Assume you have an under-determined linear system
$AX=B$
where you have more variables than constraints. It is also known that $X>0$ (element-wise). How can you determine the (scalar) lowest upper bound, $L$, such that $0 \le X \le L$ (element-wise) that will still guarantee a solution?
Example:
$A= \left( \begin{array}{ccc} 5 & 9 & 8 & 5 & 6 \\ 6 & 2 & 7 & 1 & 6 \\ 4 & 8 & 2 & 3 & 4 \end{array} \right) $
$B=[187, 154, 109]'$
Seems to have a lowest upper bound of around 7.6332, with solution
$X=[7.6332, 2.8794, 7.6332, 3.2111, 7.6332]'$
You can compute the LQ factorization of $A$, which will have the form $$ A = \begin{bmatrix}L&0\end{bmatrix}\begin{bmatrix}Q_1 \\ Q_2\end{bmatrix} $$ where $L$ is a square lower triangular matrix. In the following I will assume $L$ is full rank (if it has zeros on its diagonals at the bottom, the $Q_2$ needs to include the corresponding rows as well).
The columns of $Q_2^T$ span the nullspace of $A$, and $X_0 = Q_1^T L^{-1}B$ is the minimum norm solution. The set of all solutions is given by $X = X_0 + Q_2^T c$ for some vector $c$. You then have to solve the optimization problem $$ \min\; L \\ \text{subject to: } x_0 + Q_2^T c \le L \mathbf{1} $$ where $\mathbf{1}$ is the vector of all $1$'s. Which can be written as a linear program $$ \min\;e_1^T [L, c]^T \\ \text{subject to: } [-\mathbf{1}, Q_2^T][L, c]^T \le -x_0 $$ where $e_1$ is the vector of all zeros except a $1$ in the first entry. This is now in the form of a standard linear program.