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Given address 188.8.131.52/27, 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: 184.108.40.206 (first address)
broadcast address: 220.127.116.11 (last address)
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