From: slevin kremera (slevin.kremera@gmail.com)
Date: Wed Aug 15 2007 - 03:04:42 ART
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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