Understanding multicast destination mac addresses

Question

I'm currently learning about how to convert between a multicast ip address and the destination mac address. I fully understand the terminology, how to do the conversion, and why the 23 bits, instead of 24, are used. However, what I don't understand is what would be different if 24 bits WERE used to map between the two addresses.

I read on a website that using 24 bits would allow there to be a 1 to 1 relationship between macs and ips. However, I don't understand why. There are 5 bits that don't get mapped (28 unique ip bits - 23 mac bits = 5 bits). This means there will be 2^5 = 32 ip addresses per mac address. So, it seems to me like if the mac was 24 bits, then there would still be a 28-24=4 bit hole and, therefore, 2^4=16 ip's per mac address. Better, but still not a 1 to 1 mapping.

Kind of a complicated question, but hopefully the right person will come along and read this post!

Thank you all in advance for your answers!

Answer 1

The traditional class-D range for multicast is 224.0.0.0/4 - which is 2^28 host addresses. If you mapped this to a 24-bit namespace you would, in fact, end up with a ratio of 2^28 / 2^24 = 2^4, or 16 IP's per MAC instead of the current 32 IP's per MAC.

The reason the additional bit is used is that the most significant bit is actually a flag indicating whether the MAC is multicast. The idea was to leave a full 24 bits for OUI's while still including some kind of signifier for the frame being multicast.

It actually makes a certain amount of sense as a concession given that there used to be a lot of other protocols besides IP that used Ethernet and that these protocols didn't have any kind of universally recognizable upper-layer mechanism for network hardware to pick out whether a given frame was multicast. I suppose it might have been possible to put a multicast flag elsewhere in the header, but that would have caused other compatibility issues.