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ipv6 eui format,

sarahr202
Level 5
Level 5

Hi everybody

I was reading about how eui- 64 format is used to derive interface id in ipv6.

The book says the eui-format  requires 7 bit of mac address must be set to one.  But why?   I mean if are using burned-in mac address, it is already unique. So  why do we need to flip 7th bit  in eui-64 format ?

thanks and have a good day.

1 Accepted Solution

Accepted Solutions

cadet alain
VIP Alumni
VIP Alumni

Hi,

Cisco decided to flip the 7th bit all the time whether the MAC address from which it was derived had the U/L bit set or not.

Regards.

Alain

Don't forget to rate helpful posts.

View solution in original post

2 Replies 2

sean_evershed
Level 7
Level 7

Have you seen this reference?

http://wiki.nil.com/IPv6_EUI-64_interface_addressing

From RFC 2373, Section 2.5.1 at http://tools.ietf.org/html/rfc2373#section-2.5.1

"The motivation for inverting the "u" bit when forming the interface
   identifier is to make it easy for system administrators to hand
   configure local scope identifiers when hardware tokens are not
   available.  This is expected to be case for serial links, tunnel end-
   points, etc.  The alternative would have been for these to be of the
   form 0200:0:0:1, 0200:0:0:2, etc., instead of the much simpler ::1,
   ::2, etc."

Don't forget to rate all posts that are helpful.

Regards

Sean

cadet alain
VIP Alumni
VIP Alumni

Hi,

Cisco decided to flip the 7th bit all the time whether the MAC address from which it was derived had the U/L bit set or not.

Regards.

Alain

Don't forget to rate helpful posts.
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