From: Scott Morris (swm@emanon.com)
Date: Tue Jul 01 2008 - 18:16:26 ART
Typically we see where we may be asked to convert the 2nd and 3rd octets of
an IPv4 space to be our subnet in IPv6, but you have a little bit too much
pre-defined here. ;)
So breaking down what you are asking for:
1. Subnet is 2001:0123:3333:4444::/64 (that's 64 bits defined, 64 bits left
for host address)
2. Host part converts and uses IPv4's address. Well, the IPv4 address is
only 32-bits, so we'll be looking at something long the lines of
0000:0000:xxxx:xxxx where the x's represent IPv4 address. Remember we're
working in hex now, not dotted decimal! Each octet is 8 bits, or two hex
characters.
168 = 0xa8
100 = 0x64
10 = 0x0a
1 = 0x01
Therefore the IPv6 address completely would be:
2001:0123:3333:4444:0000:0000:a864:0a01/64
HTH,
Scott Morris, CCIE4 #4713, JNCIE-M #153, JNCIS-ER, CISSP, et al.
CCSI/JNCI-M/JNCI-ER
Senior CCIE Instructor
smorris@internetworkexpert.com
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-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
yungli2008@gmail.com
Sent: Tuesday, July 01, 2008 3:55 PM
To: ccielab@groupstudy.com
Subject: ipv6 help
Hi experts
I am from Philippines. This is my first message in groupstudy.
Currently I am working with ipv6.
Router R1's Fa0/0 address is 168.100.10.1. The question is
ipv6 address is 2001:123:3333:4444::/64
host part converts and uses ipv4's address equivalent to the interface.
I couldn't understand this question
Can any expert help me.. I really appreciate you help
Thanks
Li
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