172.16.0.0/22
1. netmask:
11111111.11111111.11111100.00000000
256-(2^y) = 256-(2^2) = 256 - 4 = 252
2. blok subnet
256 -(netmask) = 256 - 252 = 4 {0,4,8,12....252}
3. jumlah host per subnet = 2^y = 2^10 = 1024
4. Jumlah subnet = 2^x = 2^6 = 64
5. ip address
subnet 1
network :172.16.0.0
range :172.16.0.1 s.d 172.16.3.254
broadcast :172.16.3.255
pembuktian:
172.16.0.0
172.16.0.1
172.16.0.255
---------------256
172.16.1.0
172.16.1.1
172.16.1.255
---------------256
172.16.2.0
172.16.2.1
172.16.2.255
---------------256
172.16.3.0
172.16.3.1
172.16.3.255
---------------256
subnet 2:
network : 172.16.4.0
range : 172.16.4.1 s.d 172.16.7.254
broadcast : 172.16.7.255
subnet 3
network : 172.16.8.0
range : 172.16.8.1 s.d 172.16.11.254
broadcast : 172.16.11.255
tulis di buku, selanjutnya subnet 4 beserta rumus s.d subnet terakhir tulis di blogspot