Subnetting CIDR (Soal)

Soal:
1. 192.168.16.0/28. Tentukan alokasi IP untuk 4 subnet.
2. 172.122.10.0/16. Tentukan Range Network Awal dan alokasi IP untuk 9 subnet.
3. 172.186.2.0/25. Tentukan alokasi IP untuk 13 subnet.
4. 172.186.11.0/15. Tentukan alokasi IP serta Range Network Awal dengan jumlah masing-masing host sebanyak 131072 host pada tiap subnet.
5. 192.168.23.0/22. Tentukan alokasi IP serta Range Network Awal dengan jumlah masing-masing host sebanyak 1024 host pada tiap subnet.

Jawaban:
1. Jumlah host = 24 = 16
Range Network Awal: 192.168.16.0 - 192.168.16.15
Jumlah masing-masing host tiap subnetwork:
16/4 = 4 host
Masing – masing subnetwork memiliki genmask /30
Alokasi IP:
a. Subnetwork 1: 192.168.16.0/30 – 192.168.16.3/30
b. Subnetwork 2: 192.168.16.4/30 – 192.168.16.7/30
c. Subnetwork 3: 192.168.16.8/30 – 192.168.16.11/30
d. Subnetwork 4: 192.168.16.12/30 – 192.168.16.15/30

2. Jumlah host = 216 = 65536
Range Network Awal: 172.122.0.0 - 172.122.255.255
Jumlah masing-masing host tiap subnetwork:
65536/16 = 4096 host
Masing – masing subnetwork memiliki genmask /20
Alokasi IP:
a. Subnetwork 1: 172.122.0.0/20 – 172.122.15.255/20
b. Subnetwork 2: 172.122.16.0/20 – 172.122.31.255/20
c. Subnetwork 3: 172.122.32.0/20 – 172.122.47.255/20
d. Subnetwork 4: 172.122.48.0/20 – 172.122.63.255/20
e. Subnetwork 5: 172.122.64.0/20 – 172.122.88.255/20
f. Subnetwork 6: 172.122.90.0/20 – 172.122.105.255/20
g. Subnetwork 7: 172.122.106.0/20 – 172.122.121.255/20
h. Subnetwork 8: 172.122.122.0/20 – 172.122.137.255/20
i. Subnetwork 9: 172.122.138.0/20 – 172.122.153.255/20
j. Subnetwork 10: 172.122.154.0/20 – 172.122.169.255/20
k. Subnetwork 11: 172.122.170.0/20 – 172.122.185.255/20
l. Subnetwork 12: 172.122.186.0/20 – 172.122.191.255/20
m. Subnetwork 13: 172.122.192.0/20 – 172.122.207.255/20
n. Subnetwork 14: 172.122.208.0/20 – 172.122.223.255/20
o. Subnetwork 15: 172.122.224.0/20 – 172.122.239.255/20
p. Subnetwork 16: 172.122.240.0/20 – 172.122.255.255/20

3. Jumlah host = 27 = 128
Range Network Awal: 172.128.2.0 - 172.128.2.127
Jumlah masing-masing host tiap subnetwork:
128/16 = 8 host
Masing – masing subnetwork memiliki genmask /29
Alokasi IP:
a. Subnetwork 1: 172.128.2.0/29 – 172.128.2.7/29
b. Subnetwork 2: 172.128.2.8/29 – 172.128.2.15/29
c. Subnetwork 3: 172.128.2.16/29 – 172.128.2.23/29
d. Subnetwork 4: 172.128.2.26/29 – 172.128.2.31/29
e. Subnetwork 5: 172.128.2.32/29 – 172.128.2.39/29
f. Subnetwork 6: 172.128.2.40/29 – 172.128.2.47/29
g. Subnetwork 7: 172.128.2.48/29 – 172.128.2.55/29
h. Subnetwork 8: 172.128.2.56/29 – 172.128.2.63/29
i. Subnetwork 9: 172.128.2.64/29 – 172.128.2.71/29
j. Subnetwork 10: 172.128.2.72/29 – 172.128.2.79/29
k. Subnetwork 11: 172.128.2.80/29 – 172.128.2.87/29
l. Subnetwork 12: 172.128.2.88/29 – 172.128.2.95/29
m. Subnetwork 13: 172.128.2.96/29 – 172.128.2.103/29
n. Subnetwork 14: 172.128.2.104/29 – 172.128.2.111/29
o. Subnetwork 15: 172.128.2.112/29 – 172.128.2.119/29
p. Subnetwork 16: 172.128.2.120/29 – 172.128.2.127/29

4. Jumlah host = 217 = 131072
Range Network Awal: 172.186.0.0 - 172.187.255.255
Jumlah subnetwork:
131072/131072 = 1 host
Masing – masing subnetwork memiliki genmask /15
Alokasi IP:
a. Subnetwork 1: 172.186.0.0/15 – 172.187.255.255/15

5. Jumlah host = 210 = 1024
Range Network Awal: 192.168.20.0 - 192.168.23.255
Jumlah masing-masing host tiap subnetwork:
1024/1024 = 1 host
Masing – masing subnetwork memiliki genmask /22
Alokasi IP:
a. Subnetwork 1: 192.168.20.0/22 – 192.168.23.255/22

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