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IP subnet basics
Ip
addresses are 32 bits (4 octets or 4 x 8 = 32)
IP addresses
are constructed in classes based on 1st octet number
Each
class has different network id and host id fields
FIELDS 1st octet 2nd octet 3rd octet 4th octet
Class
A 1-127 net hosts hosts hosts
Class
B 128-191 net net hosts hosts
Class
C 192-223 net net net hosts
In
a typical office we may have Sales, Warehousing, Accounting, Administration,
and others.
We
may not want everyone on the same network.
TO
MANAGE OUR NETWORK WE MAY WANT TO FURTHER SUB-divide our network.
We
can use the following technique to create manageable sub-networks.
We can borrow bits from the host area and
create subnets.
We borrow from the left side of the host
bits first (Class C address shown)
(Host bits field)
EX; 192 .
101 . 114 . 00000000 (last octet is in binary form)
Borrow 3 bits - So
2 raised to 3rd (23) is 8 possible subnet address combinations
The binary
would look like this:
My 8 subnet
addresses would be:
000 00000 .0 192.101.114.0
001 “ .32 192.101.114.32
010 “ .64 192.101.114.64
011 “ .96 192.101.114.96
100 “ .128 192.101.114.128
101 “ .160 192.101.114.160
110 “ .192 192.101.114.192
111 00000 .224 192.101.114.224
The remaining 5
digits in each subnet are possible host positions. (25)
So 2 to the 5th
is 32 potential hosts. In the first potential subnet
they would look like this.
000 00000
000 00001
000 00010
000 00011
000 00100
000 00101
on to
000 11111
NOTE:
The second potential subnet would look just like the first subnet
but the first three digits in the octet would be 001 instead of 000.
Now in the original network 192.101.114.0 (without subnetting)
we couldn’t use the .0 or the .255 positions as hosts. That is because the .0 is the network ID and the .255
is the broadcast number for everyone in that IP network.
So
we have 256 - 2 = 254 actual host positions left.
After subnetting
we cannot use the first subnet or the last subnet for the same reasons. That’s right - we lose both subnets (first
& last) and all their 64 host positions.
(Possible 8 subnets – 2 = 6 usable subnets)
In each of the other subnets,
the first number is the subnet ID and the last number is it’s
own broadcast number. So we use two of
the 32 possible host positions in each subnet. (32 hosts – 2 = 30 usable hosts)
We had a
total potential of 256-2=254 hosts.
After borrowing 3 bits to subnet, we have only: (8-2) 6 subnets X (32-2) 30 hosts = 180 actual host positions.
We
pay a price for management, don’t we?
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