Kamis, 28 Maret 2019
Hitunglah jumlah Network dan Host yang akan terbentuk dari alamat IP Address dibawah ini :
- 92.168.10.1/30
- 172.168.10.1/16
- 172.168.10./22
- 10.168.5.1/8
- 10.168.5.1/17
Jawaban
Nomor 1:
Netmask = 255.255.255.252
In binary = 11111111. 11111111. 11111111.11111100
Network = 2^n
= 2^6
= 64 Network
Host = 2^h-2
= (2^2)-2
= 4-2
= 2 Host
Nomor 2 :
Netmask = 255.255.0.0
In Binary = 11111111.11111111.00000000.00000000
Network = 2^n
= 2^0
= 1 Network
Host = 2^h-2
= (2^16)-2
= 65.536 - 2
= 65.534 Host
Nomor 3 :
Netmask : 255.255.252.0
In Binary : 11111111.11111111.11111100.00000000
Network = 2^n
= 2^6
= 64 Network
Host = 2^h-2
= (2^10)-2
= 1.024-2
= 1.022 Host
Nomor 4 :
Netmask : 255.0.0.0
In Binary : 11111111.00000000.00000000.00000000
Network = 2^n
= 2^0
= 1 Network
Host = 2^h-2
= (2^24)-2
= 16.777.216-2
= 16.777.214 Host
Nomor 5 :
Netmask : 255.255.128.0
In Binary : 11111111.11111111.10000000.00000000
Network = 2^n
= 2^9
= 512 Network
Host = 2^h-2
= (2^15)-2
= 32.768-2
= 32.766 Host
Note:
Pembagian KelasIP
Address
Cara menentukan binary yang digunakan untuk
menghitung network dan host :
Tipe A : Gunakan digit ke 9 sampai digit ke 32
Contoh : 11111111.11111111.10000000.00000000
Tipe B : Gunakan digit ke 17 sampai digit ke
32
Contoh : 11111111.11111111.00000000.00000000
Tipe B : Gunakan 8 digit terakhir
Contoh : 11111111. 11111111. 11111111.11111100
