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Soran University
Faculty of Science
Mr.Ababakr I Rasul
Lecture 1 Subnetting and
VLSM
Ababakr.rasul@soran.edu.iq

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IP address classes
Class Start address Finish address
A 0.0.0.0 126.255.255.255
B 128.0.0.0 191.255.255.255
C 192.0.0.0 223.255.255.255
D 224.0.0.0 239.255.255.255
E 240.0.0.0 255.255.255.255
2

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Private IP Addresses
These IP addresses are used for internal use by company or home networks that need to
use TCP/IP
3
Class Private Start Address Private End Address
A 10.0.0.0 10.255.255.255
B 172.16.0.0 172.31.255.255
C 192.168.0.0 192.168.255.255

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Subnetting
Subnetting an IP network is to separate a big network into
smaller multiple networks for reorganization and security
purposes.

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Table :Subnet a network
Subnet mask 128 192 224 240 248 252 254 255
Bits value 128 64 32 16 8 4 2 1
2^ power number
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
Bits borrowed 1 2 3 4 5 6 - -
Subnet
prefix
Class C
Class B
Class A
/25
/17
/9
/26
/18
/10
/27
/19
/11
/28
/20
/12
/29
/21
/13
/30
/22
/14
-
/23
/15
-

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Subnet mask
Table : Default Subnet Masks for Class A, Class B and Class C Networks
IP Address
Class
Total # Of
Bits For
Network ID /
Host ID
Default Subnet Mask
First Octet
Second
Octet
Third Octet Fourth Octet
Class A 8 / 24
11111111
(255)
00000000
(0)
00000000
(0)
00000000
(0)
Class B 16 / 16
11111111
(255)
11111111
(255)
00000000
(0)
00000000
(0)
Class C 24 / 8
11111111
(255)
11111111
(255)
11111111
(255)
00000000
(0)

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How to determine host and subnet
A- (2n
) where N is equal to number of bits borrowed.
determine number of total subnet.
A- (2h
-2) where H equal to number of host bits
determine number of valid host per subnet subnet

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Example
Class C: 192.168.100.50 / 26
Subnet mask: 255.255.255.192
Number of subnets: 2n
= 22
= 4 ( n = number of borrowed bits )
Number of usable host: 2h
-2= 26
-2 = 64 – 2 = 62
To find network address Convert both IP address and subnet mask to binary number
128 64 32 16 8 4 2 1
192. 168. 100. 50
11000000. 10101000. 01100100. 00110010 IP address
255. 255. 255. 192 and
11111111. 11111111. 11111111. 11000000 subnet mask
-----------------------------------------------------------
110000000. 10101000. 01100100. 00000000
192 . 168 . 100 . 0
Network address: 192.168.100.0
First usable host: 192.168.100.1 last host =number of host or(valid host)+network address
Last host: 192.168.100.62 last host =0+62
Broadcast address: 192.168.100.63

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Example
200.10.5.68/28 what is subnet mask, network, first host, broadcast ,last host ?
How many hosts or subnets are valid?
200.10.5.68/28
255.255.255.240
200.10.5.01000100
And
255.255.255.11110000
-----------------------------------
200.10.5. 64
200.10.5.64 network address
200.10.5.65 first host
200.10.5.78 last host last host =number of host + network address =64+14
200.10.5.79 broadcast
2n
=24
= 16 subnet
2H
-2= 24
-2=14 hosts

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Variable-length subnet masking
(VLSM)

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What Variable-Length Subnet Masking
(VLSM))
VLSM amounts to "subnetting subnets” which means that VLSM
allows network engineers to divide an IP address space into a
hierarchy of subnets of different sizes, making it possible to create
subnets with very different host counts without wasting large
numbers of addresses.

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VLSM
Available subnet - 192.168.2.0/24
Solution:
In this network we have 6 networks (LAN1 – LAN2 – LAN3
– WAN link1 – WAN link2 – WAN link3)
1- Determine the class of this network 192.168.2.0/24
(Class: C , N = 24bits , H = 8bits, Default Mask = 24)
2- Order the networks from the largest size to the
smallest:
1) LAN 2 (50 hosts)
2) LAN1 (24 hosts)
3) LAN3 (8 hosts)
4) WAN link 1 - WAN link 2 - WAN link 3 (2 hosts)

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VLSM (cont….)
3-Start from the biggest network:
1) LAN 2 (50 hosts):
H = 6 bits -> 26
-2 = 62 hosts
S = 2 bits -> 22
= 4 subnets
mask = N + S = 24+2 = /26 =(255.255.255.192)
LAN2 will take the subnet ID: 192.168.2.0 /26
Subnet 192.168.2.0/26 00000000
subnetmask 255.255.255.192 11000000 and
Network address 192.168.2.0 0= 00000000
First address 192.168.2.1
Last address 192.168.2.62 last host =network address +number of host or(valid host)=0+62=62
Brodcast adress 192.168.2.63
192.168.2.64 Next network address for another host

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VLSM (cont….)
2) LAN 1 (24 hosts):
H = 5 bits -> 25
-2 = 30 hosts
S = 3 bits -> 23
= 8 subnets
/mask = N + S = 24+3 = /27 =(255.255.255.224)
subnet 192.168.2.64/27 01000000
subnet mask 255.255.255.224 11100000 and
Network address 192.168.2.64 64= 01000000
First address 192.168.2.65
Last address 192.168.2.94 last host =network address +number of host or(valid host)=64+30=94
Brodcast adress 192.168.2.95
192.168.2.96 Next network address for another host

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VLSM (cont….)
3) LAN 3 (8 hosts):
H = 4 bits -> 24
-2 = 14 hosts
S = 4 bits -> 24
= 16 subnets
/mask = N + S = 24+4 = /28 =(255.255.255.240)
subnet 192.168.2.96/28 01100000
subnet mask 255.255.255.240 11110000 and
Network address 192.168.2.96 96 = 01100000
First address 192.168.2.97
Last address 192.168.2.110 last host =network address +number of host or(valid host)=96+14=110
Brodcast adress 192.168.2.111
192.168.2.112 Next network address for another host

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VLSM (cont….)
4) WAN Links 1,2,3 (2 hosts):
H = 2 bits -> 22
-2 = 2 hosts
S = 6 bits -> 26
= 64 subnets
/mask = N + S = 24+6 = /30 =(255.255.255.252)
192.168.2.112/30
subnet 192.168.2.112/30 01110000
subnet mask 255.255.255.252 11111100 and
Network address 192.168.2.112 112= 01110000
First address 192.168.2.113
Last address 192.168.2.114 last host =network address +number of host or(valid host)=112+2=114
Brodcast adress 192.168.2.115
192.168.2.116 Next network address for another host

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VLSM (cont….)
4) WAN Links 1,2,3 (2 hosts):
H = 2 bits -> 22
-2 = 2 hosts
S = 6 bits -> 26
= 64 subnets
/mask = N + S = 24+6 = /30 =(255.255.255.252)
192.168.2.116/30
subnet 192.168.2.116/30 01110000
subnet mask 255.255.255.252 11111100 and
Network address 192.168.2.116 116= 01110000
First address 192.168.2.117
Last address 192.168.2.118 last host =network address +number of host or(valid host)=116+2=118
Brodcast adress 192.168.2.119
192.168.2.120 Next network address

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Subnetting HomeWork
Write of the subnet mask, Network address, first host address, last host address,
broadcast address, and number subnet and hosts
 192.168.5.100 /29
 192.168.4.87/25

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VLSM Homework
Find the (VLSM) and determine the range of Hosts the following term.
Available subnet = 192.168.1.0/24