SUBNETTING CALCULATION
Assume Utem Network is Class B
(172.16.0.0)
- Class
B subnet default mask is 255.255.0.0 or 11111111.11111111.0.0
- We
have 10 faculty and 1 IT Centre, therefore there need to be at least 11
subnets. 2^4 = 16. 4 bits will be borrowed from the host id.
- Mask
after subnetting will be 11111111.11111111.11110000.00000000 or
255.255.240.0/20
- Using
formula 2^h-2 = 2^4-2 = 14
- Table List of IP Address
|
Subnet |
Network
address |
Usable
address |
Broadcast
address |
|
0 |
172.16.0.0/20 |
172.16.1.0/20
to 172.16.14.0/20 |
172.16.15.0/20 |
|
1 |
172.16.16.0/20 |
172.16.17.0/20
to 172.16.30.0/20 |
172.16.31.0/20 |
|
2 |
172.16.32.0/20 |
172.16.33.0/20
to 172.16.46.0/20 |
172.16.47.0/20 |
|
3 |
172.16.48.0/20 |
172.16.49.0/20
to 172.16.62.0/20 |
172.16.63.0/20 |
|
4 |
172.16.64.0/20 |
172.16.65.0/20
to 172.16.78.0/20 |
172.16.79.0/20 |
|
5 |
172.16.80.0/20 |
172.16.81.0/20
to 172.16.94.0/20 |
172.16.95.0/20 |
|
6 |
172.16.96.0/20 |
172.16.97.0/20
to 172.16.110.0/20 |
172.16.111.0/20 |
|
7 |
172.16.112.0/20 |
172.16.113.0/20
to 172.16.126.0/20 |
172.16.127.0/20 |
|
8 |
172.16.128.0/20 |
172.16.129.0/20
to 172.16.142.0/20 |
172.16.143.0/20 |
|
9 |
172.16.144.0/20 |
172.16.145.0/20
to 172.16.158.0/20 |
172.16.159.0/20 |
|
10 |
172.16.160.0/20 |
172.16.161.0/20
to 172.16.174.0/20 |
172.16.175.0/20 |
Therefore, Ulearn System Server address is 172.16.165.0
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