Question

1. find the network address for 172.22.49.252/17

2. find the last valid assignable host address on the network that host 172.22.4.129 with mask 255.255.255.192 is a part of

3.find the first and last address of the subnet containing 192.167.25.25/16

4.what is the broadcast address of 10.75.96.0/20?

Answer 1

1. The network address for 172.22.49.252/17 is 172.22.0.0/17.

2. The last valid assignable host address of 172.22.4.129/26 is 172.22.4.190.

3. The first and last host address of 192.167.25.25/16 is 192.167.0.1 and 192.167.255.254.

4. The broadcast address of 10.75.96.0/20 is 10.75.111.255

Step-by-step explanation:

Subnetting in networking is the process of managing the use of host addresses and subnet masks of a network IP address. For example, the IP address "172.22.49.252/17" is a class B address that receives an extra bit from the third octet which changes its subnet-mask from "255.255.0.0" to "255.255.128.0". with this, only 32766 IP addresses are used, with the network address of "172.22.0.0/17".