From: Scott Morris (swm@emanon.com)
Date: Tue Aug 15 2006 - 21:52:40 ART
The 7th (second to least significant) bit in the most significant
(left-most) byte.
The 8th bit (least-significant, right-most) bit in here is that I/G one that
tells us individual (unicast) or group (multicast/broadcast)... The one
just left of that (7th bit) tells us U/L information or Universal (burned in
address) or Local (changed locally) assignment. That makes the first byte
02 instead of 00.
0000 0010 in binary in case you care! :)
HTH,
Scott Morris, CCIE4 (R&S/ISP-Dial/Security/Service Provider) #4713, JNCIE
#153, CISSP, et al.
CCSI/JNCI-M/JNCI-J
IPExpert VP - Curriculum Development
IPExpert Sr. Technical Instructor
smorris@ipexpert.com
http://www.ipexpert.com
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
uyota oyearone
Sent: Tuesday, August 15, 2006 6:45 PM
To: ccielab@groupstudy.com
Subject: EUI -64 to MAC
hey guys, am trying to understand a scenerio, and would appreciate a better
explanation. task:--------------Conf. IPv6 addre.
2001:192:10:1:1234:56ff:fe:78:9abc on r1 fa0/0. The host portion be gotten
from eui-64.solution:---------------r1#int fa0/0mac-addre
1034.5678.9abcipv6 add 2001:192:10:1::/64 eui-64 their explanation was :
Invert the 7most significant bit, then remove ff:fe, this equals
1034.5678.9abc. Understandable, but the issue i have is: :when they say
invert the 7most significant bit, do this mean the 7 most significant bit in
the first octect?? regards, Uyota
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