From: Scott Morris (smorris@ipexpert.com)
Date: Thu Feb 28 2008 - 14:28:40 ARST
Keep in mind that those values are in hex.
So "02" is 0000 0010
that "1" is in the 7th bit position. Technically, in a regular MAC address
this is known as the U/L bit for Universal (if a "0") or Local (if a "1")
address. Basically if you manipulate your own MAC address, you're supposed
to flip that bit.
We saw this a lot in token ring networks with the various functional
(alias?) addresses that controllers could assume.
As for its presence or need in IPv6, I think it's pretty goofy. If you
can't figure out that the local address has been manipulated locally when it
suddenly goes from 48 bits to 64 bits, you have issues. :) But it's the
academically correct thing to do !
(vast collection of useless knowledge)
Scott Morris, CCIE4 (R&S/ISP-Dial/Security/Service Provider) #4713, JNCIE-M
#153, JNCIS-ER, CISSP, et al.
CCSI/JNCI-M/JNCI-ER
VP - Technical Training - IPexpert, Inc.
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-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
abdul muhammed
Sent: Thursday, February 28, 2008 11:13 AM
To: Cisco certification
Subject: need help on ipv6
hi,
what does it mean to invert the 7th bit of the MAC address when calculating
ipv6 address of an interface
i.e
my interface mac address= 0030.947e.e582 when inverted gives
0230.947e.ef82and when we introduce FFFE into the mac address it gives
0230:94FF:FE7E:EF82.
how does inverting the 7th bit produce 0230.947e.e582
thank u
-- Abdul Muhammed Murtala AMerican University of Nigeria Lamido Zubairu way, Yola Adamawa +2348052001153, +2348056201237Network Manager MCSE,MCDBA,MCSA,OCPDBA,CCNA,CCIE Written.
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