# Formulation of Linear Programming problem?

I want to maximise the function:

$$l(\beta,\sigma,\alpha) = -n\log(\sigma) - \frac{1}{\sigma} A(\alpha)\vert{\bf y}-{\bf X}\beta\vert,$$

where $\vert \cdot \vert$ represents the entry-wise absolute value, $\sigma>0$, $\alpha\in {\mathbb R}$, $\beta \in{\mathbb R}^p$, ${\bf y}\in{\mathbb R}^n$, ${\bf X}$ is an $n\times p$ real matrix, and $A(\alpha)$ is a $1\times n$ vector with positive entries which depends on the parameter $\alpha$. This kind of looks like a Linear Programming problem to me but I can't figure out how to formulate it properly as such. I can optimise this function using basically any numerical software, but it would be nice to link it to Linear Programming. I would appreciate any hints in this direction.

• Two questions: (1) Is the way in which $A(\alpha)$ depends on $\alpha$ known? (2) By "entry-wise absolute value", might you mean the sum of the absolute values? (If not, then I don't understand what is meant?) ${}\qquad{}$ – Michael Hardy Mar 17 '15 at 18:32
• @MichaelHardy Yes, the functional expression of $A(\alpha)$ is known. By $\vert {\bf y} -{\bf X}\beta\vert$ I meant $(\vert y_1-x_1\beta\vert, \dots, \vert y_n-x_n\beta\vert )^{\top}$. So, it is a weighted sum, given by the product of this vector and $A$. – Gillette Mar 17 '15 at 18:35
• It is not a linear program. Far from it, as you have logarithms, products and divisions involving decision variables. – Johan Löfberg Mar 17 '15 at 19:02
• @JohanLöfberg Many thanks for the clarification. I agree with your comment. If you post it as an answer, I will accept it. – Gillette Mar 17 '15 at 19:06

• The values of $\alpha$ and $\beta$ that maximize this should be independent of $\sigma$. So if you find them, you can just plug them in and then find the maximizing value of $\sigma$.
• You have $\ell = -n\log\sigma+\dfrac B \sigma$. So $$\dfrac{d\ell}{d\sigma} = \dfrac{-n}\sigma-\frac B {\sigma^2} = \dfrac{-n\sigma-B}{\sigma^2}.$$ This $=0$ if $\sigma=\dfrac{-B}n$ and $>0$ or $<0$ according as $\sigma<\text{ or }>\text{that}$. So that is the value of $\sigma$ that maximizes $\ell$. (Notice that we must have $B<0$.)