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
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[Non-text portions of this message have been removed]
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