10-03-2008 08:38 AM - edited 03-03-2019 11:47 PM
IPv6 prefix is 2001:ABC:123::/64, remaining bits of network address should be made of 1st octet of IPv4 address.
IPv4 address : 10.1.2.3
so my ipv6 address is 2001:ABC:123:10::/64, is that correct ?
10-03-2008 08:49 AM
Hello William,
it is like IPv4 subnetting:
you start from net
2001:ABC:123::/64
you subnet adding one byte= exadecimal of IPv4 byte = 0x0A = 10
the resulting IPv6 subnet is:
2001:ABC:123:0A::/72
you moved the prefix length 8 bits to the right:
72= 64+8
be careful because IPv6 addresses use exadecimal (hex) while IPv4 are written in dotted decimal format so you need to convert from decimal to binary
Hope to help
Giuseppe
10-03-2008 08:54 AM
no, i not concern on the ipv6 subnetting. but my concern is on the question of the remaining network address.
10-03-2008 09:12 AM
Hello William,
I made an error too
the prefix is still /64 and it is built like I explained in previuos post.
2001:ABC:123:000A::/64
the host portion is the EUI-64 conversion of the PC MAC address if the process of duplicate address detection results in no one using the same EUI-64.
EUI-64 is built from the 48 bits MAC address of the NIC
Hope to help
Giuseppe
10-03-2008 11:32 PM
Your IPv6 prefix is 2001:ABC:123::/64, host portion must be made in EUI-64 format, mask on all interface must be /64, remaining bits of network address should be made of 3rd octet of IPv4 address of the interface.
IPv4 Address : 10.1.2.3
So wat is my valid IPv6 address for this question ?
Correct --> 2001:ABC:123:2:/64 eui-64 ?
10-03-2008 11:58 PM
Hello William,
I agree with 2001:ABC:123:2:/64 eui-64
the point here is to remember that each : indicate a 4 hex / 16 bits boundary so when they give you the base address IPv6 you still have 16 bits where one or even two bytes of the IPv4 address can fit.
The /64 shouldn't be passed if you want to support an EUI-64 host portion.
Hope to help
Giuseppe
10-04-2008 02:50 AM
so can i make it as 2001:ABC:123:2:3/64 eui-64 ???
but the question is "remaining bits of network address should be made of 3rd octet of IPv4 address of the interface"
so wat is the right address i should use ?
thanks in advance for the reply ! :-)
10-04-2008 03:06 AM
Hello William,
as I wrote in the previuos post
xxx : yyy : zzz:
every : counts for 16 bits and not for 8 bits.
so they have provided you the first 48 bits of the prefix leaving to be defined the last 16 bits.
To see if you are able to deal with decimal to hexadecimal conversion they ask you to set these last 16 bits to the value of the third octect in the IPv4 address on the same interface.
so the answer is
2001:ABC:123:2::/64
0000 0000 0000 0010 = 0002 = hex 2
if the ipv4 address is 10.3.2.33 and they ask you to make the remaining network bits as the second and the third byte in IPv4 address you have:
write every byte in binary then collect 4 digits and convert them in the corresponding hexadecimal digit:
0000 0011 = 03
0000 0010 = 02
the resulting 16 bits field is in hex:
0302 = 302
so in this case the answer would be:
2001:ABC:123:302::/64
Sorry if I have confused you by observing that two IPv4 bytes could fit in the 16bits to be defined.
But as you see is possible to answer to this question too.
when you write
2001:ABC:123:2:3
the notation means:
2001:0ABC:0123:0002:0003
these are 80 bits and not 64 bits
Hope to help
Giuseppe
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