
IP Address
Presentation
•
Computers
•
3rd Grade
•
Practice Problem
•
Hard
linux cvber
FREE Resource
18 Slides • 0 Questions
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Soal 1
• 192.168.0.39/28
• Total ip /28 adalah 32-28 = 4, 2^4 = 16
• Network
0
16
32
48
64
80
96
..
• Broadcast
15
31
47
63
79
95
..
• Jadi subnet yang bisa di pakai
• 192.168.0.0/28 ( 192.168.0.1-192.168.0.14 )
• 192.168.0.16/28 ( 192.168.0.17-192.168.0.30 )
• 192.168.0.32/28 ( 192.168.0.33-192.168.0.46 )
• …
• Untuk mencari subnet mask kita bisa menggunakan rumus 256-total ip
• /28 total ip = 16, jadi subnet mask (256 -16 = 240) , 255.255.255.240
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Soal 2
• 192.168.0.80/26
• Total ip /26 adalah 32-26 = 4, 2^6 = 64
• Network
0
64
128
192
256
• Broadcast
63
127
191
255
• Jadi subnet yang bisa di pakai
• 192.168.0.0/26 ( 192.168.0.1-192.168.0.63 )
• 192.168.0.64 /26 ( 192.168.0.65-192.168.0.126 )
• 192.168.0.128 /26 ( 192.168.0.129-192.168.0.190 )
• 192.168.0.192 /26 ( 192.168.0.193-192.168.0.254 )
• Untuk mencari subnet mask kita bisa menggunakan rumus 256-total ip
• /28 total ip = 16, jadi subnet mask (256 -64 = 192) , 255.255.255.192
13
Soal 3
• 192.168.0.15/29
• Total ip /29 adalah 32-29 = 3, 2^3 = 8
• Network
0
8
16
24
32
…
• Broadcast
7
15
23
31
…
• Jadi subnet yang bisa di pakai
• 192.168.0.0/29 ( 192.168.0.1-192.168.0.6 )
• 192.168.0.8 /29 ( 192.168.0.9-192.168.0.14 )
• 192.168.0.16 /29 ( 192.168.0.17-192.168.0.22 )
• 192.168.0.24 /29 ( 192.168.0.25-192.168.0.30 )
• Untuk mencari subnet mask kita bisa menggunakan rumus 256-total ip
• /29 total ip = 8, jadi subnet mask (256 -8 = 248) , 255.255.255.248
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Latihan Soal
Soal umum
• 172.17.10.20/29
• 10.10.10.56/28
• 192.168.30.150/27
• Soal Pribadi
• 10.20.30.absen+60/30
• 192.168.17.absen+130/29
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