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

Komentar

Postingan populer dari blog ini

TOPOLOGI JARINGAN

ANANDA SETIYAJI