From: Joseph Brunner (joe@affirmedsystems.com)
Date: Thu Aug 30 2007 - 16:47:50 ART
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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