Hi Daniel,
MAC address is 48 bits long. For Multicast MAC address, the first 24 bits
is 01-00-5e. The next bit is set to 0. That leave you with 23 bits.
48-24-1=23.
In your example, 224.138.8.5
1110 0000 1000 1010 0000 1000 0000 0101
XXXX XXXX X000 1010 0000 1000 0000 0101
X is not used.
000 1010 0000 1000 0000 0101 will give you 0a:08:05.
Put that together with 01:00:5e: will get 01:00:5e:0a:08:05.
Here is another example, 224.127.15.10
1110 0000 0111 1111 0000 1111 0000 0101
XXXX XXXX X111 1111 0000 1111 0000 0101\
111 1111 0000 1111 0000 0101 will give you 7f:08:0a.
Again, put it together with 01:00:5e will get you 01:00:5e:7f:08:0a.
I hope that this helps.
Albert
-----Original Message-----
From: Daniel Boutet [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, August 29, 2000 8:46 AM
To: [EMAIL PROTECTED]
Subject: Multicast adrressing: ip to Mac
I am studying for the switching exam and I am converting ip multicast
address to ethernet addresses.
What i don't get is that they state in the Cisco press book (page 294/295)
"the least 23 least significant bits of the ip multicast group are place
into the frame..."
"half of the ethernet block 01:00:5e:00:00:00 to 01:00:5e:7f:ff:ff
correspond to ip multicast...."
I really only use twenty when I am converting from binary to hex.
Scenario: 224.138.8.5 (to use their example)
1110 0000 1000 1010 0000 1000 0000 0101
01:00:5e:0A:08:05
Since the 01:00:5e:0 are always going to be, then I am only concerned with
the least significant 20. Is this right?
I did their exercise on page 319/320 and got 100% (their is an errata for
on of the address but it is a decimal to binary error) but I did not
use the 01:00:5e:0 as the base but 01:00:5e:
Thanks!
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