DRAG DROP
You are configuring a multi-subnet IPv6 network for a regional office.
The corporate network administrator allocates the 2001:0db8:1234:0800: :/54 address space for
your use.
You need to identify network IDs of the first and last subnets that you will be able to create at the
office.
Which network IDs should you identify?
To answer, drag the appropriate network IDs to the correct subnets. Each network ID may be used
once, more than once, or not at all. You may need to drag the split bar between panes or scroll to
view content.
Answer: 
why
0
0
This is not correct. We’re not even using the 64 subnet. a+e is correct
0
0
“administrator allocates the 2001:0db8:1234:0800: :/54 address space for
your use.” From there you have to subnet which is where the /64 comes from. Your first network will always be what is assigned. Thus 2001:0db8:1234:0800: :/64 is correct. Now my subnetting of IPv6 sucks but 0bff is the highest listed, so I’m going with the answers.
0
0
The network part is always the first half of an IPv6 (64 bits), and the hosts in the second half (64 bits). If we’re assigned a /54 address space, that means we don’t touch at the first 54 bits, so we have 10 bits left for our subnets.
Each IPv6 hex digit in coded on 4 bits, each IPv6 pack of 4 digits is 16 bits:
2001:0db8:1234: is 3 x 16 bits=48 (6 bits to go)
The 0800 hex part is 0000.1000.0000.0000 in binary, so the 6 bits to complete the /54 mask is 0000.10.
We have 10 bits for our subnet, starting after the 54 address space at 00 (so, at 1000 in binary, or 8 in hex).
In hex + [binary], address space is: 2001:0db8:1234:[0000.10]
After is subnet: starting at [00.00000000], ending at [11.11111111]
To convert back to hex, we regroup the first 2 bits of the subnet with the last 2 bits of the address space (that don’t change):
…:[0000.10][00.00000000] – …:[0000.10][11.11111111]
1000 binary is 8 hex, 1011 is B, 1111 is F, so: 0800 – 0BFF
0
0
what is the correct answer?
0
0
First things first… get to know your Hex to binary conversions..
A (10) = 1010
B (11) = 1011
C (12) = 1100
D (13) = 1101
E (14) = 1110
F (15) = 1111
Now as for the answer:
2001:0DB8:1234:0800:: /64 is first subnet
2001:0DB8:1234:0BFF:: /64 is last subnet
We only care about the last 4 digits in this case
when you are given a /54, count 54 bits from bit 1 in the 64-bit string they give you
Write 2001:0DB8:1234:0800 out in decimal:
0010.0000.0000.0001:0000.1101.1011.1000:0001.0010.0011.0100:0000.10|00.0000.0000:: 64 bits after this for host addreses
Where the “|” is, is where we are given our /54 network space (54 bits). We cannot touch those 54 bits. But now we have 10 more bits to play with since we are allowed the first 64 bits in an IPv6 for subnets. Hence the /64 in the answer.
If you fill all of the rest of the 10 network bits with 1s to get the last subnet [0000.10|11.1111.1111] you will get 0BFF in hex. This is the LAST subnet.
The first subnet is already in the question because all of the bits are turned off (subnet 0) [0000,10|00,0000,0000] 0800 in hex.
I know its confusing… write it out on paper…
1
0
very well explained
0
0