This document discusses subnetting and variable length subnet masking (VLSM). It explains that subnetting divides a large network into smaller networks for easier management and security. It then provides examples of calculating the subnet address, number of networks, number of hosts, broadcast address, and host range for subnet masks like /18, /24, and /30. The document also explains how VLSM allows more efficient use of IP addresses by allowing variable subnet masks tailored to the needs of different departments within a network. It provides examples of using VLSM to allocate address space from a single Class C network to three departments with different host requirements.

1. Subneting important file
Subneting
Prepared by: - FekAbe
2018

2. Subneting
Is dividing a big network in to many smaller
networks
If we have a very big network maintenance is
very difficult, and security also one more issue.

3. Cont’d
Big Network
 Identification problem is comlex
 1. Identification of the Network
 2. Identification of Subnets
 3. Identification of Hosts
 4. Identification of Procces
 Smaller Network
 1. Identification of the Network
 3. Identification of Hosts
 3. Identification of Procces
A B
C D
A

4. Cont’d
128 192 224 240 248 252 254 255
1 1 1 1 1 1 1 1
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1

5. Cont’d
Find ?
 Subnet Address ?
Number of Networks ?
Numbers of Host ?
Broadcast Address ?
Range of Host ?

6. Calculate subnet and subnet mask?
Class A
 63.24.89.21/18
Class B
172.31.254.12/24
Class C
208.100.54.209/30

7. Class A
 63.24.89.21/18
63.24.89.21 =
00111111.00011000.01011001.00010101
255.255.255.192=
11111111.11111111.11000000.00000000
No of Network= 2(No 1’s) = 210 =1024
No of Hosts = 2(No 0’s) = 214 -2
16384-2=16382
Subnet Address -we use AND operation between the
IP address and Subnet mask.
00111111.00011000.01011001.00010101
AND
11111111.11111111.11000000.00000000

8. Cont’d
Magic Number is 64
Broadcast Address ?
00111111.00011000.01
011001.00010101
AND 11111111.11111111.11
000000.00000000
00111111.00011000.01
000000.00000000
End of Mask
00111111.00011000.01111111.11111111
So, broadcast Address is =63.24.127.255
Host Range =63.24.64.1 to 63.24.127.254

9. Answers ?
 Subnet Address ? 63.24.64.0
Number of Networks ? 1024
Numbers of Host ? 16382
Broadcast Address ? 63.24.127.255
Range of Host ? 63.24.64.1 to 63.24.127.254
First usable Address ?63.24.64.1
Last Usable Address 63.24.127.254

10. VLMS
Is Variable Length Subnet Mask.
Is more efficient use of IP address.
 ISP Allocate 192.168.1.0/24 class C
Dep’t A =60 Host
Dep’t B =30 Host
Dep’t C =14 Host
Bit Value 12
8
64 32 16 8 4 2 1
Bit Borrowed 1 2 3 4 5 6 7 8
Subnet Mask 12
8
19
2
224 240 248 252 254 255
Subnet
Prefix/SIPR/
/25 /26 /27 /28 /29 /30 /31 /32

11. Cont’d
Dep’t A with 60 hosts using the above table,
borrow 2 bit with value of 64 closest we can get
60 hosts.
Dep’t A=192.168.1.0/26 Network Address
Dep’t A=192.168.1.1 Get way Address
192.168.1.2 First usable address
192.168.1.62 Last Usable Address
Total Space =192.168.1.2 to 192.168.1.62
Broadcast Address = 192.168.1.63
 (Reserve 1st and Last for Network and broadcast)
Dep’t A network mask 255.255.255.192
 We got 192=128+64Dep’t A Address Look like this 192.168.1.0/26

12. Cont’d
 Dep’t B 30 hosts using the above table, borrow 3 bit with value of 32
closest we can get 30 hosts.
 Dep’t B Address will start from 192.168.1.64 Network Address
 Now we add the 32 to the 64 we borrowed earlier =32+64=96
Dep’t B=192.168.1.65 Get way Address
192.168.1.66 First usable address
192.168.1.94 Last Usable Address
Total Space =192.168.1.66 to 192.168.1.94
Broadcast Address = 192.168.1.95
(Reserve 1st and Last for Network and broadcast)
Dep’t A network mask 255.255.255.224
We got 224=128+64+32
Dep’t A Address Look like this 192.168.1.64/27

13. Cont’d
 Dep’t C 12 hosts using the above table, borrow 4 bit with value of
16 closest we can get 12 hosts.
 Dep’t C Address will start from 192.168.1.96 Network Address
 Dep’t B=192.168.1.96/28 Network Address
 Now we add the 16 to the 96 we borrowed earlier =16+96=112
Dep’t B=192.168.1.97 Get way Address
192.168.1.98 First usable address
192.168.1.110 Last Usable Address
Total Space =192.168.1.66 to 192.168.1.94
Broadcast Address = 192.168.1.111
(Reserve 1st and Last for Network and
broadcast)
Dep’t A network mask 255.255.255.240
We got 240=128+64+32+16Dep’t C Address Look like this 192.168.1.96/28

14. Generally, in bigger network No of 0’s will be more
and No of 1’s less which means the subnet mask
will be small in number.

15. THE END