At 10:04 AM 11/1/00, Aaron wrote:
>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.
That's right. The first two bits transmitted are the Individual/Group bit
and the Universally-administered/Locally-administered bit.
If the I/G bit is a one, than it is a group address, that is, multicast or
broadcast.
If the U/L bit is a one, than it is a locally-administered address.
Locally-administered addresses aren't used much on Ethernet, with the
exception of DECnet which bases the MAC address on the network-layer
area.node ID and prepends AA-OO-04.
Since Ethernet is canonical, the I/G and U/L bits are the
"least-significant" bits, that is, the bits in the 2^0 and 2^1 positions.
>I have a question about the following whether it is right:
>when we get a MAC address, such as 0030.b6f7.3000 (Cisco),
I don't have 0030B6 listed as Cisco in the vendor list that I use, but
maybe my list is out of date.
>1. Whether the I/G and U/L bit are already set to zero?
The are set to zero in your example.
>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?
Yes, that is right. The least-significant bit is set to one. It is the
first bit transmitted. This means that all receivers recognize immediately
that this is a multicast/broadcast frame. As they continue to receive bits,
they recognize that it is a multicast address, since all bits aren't set to
zero, as they would be for broadcast.
>Thank you for your help and there may be some understanding errors in the
>questions.
Sounds like you got it!
Priscilla
>thank you very much!
>
>
>
>
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________________________
Priscilla Oppenheimer
http://www.priscilla.com
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