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
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
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
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
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
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 "a
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
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
