What would be the IPv6 address of a host using SLAAC with 2345::/64 as a network prefix and MAC address of 0300.1111.2222?
A. 2345::100:11FF:FE11:2222
B. 2345:0:0:0:0300:11FF:FE11:2222
C. 2345:0:0:0:FFFE:0300:1111:2222
D. 2345::0300:11FF:FE11:2222
How is the correct answer formulated?
the correct answer is A, because the SLAAC uses network prefix + EUI64
so Network Prefix:2345:: + EUI64 = MAC Address converted to EUI64
How that is work:
MAC Address: 0300.1111.2222
Divided into 2 pieces + Insertion of (FFFE)= 0300.11 FFFE 11.2222
Then the second rule is to flip 7th bit of 1st byte (0 to 1 or 1 to 0), you have to convert the first 2 numbers into binary in order to solve this= (03) binary 0000 0011 flip the 7th = 0000 0001, so the result it would be= 01 instead of 03
so result = 2345::100:11FF:FE11:2222
You can search on EUI64 conversion on youtube.