This inversion is bijective, so I can't figure out what is it's use.
Best Answer
RFC 4291 provides instructions on how to create the EUI64 address:
Links or Nodes with IEEE 802 48-bit MACs
[EUI64] defines a method to create an IEEE EUI-64 identifier from an
IEEE 48-bit MAC identifier. This is to insert two octets, with
hexadecimal values of 0xFF and 0xFE (see the Note at the end of
appendix), in the middle of the 48-bit MAC (between the company_id
and vendor-supplied id). An example is the 48-bit IEEE MAC with
Global scope:
|0 1|1 3|3 4|
|0 5|6 1|2 7|
+----------------+----------------+----------------+
|cccccc0gcccccccc|ccccccccmmmmmmmm|mmmmmmmmmmmmmmmm|
+----------------+----------------+----------------+
where "c" is the bits of the assigned company_id, "0" is the value of
the universal/local bit to indicate Global scope, "g" is
individual/group bit, and "m" is the bits of the manufacturer-
selected extension identifier. The interface identifier would be of
the form:
|0 1|1 3|3 4|4 6|
|0 5|6 1|2 7|8 3|
+----------------+----------------+----------------+----------------+
|cccccc1gcccccccc|cccccccc11111111|11111110mmmmmmmm|mmmmmmmmmmmmmmmm|
+----------------+----------------+----------------+----------------+
And RFC 2373 provides the 'why' behind flipping the 7th bit:
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.
But that is a bit of a mouthful. So in simpler terms... In MAC address architecture, the 7th bit signifies whether the MAC address was universally or locally assigned. A value of 0 indicates the address is universally administered. For instance, the when IANA assigns an Organizationally Unique Identifier (OUI) to a NIC card vendor, the 7th bit will be 0, indicating the OUI was universally assigned. Should a user manually change their MAC address, this 7th bit would be set to 1, indicating the Ethernet address was locally administered.
There is also some more information about this at PacketLife.
Here's a link to a Hangout with Vint Cerf (Apr. 2014) where he explains how he thought that this internet was supposed to be an experiment only:
As we were thinking about the Internet (thinking well, this is going to be some arbitrary number of networks all interconnected — we don't know how many and we don't know how they'll be connected), but national scale networks we thought "well, maybe there'll be two per country" (because it was expensive: at this point Ethernet had been invented but it wasn't proliferating everywhere, as it did do a few years later).
Then we said "how many countries are there?" (two networks per country, how many networks?) and we didn't have Google to ask, so we guessed at 128 and that would be 2 times 128 is 256 networks (that's 8 bits) and then we said "how many computers will there be on each network?" and we said "how about 16 million?" (that's another 24 bits) so we had a 32-bit address which allowed 4.3 billion terminations — which I thought in 1974/3 was enough to do the experiment!
I had already posted this as a comment to Jens Link's answer, but I felt it shoud surface a bit more.
I think one of the issues is the assumption that it looks exactly like a standard IP Packet. Most phones at this time (using standard GSM or CDMA technology) don't use an IP network but older phone technology.
The phone itself will not know anything about the end device. It will send some basic information to the carriers voice switch. That switch will handle all of the routing using a combination of SIP for calls to other phones on its network, or SS7 for calls that go to the PSTN (Public Switching Telephone Network).
Verizon Wireless, when HD calling is enabled, does do VoIP (SIP) from end to end. When these HD calling phones try to go to a non HD calling phone, they have a device that will transcode the call and do signaling for the PSTN to connect the call.
SS7 does not use IPs, it uses point codes for routing (in the format xxx-xxx-xxx). While it doesn't use IPs, it is still data and not analog. It is transferring bits across the signaling path about how the call is going to be setup. Most devices in this path will only know the next hop.
Best Answer
RFC 4291 provides instructions on how to create the EUI64 address:
And RFC 2373 provides the 'why' behind flipping the 7th bit:
But that is a bit of a mouthful. So in simpler terms... In MAC address architecture, the 7th bit signifies whether the MAC address was universally or locally assigned. A value of 0 indicates the address is universally administered. For instance, the when IANA assigns an Organizationally Unique Identifier (OUI) to a NIC card vendor, the 7th bit will be 0, indicating the OUI was universally assigned. Should a user manually change their MAC address, this 7th bit would be set to 1, indicating the Ethernet address was locally administered.
There is also some more information about this at PacketLife.