Refer to the exhibit. The IPv6 address for the LAN segment on router R1 must be configured using the EUI-64 format. When configured which Ipv6 address Is produced by the router?
Refer to the exhibit. The IPv6 address for the LAN segment on router R1 must be configured using the EUI-64 format. When configured which Ipv6 address Is produced by the router?
The IPv6 address for the LAN segment on router R1 is produced using the EUI-64 format which involves embedding the interface's MAC address into the IPv6 address. The MAC address for the E0/1 interface is 13-19-be-67-00-01. The EUI-64 process includes flipping the 7th bit of the MAC address. For the MAC address 13:19, this flipping converts the MAC address to 11:19. The full IPv6 address is then formed by inserting FF:FE in the middle of the modified MAC address. Thus, the resultant IPv6 address is 2001:db8:1006:1968:1119:BEFF:FE67:1.
The provided answer is correct. Does anyone know if the test allows the use of the computer calculator?
It’s doesn’t. You just get a pencil and paper for the in person exam.
Damn that’s suck! I was hoping it did for those subnetting questions.
If you do the exam remotely you get access to a digital whiteboard, which functions basically like a textbox you can type in, and a drawing area. I think that's no problem for subnetting; just takes a bit of practice.
It should be 2001:db8:1006:1968:1519-BEFF-FE67-1, 15 instead of 11, right? When you invert the 7th bit, 0 becomes 1, which will add another 2 to 13
The E0/1 interface is the Router interface we are looking at. The first four hex digits of that MAC address is 13-19. Convert the first 2 hex digits into binary: 1 = 0001 3 = 0011 Together: 0001-0011 Convert the 7th bit: 0001-0001 This leaves us with 11 Combine with the other 2 hex digits: 11-19 Thus, it is option B.
One trick I found out about flipping the 7th bit is that it always changes the hexadecimal value by the difference of 2 because the 7 bit is the second bit of the hexadecimal value counting from the right ( i.e.2^¹ )bit. In this case the hexadecimal value holding the 7th bit is 3. This means the 1st and second bits are both "1" or on. Flipping the second bits changes the hexadecimal value to 1. I hope this makes sense?
It should be the answer B with FE69:1 at the end of the address : 2001:db8:1006:1968:1119:BEFF:FE69:1
does anyone knows if we get to use a paper or pen for subletting ? for online exam ?
B is correct
B. 2001:db8:1006:1968:1119:BEFF:FE67:1
must be D
b is correct