192.168.1.0/22

[ArafatSultan]

How many subnets in 192.168.1.0/22 and why? I am getting confused about what is the number of borrowed bits. Please explain

Thanking in anticipation

 

[Guest]

1 network with 1022 host addresses

192.168.0.1 > 192.168.3.254


Normally 192.168.1.0 would be masked 255.255.255.0 but you have taken 2 bits from the last mask octet making 255.255.252.0 leaving the last 2 bits to be assigned to hosts.

 

[ArafatSultan]

then here how many subnets? 1 subnet (192.168.0.1-192.168.3.254) or 3 subnets? thanks for ur response

 

[Guest]

[------subnet mask /22-----][-----Host-----]
1111111.1111111.11111100.00000000

 

[ArafatSultan]

[------subnet mask /22-----][-----Host-----]
1111111.1111111.11111100.00000000

how many subnets here? thanks

 

[RealBeast]

Homework?

In CIDR notation, /22 = 11111111.11111111.11111100.00000000 so you have 10 host bits, i.e. can have 1024 total hosts (but the gateway and broadcast come out of that so 1022 useable) on one supernet with addresses from 192.168.0.0 - 192.168.3.255.

You are using a major class C address, which has a default mask of 255.255.255.0 which is /24 (i.e. 24 network bits and 8 host bits). You borrowed two bits from the network portion, hence /22, to increase the host number from 2^8 to 2^10.

 

[ArafatSultan]

thanks for ur response......