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Cisco CCNA (200-301)Network FundamentalsHard

A host has the MAC address 00:1A:2B:3C:4D:5E and receives the IPv6 prefix 2001:DB8:ACAD:1::/64 via SLAAC. Using the EUI-64 process, what is the resulting IPv6 address?

  1. A2001:DB8:ACAD:1::21A:2BFF:FE3C:4D5E
  2. B2001:DB8:ACAD:1::21A:2BFE:FF3C:4D5E
  3. C2001:DB8:ACAD:1::1A:2BFF:FE3C:4D5E
  4. D2001:DB8:ACAD:1::1A2B:3CFF:FE4D:5E00
Show answer & explanation

Correct answer: A. 2001:DB8:ACAD:1::21A:2BFF:FE3C:4D5E

EUI-64 splits the 48-bit MAC into two 24-bit halves (001A2B and 3C4D5E), inserts FFFE in the middle, and flips the 7th bit (universal/local bit) of the first byte. 00 (00000000) becomes 02 (00000010), yielding the interface ID 021A:2BFF:FE3C:4D5E, which with the leading zero dropped displays as 21A:2BFF:FE3C:4D5E appended to the prefix.

Why the other options are wrong

  • B. Reverses the byte order of the inserted FFFE value.
  • C. Fails to flip the universal/local bit, leaving the original 00 as 1A.
  • D. Incorrectly splits the MAC address at the wrong boundary.

IPv6 EUI-64

EUI-64 generates a 64-bit interface ID from a 48-bit MAC address by splitting it, inserting FFFE in the middle, and flipping the 7th bit of the first byte.

  • Split MAC into two 24-bit halves
  • Insert FFFE between the halves
  • Flip the universal/local (U/L) bit of the first byte

Memory trick: Split, stuff with FFFE, then flip the switch bit.

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