The IANA has an Ethernet address block for itself,00:00:5e. They have
allocated half of this block to mulitcast addresses. The range of multicast
addresses are 01:00:5e:00:00:00 to 01:00:5e:7f:ff:ff. With this 23 bits of
the Ethernet address can directly corrispond to the IP multicast group ID.
In this mapping the lower 23 bits of the IP address are directly mapped to
the Ethernet address. This leave 5 bits of IP address that are unused (the
first 5 bits after the initial 1110). Because of this there are 32 multicast
group ID's that can corrispond to a single Ethernet address. At this point
is up to a higher layer to do some filtering to drop unwanted packets. I
hope this helps.
Neil
"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.
>
> We all know that the first 3 bytes of the 48-bit MAC address are indicate
> the vendor. Among the 3 bytes, the first is important, because the first
2-
> bit in this byte has special meanings that are I/G bit and U/L bit.
>
> 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?
>
> Thank you for your help and there may be some understanding errors in the
> questions.
>
> thank you very much!
>
>
>
>
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