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Can someone please explain how the combination of upper and lower bounds is made here while isolating x1?
 Is there a rule for making these combination?

in * TF "Emb. Sys. and Rob." by (380 points)

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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.

by (166k points)

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