I want to show using mixed integer rounding, that
$\frac{( y_1 + 6y_2)}{4} + y_3 + 4y_4 \geq 16$
is a valid inequality for
$X = \{y \in \mathbb{Z}_{+}^{4}: y_1 + 6y_2 + 12y_3 + 48y_4 \geq 184 \}$.
If I divided it by 12 and multiply by -1 I get:
$-\frac{1}{12}y_1 - \frac{1}{2}y_2 - y_3 - 4y_4 \leq -\frac{46}{3} $ or $a_iy_1 + a_2y_2 + a_3y_3 + a_4y_4 \leq b $
now if I use the formula
$$\sum_{f_j\leq f} \left \lfloor{a_j}\right \rfloor y_j ~+~\sum_{f_j > f}( \left \lfloor{a_j}\right \rfloor + \frac{f_j - f}{1 - f})y_j \leq \left \lfloor{b}\right \rfloor$$
where $f =b - \left \lfloor{b}\right \rfloor $ and $f_j = a_j - \left \lfloor{a_j}\right \rfloor $, I will get
$f = \frac{2}{3}, \,f_1 = \frac{11}{12}, \, f_2 = \frac{1}{2}$ and $f_3 = f_4 = 0 $, so my final inequality would be:
$\frac{1}{4}y_1 + y_2 + y_3 + 4y_4 \geq 16 $
which is not what I want. Can anybody help? Thanks!