Sure, that is explained on the slides 125-130 of the related chapter and demonstrated with some examples afterwards. In the example that you mention, we find for x1 the lower bounds -1.5+x0 and 1.0 + 1.5*x0 and the upper bounds -0.75 + 0.75*x0 and 3+1.5*x0. To make sure that such an x1 can be found, we can demand that all lower bounds must be less than all upper bound, i.e.,
- -1.5+x0 <= -0.75 + 0.75*x0
- 1.0 + 1.5*x0 <= -0.75 + 0.75*x0
- -1.5+x0 <= 3+1.5*x0
- 1.0 + 1.5*x0 <= 3+1.5*x0
which is equivalent to
- 0.25*x0 <= 0.75
- 1.0 + 0.75*x0 <= -0.75
- -1.5 <= 3+0.5*x0
- 1.0 <= 3
and thus equivalent to (the last inequality is clearly true)
- x0 <= 3
- x0 <= -7/3
- -9 <= x0
Hence, we may now choose an arbitrary value for x0 between -9 and -7/3 which can then be used to compute the inequalities for x1 so that we can also choose a suitable value for x1 to come up with a solution.
