PERHITUNGAN IP ADDRESS

Menghitung Subnetting IP

Soal latihan, tentukan :
a) Alamat Subnet Mask,
b) Alamat Subnet,
c) Alamat Broadcast,
d) Jumlah Host yang dapat digunakan,
e) serta Alamat Subnet ke-3

dari alamat sebagai berikut:
1. 198.53.67.0/30
2. 202.151.37.0/26
3. 191.22.24.0/22

Penyelesaiannya:

1. 198.53.67.0/30 –> IP class C:

Subnet Mask: /30 = 11111111.11111111.11111111.11111100 = 255.255.255.252
Menghitung Subnet:
Jumlah Subnet: 26 = 64 Subnet
Jumlah Host per Subnet: 22 – 2 = 2 host
Blok Subnet: 256 – 252 = 4, blok berikutnya: 4+4 = 8, 8+4 = 12, dst…
jadi blok Subnet: 0, 4, 8, 12, dst…
Host dan broadcast yang valid:

Maka dari perhitungan diperoleh:
o Alamat Subnet Mask: 255.255.255.252
o Alamat Subnet: 198.53.67.0, 198.53.67.4, 198.53.67.8, 198.53.67.12, … , 198.53.67.252
o Alamat Broadcast: 198.53.67.3, 198.53.67.7, 198.53.67.11, 198.53.67.15 … 198.53.67.255
o Jumlah host yang dapat digunakan: 64×2 = 128
o Alamat Subnet ke-3: 198.53.67.8

Menggunakan IP Calculator/IP Subnetting

Address: 198.53.67.0 11000110.00110101.01000011.000000 00
Netmask: 255.255.255.252 = 30 11111111.11111111.11111111.111111 00
Wildcard: 0.0.0.3 00000000.00000000.00000000.000000 11
=>
Network: 198.53.67.0/30 11000110.00110101.01000011.000000 00 (Class C)
Broadcast: 198.53.67.3 11000110.00110101.01000011.000000 11
HostMin: 198.53.67.1 11000110.00110101.01000011.000000 01
HostMax: 198.53.67.2 11000110.00110101.01000011.000000 10
Hosts/Net: 2

2. 202.151.37.0/26 -> IP class C

Subnet Mask: /26 = 11111111.11111111.11111111.11000000 = 255.255.255.192
Menghitung Subnet:
Jumlah Subnet: 22 = 4 Subnet
Jumlah Host per Subnet: 26 – 2 = 62 host
Blok Subnet: 256 – 192 = 64, blok berikutnya: 64+64 = 128, 128+64 = 192
Jadi blok Alamat Subnet: 0, 64, 128, 192
Host dan broadcast yang valid:



Maka dari perhitungan diperoleh:
o Alamat Subnet Mask: 255.255.255.192
o Alamat Subnet: 202.151.37.0, 202.151.37.64, 202.151.37.128, 202.151.37.192
o Alamat Broadcast: 202.151.37.63, 202.151.37.127, 202.151.37.191, 202.151.37.255
o Jumlah host yang dapat digunakan: 4×62 = 248
o Alamat Subnet ke-3: 202.151.37.128

Menggunakan IP Calculator/IP Subnetting

Address: 202.151.37.0 11001010.10010111.00100101.00 000000
Netmask: 255.255.255.192 = 26 11111111.11111111.11111111.11 000000
Wildcard: 0.0.0.63 00000000.00000000.00000000.00 111111
=>
Network: 202.151.37.0/26 11001010.10010111.00100101.00 000000 (Class C)
Broadcast: 202.151.37.63 11001010.10010111.00100101.00 111111
HostMin: 202.151.37.1 11001010.10010111.00100101.00 000001
HostMax: 202.151.37.62 11001010.10010111.00100101.00 111110
Hosts/Net: 62


3. 191.22.24.0/22 –> IP class B
Subnet Mask: /22 = 11111111.11111111.11111100.00000000 = 255.255.252.0
Menghitung Subnet:
Jumlah Subnet: 26 = 64 Subnet
Jumlah Host per Subnet: 22– 2 = 2 host
Jumlah Blok Subnet: 256 – 252 = 4, blok berikutnya: 4+4 = 8, 8+4 = 12, dst…
Jadi blok Alamat Subnet: 0, 4, 8, 12, 16, dst…

Alamat host yang valid:


o Alamat Subnet Mask: 255.255.252.0
o Alamat Subnet: 191.22.24.0, 191.22.24.4, 191.22.24.8, …, 191.22.24.252
o Alamat Broadcast: 191.22.24.3, 191.22.24.7, 191.22.24.11, …, 191.22.24.255
o Jumlah host yang dapat digunakan: 2×64 = 128
o Alamat Subnet ke-3: 191.22.24.8

Menggunakan IP Calculator/IP Subnetting

Address: 191.22.24.0 10111111.00010110.000110 00.00000000
Netmask: 255.255.252.0 = 22 11111111.11111111.111111 00.00000000
Wildcard: 0.0.3.255 00000000.00000000.000000 11.11111111
=>
Network: 191.22.24.0/22 10111111.00010110.000110 00.00000000 (Class B)
Broadcast: 191.22.27.255 10111111.00010110.000110 11.11111111
HostMin: 191.22.24.1 10111111.00010110.000110 00.00000001
HostMax: 191.22.27.254 10111111.00010110.000110 11.11111110
Hosts/Net: 1022