IPv6 - EUI-64 to IPv6 Identifier

Unanswered Question
Nov 7th, 2007

Hi all

I'm studying 8th Module of BSCI curriculum(IPv6)

I'm confused by mistakes in the book concerning the seventh bit in an IPv6 interface identifier (Universal/Local and eight bit (Individual/Group).

I think that:

U/L set to 0, the address is locally administered

U/L set to 1, the address is universally administered

I/G set to 0, the address is an unicast address

I/G set to 1, the address is an multicast address

That is correct?

Thanks

Miquel

I have this problem too.
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bvsnarayana03 Sat, 11/10/2007 - 03:21

U r perfect.

Here is how to remember:

Universally administered / Locally

administered (U/L)

U/L = 0 means U=0 & L=0

So it is universally administered

U/L = 1 means U=0 & L=1

which means locally administered.

Similarly,

Unicast add / group add (U/G)

U/G = 0 means U=0 & G=0

Thus unicast address

U/G = 1 means U=0 & G=1

this is a multicast addres

swmorris Sat, 11/10/2007 - 08:34

Being that each bit is independently modified I think you may be making that more complicated that it needs to be!

Yes, the least significant bit (1-bit position) of the most significant byte of a MAC address is the I/G or Individual/Group address. Keep in mind that this bit as a 1 indicated multicast OR broadcast. (Any group address)

The U/L bit is in the 2-bit position, or the next one to the left. 0 = universal/assigned. 1 = locally modified.

We used this a lot back in the token ring days with functional addresses.

Personally, I never understood the logic of modifying it in an EUI-64 address. Yes, it's derived from a 48-bit MAC, but if you can't figure out that it's been modified since it's now 64-bits long instead of 48, you have issues. :) And the underlying MAC is NOT modified. *shrug*

Designed by academics.

But yes, the underlying bits mean what you noted in the original post here.

HTH,

Scott

[email protected]

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