Latihan Subnetting

1. Sebuah network di alokasikan IP Address 100.100.0.0/24. Tentukan alokasi IP address untuk subnetwork yang masing-masing terdiri atas 30 IP address yang digunakan untuk pengalamatan hostnya!

2.Network 100.100.0.0/24. Tentukan alokasi atau range subnetwork yang masing-masing terdiri dari 200 host!

3. Tentukan range network dari 222.168.0.0/24 apabila akan dibagi menjadi 4 subnetwork!

4.192.168.0.0/24 akan dibuat distribusi alamat untuk beberapa subnetwork dengan rincian sebagai berikut:

* Network A = 5 PC

* Network B = 15 PC

* Network C = 25 PC

* Network C = 35 PC

Tentukan range setiap subnetworknya.!

Jawab :

1. Jumlah host  : 2ⁿ - 2

                  30 : 2⁵ - 2 = 5 -> 5 bit terakhir

Subnetwork : 100.100.0.0 /27     -> 100.100.0.1 /27     - 100.100.0.30 /27

                     100.100.0.32 /27   -> 100.100.0.33 /27   - 100.100.0.62 /27

                     100.100.0.64 /27   -> 100.100.0.65 /27   - 100.100.0.94 /27

                     100.100.0.96 /27   -> 100.100.0.97 /27   - 100.100.0.126 /27

                     100.100.0.128 /27 -> 100.100.0.129 /27 - 100.100.0.158 /27

                     100.100.0.160 /27 -> 100.100.0.161 /27 - 100.100.0.190 /27

                     100.100.0.192 /27 -> 100.100.0.193 /27 - 100.100.0.222 /27

                     100.100.0.224 /27 -> 100.100.0.225 /27 - 100.100.0.254 /27

2. Jumlah host  : 2ⁿ- 2

               200  : 2⁸- 2

               8 bit terakhir

Subnetwork : 100.100.0.0 /24  ->  100.100.0.1 /24  -  100.100.0.254 /24

                     100.100.1.0 /24  ->  100.100.1.1 /24  -  100.100.1.254 /24

                     100.100.2.0 /24  ->  100.100.2.1 /24  -  100.100.2.254 /24

                     100.100.3.0 /24  ->  100.100.3.1 /24  -  100.100.3.254 /24

3. Jumlah subnet : 2ⁿ 

                               4 : 2²  -> 2 bit

Subnetwork  : 222.168.0.0 /26      -> 222.168.0.1 /26     -   222.168.0.61 /26

                      222.168.0.64 /26    -> 222.168.0.65 /26   -   222.268.0.125 /26

                      222.168.0.128 /26  -> 222.168.0.129 /26 -   222.168.0.189 /26

                      222.168.0.192 /26  -> 222.168.0.193 /26 -   222.168.0.253 /26

4. A = 5 + 2 = 7           IP -> 8 IP address         Mask : /29

    B = 15 + 2 = 17       IP -> 32 IP address       Mask : /27

    C = 25 + 2 = 27       IP -> 32 IP address       Mask : /27

    D = 35 + 2 = 37       IP -> 64 IP address       Mask : /26

D = 192.168.0.0 /26     -    192.168.0.63 /26

C = 192.168.0.64 /27   -    192.168.0.95 /27

B = 192.168.0.96 /27   -    192.168.0.127 /27

A = 192.168.0.128 /29 -    192.168.0.135 /29

Sisa = 192.168.0.136 - 192.168.0.255

Soal

1.Diketahui : Network 192.168.11.0/23. Buatlah menjadi 16 subnetwork

2. Diketahui : Network 192.168.10.14/26. Buatlah menjadi 7 subnetwork

3.Diketahui : Network 192.168.112.0/20. Tentukan alokasi subnetwork yang masing-masing terdiri atas 600 host

4. Diketahui : Network 100.100.10.10/28. Tentukan alokasi subnetwork yang masing-masing terdiri atas 8 host

Jawab :

1. 32 -23 = 9

    2⁹ = 512

    512 - 2 =510

    Range total : 192.168.11.0  -> 192.168.12.255

    2ⁿ = 16, n = 4

    23+4 = 27

    32-27 = 5

    2⁵-2 = 32-2 = 30

192.168.11.0       /27         –              192.168.11.31     /27

192.168.11.32     /27         –              192.168.11.63     /27

192.168.11.64     /27         –              192.168.11.95     /27

192.168.11.96     /27         –              192.168.11.127   /27

192.168.11.128   /27         –              192.168.11.159   /27

192.168.11.160   /27         –              192.168.11.191   /27                       

192.168.11.192   /27         –              192.168.11.223   /27

192.168.11.224   /27         –              192.168.11.255   /27

192.168.12.0       /27         –              192.168.12.31     /27

192.168.12.32     /27         –              192.168.12.63     /27

192.168.12.64     /27         –              192.168.12.95     /27

192.168.12.96     /27         –              192.168.12.127   /27

192.168.12.128   /27         –              192.168.12.159   /27

192.168.12.160   /27         –              192.168.12.191   /27                       

192.168.12.192   /27         –              192.168.12.223   /27

192.168.12.224   /27         –              192.168.12.255   /27

2. Network awal 192.168.10.14 /26

    Range awal  : 32 - 26 = 6

    2⁶ = 64

2ⁿ = 7, n = 3

    26+3 = 29

    32-29 = 3

    2³= 8

192.168.10.0 /29

192.168.10.8 /29

192.168.10.16 /29

192.168.10.24 /29

192.168.10.32 /29

192.168.10.40 /29

192.168.10.48 /29

192.168.10.56 /29

3.600 + 2 = 2n
   n = 10
  32 - 10 = 22
192.168.112.0 /22       -> 192.168.112.1           -        192.168.114.92
192.168.116.0 /22       -> 192.168.116.1           -        192.168.118.92
192.168.120.0 /22       -> 192.168.120.1           -        192.168.122.92
192.168.124.0 /22       -> 192.168.124.1           -        192.168.124.92

4. 8 + 2 = 2n

      n = 4

32 - 4 = 28

NA = 100.100.10.0

BA = 100.100.10.255

100.100.10.0/28

100.100.10.16/28

100.100.10.32/28

100.100.10.48/28

100.100.10.64/28

100.100.10.80/28

100.100.10.96/28

100.100.10.112/28

100.100.10.128/28

100.100.10.144/28

100.100.10.160/28

100.100.10.176/28

100.100.10.192/28

100.100.10.208/28

100.100.10.224/28

100.100.10.248/28 - 100.100.10.256/28

Latihan 1

192.168.10.10 /27 buat menjadi 6 subnet

NA = 192.168.10.0

BA = 192.168.10.31

192.168.10.0 /30

192.168.10.4 /30

192.168.10.8 /30

192.168.10.12 /30

192.168.10.16 /30

192.168.10.20 /30

Latihan 2

Network 172.16.0.0 /22 buat alokasi IP address untuk subnetwork

A = 80 pc + 2 = 82       IP = 128       Mask /25

B = 100 pc + 2 = 102   IP = 128       Mask /25

C = 200 pc + 2 = 202   IP = 256       Mask /24

D = 150 pc + 2 = 152   IP = 256       Mask /24

E = 50 pc + 2 = 52       IP = 64         Mask /26

C = 172.16.0.0 /24       -    172.16.0.255 /24

D = 172.16.1.0 /24       -    172.16.1.255 /24

B = 172.16.2.0 /25       -    172.16.2.127 /25

A = 172.16.2.128 /25   -    172.16.2.255 /25

E = 172.16.3.0 /26       -     172.16.3.63 /16

Sisa = 172.16.3.64 - 172.16.3.155

Latihan 3

Bank mandiri bandung raya memiliki alokasi network 60.60.60.0 /23 tetukan alokasi IP address untuk cabang-cabang bank mandiri yang pengguna jaringannya :

A = Bandung 60 pc

B = Cimahi 120 pc

C = Baros 40 pc

D = Padalarang 100 pc

E = Soreang 120 pc

NA : 60.60.60.0 /23

BA : 60.60.61.255 /23

60 + 2 = 62       IP = 64        Mask /26

120 + 2 =122    IP = 128      Mask /25

40 + 2 = 42       IP = 64        Mask /26

100 + 2 = 102   IP = 128      Mask /25

120 + 2 = 122   IP = 128      Mask /25

B = 60.60.60.0 /25      - 60.60.60.127 /25

E = 60.60.60.128 /25  - 60.60.60.255 /25

D = 60.60.61.0 /25     - 60.60.61.127 /25

A = 60.60.61.128 /26 - 60.60.61.191 /26

C = 60.60.61.192 /26 - 60.60.61.255 /26

Latihan 4

Network 111.111.11.0 /28

A = 100 pc + 2 = 102            IP = 128        Mask /25

B = 60 pc + 2 = 62                IP = 64          Mask /26

C = 150 pc + 2 = 152            IP = 256        Mask /24

D = 25 pc + 2 =27                 IP = 32          Mask /27

E = 3 pc + 2 = 5                    IP = 8            Mask /29

C = 111.111.11.0 /24       - 111.111.11.255 /24

A = 111.111.12.0 /25       - 111.111.12.127 /25

B = 111.111.12.128 /26   - 111.111.12.191 /26

D = 111.111.12.192 /27   - 111.111.12.223 /27

E = 111.111.12.224 /29   - 111.111.12.231 /29

Sisa = 111.111.12.232 - 111.111.12.255

Latihan 5

IP = 13.13.3.3 /20

A = 62 pc + 2 = 64            IP = 64        Mask /26

B = 100 pc + 2 + 102        IP = 128      Mask /25

C = 500 pc + 2 = 502        IP = 512      Mask /23

D = 1000 pc + 2 = 1002    IP = 1024    Mask /22

E = 1022 pc + 2 = 1024     IP = 1024   Mask /22

NA : 13.13.3.3

BA : 13.13.15.255

Host awal : 2^32 - 20 = 2^12 = 4096

E = 13.13.0.0 /22       - 13.13.3.255 /22

D = 13.13.4.0 /22       - 13.13.7.255 /22

C = 13.13.8.0 /23       - 13.13.9.255 /23

B = 13.13.10.0 /25     - 13.13.10.127 /25

A = 13.13.10.128 /26 - 13.13.10.191 /26

Sisa = 13.13.10.192 - 13.13.10.255

Latihan 6

IP 6.6.6.0 /21

A = 6 pc + 2 = 8              IP = 8        Mask /29

B = 666 pc + 2 = 668      IP = 1024  Mask /22

C = 66 pc + 2 = 68          IP = 128    Mask /25

D = 13 pc + 2 = 15          IP = 16      Mask / 28

E = 60 pc + 2 = 62          IP = 64      Mask /26

NA : 6.6.0.0

BA : 6.6.7.255

Host awal :2^32 - 21 = 2^11 = 2048

B = 6.6.0.0 /22       - 6.6.3.255 /22

C = 6.6.4.0 /25       - 6.6.4.127 /25

E = 6.6.4.128 /26   - 6.6.4.191 /26

D = 6.6.4.192 /28  - 6.6.4.207 /28

A = 6.6.4.208 /29  - 6.6.4.215 /29

Sisa = 6.6.4.216 - 6.6.4.255