Explanation : No. of host = 1500 No. of host bits = [log2 1500] = 11 bits ∴ Total possible hosts = 211 = 23 × 28 n is the netmask bits, Range of addresses is = 232 – n 211 = 232-n Available IP address ⇒ 202.61.0.0/17
So the IP address follow the pattern
0.0 to 7.255
8.0 to 15.255
16.0 to 23.255
:
64.0 to ...
:
104.0 to ...
∴ The possible IP addresses are 202.61.64.0/21 & 202.61.104.0/21