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I need help understanding MAC address convention as described here. If I understand it correctly, then MAC address 00-27-C7-38-42-11:

  1. Has 00 as the most significant byte.
  2. Has 0 (zero) set in its least significant bit of the most significant byte -- 00 Hex is 0000 0000 in binary.
  3. And therefore it's universally administered.

While MAC address A9-5E-4C-22-AF-17:

  1. Has A9 as the most significant byte.

  2. Has 1 (one) set in its least significant bit of the most significant byte -- A9 Hex is 1010 1001 in binary.

  3. And therefore it's locally administered.

With that;

  1. Is my understanding correct?

  2. The the above MAC addresses are all 6 bytes and contain no unicast/multicast bit field. Is this correct?

Simon Sheehan
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Larssend
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1 Answers1

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If I understand it correctly, then MAC address 00-27-C7-38-42-11

  1. Has 00 as the most significant byte.
  2. Has 0 (zero) set in its least significant bit of the most significant byte -- 00 Hex is 0000 0000 in binary.
  3. And therefore it's universally administered.

It's the second least significant bit, not the least significant bit. It's still 0. So it is indeed universally administered (globally unique).

While MAC address A9-5E-4C-22-AF-17:

  1. Has A9 as the most significant byte.
  2. Has 1 (one) set in its least significant bit of the most significant byte -- D9 Hex is 1010 1001 in binary.
  3. And therefore it's locally administered.

You mean A9, not D9. Yes, A9 is 101010*0*1 in binary.

Again, it's the second least significant bit. Which is still 0 (I marked it between asterisks above). It's universally administered (globally unique).

The the above MAC addresses are all 6 bytes and contain no unicast/multicast bit field. Is this correct?

They both contain the Unicast/Multicast bit, naturally. This is where the least significant bit of the most significant octet (byte) enters into play.

  • The first address is unicast. The least significant bit of the most significant octet is 0.
  • The second address is multicast. 1010100*1* has 1 as the least significant bit.

Hope this helps.

A Dwarf
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