqrt{B^2-C^2}} + \frac{\left| D \right|}{a\delta} $$ Since $\delta$ is small, the final term is large, and this approximate form of $f$, with domain $x \leq \left| B \right|$ $-x/\sqrt{B^2-x^2}$, diverges for $x$ near $B$. So, in practice this constraint for $\theta$ not too far below $\pi/2$ would tend to just restrict $A$ to not have a magnitude too close to $B$, unless $\left| D \right|/a$ is small enough to make it a tighter constraint.