Subnet Question Broadcast Valid Host Range - Class A

Example: 1

Calculate Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.0.0 (/16) subnet.

Solution

Class A courtship use a default mask of 255.0.0.0.
Given subnet mask = 255.255.0.0
                            = 255. 11111111. 00000000. 00000000
Subnets = 2^8 - 2 = 254.
Hosts = 2^16 – 2 = 65,534.
Block Size = 2^0 = 1.
Valid subnets = 256 – 255 = 1, that means we have a block size of 1 in the second octet. 
Subnets are 0, 1, 2, 3, … , 255.
Address 0 falls in the 0 subnet.
Subnet = 10. 0. 0. 0

The subnets would be 10.0.0.0, 10.1.0.0, 10.2.0.0, 10.3.0.0, etc., up to 10.255.0.0.

NetworkAssignable HostBroadcast 10.0.0.0

10.0.0.1-10.0.255.254

10.0.255.255

10.1.0.0

10.1.0.1-10.1.255.254

10.1.255.255

10.2.0.0

10.2.0.1-10.2.255.254

10.2.255.255

10.3.0.0

10.3.0.1-10.3.255.254

10.3.255.255

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10.254.0.0

10.254.0.1-10.254.255.254

10.254.255.255

10.255.0.0

10.255.0.1-10.255.255.254

10.255.255.255

The subnet is: 10.0.0.0.
The broadcast consign is:10.0.255.255.
The conclusive host range: 10.0.0.1-10.0.255.254.

Example: 2

Calculate Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.255.192(/26) subnet.

Solution

Class A courtship use a default mask of 255.0.0.0.
Given subnet pretence = 255.255.255.192
                            = 255. 11111111. 11111111. 11000000
Subnets = 2^18 - 2 = 262142.
Hosts = 2^6 – 2 = 62.
Block Size = 2^6 = 64.
Valid subnets = 256 – 192 = 64, that means we have a block bulk of 64 in the forth octet.
Subnets are 0, 64, 128, 192.
Address 0 falls in the 0 subnet.
Subnet = 10. 0. 0. 0

The subnets would have existence 10.0.0.0, 10.0.0.64, 10.0.0.128, 10.0.0.192.

NetworkAssignable HostBroadcast 10.0.0.0

10.0.0.1-10.0.0.62

10.0.0.63

10.0.0.64

10.0.0.65-10.0.0.126

10.0.0.127

10.0.0.128

10.0.0.129-10.0.0.190

10.0.0.191

10.0.0.192

10.0.0.193-10.0.0.254

10.0.0.255

The subnet is: 10.0.0.0.
The broadcast discourse is: 10.0.0.63.
The strong host range: 10.0.0.1-10.0.0.62.

Example: 3

Calculating Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.240.0(/20) subnet.

Solution

Class A courtship use a default mask of 255.0.0.0.
Given subnet pretence = 255.255.240.0
= 255. 11111111. 11110000. 00000000
Subnets = 2^12 - 2 = 4094.
Hosts = 2^12 – 2 = 4094.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, which means we have a block weak glue of 16 in the third octet. 
Subnets are 0, 16, 32, 48, … , 240.
Address 0 falls in the 0 subnet.
Subnet = 10. 0. 0. 0 
The subnets would have existence 10.0.0.0, 10.0.16.0, 10.0.32.0, 10.0.48.0, … , 10.0.240.0.

NetworkAssignable HostBroadcast 10.0.0.0