A Network of 220.127.116.11 / 20 would give you 2^12-2= valid 4094 hosts and 2^(8+4)-2= 4094 networks, thought you could also use the 2 networks we subtracked in the equasion which would give you 4096 networks.
I think you got mixed up with the subnet bits as you counted 4 subnet bitspossibly because you saw the network as a Default Class C network, where as its 8(default class C)+4 which is a total of 12.
To confirm this number all you simply need to do is the following: The /20 means 20 subnetmask bits from a total of 32, therefore, 32-20=12 This is the number you need to put in the power of 2 in order to find out your networks.
Chris i have a small doubt in this...
202.X.X.X is a class c address, by default with which u can get 255 host ids. Now when u give a mask of 20(i.e 255.255.240.0), then does the 3rd octet account for both the host ids and network ids.
First of all let us looking at the first octet in the ip address it is a class C address & so should have 8 host bits meaning 2^8-2 hosts ignoring the first & the last address.
I am not sure if your question as such is based on Class C network or on CIDR (Classless Inter Domain Routing) addresses which are normally written in the /20 or /x format & has no specific classed defined.
In simple language when u say 18.104.22.168/20 means the first 20 bits are for network & the remaining 12 bits are for hosts.
Further analyzing the above address basically the network portion would be defined as 22.214.171.124 network & would have hosts starting with 126.96.36.199 to 188.8.131.52
Please confirm what exactly are u talking abt if not CIDR then we should have a proper class C address with the right subnet mask.