I am stuck solving the following:
$$ S =\lfloor\lfloor(2*B+I+E) * L / 100 + 5\rfloor * N \rfloor $$ I am trying to isolate $I$ in the equation (like $I = $ {everything else}), but I do not know how to deal with the floors in the equation. Any help and explanation would be much appreciated. Thank you!

  • $\begingroup$ Is everything an integer? $\endgroup$ – K Split X Aug 1 '17 at 16:09
  • $\begingroup$ Yes, everything except N is an integer, but N is always 0.9, 1.0 or 1.1. $\endgroup$ – Kyle Fenole Aug 1 '17 at 16:12
  • $\begingroup$ you won't be able to find something exact, since the floor function is not bijective there is no inverse function $\endgroup$ – Dando18 Aug 1 '17 at 16:14
  • $\begingroup$ Oh, ok, thanks, if all the other variables besides I were known, would any kind of range or estimation be able to be found? $\endgroup$ – Kyle Fenole Aug 1 '17 at 16:21

The floor function doesn't have an inverse, since it's not bijective, as Dando18 pointed out, so you won't be able to invert it directly. At best, you'll be able to get an inequality for $I$, since you can use the fact that $\lfloor x \rfloor \le x$ - so $S \le\lfloor (2*B+I+E)*L/100 + 5\rfloor * N$. If any of your variables can be negative, then be careful with the direction of your inequality.

I get $I \ge (\frac{S}{N} - 5)*\frac{100}{L} - (2B+E)$, provided $L>0$, and you already state $N \in \{0.9,1,1.1\}$. If $L<0$, reverse the inequality. If $L=0$, you can't make $I$ the subject since the whole expression reduces to $S=\lfloor 5N \rfloor$.

  • $\begingroup$ Thank you for explaining everything! Good Day! $\endgroup$ – Kyle Fenole Aug 1 '17 at 16:28

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