Information Communication Technology Module ICTBVF8 1 Computer Networks

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

Information & Communication Technology 1 128 Module ICT-BVF-8. 1 Computer Networks Unit ICT-BVF-8. 1 Subnetting 1 0 0 0 64 32 16 8 4 2 1 192. 168. 10. 0/24 11000000. 10101000. 00001010. 0000 11111111. 0000 Network portion 255. 0 Host portion

Information & Communication Technology Module ICT-BVF-8. 1 Computer Networks Unit ICT-BVF-8. 1 Subnetting

Information & Communication Technology Module ICT-BVF-8. 1 Computer Networks Unit ICT-BVF-8. 1 Subnetting 192. 168. 1. 0/25 255. 128 1 -How many networks I need? 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. 128 2 -How many hosts in each networks? 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. 128 11111111. 10000000 3 - What are the valid subnets? 256 – 128 = 128 Network 1 : 192. 168. 1. 0 Network 2 : 192. 168. 1. 128 +128

Information & Communication Technology Module ICT-BVF-8. 1 Computer Networks Unit ICT-BVF-8. 1 Subnetting 192. 168. 1. 0/25 255. 128 11111111. 10000000 3 - What are the valid subnets? 256 – 128 = 128 Network 1 : 192. 168. 1. 0 Network 2 : 192. 168. 1. 128 +128

Information & Communication Technology Module ICT-BVF-8. 1 Computer Networks Unit ICT-BVF-8. 1 Subnetting 192. 168. 1. 0/25 255. 128 4 -What is the broadcast address? 5 - What the valid hosts in each subnet? Network First Host Last Host Broadcast 192. 168. 1. 0 192. 168. 1. 126 192. 168. 1. 127 192. 168. 1. 128 192. 168. 1. 129 192. 168. 1. 254 192. 168. 1. 255

Information & Communication Technology Module ICT-BVF-8. 1 Computer Networks Unit ICT-BVF-8. 1 Subnetting 192. 168. 1. 0 255. 128 Is the same as 192. 168. 1. 0/25 11111111. 10000000

Information & Communication Technology Module ICT-BVF-8. 1 Computer Networks Unit ICT-BVF-8. 1 Subnetting 172. 10. 0. 0/22 255. 252. 0 1 -How many networks I need? 1111111100. 0000 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. 252. 0 2 -How many hosts in each networks? 1111111100. 0000 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. 252. 0 3 - What are the valid subnets? 256 – 252= 4 Network 1 : 172. 10. 0. 0 +4 Network 2 : 172. 10. 4. 0 +4 Network 3 : 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. 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 First Host 172. 10. 8. 0 172. 10. 8. 1 172. 10. 252. 0 172. 10. 252. 1 Last Host 172. 10. 11. 254 172. 10. 255. 254 Broadcast 172. 10. 11. 255 172. 10. 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. 128 2 - 10. 0. 0. 1 /18. 255. 192. 0 3 - 172. 10. 0. 0 / 19. 255. 224. 0 4 - 192. 168. 0. 2 / 24. 255. 0 *Please identify the host and the network portion of each IP address: 1 - 192. 168. 0. 1 /24 11111111. 0000 2 - 10. 0. 0. 5 /26 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. 1010100001100. 00000011 255. 254. 0 Subnet mask : 111111110. 0000 Network : 2^7 = 128 Hosts : 2^9 – 2 = 512 – 2 = 510 Network 192. 168. 0. 0 192. 168. 254. 0 First Host 192. 168. 0. 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

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. 028 11111111. 11110000 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. 028 11111111. 11110000 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. 028 11111111. 11110000 255. 240 Network 172. 16. 0. 0 172. 16. 0. 16 First Host 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 Technology Module ICT-BVF-8. 1 Computer Networks Unit ICT-BVF-8. 1 Subnetting