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• Valid subnets? 256 – 255 = 1. 0, 1, 2, 3, etc. (all in the second octet). 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.

• Valid hosts?

The following table shows the first two and last two subnets, valid host range, and broadcast addresses for the private Class A 10.0.0.0 network:

Subnet

10.0.0.0

10.1.0.0

10.254.0.0

10.255.0.0

First host

10.0.0.1

10.1.0.1

10.254.0.1

10.255.0.1

Last host

10.0.255.254

10.1.255.254

10.254.255.254

10.255.255.254

10.0.255.255

10.1.255.255

10.254.255.255

10.255.255.255

Subnet

10.0.0.0

10.0.16.0

10.0.32.0

10.0.240.0

First host

10.0.0.1

10.0.16.1

10.0.32.1

10.0.240.1

Last host

10.0.15.254

10.0.31.254

10.0.47.254

10.0.255.254

10.0.15.255

10.0.31.255

10.0.47.255

10.0.255.255

Practice Example #2A: 255.255.240.0 (/20) 255.255.240.0 gives us 12 bits of subnetting and leaves us 12 bits for host addressing.

• Subnets? 212 = 4096.

• Hosts? 212 – 2 = 4094.

• Valid subnets? 256 – 240 = 16. The subnets in the second octet are a block size of 1 and the

subnets in the third octet are 0, 16, 32, etc.

• Valid hosts?

The following table shows some examples of the host ranges—the first three and the last subnets:

Practice Example #3A: 255.255.255.192 (/26) Let’s do one more example using the second, third, and fourth octets for subnetting.

• Subnets? 218 = 262,144.

• Hosts? 26 – 2 = 62.

• Valid subnets? In the second and third octet, the block size is 1 and in the fourth octet the

block size is 64.