TUGAS KELOMPOK PERTEMUAN KE-2
Nama: Ananda Setiyaji
NIM : 17220708
Kelas : 17.2B.05
1.) 192.168.10.1/30
Netmask = 255.255.0.0
Subnet = 11111111.11111111.11111111.11111100
Network = 2⁶ = 64
Host = 2²-2 = 2
Alokasi IP
Network : 192.168.10.0
Host : 192.168.10.1 - 192.168.10.2
Broadcast : 192.168.10.3
2.) 172.168.10.1/16
Netmask = 255.255.000.000
Subnet = 11111111.11111111.00000000.00000000
Network = 2n = 1
Host = 2h - 2 = 216 - 2 = 65.534
Alokasi ip :
Network : 172.168.0.0
Host : 172.168.0.1 - 192.168.255.254
Broadcast : 192.168.255.255
3.) 172.168.10.1/22
Netmask = 255.255.252.0
Subnetmask = 11111111.11111111.11111100.00000000
Network = 2⁶ = 64
Host = 2^10-2 = 1024 - 2 = 1022
Alokasi ip :
Network : 172.168.10.0
Host 1 : 172.168.10.1
Host 2 : 172.168.10.2
Host akhir : 172.168.255.255
4.) 10.168.5.1/8
Netmask = 255.0.0.0
Subnet = 11111111.00000000.00000000.00000000
Network = 2⁰ = 1
Host = 2²⁴ - 2 = 16.777.214
Alokasi ip :
- Network Address : 10.168.5.0/8
- Host Address : 10.0.0.1 - 10.0.0.4
Host address terakhir : 10.255.255.254
Broadcast : 10.255.255.255
5.) 10.168.5.1/17
Netmask : 255.255.128.0
Subnet : 11111111.11111111.10000000.00000000
Network : 2^n = 2⁹ = 512
Host = 2^h - 2 = 2¹⁵ - 2 = 32.768 - 2 = 32.766
Alokasi ip :
Network : 10.0.0.0
Host address pertama : 10.0.0.1
Host address terakhir : 10.0.127.254
Broadcast : 10.0.127.255
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