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A CIDR block contains the following subnets with the IP addresses of 192.168.68.0/22 192.168.69.0/22 192.168.70.0/22 192.168.71.0/22

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

Why?

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

2 Answers

1 vote

Final answer:

The student's question about potential issues with a CIDR block of subnets reveals an overlap problem. The four /22 subnets provided share the same address space, which is not permitted in subnetting. A single /22 subnet covers the IP address range indicated by all the subnets listed, which demonstrates the issue of improper configuration and overlap.

Step-by-step explanation:

The student has inquired about potential issues with a group of subnets within a CIDR block. The subnets mentioned are 192.168.68.0/22, 192.168.69.0/22, 192.168.70.0/22, and 192.168.71.0/22. Upon examination, we find that these subnets do indeed have a problem. The notation /22 indicates that the first 22 bits of the IP address are reserved for the network portion, which leaves 10 bits for host addresses. A subnet with a /22 prefix has 2^10, or 1024, possible IP addresses.

When we calculate the address range for the first subnet, 192.168.68.0/22, it spans from 192.168.68.0 to 192.168.71.255, encompassing all addresses within that range. However, the subsequent subnets, such as 192.168.69.0/22, overlap with this range. This means there is an improper configuration as multiple subnets cannot occupy the same IP address space. Therefore, there should only be one /22 subnet block instead of four overlapping ones.

The correct address range for a single /22 subnet starting at 192.168.68.0 would include all subnets and IP addresses from 192.168.68.0 through 192.168.71.255.=

User Yaroslav Gaponov
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4 votes

Answer:

Yes there is a problem.

Step-by-step 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.

User Anol
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