In message <CC96542306D7D2119E0B080009EB58FE9582BA(_at_)MERCURY>, Timothy Behne
Also, I dont see how you got 25*10^9 * 1000 * 10 = 25*10^9. Should be
25*10^13. This requires log(25*10^13)/log(2), or 48 bits, to use every
address. So the original answer (80 bits left over) was correct, and using
25*10^9 in the calculation must have been a typo. I just had to satisfy
Yes, that was in fact a typo, and my calculations used 10^13.
Anyway, even if we REALLY only have 64 bits, we still have 16 left over, and
given the assumptions we made about the number of addresses each person will
require, it still seems to be more than enough.
As I indicated, the real issue is innovative ways to use the address space.