SUBNETTING CALCULATION

Subnetting calculation

Default Class B mask = 255.255.0.0 (16 bits)

There will be 8 subnet = 7 faculties + 1 IT Center

IP address = 172.16.0.0 / 16

As you can see that we can use 8 subnets because it has the same number of networks which is 8 and it fulfill the requirement.

Calculate the required host bits to be borrowed

 Number of subnets: 2^N 

 N = number of bits borrow

2^3 = 8

 

New subnet mask

From the calculation of borrowed host bits above, a new subnet mask is created

Subnet mask prefix = 16 bits (before) + 3 bits (borrowed) = 19

The new subnet mask prefix is = /19

From

/16 = 255.255.0.0 or 1111 1111. 1111 1111. 0000 0000. 0000 0000 – old subnet mask

/19 = 255.255.224.0 or 1111 1111. 1111 1111. 1110 0000. 0000 0000 – new subnet mask

3 bits = 128 + 64 + 32 = 224

 

Find the total number of hosts

  Number of hosts: 2^H

  H = number of hosts on the subnet

  2^13 = 8192

 

Find the number of subnet range

 Number of range:

 256 - 224 = 32

 

Subnet	Subnet Address	Host Address Range	Broadcast Address
1	172.16.0.0 /19	172.16.0.1 /19 – 172. 16.31.254 /19	172.16.31.255 /19
2	172.16.32.0 /19	172.16.32.1 /19 – 172.16.63.254 /19	172.16.63.255 /19
3	172.16.64.0 /19	172.16.64.1 /19 – 172.16.95.254 /19	172.16.95.255 /19
4	172.16.96.0 /19	172.16.96.1 /19 – 172.16.127.254 /19	172.16.127.255 /19
5 (IT Server)	172.16.128.0 /19	172.16.128.1 /19 – 172.16.159.254 /19	172.16.159.255 /19
6	172.16.160.0 /19	172.16.160.1 /19 – 172.16.191.254 /19	172.16.191.255 /19
7	172.16.192.0 /19	172.16.192.1 /19 – 172.16.223.254 /19	172.16.223.255 /19
8	172.16.224.0 /19	172.16.224.1 /19 – 172.16.255.254 /19	172.16.255.255 /19

The IT center is assigned the 5th subnet, which is 172.16.128.0/19

The U-learn System Server is assigned the 5th available address, according to the question. As a result, we can assign the learn System server's address as 172.16.128.6 since the IT Center valid host is 172.16.128.1 until 172.16.159.254