172.16.0.0 s.d 172.31.255.255
jumlah : 256^2 = 65.536
contoh
172.16.0.0/22
11111111.11111111.11111100.00000000
1. netmask : 256 - 2^y = 256 - 2^2 = 256 - 4 = 252
(255.255.252.0)
2. Jumlah Host All = 2^10 = 1024 /subnet
3. Jumlah Host Valid = (2^10)-2 = 1022
4. Jumlah Subnet = 2^6 = 64 subnet
5. Blok subnet = 256 - 252 = 4 {0,4,8,12,16 .....
252}
6. Ip Address
subnet 1:
network : 172.16.0.0
IP 1 : 172.16.0.1
Ip end : 172.16.3.254
IP BC : 172.16.3.255
subnet 2
network : 172.16.4.0
IP 1 : 172.16.4.1
IP end : 172.16.7.254
IP BC : 172.16.7.255
Subnet 3
network : 172.16.8.0
IP 1 : 172.16.8.1
IP End : 172.16.11.254
IP BC : 172.16.11.255
Lanjutkan di blogspot s.d subnet 64