Given address 22.214.171.124/27, what is the network and broadcast address?
consider only the 4th octet, where the subnet mask (27bits) boundary occurs.
step1. how many host bits? number of host bits = 5
step2. what is the starting address for the address block containing given address?
possible starting addresses in the 4th octet are: 0, 32, 64, 96, 128, 160,....
Ans: network address: 126.96.36.199 (first address)
broadcast address: 188.8.131.52 (last address)
on the question above I can understand step1 but I'm kind of confused with step2 where it says "possible starting addresses in the 4th octet are: 0, 32, 64, 96, 128, 160,...."
Need help with understanding step2.