# Why are the bounds 3 and 6 instead of 7, in this binary expansion of a slack variable in this QUBO problem?

I've recently started to study how to formulate optimization problems as QUBO models through this paper/tutorial: https://arxiv.org/pdf/1811.11538.pdf

One of the steps is to transform the inequalities of the original optimization problem (in an LP formulation for example) into equalities by adding a slack variable. In the example below (taken from page 17) is quite trivial how the bound was found to be 7, assuming x1 = x2 = 0 and x3 = 1, which is the "worst case" or the "furthest" we can be from 6 in the right hand side.

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On the other hand, in page 25 there's another example which is making me a bit puzzled:

I'm failing to understand why 3 and 6 were chosen as the bounds here.

For example, for the first constraint, why isn't the bound 7? (assuming x1=x2=x3=x4=x5=0)

As for the third constraint, why isn't the bound 11? (assuming x1=x2=x3=x4=x5=1, so we'd have -16 + s <= -5)

Any help will be much appreciated! Maybe there's something obvious which I haven't realized :)

According to the equality constraint (the equal to 4 one), at least two of the $$x_{?}$$ are 1. Therefore, the slack for the 1st constraint is at most 3.
According to the equality constraint, the worst case is $$x_1$$, $$x_4$$, $$x_5$$ are 1 while other xs are 0. The slack is 6 in that case for the 3rd constraint.