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Septian Adji Pamungkas XI TKJ 3

Septian Adji Pamungkas XI TKJ 3. Latihan. 192.168.0.10/28 192.168.0.15/netmask 255.255.255.240 192.168.0.16/28 192.168.0.20 netmask 255.255.255.240 192.168.0.20/28 192.168.0.9/30 192.168.0.11/255.255.255.250. Pertanyaan :.

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Septian Adji Pamungkas XI TKJ 3

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  1. Septian Adji Pamungkas XI TKJ 3

  2. Latihan • 192.168.0.10/28 • 192.168.0.15/netmask 255.255.255.240 • 192.168.0.16/28 • 192.168.0.20 netmask 255.255.255.240 • 192.168.0.20/28 • 192.168.0.9/30 • 192.168.0.11/255.255.255.250

  3. Pertanyaan : • Manakah IP yang dapat saling berhubungan (berada dalam segmen yang sama) ? • Berapakah netmask untuk IP 1,3,5,6 ? • Berapakah nilai prefix untuk IP 2,4,7 ? • Manakah IP yang tidak dapat digunakan dalam jaringan, dan apa sebabnya ? • Berapa range untuk masing-masing IP ? • Bagaimana cara menguji konektivitas masing-masing IP ?

  4. Jawab !!! 1. 192.168.0.10/28 192.168.0.16/28 192.168.0.20/28 2.- 192.168.0.16/28 netmask : 255.255.255.240 = 11111111.11111111.11111111.11110000 (1x2)7 + (1x2)6 + (1x2)5 + (1x2)4 + (1x2)3 + (1x2)2 + (1x2)1 + (1x2)0 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1= 255 (1x2)7 + (1x2)6 + (1x2)5 + (1x2)4 + (0x2)3 + (0x2)2 + (0x2)1 +(0x2)0 128 + 64 + 32 + 16 + 0 + 0 + 0 + 0= 240 Jadi subnetmask dari 192.168.0.10/29 adalah 255.255.255.240

  5. - 192.168.0.16/28 Netmask : 255.255.255.240 = 11111111.11111111.11111111.11110000 (1x2)7 + (1x2)6 + (1x2)5 + (1x2)4 + (1x2)3 + (1x2)2 + (1x2)1 + (1x2)0 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 (1x2)7 + (1x2)6 + (1x2)5 + (1x2)4 + (0x2)3 + (0x2)2 + (0x2)1 + (0x2)0 128 + 64 + 32 + 16 + 0 + 0 + 0 + 0 = 240 Jadi subnetmask dari 192.168.0.16/28 adalah 255.255.255.240

  6. - 192.168.0.20/28 Netmask : 255.255.255.240 = 11111111.11111111.11111111.11110000 (1x2)7 + (1x2)6 + (1x2)5 + (1x2)4 + (1x2)3 + (1x2)2 + (1x2)1 + (1x2)0 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 (1x2)7 + (1x2)6 + (1x2)5 + (1x2)4 + (0x2)3 + (0x2)2 + (0x2)1 + (0x2)0 128 + 64 + 32 + 16 + 0 + 0 + 0 + 0 = 240 Jadi subnetmask dari 192.168.0.20/28 adalah 255.255.255.240

  7. - 192.168.0.9/30 Netmask : 255.255.255.252 = 11111111.11111111.11111111.11111100 (1x2)7 + (1x2)6 + (1x2)5 + (1x2)4 + (1x2)3 + (1x2)2 + (1x2)1 + (1x2)0 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 (1x2)7 + (1x2)6 + (1x2)5 + (1x2)4 + (1x2)3 + (1x2)2 + (0x2)1 + (0x2)0 128 + 64 + 3 + 16 + 8 + 4 + 0 + 0= 252 Jadi subnetmask dari 192.168.0.9/30 adalah 255.255.255.252

  8. 3. Prefix Adalah penulisan singkat dari sebuah Netmask. - 192.168.0.15/netmask 255.255.255.240 prefix : /28 - 192.168.0.20/ netmask 255.255.255.240 prefix : /28 - 192.168.0.11/ netmask 255.255.255.250 prefix :

  9. 4. 192.168.0.11/ netmask 255.255.255.250 sebab netmask dan prefix tidak diketahui. 5. A. 192.168.0.10/28 Range ip : 192.168.0.1 - 192.168.0.14 B. 192.168.0.15/netmask 255.255.255.240 Range ip : 192.168.0.1 - 192.168.0.14 C. 192.168.0.16/28 Range ip : 192.168.0.17 - 192.168.0.30 D. 192.168.0.20 netmask 255.255.255.240 Range ip : 192.168.0.17 - 192.168.0.30 E. 192.168.0.20/28 Range ip : 192.168.0.17 - 192.168.0.30 F. 192.168.0.9/30 Range ip : 192.168.0.9 - 192.168.0.10 G. 192.168.0.11/255.255.255.250 Range ip :

  10. 6. Dengan cara di PING (Packet InterNet Gropher)

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