Subnetting Practice

Ask Scott Boyer to check my work.

Given 10.20.97.50/22

Answer:

Class A

255.0.0.0. = default subnet mask

11111111.11111111.11111100.00000000 = /22 subnet mask in binary

255.255.252.0 = /22 subnet mask in dotted decimal

"The Line" falls between the last consecutive 1 and the first 0 in the third octet.

"Network Separation" is the value of the bit immediately to the left of "The Line", in this case 4.

The "Network ID" is the IP address ANDED with the subnet mask.

All octets before "the interesting octet" remain the same.

All octets after "the interesting octet" go to zero.

So we only have to AND the interesting octets.

10.20.97.50 ANDED with
255.255.252.0


The third octet is the where the line is.

So we only have to convert the third octet to binary.

01100001 = 97
11111100 = 252
---------
01100000 = 96

10.20.96.0 = The Network ID
10.20.96.1 = The first usable IP address
10.20.99.254 = The last usable IP address
10.20.99.255 = The broadcast address
10.20.100.0 = The Next Network ID