Oleh: Ahmad Sachowi, M.Kom.
Contoh Soal 1:
192.168.1.50 /28
Prefix /28
28 = 8.8.8.4 = 1111111.1111111.1111111.11110000 = 255.255.255.240
jadi subnetmasknya adalah 255.255.255.240
perlu diketahui bahwa:
x = 1
y=0
Sekarang kita mencari jumlah Subnet dan Hostnya yaitu menggunakan rumus:
1. Jumlah Subnet 2 pangkat x = 2 pangkat 4 = 16
2. Jumlah Host per subnet pangkat 4 - 2 = 16 - 2 = 14
3. Broadcast 256 - 240 = 16 (0,16,32,48,64...)
Network 192.168.1.0 192.168.1.16 192.168.1.32 192.168.1.48
IP Awal 192.168.1.1 192.168.1.17 192.168.1.33 192.168.1.49
IP Akhir 192.168.1.14 192.168.1.30 192.168.1.46 192.168.1.62
Broadcast 192.168.1.15 192.168.1.31 192.168.1.47 192.168.1.63
Ip Address di contoh kasus diatas adalah 192.168.1.52 /28 sehingga, Network Addresnya adalah 192.168.1.48 dan Broadcast Address nya adalah 192.168.1.63
=========================================
Jumlah IP = 256 (0-255)
0 = IP Network
255 = IP Broadcast
Contoh Soal 2:
192.168.1.52 /28
Penyelesaian:
/28 Jumlah IP = 16
52 : 16 = 3,25
3 x 16 = 48
48 + 16 - 1 = 64 - 1 = 63
RENTANG IP : 192.168.1.48 - 192.168.1.63
IP NETWORK : 192.168.1.48
IP BROADCAST : 192.168.1.63
IP HOST : 192.168.1.49 - 192.168.1.62
SUBNETMASK : 255.255.255.240
Contoh Soal 3:
23.51.120.155/27
155 : 32= 4,843677
4 x 32=128
128 + 32 – 1 = 160 – 1 = 159
RENTANG IP : 23.51.120.128 - 23.51.120.159
IP NETWORK : 23.51.120.128
IP BROADCAST : 23.51.120.159
IP HOST : 23.51.120.129 - 23.51.120.158
SUBNETMASK : 255.255.255.224
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