It is known that the MAC address of an interface is 0A-04-3B-45-1A-03, then the interface ID generated according to the IEEE EUI-64 specification is: ______.

Question

Options

• A08-04-3B-FF-FE-45
• B
• C
• D

• A
  70% (31)
• B
  11% (5)
• C
  14% (6)
• D
  5% (2)

Explanation

Option A is correct because IEEE EUI-64 transforms a 48-bit MAC address into a 64-bit interface identifier through three steps: (1) split the MAC address in the middle - 0A-04-3B | 45-1A-03, (2) insert FF-FE between the two halves - 0A-04-3B-FF-FE-45-1A-03, and (3) flip the U/L bit (bit 1, the second-least-significant bit of the first octet). Since 0A = 0000 1010 in binary, flipping bit 1 changes it to 0000 1000 = 08, yielding the final result 08-04-3B-FF-FE-45-1A-03. Options B, C, and D are not provided as distractors in this question, so A stands as the only valid answer.

Memory tip: Use the phrase "Split → FFFE → Flip" - split the MAC in half, inject FF-FE in the middle, then flip the U/L bit (bit 1) of the first byte. The U/L flip is the most commonly missed step: if the first octet ends in A (binary ...10), flipping bit 1 drops it to ...00, which is why 0A becomes 08.

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