A company is planning to open two new branches and allocate the 2a01:c30:16:7009::3800/118 IPv6 network for the region.
Each branch should have the capacity to accommodate a maximum of 200 hosts.
Which two networks should the company use? (Choose two.)
Click on the arrows to vote for the correct answer
A. B. C. D. E.AE.
The given IPv6 network is 2a01:c30:16:7009::3800/118, which means that the network has 118 bits for the network prefix and the remaining 46 bits for the host address.
To accommodate a maximum of 200 hosts per branch, we need to have at least a /25 prefix, which provides 128 addresses. However, we also need to consider the network and broadcast addresses, which reduces the available addresses to 126.
Therefore, we need to allocate two /25 subnets from the given /118 network, with each subnet having a capacity of 126 addresses.
To find the two subnets, we can start by borrowing 7 bits from the host portion of the /118 network, which gives us a /125 network prefix. This creates 128 subnets, each with 122 usable addresses.
Next, we can assign the first subnet to the first branch by taking the first 7 bits of the remaining 39 bits in the network address (after the first 118 bits), which gives us 2a01:0c30:0016:7009::3a00/125.
For the second subnet, we can take the next available 7 bits, which gives us 2a01:0c30:0016:7009::3b00/125.
Now we need to choose two of the given options that match the calculated subnets. Let's go through the options:
A. 2a01:0c30:0016:7009::3a00/120 - This matches the first calculated subnet, so it is a valid option.
B. 2a01:0c30:0016:7009::3b00/121 - This does not match the calculated subnet, as it has a /121 prefix instead of the required /125 prefix.
C. 2a01:0c30:0016:7009::3a80/121 - This does not match the calculated subnet, as it has a different host portion than the first calculated subnet.
D. 2a01:0c30:0016:7009::3c00/120 - This does not match either of the calculated subnets, as it has a different prefix length and host portion.
E. 2a01:0c30:0016:7009::3b00/120 - This matches the second calculated subnet, so it is a valid option.
Therefore, the two valid options are A and E.