Answer:
Following subnet mask we are going to used
.
Explanation:
Subnet Mask:255.255.255.224
Its class C Subnet mask hence we are considering last octal of binary representation. (binary representation 11111111.11111111.11111111.11100000)
last octal cosist 3 bits as 1's hence total number of subnet can be form is 2^3=8. and each subnet consist 2^5=32 (32-2=30,Usable host in each subnet)
All 8 of the Possible /27 Networks for 192.168.162.*
Sr. No Network Address Usable Host Range Broadcast Address:
1st 192.168.162.0 192.168.162.1 - 192.168.162.30 192.168.162.31
2nd 192.168.162.32 192.168.162.33 - 192.168.162.62 192.168.162.63
3rd 192.168.162.64 192.168.162.65 - 192.168.162.94 192.168.162.95
4th 192.168.162.96 192.168.162.97 - 192.168.162.126 192.168.162.127
5th 192.168.162.128 192.168.162.129 - 192.168.162.158 192.168.162.159
6th 192.168.162.160 192.168.162.161 - 192.168.162.190 192.168.162.191
7th 192.168.162.192 192.168.162.193 - 192.168.162.222 192.168.162.223
8th 192.168.162.224 192.168.162.225 - 192.168.162.254 192.168.162.255
