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by. oggix.com

21.02 |

Tugas jaringan computer

NPM : 43A87006060018

Nama : lina octaviani

Jurusan : S1/TI/6 A Pagi

Soal:

172.10.1.0/20
200.200.1.0/27
10.1.0.0/7

Jawab:

172.10.1.0/20 (IP address di class B)
11111111.11111111.11110000.00000000(255.255.240.0)
Penghitungan :

Jumlah subnet= 2⁴= 16 subnet
Jumlah host persubnet =2¹² – 2 = 4094 host
Blok subnet = 256 – 240 = 16
Bila di gambarkan

Subnet	172.10.1.0	172.10.1.16	.	.	.	.	.	.	.	.	.	.	.	.	172.10.1.224
Host pertama	172.10.1.1	172.10.1.17	.	.	.	.	.	.	.	.	.	.	.	.	172.10.1.225
Host terakhir	172.10.1.14	172.10.1.30													172.10.1.254
boardcast	172.10.1.15	172.10.1.31													172.10.1.255

200.200.1.0/27 (IP address di class C)
11111111.11111111.11111111.11100000(255.255.255.224)
Penghitungan :

Jumlah subnet= 2³= 8 subnet
Jumlah host persubnet =2⁵ – 2 = 30 host
Blok subnet = 256 – 224 = 32
Bila di gambarkan

Subnet	200.200.1.0	200.200.1.32	..	..	..	..	200.200.1.224
Host pertama	200.200.1.1	200.200.1.33					200.200.1.225
Host terakhir	200.200.1.30	200.200.1.62					200.200.1.254
boardcase	200.200.1.31	200.200.1.63					200.200.1.255

10.1.0.0/7 (IP address class A)
11111110.00000000.00000000.00000000(254.0.0.0)
Penghitungan :

Jumlah subnet= 2⁷= 128 subnet
Jumlah host persubnet =2²⁴ – 2 = 16777214 host
Blok subnet = 256 – 254= 2
Bila di gambarkan

Subnet	10.1.0.0	10.1.0.2	..	..	10.1.255.0.2
Host pertama	10.1.0.1	10.1.0.3			10.1.255.0.3
Host terakhir	10.1.255.254	10.1.255.254			10.1.255.254
boardcase	10.1.255.255	10.1.255.254			10.1.255.254

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