This document discusses subnetting and variable length subnet masking (VLSM). It provides examples of calculating subnet masks using fixed length subnet masking (FLSM) and VLSM. For FLSM, it shows how to determine the number of host bits needed based on required hosts and how to calculate the subnet mask and number of subnets. For VLSM, it demonstrates calculating subnet masks and IP ranges to support different numbers of hosts in the same class C network.

1. Subnetting (FLSM & VLSM)
with examples
by
Krishna Mohan
Krishna.mohan.ec@gmail.com

2. As IP addresses are limited, it is not advisable to waste it, weather it is private IP or public IP.
A large network can be divide into various small networks using subnetting.
Subnetting involves conversion of host bits into network bits.
Methods of Subnetting:
a) FLSM (Fixed Length Subnet Mask)
b) VLSM (Variable Length Subnet Mask)
Division of network based on requirement:
a) Based on host bits : 2^h-2>=required address where h= no. of host bits
b) Based on network bits : 2^n>=required address where n=no of network bits
Addressing format with network & host bits as
Class & format Network Bits (N) Host bits (H)
A: N.H.H.H 8 24
B: N.N.H.H 16 16
C: N.N.N.H 24 8

3. Calculation of Subnet mask with FLSM
Example. Calculate subnet mask using Host Bits , each having 12 hosts or nodes or devices. Also provide IP
addresses of all devices. How many subnets available. Given network id as 199.1.1
Solution:
Step I: Calculation of host bits
For h host bits, 2^ℎ-2≥requirement
Here requirement=12
Hence h=4 (as 2^4-2=16-2=14 ≥ 12
Hence host bits required is 4
Step II: Calculation of subnet mask
Given class C address, which has format N.N.N.H or
11111111.11111111.11111111.00000000 (24 network bits & 8 host bits)
From the host bit calculation h=4, hence 4 LSB bits are host bits & remaining 4 bits of are for network bits.
On rearranging 11111111.11111111.11111111.11110000 (24+4=28 total network bits & 4 host bits), which is
subnet mask for the designed network, can be given as 28/4 (network bits/host bits).
Converting into decimal, subnet mask is 255.255.255.240
Step III: Calculation of no of sub-networks
No of networks=2^(no of network bits taken from host bits)= 2^(8-4)=2^4=16 subnets

4. Step IV: IP Range
Given IP is 199.1.1
Range=2^4= 16 (group of 16 is one subnetwork) & each having 16 devices or nodes.
199.1.1.0, 199.1.1.1, 199.1.1.2………………………………………………………………………………199.1.1.15
199.1.1.16, 199.1.1.17………………………………………………………………………………………...199.1.1.31
199.1.1.32…………………………………………………………………………………………………………………………….
199.1.1.48…………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………………………………….
199.1.1.240, 199.1.1.241……………………………………………………………………………………….199.1.1.255
First address of each subnet is network ID (here 199.1.1.0, 199.1.1.16…….199.1.1.240)
Last address of each subnet is broadcast ID (here 199.1.1.15, 199.1.1.31……..199.1.1.255)
Network ID & Broadcast ID cannot be assigned to any device.
IPs other than network & broadcast ID may be assigned to any device
Above IPs are for subnet mask 255.255.255.240
Total IPs available for devices are 16*16-16*2=224

5. What is 192.168.1.10/26
It is a class C IP address with 26 network bits. The network bits used to calculate subnet mask.
Since class C has format N.N.N.H, where N is 8 bit network octet, & H is host bits
Given 26 network bits , hence the format 11111111.11111111.11111111.11000000
Host bits can be calculated as h=32-network bits. Here h=32-26=6
Thus subnet mask is 255.255.255.192 with IP 192.168.1.10

6. Calculation of subnet mask using VLSM (Variable Length Subnet Mask)
Here the length of block is variable
Example: Given the no of hosts as 100, 50 20 & 6. find IP addresses & subnet mask. Using Class C (192.168.1
network)
Solution: For 100 hosts:
2^h-2>=100, hence h=7 (host bits)
Hence value is 8+8+8+(8-7)=25
Hence subnet mask is 11111111.11111111.11111111.10000000 i.e.255.255.255.128
Block size is 2^7=128 hence the range is
192.168.1.0/25 to 192.168.1.127/25 for 100 hosts
For 50 hosts:
2^h-2>=50, hence h=6 (host bits)
Hence value is 8+8+8+(8-6)=26
Hence subnet mask is 11111111.11111111.11111111.11000000 i.e.255.255.255.192
Block size is 2^6=64 hence the range is
192.168.1.128/26 to 192.168.1.191/26 for 50 hosts

7. Solution: For 20 hosts:
2^h-2>=20, hence h=5 (host bits)
Hence value is 8+8+8+(8-5)=27
Hence subnet mask is 11111111.11111111.11111111.11100000 i.e.255.255.255.224
Block size is 2^5=32 hence the range is
192.168.1.192/27 to 192.168.1.223/27 for 20 hosts
For 6 hosts:
2^h-2>=6, hence h=3 (host bits)
Hence value is 8+8+8+(8-3)=29
Hence subnet mask is 11111111.11111111.11111111.11111000 i.e.255.255.255.248
Block size is 2^3=8 hence the range is
192.168.1.224/29 to 192.168.1.231/29 for 6 hosts
Hence complete range:
192.168.1.0/25 to 192.168.1.127/25 for 100 hosts
192.168.1.128/26 to 192.168.1.191/26 for 50 hosts
192.168.1.192/27 to 192.168.1.223/27 for 20 hosts
192.168.1.224/29 to 192.168.1.231/29 for 6 hosts
Any doubt mail at: Krishna.mohan.ec@gmail.com