lengkap... On Jun 27, 2011 8:53 AM, "priyanto uki" <ukay0...@yahoo.com> wrote: **
jawaban no 1 subnetting: membagi sebuah jaringan/network menjadi beberapa sub-jaringan/subnet caranya: meminjam host bit utk dijadikan network bit rumus: 2^N - 2 >= subnet yg dibutuhkan N = subnet bit (host bit yg dipinjam) Contoh soal: jaringan 172.16.0.0/16 dibagi menjadi minimal 5 subnet langkah 1: mencari jumlah subnet bit (N) 2^N - 2 >= 5 N = 3 subnet bit langkah 2: menghitung subnet mask utk setiap subnet yg baru subnet mask = NW bit yg ada + N = 16 + 3 = /19 --> /8./8./3./0 = 255.255.224.0 langkah 3: menghitung interval antar subnet question --> di octet ke berapa, NW bit terakhir? octet ke-3 interval = 256 - 224 = 32 interval = 2^hostBit di octet tsb = 2^5 = 32 BC address = next subnet - 1 jumlah subnet yg valid = 2^N - 2 = 2^3 - 2 = 6 subnet valid dikurangi 2 karena ada subnet zeroes & ones khusus subnetting, ada 2 invalid subnet, yaitu: 1. subnet zeroes --> subnet paling pertama (172.16.0.0) 2. subnet ones --> subnet paling terakhir jumlah host bit di subnet /19 ==> 32 bit - 19 NW bit = 13 host bit pada subnet /19, jumlah host per subnet = 2^13 - 2 = 8190 host per subnet /19 subnetting = FLSM (fixed-length subnet mask) kekurangan subnetting: alokasi ip kurang efisien (boros) ------------------------- VLSM (variable-length subnet mask) menghitung pembagian subnet berdasarkan kebutuhan host per subnet rumus: 2^H - 2 >= host yg dibutuhkan H = host bit yg digunakan note: mulai dari subnet dengan kebutuhan host terbanyak -------- 2 subnet @ min 100 host 2^H - 2 >= 100 ; H = 7 host bit subnet mask = total bit IPv4 address - H = 32 - 7 = /25 --> /8./8./8./1 255.255.255.128 interval = 256 - 128 = 128 (di octet ke-4) interval = 2^7 = 128 jumlah host per subnet /25 = 2^7 - 2 = 126 host LAN JKT: 172.16.0.0/25 BC: 172.16.0.127/25 172.16.0.128/25 BC: 172.16.0.255/25 172.16.1.0 ---------- 3 subnet @ min 50 host 2^H - 2 >= 50 ; H = 6 host bit subnet mask = 32 - 6 = /26 --> /8./8./8./2 255.255.255.192 interval = 256 - 192 = 64 (interval ke 4) interval = 2^6 = 64 jumlah host per subnet /26 = 2^6 - 2 = 62 host LAN SBY: 172.16.1.0/26 BC: 172.16.1.63/26 172.16.1.64/26 BC: 172.16.1.127/26 172.16.1.128/26 BC: 172.16.1.191/26 jawaban no 2 : ip local yang di subnetting mohon jawaban yang lainnya [Non-text portions of this message have been removed] [Non-text portions of this message have been removed] ------------------------------------ -- www.itcenter.or.id - Komunitas Teknologi Informasi Indonesia Gabung, Keluar, Mode Kirim : itcenter-h...@yahoogroups.com Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/ITCENTER/ <*> Your email settings: Individual Email | Traditional <*> To change settings online go to: http://groups.yahoo.com/group/ITCENTER/join (Yahoo! ID required) <*> To change settings via email: itcenter-dig...@yahoogroups.com itcenter-fullfeatu...@yahoogroups.com <*> To unsubscribe from this group, send an email to: itcenter-unsubscr...@yahoogroups.com <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/