Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting Welcome

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting Subnetting

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 1 128 64 32 16 8 4 2 192.168.10.0/24 11000000.10101000.00001010.00000000 11111111.11111111.11111111.00000000 255.255.255.0 Host portion Network portion

1-How many networks I need? 2-How many hosts in each networks? Information & Communication Technology Module 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 Technology Module 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 Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 192.168.1.0/25 255.255.255.128 2-How many hosts in each networks? 11111111.11111111.11111111.10000000 2^N – 2 = X 2^7 – 2 = 128 – 2 = 126 N = 0’s X = Hosts

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 192.168.1.0/25 255.255.255.128 11111111.11111111.11111111.10000000 3- What are the valid subnets? 256 – 128 = 128 Network1 : 192.168.1.0 +128 Network2 : 192.168.1.128

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 192.168.1.0/25 255.255.255.128 11111111.11111111.11111111.10000000 3- What are the valid subnets? 256 – 128 = 128 Network1 : 192.168.1.0 +128 Network2 : 192.168.1.128

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 192.168.1.0/25 255.255.255.128 4-What is the broadcast address? 5- What the valid hosts in each subnet? Network 192.168.1.0 192.168.1.128 First Host 192.168.1.1 192.168.1.129 Last Host 192.168.1.126 192.168.1.254 Broadcast 192.168.1.127 192.168.1.255

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 192.168.1.0 255.255.255.128 Is the same as 192.168.1.0/25 11111111.11111111.11111111.10000000

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 172.10.0.0/22 255.255.252.0 1-How many networks I need? 11111111.11111111.11111100.00000000 2^N = X 2^6 = 64 N = borrowed bits X = networks

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 172.10.0.0/22 255.255.252.0 2-How many hosts in each networks? 11111111.11111111.11111100.00000000 2^N – 2 = X 2^10 – 2 = 1024 – 2 = 1022 N = 0’s X = Hosts number

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 172.10.0.0/22 255.255.252.0 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 Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 172.10.0.0/22 255.255.252.0 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 Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting TO Network 172.10.8.0 172.10.252.0 First 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

Information & Communication Technology Module 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. 255.255.224.0 4- 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 Technology Module 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? 192.168.12.3/23 IP address : 11000000.101010000.00001100.00000011 255.255.254.0 Subnet 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 Technology Module 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\28 11111111.11111111.11111111.11110000 255.255.255.240 Networks : 2^12= 4096 Hosts : 2^4 – 2 = 16 – 2 = 14

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 172.16.0.0\28 11111111.11111111.11111111.11110000 255.255.255.240 256 – 240 = 16 Networks : 1- 172.16.0.0 2- 172.16.0.16 To 4096 - 172.16.255.240

Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting 172.16.0.0\28 11111111.11111111.11111111.11110000 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

Thanks for the attendance Information & Communication Technology Module ICT-BVF-8.1 Computer Networks Unit ICT-BVF-8.1 Subnetting Thanks for the attendance Preparing by Sayfuldeen & Abdulaziz