I'm implementing on Xpress a problem with different solution proposed on a paper. The idea is to decompose a matrix $X$ into a convex sum $\sum_{t}\lambda_t M^{(t)}$, where each $M^{(t)}$ has only either $0$ or $1$, better stated we have that $M^{(t)} = (m^{(t)}_{ij})_{ij}$ and each $m_{ij}$ is a binary variable. In order to find such decomposition, we consider another continuous variable $\alpha$ and resolve the problem:
$$\begin{align} &\min\; \alpha \tag{1}\\ &\text{s.t.}\\ &\sum_{t}\lambda_t = 1 \tag{2}\\ &\sum_{j} m^{(t)}_{ij}\leq 1 \quad \forall i\;\forall t \tag{3}\\ &\sum_{i}m^{(t)}_{ij}\leq q_{i} \quad \forall j\;\forall t \tag{4}\\ &\sum_{t}\lambda_t M^{(t)} - X \geq 0 \tag{5}\\ &\sum_{t}\lambda_t M^{(t)} - X \leq \alpha\tag{6}\\ &\lambda_t,\alpha\geq 0 \end{align}$$
Equations $(5)$ and $(6)$ are problematic because they are quadratic and would yield a non convex region. We can linearize it by introducing a variable $z^{(t)}_{ij}=\lambda_{t}m^{(t)}_{ij}$ where $Z^{(t)} = (z^{(t)}_{ij})_{ij}$ which ultimately changes the equations to: $$\sum_t Z^{(t)} - X \geq 0\qquad\text{and}\qquad \sum_{t}Z^{(t)} - X \leq\alpha$$
which is now linear but requires the additional constraints:
$$\forall i, j,t\hspace{2mm}: 0\leq z_{ij}^{(t)}\leq \lambda_t \qquad z_{ij}^{(t)}\geq \lambda_t + m_{ij}^{(t)}-1$$
The code runs perfectly with a non trivial output, meaning that the problem is feasible. Nontheless the new solution looses the meaning of the initial implementetion: the non linearized equation would have given as a result, $0-1$ matrixes, which is my aim actually. By linearizing, the solution looses any interpretation (as the $z$'s need now to be continuous variables). What's the best idea to set back now? Is dividing each $Z^{(t)}$ by its $\lambda_t$ and rounding the numbers a meaningful step?