MAPE (Mean Average Percentage Error):Let $\left\{p_{1},p_{2},p_{3},p_{4}\right\} $ be the numbers $1,-4,4,-5$ respectively. Find the number $x$ that minimizes the Mean Average Percentage Error, \begin{equation*} f\left( x\right) =\frac{1}{4}\sum_{i=1}^{4}\left\vert \frac{x-p_{i}}{p_{i}} \right\vert \end{equation*} by minimizing the sum $\mu _{1}+\mu _{2}+\mu _{3}+\mu _{4}$ subject to \begin{equation*} \left\vert \frac{x-p_{i}}{p_{i}}\right\vert \leq \mu _{i}\quad or\ equivalently\quad \left\vert x-p_{i}\right\vert \leq \left\vert p_{i}\right\vert \mu _{i} \end{equation*} What is the dual program (and its solution) for this problem. Complementary slackness has a natural intepretation for this problem. What is it?
SOLUTION :WHAT I HAVE DONE SO FAR
Minimize $z=\mu_{1}+\mu_{2}+\mu_{3}+\mu_{4}$ subject to \begin{equation*} \left\vert \frac{x-1}{1}\right\vert \leq \mu _{1}\quad ,\\ \left\vert \frac{x+4}{-4}\right\vert \leq \mu _{2}\quad ,\\ \left\vert \frac{x-4}{4}\right\vert \leq \mu _{3}\quad ,\\ \left\vert \frac{x+5}{-5}\right\vert \leq \mu _{4}\quad \end{equation*} Continuing in this way , we have the following constraints
$x-\mu_{1}\leq1$ , $x+\mu_{1}\geq1$
$x-4\mu_{2}\leq -4$ , $x+4\mu_{2}\geq -4$
$x-4\mu_{3}\leq 4$ , $x+4\mu_{3}\geq 4$
$x-5\mu_{4}\leq -5$ , $x+\mu_{4}\geq -5$
However, $x$ is free, so we replace it by $x=u-v$ to obtain
$u-v-\mu_{1}\leq1$ , $u-v+\mu_{1}\geq 1$
$-u+v+4\mu_{2}\geq 4$ , $-u+v-4\mu_{2}\leq 4$
$u-v-4\mu_{3}\leq 4$ , $u-v+4\mu_{3}\geq 4$
$-u+v+5\mu_{4}\geq 5$ , $-u+v-\mu_{4}\leq 5$
WHAT DO I DO NEXT, IS MY HEADACHE ???