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exis [7]
3 years ago
10

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?
Computers and Technology
1 answer:
Dafna1 [17]3 years ago
4 0

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.

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Explanation:

The class was created called Solution. The second line define the main function. Inside the main function; we initialized and assign an array called allNames to hold the list of all name. Then a String called searchString was also defined. The searchString is the string to search for in each element of allNames array. The two variables (allNames and searchString) are passed as argument to the findNames method when it is called.

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