I wanted to add that there is also an errata to the figure 8-5 page 294 of
the BCMSN book by Karen Webb
The 48-bit ethernet address should read:
0000 0001 0000 0000 0101 1110 0 which converts to 01:00:5e This leaves
23 bit that can be matched to the Ip Multicast least-significant
23 bits.
""Daniel Boutet"" <[EMAIL PROTECTED]> wrote in message
8ogm9b$bsb$[EMAIL PROTECTED]">news:8ogm9b$bsb$[EMAIL PROTECTED]...
> 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
> one 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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