Buat 5 soal bagaimana cara menghitung dari Subnet

Tentukan

1.Alamat Subnet Mask

2.Alamat Subnet

3.Alamat broadcast

4.Jumlah host yg dapat di gunakan

5.Alamat Subnet ke 3

Jawab :

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:

§ Alamat Subnet Mask: 255.255.255.252

§ Alamat Subnet: 198.53.67.0, 198.53.67.4, 198.53.67.8, 198.53.67.12, … , 198.53.67.252

§ Alamat Broadcast: 198.53.67.3, 198.53.67.7, 198.53.67.11, 198.53.67.15 … 198.53.67.255

§ Jumlah host yang dapat digunakan: 64×2 = 128

§ Alamat Subnet ke-3: 198.53.67.8

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:

§ Alamat Subnet Mask: 255.255.255.192

§ Alamat Subnet: 202.151.37.0, 202.151.37.64, 202.151.37.128, 202.151.37.192

§ Alamat Broadcast: 202.151.37.63, 202.151.37.127, 202.151.37.191, 202.151.37.255

§ Jumlah host yang dapat digunakan: 4×62 = 248

§ Alamat Subnet ke-3: 202.151.37.128

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:

§ Alamat Subnet Mask: 255.255.252.0

§ Alamat Subnet: 191.22.24.0, 191.22.24.4, 191.22.24.8, …, 191.22.24.252

§ Alamat Broadcast: 191.22.24.3, 191.22.24.7, 191.22.24.11, …, 191.22.24.255

§ Jumlah host yang dapat digunakan: 2×64 = 128

§ Alamat Subnet ke-3: 191.22.24.8