172.16.0.0/23 ( XI TKJ 2 dan XI TKJ 1 ) yadika

 172.16.0.0/23

11111111.11111111.11111110.00000000

netmask : 256 - 2^y= 256 - 2^1

256 - 2 = 254 {255.255.254.0}

1. berapa blok subnet:

   256 -(netmask) = 256 -254 = 2 

   jadi blok (0,2,4,6 ...... 254)

2. Jumlah Host /subnet= 2^y = 2^9 = 512 Per subnet 

3. Jumlah subnet = 2^x = 2^7 = 128 Subnet

4. ip per subnet :

subnet 1

network :172.16.0.0

range :172.16.0.1 s.d 172.16.1.254

broadcast :172.16.1.255

pembuktian subnet 1:

172.16.0.1

172.16.0.2

172.16.0.255

------------------256

172.16.1.0

172.16.1.1

172.16.1.255

------------------256

subnet ke 2

network : 172.16.2.0

range : 172.16.2.1 s.d 172.16.3.254

broadcast : 172.16.3.255


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