1. 172.10.1.0/20
2. 200.200.1.0/27
3. 10.1.0.0/7
Jawab:
1. 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
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
2. 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
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
3. 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
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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