Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

Welcome

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

Subnetting

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

192.168.10.0/2411000000.10101000.00001010.0000000011111111.11111111.11111111.00000000

255.255.255.0

1 1 0 0 0 0 0 0

128 64 32 16 8 4 2 1

Network portion Host portion

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

If you want to do subnetting you must answer these questions :

1-How many networks I need?2-How many hosts in each networks?3-What are the broadcasts address?4-What are the valid hosts in each subnets?

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

192.168.1.0/25

255.255.255.128

1-How many networks I need?11111111.11111111.11111111.10000000

2^N = X 2^1 = 2

N = borrowed bits X = networks number

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

192.168.1.0/25

255.255.255.1282-How many hosts in each networks?

11111111.11111111.11111111.10000000

2^N – 2 = X2^7 – 2 = 128 – 2 = 126

N = 0’sX = Hosts

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

192.168.1.0/25

255.255.255.12811111111.11111111.11111111.10000000

3- What are the valid subnets?256 – 128 = 128

Network1 : 192.168.1.0 +128 Network2 : 192.168.1.128

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

192.168.1.0/25

255.255.255.12811111111.11111111.11111111.10000000

3- What are the valid subnets?256 – 128 = 128

Network1 : 192.168.1.0 +128 Network2 : 192.168.1.128

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

192.168.1.0/25

255.255.255.1284-What is the broadcast address?

5- What the valid hosts in each subnet?

Network 192.168.1.0 192.168.1.128First Host 192.168.1.1 192.168.1.129Last Host 192.168.1.126 192.168.1.254Broadcast 192.168.1.127 192.168.1.255

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

192.168.1.0

255.255.255.128

Is the same as

192.168.1.0/2511111111.11111111.11111111.10000000

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

172.10.0.0/22255.255.252.0

1-How many networks I need?11111111.11111111.11111100.00000000

2^N = X 2^6 = 64

N = borrowed bits X = networks

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

172.10.0.0/22255.255.252.0

2-How many hosts in each networks? 11111111.11111111.11111100.00000000

2^N – 2 = X2^10 – 2 = 1024 – 2 = 1022

N = 0’sX = Hosts number

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

172.10.0.0/22255.255.252.0

3- What are the valid subnets?256 – 252= 4

Network1 : 172.10.0.0 +4 Network2 : 172.10.4.0 +4 Network3 : 172.10.8.0 to Network 64 : 172.10.252.0

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

172.10.0.0/22255.255.252.0

4-What is the broadcast address?5- What the valid hosts in each subnet?

Network 172.10.0.0 172.10.4.0

First Host 172.10.0.1 172.10.4.1

Last Host 172.10.3.254 172.10.7.254

Broadcast 172.10.3.255 172.10.7.255

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

Network 172.10.8.0 172.10.252.0First Host 172.10.8.1 172.10.252.1

Last Host 172.10.11.254 172.10.255.254

Broadcast 172.10.11.255 172.10.255.255

TO

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

Please identify the right network mask for the following IP address:* 1 -192.168.0.1/ 25 .255.255.255.128

2 -10.0.0.1/ 18 . 255.255.192.0 3 -172.10.0.0 / 19 .255.255.224.04 -192.168.0.2 / 24 .255.255.255.0

*Please identify the host and the network portion of each IP address: 11111111.11111111.11111111.00000000 1 -192.168.0.1/ 24

2 -10.0.0.5/ 26 11111111.11111111.11111111.11000000

* Red 1 : Network portion* Black 0 : Host portion

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

* Please convert this IP address 192.168.12.3 / 23 to binary number and Identify the first host, last host, broadcast, and network?

192.168.12.3/23IP address : 11000000.101010000.00001100.00000011

255.255.254.0Subnet mask : 11111111.11111111.11111110.00000000

128Network : 2^7 = Hosts : 2^9 – 2 = 512 – 2 = 510

Network 192.168.0.0 192.168.2.0 192.168.254.0

First Host 192.168.0.1 192.168.2.1 192.168.254.1

Last Host 192.168.1.254 192.168.3.254 192.168.255.254

Broadcast 192.168.1.255 192.168.3.255 192.168.255.255

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

*Create two sub-networks from this address 172.16.0.0 and each sub-network has only fourteen hosts?

172.16.0.0\2811111111.11111111.11111111.11110000

255.255.255.240 Networks : 2^12= 4096

Hosts : 2^4 – 2 = 16 – 2 = 14

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

172.16.0.0\2811111111.11111111.11111111.11110000

255.255.255.240 256 – 240 = 16

Networks: 1 -172.16.0.0

2 -172.16.0.16To

4096 - 172.16.255.240

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

172.16.0.0\2811111111.11111111.11111111.11110000

255.255.255.240

Network 172.16.0.0 172.16.0.16

First Host 172.16.0.1 172.16.0.17

Last Host 172.16.0.14 172.16.0.30

Broadcast 172.16.0.15 172.16.0.31

Information & Communication TechnologyModule ICT-BVF-8.1 Computer Networks

Unit ICT-BVF-8.1 Subnetting

Thanks for the attendance

Preparing by Sayfuldeen & Abdulaziz