Subnetting Class B Addresses
Subnetting Class B
network is much more similar to subnettins Class
C, the only difference is that when subnetting class B, you will be working
on the third octect; while Class C,
you will work on the fouth octect.
Look at this:
To enable you subnet Class
B, use the same subnet numbers for the third octect just as in Class C. All
you need to do is just to add zero (0)
to the network portion and a 255 to
the broadcast section in the fourth octect. Remember we have more possible
subnet mask in Class B than Class C.
I will bring in the cram table once more, only this time
we are applying it on the THIRD
octect;
Class B cram
table:
Class B
network address has 16 bits available for host addressing (14 bits for
subnetting, 2 bits for host addressing).
Example 1
Let’s look at some examples, using the table above,
remember we are working on the THIRD
octect of Class B. Given network address: 172.16.0.0 /20
From the above network IP address, the mask will be
255.255.240.0 which means we are
using the bit value or block size of 16.
We are going to subnet it to three different networks
with equal host IP addresses; remember we are working on the THIRD octect with
the block size of 16.
Network A
Network address:
172.16.16.0
First Host address: 172.16.16.1
Last host
address: 172.16.31.254
Broadcast
address: 172.16.31.255
What we did above is to add the bit value or size (16+16=32) to obtain the next network
address which is 172.16.32.0
Network B
Network address: 172.16.32.0
First Host address: 172.16.32.1
Last host
address: 172.16.47.254
Broadcast
address: 172.16.47.255
We carried out the same addition here to get the next
network address (32+16=48)
Network C
Network
address : 172.16.48.0
First Host
address : 172.16.48.1
Last host
address: 172.16.63.254
Broadcast address:
172.16.63.255
Same addition before for the next network.
For the WAN (serial links) We need only 4 bits value or block size here due to the
number of network and hosts involved so as not to waste much address space.
looking at the cram table, 4 bit value gives us /30 which results to mask 255.255.252.0 (just like Class C) so we continue from the next network which
is (48+16=64)
WAN 1
Connection from Router A to Router B
Network
address: 172.16.64.0
Network A to B address: 172.16.64.1 255.255.252.0
Network B to A address: 172.16.64.2 255.255.252.0
Next network will also have 4 bits value added to the
last network; (64+4=68)
Same four bit value is used. The next network is:
WAN 2
Connections from
Router A to Router C
Network
address: 172.16.68.0
Network A to C address: 172.16.68.1 255.255.252.0
Network C to A address: 172.16.68.2 255.255.252.0
There are different ways to subnet; you have to device a
way to make it simple for yourself! I think by using the cram table saves you a
lot of time from all the equation of all sort. Lets apply it to a topology:
