SUBNETTING CALCULATION

In this section, you are expecting to show the calculation on how to identify the IP address that was assigned for the Online Learning System Server based on the information given in the project scenario. Explain the concept and show the calculation to identify the IP address.

For the following scenario, a class A IP addressing scheme 60.0.0.0/12 has been proposed and 4 subnets have been identified. 

The first subnet which holds a total number of 500 hosts are made for the students followed by 100 hosts in the second subnet for the lecturers followed by 30 hosts in the third subnet for the management and 10 hosts in the fourth subnet for the administrators.

                                        Subnet 1 – Student: 500 hosts 

                                        Subnet 2 – Lecturer: 100 hosts 

                                        Subnet 3 – Management: 30 hosts 

                                        Subnet 4 – Administrator: 10 hosts

Step 1 – Do the largest network: 

Largest subnet = 500 host 

Quantity of host = 2H – 2 ≈ 500 2H = 29 (more than 500) 

 When H = 9, 29 – 2 = 512 – 2 = 510 > 500 

 This is the situation: 60 . 0 . 0000 000X . XXXX XXXX 

 Next, Subnet 0: 60 . 0 . 0000 000X . XXXXXXXX 

Subnet 1: 60 . 0 . 0000 001X . XXXXXXXX 

Subnet 2: 60 . 0 . 0000 010X . XXXXXXXX 

Subnet 3: 60 . 0 . 0000 011X . XXXXXXXX 

                     . 

                     . 

                     . 

Subnet Broadcast: 60 . 0 . 1111 111X . XXXXXXXX 

So, the IP for 1st subnet: 

For 500 host, 60.0.0000 000X . XXXX XXXX 

Then, from subnet 0: 

Network Address: 60 . 0 . 0000 0000 . 0000 0000 

IP Range: 60 . 0 . 0000 0000 . 0000 0001 – 60 . 0 . 0000 0001 . 1111 1110 

Broadcast Address: 60 . 0 . 0000 0001 . 1111 1111 

Decimal Form of Network Address for Subnet 0: 60.0.0.0/23

Step 2 – Do the next largest network: 

So, the IP for 2nd subnet: 

For 100 host, 

Take subnet 1 from previous subnetting: 60 . 0 . 0000 001X . XXXX XXXX 

Quantity for the host: 2H – 2 ≈ 100 So, 2H = 27 (more than 100) H = 7, 27 – 2 = 126 > 100 

So, this is the situation: 60 . 0 . 0000 001X . XXXX XXXX 

Subnet 0: 60 . 0 . 0000 0010 . 0XXX XXXX 

Subnet 1: 60 . 0 . 0000 0010 . 1XXX XXXX 

Subnet 2: 60 . 0 . 0000 0011 . 0XXX XXXX 

Subnet Broadcast: 60 . 0 . 0000 0011 . 1XXX XXXX 

Then, from subnet 0: 

Network Address: 60 . 0 . 0000 0010 . 0000 0000 

IP Range: 60 . 0 . 0000 0010 . 0000 0001 – 60 . 0 . 0000 0010 . 0111 1110 

Broadcast Address: 60 . 0 . 0000 0010 . 0111 1111 

Decimal form of Network Address for Subnet 0: 60.0.2.0/25

Step 3 – Do the next largest network: 

For the IP for 3rd subnet: 

For 30 host, 

Take subnet 1 from previous subnetting: 60 . 0 . 0000 0010 . 1XXXXXXX 

Quantity of host: 2H – 2 ≈ 30 So, 2H = 25 (more than 30) H = 5, 25 – 2 = 30 >= 30 

So, this is the situation: 60 . 0 . 0000 0010 . 1XXX XXXX 

Subnet 0: 60 . 0 . 0000 0010 . 100X XXXX 

Subnet 1: 60 . 0 . 0000 0010 . 101X XXXX 

Subnet 2: 60 . 0 . 0000 0010 . 110X XXXX 

 .

 .

 . 

Subnet Broadcast: 60 . 0 . 0000 0010 . 111X XXXX 

Then, from subnet 0: 

Network Address: 60 . 0 . 0000 0010 . 1000 0000 

IP Range: 60 . 0 . 0000 0010 . 1000 0001 - 60 . 0 . 0000 0010 . 1001 1110 

Broadcast Address: 60 . 0 . 0000 0010 . 1001 1111 

Decimal Form of Network Address for subnet 0: 60.0.2.128/27

Step 4 – Do the next largest network: 

For the IP for 4th subnet: 

For 10 host: 

Take subnet 1 from previous subnetting: 60 . 0 . 0000 0010 . 101X XXXX 

Quantity of host: 2H – 2 ≈ 10 So, 2H = 24 (equal or more than 10) H = 4, 24 – 2 = 14 > 10 

So, this is the situation: 60 . 0 . 0000 0010 . 101X XXXX 

Subnet 0: 60 . 0 . 0000 0010 . 1010 XXXX 

Subnet Broadcast: 60 . 0 . 0000 0010 . 1011 XXXX 

Then, from subnet 0: 

Network Address: 60 . 0 . 0000 0010 . 1010 0000 

IP Range: 60 . 0 . 0000 0010 . 1010 0001 - 60 . 10 . 0000 0010 . 1010 1110 

Broadcast Address: 60 . 0 . 0000 0010 . 1010 1111 

 Decimal form of Network Address for subnet 0: 60.0.2.160/28