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I am trying to solve problem 6.3(b) of Boyd & Vandenberghe's Convex Optimization:

Express the following minimization as a QP, LP, SOCP or SDP $$ \min_{\mathbf{x}}\sum\limits_{i=0}^K \phi(\mathbf{a}_i^T \mathbf{x}-\mathbf{b})$$ where $$\phi(u) = \begin{cases} u^2 & |u| \leq M \\ M(2|u|-M) \quad & |u| > M \end{cases}$$


I have no idea on how to start this. Any help would be greatly appreciated.

Karthik Upadhya
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1 Answers1

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One of the reasons we like the Huber penalty is that it is the "Moreau-Yosida regularization" of the absolute value function, which means that \begin{equation*} \phi(y) = \inf_u \quad | u | + \frac{1}{2M} (u - y)^2. \end{equation*}

So, your optimization problem can be written as \begin{equation*} \operatorname*{minimize}_{x} \quad \sum_i \inf_{u_i} \, |u_i | + \frac{1}{2M} (u_i - a_i^T x + b)^2 \end{equation*} which is equivalent to \begin{equation*} \operatorname*{minimize}_{x,u_1,\ldots,u_K} \quad \sum_i |u_i| + \frac{1}{2M} (u_i - a_i^T x + b)^2. \end{equation*} Now you can deal with the absolute value terms using the standard trick of introducing variables $t_i$ that satisfy the constraints $|u_i| \leq t_i$ (or in other words, $u_i \leq t_i$ and $-u_i \leq t_i$).

littleO
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  • Thanks a lot for the advice. I managed to reduce it to the quadratic program $$min \sum\limits_{i} 2M t_i^2 + v_i$$ subject to $$t\geq 0 , \ -t \leq u \leq t ,\ -(u-v)\leq Ax-b \leq(u-v)$$.

    Can you suggest any other approach other than the Moreau-Yosida regularization? I am looking for the following answer.

    $$min \sum_i {u_i^2 + 2M v_i} \ \text{subject to} \ -(u+v) \leq Ax-b \leq u+v \ 0 \leq u \leq M\mathbf{1} \ v \geq 0 $$

    I cant seem to reduce it to this form. Can you please help.

    – Karthik Upadhya Nov 21 '14 at 18:58