At 12:41 PM 11/1/00, Jim Erickson wrote:
>"Aaron" <[EMAIL PROTECTED]> wrote in message
>[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > Hi, everyone!
> >
> > I have a question about the MAC layer address, and I use the Ethernet for
> > making an example.
> >
> > I have a question about the following whether it is right:
> > when we get a MAC address, such as 0030.b6f7.3000 (Cisco),
> > 1. Whether the I/G and U/L bit are already set to zero?
> > 2. When a multicast packet shoud be sent to this address, the destination
> > address in the MAC packet header should be set to 0130.b6f7.3000?
>
>Wouldn't a multicast packet be sent to the MAC address of the multicast
>group, as translated from the multicast IP address, not to the individual
>host's MAC address (or an alteration thereof)? If so, the first three bytes
>are 0100.5e, with the remainder determined by the translation of the
>multicast IP address. Then, if it was, say, an all routers multicast, the
>destination MAC would be 0100.5e00.0002
What if it's not IP? &;-) Seriously, you're up a layer from the original
question. He was just trying to understand the MAC-layer multicast bit.
Notice that the multicast bit is set in your example. The first byte is 01
in HEX, or 0000 0001 in binary. The first bit transmitted will be a one.
It's on the right because we're using canonical Ethernet in the example.
For IP, when a station sends a frame to an IP group that is identified with
a Class-D address, the station inserts the right-most 23 bits of the
Class-D address into the right-most 23 bits of the MAC-layer destination
address. The left-most 9 bits of the Class-D address are not used. The
left-most bits of the Mac address are 00000001 00000000 01011110 0. That's
starts with 01:00:5E in HEX.
Priscilla
>Or am I missing something important in the question?
>
>---JRE---
>
>
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Priscilla Oppenheimer
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