Tugas Pertemuan II
Hitunglah jumlah Network dan Host yang akan terbentuk 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./22
4. 10.168.5.1/8
5. 10.168.5.1/17
Berikut merupakan jawaban dari soal diatas:
1. 192.168.10.1/30
(Kelas C)
a. IP
Address: 11000000.10101000.00001010.00000001
b. Netmask:
11111111.11111111.11111111.11111100 = 255.255.255.252
c. Network:
26 = 64
d. Host:
22 = 4 - 2 = 2
e. Network
ID & Broadcast ID: 192.168.10.0 & 192.168.10.3
f.
IP Range: 192.168.10.1 - 192.168.10.2
2. 172.168.10.1/16
(Kelas B)
a. IP
Address: 10101100.10101000.00001010.00000001
b. Netmask:
11111111.11111111.00000000.00000000 = 255.255.0.0
c. Network:
20 = 1
d. Host:
216 = 65,536 – 2 = 65,534
e. Network
ID & Broadcast ID: 172.168.0.0 & 172.168.255.255
f.
IP Range: 172.168.0.1 - 172.168.255.254
3. 172.168.10.1/22
(Kelas B)
a. IP
Address: 10101100.10101000.00001010.00000001
b. Netmask:
11111111.11111111.11111100.00000000 = 255.255.252.0
c. Network:
26 = 64
d. Host:
210 = 1,024 – 2 = 1,022
e. Network
ID & Broadcast ID: 172.168.8.0 & 172.168.11.255
f.
IP Range: 172.168.8.1 - 172.168.11.254
4. 10.168.5.1/8
(Kelas A)
a. IP
Address: 00001010.10101000.00000101.00000001
b. Netmask:
11111111.00000000.00000000.00000000 = 255.0.0.0
c. Network:
20 = 1
d. Host:
224 = 16,777,216 – 2 = 16,777,214
e. Network
ID & Broadcast ID: 10.0.0.0 & 10.255.255.255
f.
IP Range: 10.0.0.1 - 10.255.255.254
5.
10.168.5.1/17 (Kelas A)
a. IP
Address: 00001010.10101000.00000101.00000001
b. Netmask:
11111111.11111111.10000000.00000000 = 255.255.128.0
c. Network:
29 = 512
d. Host:
215 = 32,768 – 2 = 32,766
e. Network
ID & Broadcast ID: 10.168.0.0 & 10.168.127.255
f.
IP Range: 10.168.0.1 - 10.168.127.254
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