A CIDR block contains the following subnets with the IP addresses of 192.168.68.0/22 192.168.69.0/22 192.168.70.0/22 192.168.71.

Question

4 years ago

10

0/22
Are there any problems with this group of subnets in the CIDR block?

Why?

Be specific Which subnet(s) are the given addresses in?

1 answer:

Dafna1 [17] · Answer 1 · 2022-01-13T03:16:15+03:00

Answer:

Yes there is a problem.

Explanation:

Yes there is a problem.

Looking at /22, the host part has 10 bits(right most 10 bits). The subnet mask is:

1111111.11111111.11111100.00000000

The first IP address which is 192.168.68.0 in binary form:

11000000.10101000.1000100.00000000

Here all the 10 bits of host part are 0's. Hence it is a valid network IP address.

192.168,68.0 - This is avalid subnet id

The Next IP address 192.168.69.0 in binary form:

11000000.10101000.1000101.00000000

Here all the 10 bits of host part are not 0's. Hence it is not a valid network IP address for /22 cidr.

Moreover this address lies in the first subnet - 192.168.68.0, which has host addresses in the range of 192.168.68.1 to 192.168.71.254

Hence other IP addresses 192.168.69.9/22, 192.168.70.0/22 and 192.168.71.0/22 lies n the subnet 192.168.68.0/22.