0
$\begingroup$

For any given real numbers such that $\lambda_1\geq\lambda_2\geq\lambda_3\geq\lambda_4\geq\lambda_5\geq\lambda_6$, show that the optimal solution of the problem \begin{align} \mbox{maximize}& -(\lambda_1^+ + \lambda_1^-) + \sqrt{(\lambda_1^+-\lambda_1^-)^2+(\lambda_3^+-\lambda_3^-)^2}\\ \mbox{subject to }& \{\lambda_1^+,\lambda_1^-,\lambda_2^+,\lambda_2^-,\lambda_3^+\lambda_3^-\} = \{\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5,\lambda_6\}\quad\mbox{and}\quad\lambda_i^+\geq \lambda_i^-\quad \mbox{for every}\quad i\in\{1,2,3\} \end{align} is given by \begin{equation} (\lambda_1^+,\lambda_1^-,\lambda_2^+,\lambda_2^-,\lambda_3^+,\lambda_3^-)=(\lambda_4,\lambda_6,\lambda_2,\lambda_3,\lambda_1,\lambda_5) \end{equation}

OBS 1: The proposed solution was obtained numerically by enumerating more than a million instances of sets $\{\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5,\lambda_6\}$ and, for each of them, explicitly evaluating the figure of merit for every possible permutation of the six elements and picking the largest one.

OBS2: Although the result seems to hold for any real values of $\lambda_j$, I am only interested in the case where $\lambda_1\geq\lambda_2\geq\lambda_3\geq\lambda_4\geq\lambda_5\geq\lambda_6\geq 0$ and $\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5+\lambda_6=1$. So, feel free to use these restrictions if that makes the proof any easier.

$\endgroup$
  • $\begingroup$ Can you try to clarify what the pluses and minuses are about? Are you actually dealing with just three numbers and their negations? If so, what's supposed to match up to what? $\endgroup$ – dfeuer Jun 3 '15 at 13:25
  • $\begingroup$ Sorry, it's probably confusing notation that only makes sense within the physical context where the problem arises. Mathematically, these are just independent variables that you could rename as follows: $\lambda_1^+\to x$, $\lambda_1^-\to y$, $\lambda_2^+\to u$, $\lambda_2^-\to v$, $\lambda_3^+\to z$, $\lambda_3^-\to w$. $\endgroup$ – AquilaXi Jun 3 '15 at 14:03
  • $\begingroup$ What do you mean, then, by $\{\lambda_1^\pm,\lambda_2^\pm,\lambda_3^\pm\} = \{\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5,\lambda_6\}$? That these are equal as sets, where $\lambda_k^\pm$ is expanded textually to $\lambda_k^+,\lambda_k^-$? $\endgroup$ – dfeuer Jun 3 '15 at 15:21
  • $\begingroup$ That's right. It's a equality between sets, the elements are the same without respect to the order. I've edited the question for clarifying this point. $\endgroup$ – AquilaXi Jun 3 '15 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.