CIDR 23 dan 22

 Contoh :

172.16.0.0/23

1. Netmask :

11111111.11111111.11111110.00000000

255.255.254.0

2. Jumlah Blok subnet 

   256 - netmask = 256-254 = 2 {0,2,4,6,8 ....... 254}

3. Jumlah subnet = 2^7 = 128 subnet

4. Jumlah Host Per subnet = 2^9 = 512 host Per subnet

5. Jumlah Host Per subnet Valid = (2^9)-2 = 510 Valid

6. ip address sesuai subnet:

subnet 1:

network : 172.16.0.0

ip 1 : 172.16.0.1

ip end : 172.16.1.254

ip bc : 172.16.1.255

subnet 2:

network : 172.16.2.0

ip 1 : 172.16.2.1

ip end : 172.16.3.254

ip bc : 172.16.3.255

subnet 3

network : 172.16.4.0

ip 1 : 172.16.4.1

ip end : 172.16.5.254

ip bc : 172.16.5.255 Tuliskan 22 subnet ke depan

=========================

172.16.0.0/22

netmask = 255.255.252.0

berapa subnet: 11111111.11111111.11111100.00000000

2^6 = 64 subnet

berapa host per subnet: 2^10 = 1024

berapa host per subnet valid = 1022

blok subnet : 256-252 = 4 {0,4,8,......252}

ip address:

subnet 1:

172.16.0.0

172.16.0.1

172.16.3.254

172.16.3.255

subnet 2:

172.16.4.0

172.16.4.1

172.16.7.254

172.16.7.255

subnet 3:

172.16.8.0

172.16.8.1

172.16.11.254

172.16.11.255


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