Belajar Subnetting
subnetting: membagi sebuah jaringan/network menjadi beberapa sub-jaringan/subnet yg lebih kecil
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
2^N >= 5 + 2
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
/0 = 0000 0000 = 0
/1 = 1000 0000 = 128
/2 = 1100 0000 = 192
/3 = 1110 0000 = 224
/4 = 1111 0000 = 240
/5 = 1111 1000 = 248
/6 = 1111 1100 = 252
/7 = 1111 1110 = 254
/8 = 1111 1111 = 255
langkah 3: menghitung interval antar subnet
question --> di octet ke berapa, NW bit terakhir? octet ke-3
interval = 256 - 224 = 32 (octet 3)
interval = 2^hostBit di octet tsb = 2^(8-3) = 2^5 = 32
langkah 4:
NW address BC address host address
172.16.0.0/19 172.16.31.255/19 172.16.0.1-172.16.31.254/19
172.16.32.0/19 172.16.63.255/19 172.16.32.1-172.16.63.254/19
172.16.64.0/19 172.16.95.255/19 172.16.64.1-172.16.95.254/19
172.16.96.0/19 172.16.127.255/19 172.16.96.1-172.16.127.254/19
172.16.128.0/19 172.16.159.255/19 172.16.128.1-172.16.159.254/19
172.16.160.0/19 172.16.191.255/19 172.16.160.1-172.16.191.254/19
172.16.192.0/19 172.16.223.255/19 172.16.192.1-172.16.223.254/19
172.16.224.0/19 172.16.255.255/19 172.16.224.1-172.16.255.254/19
_______________
172.17.0.0
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
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
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
2^N >= 5 + 2
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
/0 = 0000 0000 = 0
/1 = 1000 0000 = 128
/2 = 1100 0000 = 192
/3 = 1110 0000 = 224
/4 = 1111 0000 = 240
/5 = 1111 1000 = 248
/6 = 1111 1100 = 252
/7 = 1111 1110 = 254
/8 = 1111 1111 = 255
langkah 3: menghitung interval antar subnet
question --> di octet ke berapa, NW bit terakhir? octet ke-3
interval = 256 - 224 = 32 (octet 3)
interval = 2^hostBit di octet tsb = 2^(8-3) = 2^5 = 32
langkah 4:
NW address BC address host address
172.16.0.0/19 172.16.31.255/19 172.16.0.1-172.16.31.254/19
172.16.32.0/19 172.16.63.255/19 172.16.32.1-172.16.63.254/19
172.16.64.0/19 172.16.95.255/19 172.16.64.1-172.16.95.254/19
172.16.96.0/19 172.16.127.255/19 172.16.96.1-172.16.127.254/19
172.16.128.0/19 172.16.159.255/19 172.16.128.1-172.16.159.254/19
172.16.160.0/19 172.16.191.255/19 172.16.160.1-172.16.191.254/19
172.16.192.0/19 172.16.223.255/19 172.16.192.1-172.16.223.254/19
172.16.224.0/19 172.16.255.255/19 172.16.224.1-172.16.255.254/19
_______________
172.17.0.0
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
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
contoh: carilah network address dari tiap host dalam topology diatas
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 (octet 4)
interval = 2^7 = 128 (octet 4)
jumlah host per subnet /25 = 2^7 - 2 = 126 host per network
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 (octet 4)
interval = 2^6 = 64 (octet 4)
jumlah host per subnet /26 = 2^6 - 2 = 62 host per network
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
172.16.1.192
-----------
1 subnet @ min 25 host
2^H - 2 >= 25 H = 5 host bit
subnet mask = 32 - 5 = /27 --> /8./8./8./3
255.255.255.224
interval = 256 - 224 = 32 (octet 4)
interval = 2^5 = 32 (octet 4)
jumlah host per subnet /27 = 2^5 - 2 = 30 host per subnet
LAN MDN:
172.16.1.192/27 BC=172.16.1.223/27
172.16.1.224
------------
2 subnet @ min 2 host
2^H - 2 >= 2 H = 2
subnet mask = 32 - 2 = /30 --> /8./8./8./6
255.255.255.252
interval = 256 - 252 = 4 (octet 4)
interval = 2^2 = 4 (octet 4)
jumlah host per subnet /30 = 2^2 - 2 = 2 host per subnet
Antar router (Link Serial):
172.16.1.224/30 BC=172.16.1.227/30
172.16.1.228/30 BC=172.16.1.231/30
=================================
manajemen IP: subnetting, VLSM, summarization
summarization: meringkas jaringan-jaringan/subnet-subnet menjadi 1 subnet (bisa juga lebih)
tujuan:
1. meringkas isi routing table
2. meringkas routing update ke router neighbor
contoh 1:
A punya 4 subnet:
10.1.1.0/26 = 00001010.00000001.00000001.00000000
10.1.1.64/26 = 00001010.00000001.00000001.01000000
10.1.1.128/26 = 00001010.00000001.00000001.10000000
10.1.1.192/26 = 00001010.00000001.00000001.11000000
-------------------------------------
00001010.00000001.00000001.00000000
summary: 10.1.1.0/24 (10.1.1.0 s.d. 10.1.1.255)
contoh 2:
A punya 2 subnet:
10.1.1.0/26 = 00001010.00000001.00000001.00000000
10.1.1.64/26 = 00001010.00000001.00000001.01000000
-------------------------------------
00001010.00000001.00000001.00000000
summary 10.1.1.0/25 (10.1.1.0 s.d. 10.1.1.127)
C punya 2 subnet:
10.1.1.128/26 = 00001010.00000001.00000001.10000000
10.1.1.192/26 = 00001010.00000001.00000001.11000000
-------------------------------------
00001010.00000001.00000001.10000000
summary: 10.1.1.128/25 (10.1.1.128 s.d. 10.1.1.255)
10.1.1.64/26 = 00001010.00000001.00000001.01000000
10.1.1.128/26 = 00001010.00000001.00000001.10000000
-------------------------------------
00001010.00000001.00000001.00000000
summary: 10.1.1.0/24 (not true)
cara cepat:
1. apakah kelipatan 2^n
2. ip terakhir - ip pertama = wildcard mask
3. 255.255.255.255 - wildcard mask = subnet mask
4. hasil summary = ip pertama & subnet mask (dari langkah 3)
contoh 3:
/24 => /8./8./8./0 = 255.255.255.0
interval = 256 - 255 = 1
2^0 = 1
langkah 1:
172.16.0.0/24
172.16.1.0/24
172.16.2.0/24
172.16.3.0/24
langkah 2: 172.16.3.255 - 172.16.0.0 = 0.0.3.255 (wildcard mask)
ip terakhir = BC address nya subnet terakhir
langkah 3: 255.255.255.255 - 0.0.3.255 = 255.255.252.0
/8./8./6./0 = /22
langkah 4: summary: 172.16.0.0/22 = 172.16.0.0 255.255.252.0
===============================================================
172.16.4.0/24
172.16.5.0/24
langkah 2: 172.16.5.255 - 172.16.4.0 = 0.0.1.255
langkah 3: 255.255.255.255 - 0.0.1.255 = 255.255.254.0
/8./8./7./0 = /23
langkah 4: summary = 172.16.4.0/23
contoh 4:
192.168.160.0/24
192.168.161.0/24
192.168.162.0/24
192.168.163.0/24
192.168.164.0/24
192.168.165.0/24
192.168.166.0/24
192.168.167.0/24
------------------------> summary: 192.168.160.0/21
192.168.168.0/24
contoh 5:
202.10.4.0/24
202.10.5.0/24
202.10.6.0/24
202.10.7.0/24 ==> 202.10.4.0/22
202.10.8.0/24
contoh 6:
10.1.0.0/25
10.1.0.128/25
10.1.1.0/26
10.1.1.64/26
10.1.1.128/26
10.1.1.192/26
==============> 10.1.0.0/23
langkah 2: 10.1.1.255 - 10.1.0.0 = 0.0.1.255
langkah 3: 255.255.255.255 - 0.0.1.255 = 255.255.254.0
/23
contoh 7:
10.2.0.0/24
10.2.1.0/25
10.2.1.128/25
langkah 2: 10.2.1.255 - 10.2.0.0 = 0.0.1.255
langkah 3: 255.255.255.255 - 0.0.1.255 = 255.255.254.0
/23
langkah 4: summary 10.2.0.0/23
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