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IP Address

IP Address

Assessment

Presentation

Computers

3rd Grade

Practice Problem

Hard

Created by

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

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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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