Question

One of the addresses in a block is 165.199.170.83/27. find the number of addresses, the first address, last address in the block, network address, and the broadcast address

Answer 1

Given IP address : 165.199.170.83

Network Mask : 255.255.255.224

Convert IP address in Binary Format :

165.199.170.83

10100101.11000111.10101010.01010011

Convert Network Mask in Binary Format :

255.255.255.224

11111111.11111111.11111111.11100000

Network Address (First Address) =
(IP Address) Logical AND (Network Mask)

10100101.11000111.10101010.01010011
+ 11111111.11111111.11111
111.11100000
----------––-----------------–––––---------–----–
10100101.11000111.10101010.01000000

i.e. 165.199.170.64

Network Broadcast Address (Last Address) =
(IP Address) Logical OR [Logical NOT(Network Mask)]

10100101.11000111.10101010.01010011
- 00000000.00000000.00000000.00011111
----------–-----–-------–------–-------–-----–-----–-------
10100101.11000111.10101010.01011111

i.e. 165.199.170.95

n=27 (Given)

Total no. of Hosts bits = 32-n = 32-27 = 5

Number of Hosts (N) = 2^(32-n) = 2^5 = 32

First Host Address = Network Address + 1
= 165.199.170.64 + 1
= 165.199.170.65

Last Host Address
= Network Broadcast Address - 1
= 165.199.170.95 - 1
= 165.199.170.94