Computer Science - Networking - Discussion

Thanks @Rohit Singh.

Total 16 bits available for representing subnets and hosts since Class B address has 2 octets available.
No. of hosts = 600.
No. of bits needed to represent 600 hosts = round( log(600) to the base (2) ) = 10.
Hence, no. of bits available to represent subnets = 16 - 10 = 6.
From the right end, first host bits are taken then subnet bits.
The subnet mask is when all subnet bits are set to 1 and remaining host bits are set to 0 (in the available octets).

Hence, subnet mask = 11111100.00000000 = 252.0 (Remaining portion will remain the same, i.e., 255.255).

Final mask = 255.255.252.0.

No. of hosts=2^n-2.

So, 2^n=602, Approx. n=10.

The subnet mask will be:
11111111.11111111.11111100.00000000 (255.255.252.0).

Ans: Its a class B IP.
No. of host= 2^n-1=600 (where n=numbers of zeros)
So, 2^n=601. Approx. n=10

the subnet mask will be:
11111111.11111111.11111100.00000000 (255.255.252.0)