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 :)
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.
-----Original Message-----
From: John Day [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, April 25, 2000 10:02 AM
To: Steven M. Bellovin; Graham Klyne
Cc: Richard Shockey; [EMAIL PROTECTED]
Subject: Re: How many IP addresses?
At 9:41 -0400 4/25/00, Steven M. Bellovin wrote:
>In message <[EMAIL PROTECTED]>, Graham
>Klyne wri
>tes:
>>At 11:06 PM 4/23/00 -0500, Richard Shockey wrote:
>>>With "always on" IP and IP on anything this is closer to reality than we
>>>might think. In order to permit a reasonable allocation of addresses with
>>>room for growth the idea of 25 IP address per household and 10 person
>>>actually seems conservative.
>>
>>Following this line of thought, I'd suggest taking the number of
electrical
>>outlets and multiplying by some suitable constant (say, 10, or 1000).
>>
>>And what about all those little wireless-connected gadgets; an IP address
>>for each TV remote-control (where everyone has their own, of course, for
>>personalized access and prioritizing control conflicts...)
>
>I've been in rooms where people have walked through exactly calculation.
>Let's throw a few numbers around.
>
>Assume that the average person in the world has 1000 outlets. That's
>preposterously large for even Bill Gates' house, I suspect, and it doesn't
>even account for dividing by the number of people per house. But let's
stick
>with 1000. Assume that there are 25*10^9 people in the world -- 4x the
>current population. And allocate 10 IP addresses for each of those
outlets.
>That means that we need a minimum of 25*10^9 IP addresses, plus allowances
>for
>delegation on TLA boundaries, smaller provider chunks, homes, etc.
>
>So -- when I divide 2^128 by 25*10^9, I get ~2^80. That's right -- 80 bits
>worth of address space for allocation inefficiencies. If, at each of three
>levels, we really use just one address out of every 2^16, we *still* have
>32 bits left over.
Doesn't this leave out a few pieces of data? Given the current IPv6
address format, which includes a globally unique 64 bit interface ID and 64
bits of globally unique routing goop. My calculation is that you only have
2^64 addresses to work with which leaves roughly 12 bits, maybe 14 to work
with.
But still, should be more than enough, don't you think?
