Re: How many IP addresses?

[Keith Moore]

> 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.

right.  but for privacy reasons you really don't want to transmit the 
64 bit interface ID around anyway.  so at some point you reclaim
those bits for use in routing.

Keith

Re: How many IP addresses?

[Steven M. Bellovin]

In message , 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

RE: How many IP addresses?

[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
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?

RE: How many IP addresses?

[Tripp Lilley]

On Tue, 25 Apr 2000, Michael B. Bellopede wrote:

> >wrong idea -- big iron routers don't use "code" to do forwarding, they use
> >silicon, and very fast silicon at that.  There are routers in production
> >--Steve Bellovin
> 
> Software, firmware, its a matter of semantics.  Do you think that silicon
> magically performs routing and logical functions.  Hmmm...

Uh, "hardware" != "EEPROM + CPU" here. "hardware" == "custom routing
ASICs". In the "no options, entry in cache" scenario Steve mentions, the
processor "don't enter into it", as it all goes through the raw silicon.

Think gates. Not bill gates, logic gates.

-- 
   Tripp Lilley  *  [EMAIL PROTECTED]  *  http://stargate.sg505.net/~tlilley/
--
  "I get a lot of letters like, 'Dear John, I've got a dead alien. What
   should I do with it?'  One word: barbecue!"

   - John Lovitz, A Yellow Pages commercial

Re: How many IP addresses?

[John Day]

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?

Re: How many IP addresses?

[Steven M. Bellovin]

In message <[EMAIL PROTECTED]>, "Mi
chael B. Bellopede" writes:
>
>
>
>>  But we have to engineer this in some fashion that
>>permits efficient use of these addresses by hosts and (especially) routers.
>>(An earlier poster wrote that you "just have to write the code".  That's
>the
>>wrong idea -- big iron routers don't use "code" to do forwarding, they use
>>silicon, and very fast silicon at that.  There are routers in production
>use
>>today that are handling OC-192 -- 10 Gbps -- links.  You can't do that in
>>software.)
>>--Steve Bellovin
>
>Software, firmware, its a matter of semantics.  Do you think that silicon
>magically performs routing and logical functions.  Hmmm...

Well, yes, magic might help the routing problem, if you have suitable pixie 
dust...  

More seriously, my point is that the techniques, optimizations, and costs are 
very different for software versus hardware.  If you're accustomed to thinking 
in software terms, you're not likely to make the proper design tradeoffs.

--Steve Bellovin

RE: How many IP addresses?

[Michael B. Bellopede]

>  But we have to engineer this in some fashion that
>permits efficient use of these addresses by hosts and (especially) routers.
>(An earlier poster wrote that you "just have to write the code".  That's
the
>wrong idea -- big iron routers don't use "code" to do forwarding, they use
>silicon, and very fast silicon at that.  There are routers in production
use
>today that are handling OC-192 -- 10 Gbps -- links.  You can't do that in
>software.)
>--Steve Bellovin

Software, firmware, its a matter of semantics.  Do you think that silicon
magically performs routing and logical functions.  Hmmm...

-Michael B. Bellopede
 [EMAIL PROTECTED]

Re: How many IP addresses?

[Steven M. Bellovin]

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.

Conclusion:  if we don't do things differently, we have more than enough 
addresses for any conceivable size of the Internet.  (In fact, if we assume 
that there are 300*10^9 stars in the Milky Way, we're not very far from being 
able to handle interstellar networking, at least in terms of the number of 
hosts.  But we may have to redefine the TTL field to let packets last for 50K 
years or so...)

OK -- if 2^128 is that large, why did I (among others) conclude that we needed 
some headroom?  There's a lot of history showing that address space *always* 
runs out, and that the consumption often comes from changing your paradigm.  A 
routing/addressing scheme like 8+8 is an example of what I mean.  In another 
vein, I have a project going that requires 2^16 addresses per host, and maybe 
2^32.  There are undoubtedly other new ideas we can play with *if* we have 
enough address space.  But we have to engineer this in some fashion that 
permits efficient use of these addresses by hosts and (especially) routers.  
(An earlier poster wrote that you "just have to write the code".  That's the 
wrong idea -- big iron routers don't use "code" to do forwarding, they use 
silicon, and very fast silicon at that.  There are routers in production use 
today that are handling OC-192 -- 10 Gbps -- links.  You can't do that in 
software.)

--Steve Bellovin