Which two statements about EUI-64 addressing are true? (Choose two.)

Question:
Which two statements about EUI-64 addressing are true? (Choose two.)

• The address includes the hex digits FFFE after the last 24 bits of the interface MAC address.
• The address includes the hex digits FFFE after the first 24 bits of the interface MAC address
• A 64-bit interface identifier is derived from the interface MAC address.
• A locally administrated address has the universal/local bit set to 0.
• A 96-bit interface identifier is derived from the interface MAC address.
• The address includes the hex digits FFFE after the first 14 bits of the interface MAC address.

Explanation: Extended Unique Identifier (EUI) allows a host to assign itself a unique 64-Bit IPv6 interface identifier (EUI-64). This feature is a key benefit over IPv4 as it eliminates the need of manual configuration or DHCP as in the world of IPv4. The IPv6 EUI-64 format address is obtained through the 48-bit MAC address. The MAC address is first separated into two 24-bits, with one being OUI (Organizationally Unique Identifier) and the other being NIC specific. The 16-bit 0xFFFE is then inserted between these two 24-bits for the 64-bit EUI address. IEEE has chosen FFFE as a reserved value which can only appear in EUI-64 generated from the an EUI-48 MAC address.

For example, suppose we have the MAC address of C601.420F.0007. It would be divided into two 24-bit parts, which are “C60142” (OUI) and “0F0007” (NIC). Then “FFFE” is inserted in the middle. Therefore we have the address: C601.42FF.FE0F.0007.

Then, according to the RFC 3513 we need to invert the Universal/Local bit (“U/L” bit) in the 7th position of the first octet. The “u” bit is set to 1 to indicate Universal, and it is set to zero (0) to indicate local scope.

Therefore with the subnet of 2001:DB8:0:1::/64, the full IPv6 address is 2001:DB8:0:1:C601:42FF:FE0F:7/64

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