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Given $A \in \mathbb{R}^{m \times n}$. Is there any way to find a closed form solution for $$ \max_{\sum_{i=1}^n x_i=1, \quad x_i\geq 0} \sum_{j=1}^m\sum_{i=1}^n A_{j i}x_i? $$

The max can be changed to $\min$ as follows: $$ \min_{\sum_{i=1}^n x_i=1, \quad x_i\geq 0} -\sum_{j=1}^m\sum_{i=1}^n A_{j i}x_i? $$ Then one can write the Lagrangian as follows:

$$ L(x, \lambda_0, \lambda)=-\sum_{j=1}^m\sum_{i=1}^n A_{j i} x_i +\lambda_0(\sum_{i=1}^n x_i=1)+\lambda_i(-x_i) $$ where $\lambda_0 \in \mathbb{R}$ and $0 \leq \lambda \in \mathbb{R}^n$. Then stationary and KKT conditions are as follows:

$$ -\sum_{j=1}^m A_{j i}+\lambda_0-\lambda_i=0, $$ and $\lambda_i(x_i)=0$ and $\lambda_i \geq 0$.

Saeed
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1 Answers1

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Are you sure your matrix $A$ is size $n \times m$? From the expression $\sum_{j = 1}^m \sum_{i = 1}^n A_{ji} x_i$, it seems like $A$ is size $m \times n$ instead. For now, I assume you meant $\sum_{j = 1}^m \sum_{i = 1}^n A_{ij} x_i$.

Assuming I interpreted the question correctly, the solution seems to be trivial.

You are maximizing $x^T f$, where $f = A\mathbf{1}$, and $\mathbf{1} = \begin{pmatrix} 1\\ \vdots\\ 1\end{pmatrix} \in \mathbb{R}^m$.

Write $f = \begin{pmatrix} f_1\\ \vdots\\ f_n \end{pmatrix} \in \mathbb{R}^n$. Suppose $f_k$ is the largest among the entries $f_1, \dots, f_n$. Then the optimal value of your problem is $f_k$, achieved by setting $x_k = 1$, and $x_i = 0$ for $i \neq k$.

nowhere
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