From: ccie1101 (ccie1101@gmail.com)
Date: Wed Aug 15 2007 - 04:42:28 ART
Thanks mate .... :)
Cheers,
ccie1101.
On 8/15/07, slevin kremera <slevin.kremera@gmail.com> wrote:
>
>
> here goes,the expansion of address is
>
> 2001:0141:0001:0012:0:0:0:0
> 2001:0141:0001:0025:0:0:0:0
>
>
> now ip ipv6 each digit is hex...if u count u will get /58 becos til
> 2001:0141:0001:00 u get 56 bits same now if u convert 12 and 25 into hex
> they are 00010010
> 0010 0101.Here only the first two bits are same..so 58
>
>
> well even i never knew..i just figured this now..just to answer this
> question...
> well keep those questions coming...who knows u may touch a soft
> spot.........
>
> SK
>
> On 8/15/07, ccie1101 <ccie1101@gmail.com> wrote:
>
> > Hi Experts,
> > How do you summarize between 2001:141:1:12::/64 and
> > 2001:141:1:25::/64
> > ? I don't seem to be able to get the answer 2001:141:1::/58.
> >
> > Is the summarization the same as in ipv4 but of course each octet has
> > a
> > value of 16 ? Coz i am doing it as in ipv4 but it does not come
> > to 2001:141:1::/58 ....
> >
> > Pls advice on the way to do this ...
> >
> > Thank you,
> >
> > Cheers,
> >
> > ccie1101.
> >
> > _______________________________________________________________________
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