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## RE: How many IP addresses?

2000-04-25 09:30:06
```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
require, it still seems to be more than enough.

-----Original Message-----
From: John Day [mailto:day(_at_)bbn(_dot_)com]
Sent: Tuesday, April 25, 2000 10:02 AM
To: Steven M. Bellovin; Graham Klyne
Cc: Richard Shockey; ietf(_at_)ietf(_dot_)org
Subject: Re: How many IP addresses?

At 9:41 -0400 4/25/00, Steven M. Bellovin wrote:
```
```In message
<4(_dot_)2(_dot_)2(_dot_)20000425090404(_dot_)00a3c3f0(_at_)pop(_dot_)dial(_dot_)pipex(_dot_)com>,
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).

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?

```
 Current Thread Re: How many IP addresses?, Steven M. Bellovin Re: How many IP addresses?, John Day Re: How many IP addresses?, Keith Moore RE: How many IP addresses?, Michael B. Bellopede RE: How many IP addresses?, Tripp Lilley Re: How many IP addresses?, Steven M. Bellovin RE: How many IP addresses?, Timothy Behne <= Re: How many IP addresses?, Steven M. Bellovin