If an Ethernet port on a router was assigned an IP address of 172.16.112.1/20, what is the
maximum number of hosts allowed on this subnet?
A.
1024
B.
2046
C.
4094
D.
4096
E.
8190
Explanation:
Each octet represents eight bits. The bits, in turn, represent (from left to right): 128, 64, 32 , 16 , 8,
4, 2, 1
Add them up and you get 255. Add one for the all zeros option, and the total is 256.
Now take away one of these for the network address (all zeros) and another for the broadcast
address (all ones). Each octet represents 254 possible hosts. Or 254 possible networks. Unless
you have subnet zero set on your network gear, in which case you could conceivably have 255.
The CIDR addressing format (/20) tells us that 20 bits are used for the network portion, so the
maximum number of networks are 2^20 minus one if you have subnet zero enabled, or minus 2 if
not.
You asked about the number of hosts. That will be 32 minus the number of network bits, minus
two. So calculate it as (2^(32-20))-2, or (2^12)-2 = 4094
I feel like the answer should be 4093, because the question implies that one of the host addresses is already taken and you’d expect a terrible answer like that to be correct for this exam.