SUBNET
192 .168 .10.
0/26
1111111.1111111.1111111.11000000
1. Network / Subnet = 2x = 4 network / subnet
2. Host = 2y-2 = 2⁶-2 = 64-2 = 62host
3.Blok Subnet = 256-netmask 256-192 =64
Network
/Subnet
|
0
|
64
|
128
|
192
|
HOST
|
1-62
|
65-126
|
129-190
|
193-254
|
Broadcast
|
63
|
127
|
191
|
255
|
Diket 30 Host 192.168.50.0
Ditanta berpa prefix ?
Jawab host =
2y-2
32 = 2y
Y
= 5 = 32
8 – 5 = 3 x = 3 soalnya kan ada 8
11111111 kalau ynya 5 berarti 1nya 3
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
|
Y=
|
7
|
6
|
5
|
4
|
3
|
2
|
1
|
0
|
Contoh Soal
2. Berapa prefix dari IP 192.168.25.0/ jika mempunyai 126 host
Jawab : host = 126 +2 =128 (rumus nyari host 2y-2=host)
= 128 (punya 0 = 7)
8 - 7 = 1 x = 1
= 11111111.11111111.11111111.10000000/25
172.16.10.0/18
11111111.1111111.11000000.0000000
255.255.192.0
1. Network / Subnet 2x = 2² =4 subnet
2. Host 2y-2 =2pangkat14-2
= 16384-2
= 16382 host
3. Blok Subnet = 256-192 = 64
Subnet
|
0.0
|
64.0
|
128.0
|
192.0
|
HOST
|
0.1-63.254
|
64.254
|
128.254
|
192.254
|
Broadcast
|
63.255
|
127.255
|
191.255
|
255.255
|
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