From: Narbik Kocharians (narbikk@gmail.com)
Date: Thu Aug 30 2007 - 18:54:47 ART
YES
On 8/30/07, Usankin, Andrew <Andrew.Usankin@twtelecom.com> wrote:
>
> :) In the light of Joseph's mistake I have a question. Can we use
> windows calculator in the lab?
>
> 12(hex) = 00010010
> 25(hex) = 00110101
>
> Andrew
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
> Joseph Brunner
> Sent: Thursday, August 30, 2007 1:48 PM
> To: 'Sadiq Yakasai'; 'Cisco certification'
> Subject: RE: IPV6 Address Summarization
>
> The first 48 bits wont change, agreed?
> 2001:141:1
> 2001:141:1
>
> Lets look at the 16 bits of the 128 where have a different value...
>
> 12 in hex is shorted from 0012
> 25 in hex is shorted from 0025
>
> Each XX equals 8 bits
>
> So
>
> 00000000
> 00000000
>
> That means we already have a /56 up to midway through the 4th 16 bit
> block...
>
> Now convert each nibble to binary and pad up to /64...
>
> 1 2
> 00000110
>
> 2 5
> 00010101
>
> 00000000000 (11 zero's of similarity in the 4th 16th bit block)
>
> 11 + 48 = 59
>
> You are correct it's a /59
>
> Who told you other wise?
>
> -Joe
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
> Sadiq Yakasai
> Sent: Thursday, August 30, 2007 2:56 PM
> To: Cisco certification
> Subject: IPV6 Address Summarization
>
> Hi Guys,
>
> Please could someone help out a guy here:
>
> I need to summarize these two IPV6 addresses:
> 2001:141:1:12::/64
> 2001:141:1:25::/64
>
> I have done it many times over, and what I find to be the summarized
> address
> is:
> 2001:141:1::/59
>
> However, an excercise I am doing here says is
> 2001:141:1::/58
>
> Please can someone confirm if I am right or wrong here?
>
> Thanks!!
>
> Sadiq
>
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-- Narbik Kocharians CCIE# 12410 (R&S, SP, Security) CCSI# 30832 www.Net-WorkBooks.com
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