Tugas pertemuan 2 IP Adress

 Nama : Muhammad Ilham pernanda

Kelas : 17.3A.03

NIM : 17220179

Tugas Pertemuan 2 Jaringan Komputer

September 24, 2021

Hitunglah jumlah Network dan Host yang akanterbentuk dari alamat IP Address dibawah ini, kemudian buat rincian alamat (Network, Broadcast) beserta alamat yang dapat diberikan kepada setiap host (minimal untuk 5 host/komputer):

1. 192.168.10.1/30

2. 172.168.10.1/16

3. 172.168.10.0/22

4. 10.168.5.1/8

5. 10.168.5.1/17

Jawab :

1) 192.168.10.1/30

Netmask = 255.255.255.252

Subnet = 11111111.11111111.11111111.11111100 : /30

Network = 2n = 26 = 64 network

Host = 2h-2 = 22-2 = 2 host

Alokasi IP :

Network : 192.168.10.0

Host : 192.168.10.1 - 192.168.10.2

Broadcast : 192.168.10.3

Network :192.168.10.4

Host : 192.168.10.5 - 192.168.10.6

Broadcast : 192.168.10.7

Network :192.168.10.8

Host : 192.168.10.9 - 192.168.10.10

Broadcast : 192.168.10.11

2) 172.168.10.1/16

Netmask = 255.255.000.000

Subnet = 11111111.11111111.00000000.00000000 : /16

Network = 2n = 20 = 1 network

Host = 2h-2 = 216-2 = 65.534 host

Alokasi IP :

Network :172.168.0.0

Host : 172.168.0.1 - 192.168.255.254

Broadcast : 192.168.255.255

3) 172.168.10.0/22

Netmask = 255.255.252.000

Subnet = 11111111.11111111.11111100.00000000 : /22

Network = 2n = 26 = 64 network

Host = 2h-2 = 210-2 = 1.022 host

Alokasi IP :

Network :172.168.10.0

Host : 172.168.10.1 - 192.168.255.254

Broadcast : 192.168.255.255

4) 10.168.5.1/8

Netmask = 255.000.000.000

Subnet = 11111111.00000000.00000000.00000000 : /8

Network = 2n = 20 = 1 network

Host = 2h-2 = 224-2 = 167.772.614 host

Alokasi IP :

Network :10.168.5.0

Host : 10.0.0.1 - 10.255.255.254

Broadcast : 10.255.255.255

5) 10.168.5.1/17

Netmask = 255.255.128.000

Subnet = 11111111.11111111.10000000.00000000 : /17

Network = 2n = 21 = 2 network

Host = 2h-2 = 215-2 = 32.766 host

Alokasi IP :

Network :10.168.0.0

Host : 10.168.0.1 - 10.168.127.254

Broadcast : 10.168.127.255