1

How to solve the follow optimization problem?

$$\begin{array}{ll} \text{minimize} &\displaystyle\sum_{i=1}^{T} \| \mathbf{A}_i\mathbf{x} - \mathbf{b}_i \|^2\\ \text{subject to} & \mathbb{1}^\top {\mathbf{x}} = 1\\ & \mathbf{x} \geq 0\end{array}$$

where $\mathbf{x} \in \mathbb R^{n}$ is a vector and $\mathbf{b}_i \in \mathbb R^{m}$ are a sequence of vectors and $\mathbf{A}_i \in \mathbb R^{m \times n}$ is a sequence of matrices. $T$ is the number of equation. The constraints mean that all element of vector $\mathbf{x}$ are non-negative and that the sum of its elements is equal to $1$.

A simple reduction is solve in How to optimize $\|Ax - b\|^2$ subject t0 $x1 = 1$, $x\geq 0$

This time, my goal is to jointly optimize those $T$ equation.

My second question is the following. How to show constraint $\mathbf{x} $ is in compact convex set?

Rodrigo de Azevedo
  • 1
jason
  • 829
  • What do you mean by $\boldsymbol{A}_i$ and $\boldsymbol{x}1$? Your notation is not clear. – MrYouMath May 10 '18 at 13:35
  • @MrYouMath, I modify it. $x1$ means that sum of all element in $x$. – jason May 10 '18 at 13:38
  • The sum of all the elements of $\boldsymbol{x}$ or the sum of the absolute values of all the elements of $\boldsymbol{x}$? – MrYouMath May 10 '18 at 13:40
  • @MrYouMath since all element of $x$ is nonnegative. so they are the same. Thanks – jason May 10 '18 at 13:43
  • No, they are not the same. they become the same because of the constraint. Please reformulate your question. It is very unlikely that people will invest time in answering your question when you do not take some time to formulate it well. – MrYouMath May 10 '18 at 13:46
  • @MrYouMath Harsh? – measure_theory May 10 '18 at 13:50
  • @measure_theory It is not harsh. The OPs question is still not readable. The OP uses non-standard notation without explaining it. A question should be self-explanatory and at the moment it is not self-explanatory. – MrYouMath May 10 '18 at 13:53
  • 1
    @MrYouMath It just came off overly rude. There's a difference between providing constructive comments that help the OP get their question answered and trying to make them feel stupid. Seems to me like he put a good amount of effort into thinking about the question before posting, which is much more than we can say for a lot of other posts. – measure_theory May 10 '18 at 15:20

2 Answers2

4

$||z||_2^2 + ||w||_2^2$ is equal to $\left|\left|\begin{matrix}z\\w\end{matrix}\right|\right|_2^2$ so you already have the answer as all you have to do is to stack your expressions (i.e. create a large $A$ and $b$ matrix from the given ones)

Johan Löfberg
  • 9,497
  • 1
  • 15
  • 15
1

I assume $\| \cdot \|$ is the $\ell_2$-norm. Let $$ A = \begin{bmatrix} A_1 \\ A_2 \\ \vdots \\A_T \end{bmatrix}, \qquad b = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_T \end{bmatrix}. $$ Then $$ \sum_{i=1}^T \| A_i x - b_i \|^2 = \| Ax - b \|^2. $$ So you can use the solution in the question you linked to.

littleO
  • 51,938