Subnetting
165
Practice Example #1A: 255.255.0.0 (/16)
Class A addresses use a default mask of 255.0.0.0, which leaves 22 bits for
subnetting since you must leave two bits for host addressing. The 255.255.0.0
mask with a Class A address is using eight subnet bits.
Subnets? 2
8
- 2 = 254.
Hosts? 2
16
- 2 = 65,534.
Valid subnets? 256
- 255 = 1, 2, 3, etc. (all in the second octet). The
subnets would be 10.1.0.0, 10.2.0.0, 10.3.0.0, etc., up to 10.254.0.0.
Broadcast address for each subnet?
Valid hosts?
The following table shows the first and last subnet, valid host range, and
broadcast addresses:
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? 2
12
- 2 = 4094.
Hosts? 2
12
- 2 = 4094.
Valid subnets? 256
- 240 = 16. And since the second octet is 255,
or all subnet bits on, we can start the third octet with 0 as long as a
subnet bit is turned on in the second octet. So, the first valid subnets
are 10.0.16.0, 10.0.32.0, and 10.0.48.0, all the way to 10.0.224.0.
The next set of subnets would be 10.1.0.0, 10.1.16.0, 10.1.32.0,
10.1.48.0, all the way to 10.1.240.0. Notice that we can use 240
in the third octet as long as the subnet bits in the second octet are
not all on (1s). In other words, 10.255.240.0 is invalid because all
subnet bits are turned on. The last valid subnet would have to be
10.255.224.0.
Subnet
10.0.16.0
...
10.254.0.0
First host
10.1.0.1
...
10.254.0.1
Last host
10.1.255.254
...
10.254.255.254
Broadcast
10.1.255.255
...
10.254.255.255
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