no, i not concern on the ipv6 subnetting. but my concern is on the question of the remaining network address.

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

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 ?

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

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 ! :-)

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