Hi I am obstructed by the following problem regarding the integer linear inequality in the following form:
$\mathbf{A}\cdot\mathbf{x}\geq\mathbf{b}$
Where $\mathbf A$ is $m\times n$ integer $(0, \pm1, \pm2, \ldots)$ matrix, $\mathbf b$ is $m$-dimensional integer vector $(0, \pm1, \pm2, \ldots)$, and $\mathbf x$ is $n$-dimensional integer nonnegative vector ($\mathbf x\geq\mathbf 0$) to be determined.
Let $S$ be the set of all integer nonnegative solutions of the above linear inequality, and let $S_{min}$ be the set of minimal elements in $S$. My problem is: what is the maximal cardinality of $S_{min}$? Clearly, if $\mathbf b=\mathbf 0$ then the answer is 1: $S_{min}=\{\mathbf{0}\}$. But what if $\mathbf b\neq \mathbf 0$?
Note that I am not trying to compute $S_{min}$ but just want to find a bound of the size of $S_{min}$.
I feel that this problem might be related to the properties of Hilbert basis, but I did not find any result. Maybe this clue is a wrong lead.
Does any one has some ideas on solving this or any existing results? Even if some approximation is fine. Many thanks.
For example:
$\left[\begin{array}{c}-1&0&0\\0&-1&0\\1&1&-1\end{array}\right]\cdot\mathbf{x}=\left[\begin{array}{c}-2\\-2\\1\end{array}\right]$
This system has six integer solutions among which two are minimal: $\mathbf{x}_1=\left[\begin{array}{c}1\\0\\0\end{array}\right], \mathbf{x}_2=\left[\begin{array}{c}0\\1\\0\end{array}\right]$ So $|S_{min}|=2$