172.16.0.0/12
1. netmask
11111111.11110000.00000000.00000000
256 - 2^y = 256-2^4 = 256-16=240
{255.240.0.0}
2. Blok subnet
256-(netmask)= 256-240 =16{0,16,32,....240}
3. Jumlah host per subnet = 2^y=2^20 =1.048.576
4. Jumlah Subnet = 2^x= 2^4=16 subnet
5. IP address
subnet 1
network :172.0.0.0
range :172.0.0.1 s.d 172.15.255.254
broadcast :172.15.255.255
subnet 2 :
network : 172.16.0.0
range : 172.16.0.1 s.d 172.31.255.254
broadcast : 172.31.255.255
pembuktian
16 x 256 x 256 = 1.048.576 x 16 = 16.777.216
tulis subnet selanjutnya di buku tulis