# Ha: [address-policy-wg] RE: an arithmetic lesson

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**Marshall Eubanks**tme at multicasttech.com

*Thu Dec 3 15:20:22 CET 2009*

On Dec 3, 2009, at 7:55 AM, Dmitriy V Menzulskiy wrote: >>>>> > On 3 Dec 2009, at 10:00, <michael.dillon at 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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