Answer:
See the explanation for the answer
Explanation:
Given, network address is: 192.168.0.0/24
The requirement is to create 3 subnets from this network.
3 is near to 4 in powers of 2 => 22
We need 2 bits for subnetting.
So, total number of network bits are: 24 + 2 = 26 bits
Number of host bits are: 32 - 26 = 6 bits [since total number of bits in a address is 32]
Total number of hosts in a subnet: 26 = 64 hosts
Total number of valid hosts: 64 - 2 = 62 hosts [2 => 1 is for network address and 1 is subnet mask]
Subnet Network ID Subnet Mask
1 192.168.0.0/26 255.255.255.192
2 192.168.0.64/26 255.255.255.192
3 192.168.0.128/26 255.255.255.192
4 192.168.0.192/26 255.255.255.192
Subnet Host ID range #of usable host IDs Broadcast ID
1 192.168.0.1 - 62 192.168.0.63
192.168.0.62
2 192.168.0.65 - 62 192.168.0.127
192.168.0.126
3 192.168.0.129 - 62 192.168.0.191
192.168.0.190
4 192.168.0.193 - 62 192.168.0.255
192.168.0.254
