video installasi debian dan konfigurasi DNS server

tugas ulfahemil

TUGAS PRODUKTIF !

Membuat perhitungan subnetting kelas A,B,C masing-masing 2 perhitungan

1)KELAS A

a)12.10.8.2/10

b)13.12.10.9/9

jawab:

a)12.10.8.2/10

255.192.0.0

11111111.11000000.00000000.00000000

-Jumlah subnet =2^x=2²=4

-Jumlah host persubnet=2^y-2=2²²-2=4194304-2=4194302

-blok subnet=256-192=64(0,64,128,192)

subnet	Host pertama	Host terakhir	Broadcast
12.0.0.0	12.0.0.1	12.63.255.254	12.63.255.255
12.64.0.0	12.64.0.1	12.127.255.254	12.127.255.255
12.128.0.0	12.128.0.1	12.191.255.254	12.191.255.255
12.192.0.0	12.192.0.1	12.255.255.254	12.255.255.255

b)13.12.10.9/9

          255.128.0.0

          11111111.10000000.00000000.00000000

-Jumlah subnet = 2^x=2¹=2

-Jumlah host persubnet=2^y-2=2²³-2=8388608-2=8388606

-blok subnet=256-128=128

(0,128)

subnet	Host pertama	Host terakhir	Broadcast
255.0.0.0	255.0.0.1	255.127.255.254	255.127.255.255
255.128.0.0	255.128.0.1	255.255.255.254	255.255.255.255

2.kelas b

a)182.128.5.9/18

b)200.18.20.4/19

Jawab:

a)182.128.5.9/18

            255.255.192.0

11111111.11111111.11000000.00000000

-jumlah subnet  =2^x=2²=4

-jumlah host persubnet=2^y-2=2¹⁴-2=16384-2=16382

-blok subnet =256-192=64

(0,64,128,192)

subnet	Host pertama	Host terakhir	Broadcast
182.128.0.0	182.128.0.1	182.128.63.254	182.128.63.255
182.128.64.0	182.128.64.1	182.128.127.254	182.128.127.255
182.128.128.0	182.128.128.1	182.128.191.254	182.128.191.255
182.128.192.0	182.128.192.1	182.128.255.254	182.128.255.255

b)200.18.20.4/19

            255.255.224.0

            11111111.11111111.11100000.00000000

-jumlah subnet =2^x=2³=8

-jumlah host persubnet=2^y-2=2¹³-2=8192-2=8190

-blok subnet=256-224=32

(0,32,64,96,128,160,192,224)

Subnet	Host pertama	Host terakhir	Broadcast
200.18.0.0	200.18.0.1	200.18.31.254	200.18.31.255
200.18.32.0	200.18.32.1	200.18.63.254	200.18.63.255
200.18.64.0	200.18.64.1	200.18.95.254	200.18.95.255
200.18.96.0	200.18.96.1	200.18.127.254	200.18.127.255
200.18.128.0	200.18.128.1	200.18.159.254	200.18.159.255
200.18.160.0	200.18.160.1	200.18.191.254	200.18.191.255
200.18.192.0	200.18.192.1	200.18.223.254	200.18.223.255
200.18.224.0	200.18.224.1	200.18.255.254	200.18.255.255

3)KELAS C

          a)192.168.10.4/25

          b)192.168.50.1/26

jawab:

a)192.168.10.4/25

          255.255.255.128

          11111111.11111111.11111111.10000000

-jumlah subnet=2^x=2¹=2

-jumlah host persubnet= 2^y-2=2⁷-2=128-2=126

-blok subnet =156-128=128

(0.128)

Subnet	Host pertama	Host terakhir	Broadcast
192.168.10.0	192.168.10.1	192.168.10.126	192.168.10.127
192.168.10.128	192.168.10.129	192.168.10.254	192.168.10.255

2)192.168.50.1

          255.255.255.192

11111111.11111111.11111111.11000000

-jumlah subnet=2^x=2²=4

-jumlah host persubnet=2^y-2=2⁶-2=64-2=62

-blok subnet=256-192=64

(0.64,128,192)

Subnet	Host pertama	Host terakhir	Broadcast
192.168.50.0	192.168.50.1	192.168.50.62	192.168.50.63
192.168.50.64	192.168.50.65	192.168.50.126	192.168.50.127
192.168.50.128	192.168.50.129	192.168.50.190	192.168.50.191
192.168.50.192	192.168.50.193	192.168.50.254	192.168.50.255