In message <CC96542306D7D2119E0B080009EB58FE9582BA@MERCURY>, Timothy Behne writ
es:
>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
>myself :)
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.
--Steve Bellovin
