ok ok ok!!! some of you might think that I was in left field. I think I know
where I went wrong.
I am mixing the 48-bit ethernet address with the conversion process. Let me
explain:
I was adding a full "half octet field" 01:00:5e:0 when it is actually just
one bit (0) from the ethernet address. Group of four bits to convert from
binary to hex.
I actually discover this by replying to my own question.
Sorry for using valuable thread!
Should of tried to understand before asking!!!!!!!!!!!!!
""Daniel Boutet"" <[EMAIL PROTECTED]> wrote in message
8ogo5a$l3h$[EMAIL PROTECTED]">news:8ogo5a$l3h$[EMAIL PROTECTED]...
> 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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>
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