To upset some more sentiments; compare v4 /24s with the available v4 unicast; do the same with v6 /24s and current v6 unicast space. Rough arithmetic shows then that in that line of reasoning, a v6 /24 is more comparable to a v4 /20. Remco ----- Original Message ----- From: address-policy-wg-admin@ripe.net <address-policy-wg-admin@ripe.net> To: Dmitriy V Menzulskiy <DMenzulskiy@beeline.ru> Cc: michael.dillon@bt.com <michael.dillon@bt.com>; address-policy-wg@ripe.net <address-policy-wg@ripe.net> Sent: Thu Dec 03 14:20:22 2009 Subject: Re: Ha: [address-policy-wg] RE: an arithmetic lesson On Dec 3, 2009, at 7:55 AM, Dmitriy V Menzulskiy wrote:
On 3 Dec 2009, at 10:00, <michael.dillon@bt.com> wrote:
an IPv6 /24 and an IPv4 /24 use up the same percentage of the
total
address space.
How do you work that out? Please enlighten me. 2^24/2^128 x 100 is many orders of magnitude smaller than 2^24/2^32 x 100: gromit% bc scale=50 2^24/2^128*100 .00000000000000000000000000000493038065763132378300 2^24/2^32*100 .39062500000000000000000000000000000000000000000000
There are of course the same number of IPv4 and IPv6 /24s.
Percentage is calculated by dividing the number of things under consideration by the total number of things. When I used the word "an", I meant one thing.
Assuming that the number of IPv4 and IPv6 /24s is 10
1/10 = 1/10
Assuming that the number of IPv4 and IPv6 /24s is 8192
1/8192 = 1/8192
Assuming that the number of IPv4 and IPv6 /24s is 2882873787
1/2882873787 = 1/2882873787
Do you see a pattern forming?
--Michael Dillon
As I understand:
IPv4 /24 is (Total IPv4)/(2^24) IPv6 /24 is (Total IPv6)/(2^24)
Or not ?
Not. The ratio you want, using your formalism, is (2^(size of address space - 24)) / (Total IPvX) which is 2^(N - 24) / 2^N = 1 / 2^24 (where N is the number of bits in the address space). Regards Marshall
WBR,
Dmitry Menzulskiy, DM3740-RIPE
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