Doubt regarding Fourier-Motzkin

Question

I have doubt regarding  Fourier-Motzkin algorithm. When I was reading the slides, i have seen the below:

How it's concluded that x1 does not have lower bounds? If the lower bound is negative infinity, then is it applicable for x0 and x2? How x2 does not have upper bound? Could you please clarify these?

Answer 1

If you solve the inequalities for x1, you will see that the left hand sides are minus infinity, so that x1 does not have lower bounds. That is a preferable case since then we can remove all equations where x1 occurs (see slides 85-90).

Comments

I understood the point. But if i try to solve for x0 also i will get the same.so can we take any of the variable from x0 and x1 ?

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Sure, and also that is just an optimization. In general, you an pick any variable for elimination, just to save some work, those without upper/lower bounds are preferable.

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Thank you, Professor.