11100000 : 1000 1010 : 0000 0000 : 0000 0001= 224.138.0.1

Nice explanation but a couple of things:

1) This is wrong (you made a typo)
1110 0000 : 1000 1010 : 0000 0000 : 0000 0001 = 224.10.0.1

It should be
1110 0000 : 1000 1010 : 0000 0000 : 0000 0001 = 224.138.0.1

2) (To the readers), the book assumes you went over the CCNA and assumes you know how to do (bin/dec/hex) to (bin/dec/hex) conversions plus you know the meaning of *high order* and *low order* bits. My advice, if you can’t explain these two assumptions to a non tech guy go learn them until you can. There will be more cases when you need to understand these concepts very well (IPv6 comes to mind).

With that assumption the book explanation/example should be sufficient to allow you to draw a similar recipe as Aragoen did. It might force you to take sometime to think about it but that’s exactly the goal of the book in this case.

3) Now why do we reduce a perfectly normal 24 bits space, for the MAC conversion it would make a perfect map to the last 3 IP octets, to a 23bits space and replace the 24th bit with a 0? Doesn’t seem to make much sense does it?

By replacing that 24th bit to 0 we are splitting the 2nd octet into one of two possibilities. Either will be a constant difference from 0 or 128, so it’s easy to do a quick conversion:

e.g.
if the 2nd octet is 18 the “higher twin” octet, that will be mapped to the same MAC, will be 128+18=146
if the 2nd octet is 178 the “lower twin” octet, that will be mapped to the same MAC, will be 178-128=50

4) The “lost” 1st octet makes sense since after applying the OUI in the construction of the MAC address we only have 24bits left, but it’s interesting as the last 4 bits of the OUI map to the first 4 bits of the Multicast IP address (1110)

Hex 53 = 0101 (1110)
IP 224 = (1110) 0000

Thank you for your explanation, now its clear how multicast MAC-to-IP conversion works. There will be no problem on the BSCI exam )

Ciao!

thanks again