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 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
>