Subnetting a Class c address Example

 Date 08.06.2021 Size 21.87 Kb.

Subnetting a Class C Address Example
You have been tasked with producing an addressing scheme to support five networks (subnets) within your LAN. You will be using the private class C address 192.168.10.0 /24.
Step 1 – Determine how many network bits you will need to borrow from the available host bits.
Use the formula 2n where n >= the number of networks needed.
2n >= 5 n=3
We will actually have 8 networks even though we only need 5 because 23 = 8
Step 2 – Change the subnet mask to reflect the value for n in step 1. Since n=3 add replace the first 3 zeros in the last octet to ones.
The original subnet mask was 11111111.11111111.11111111.00000000 or 255.255.255.0
The new subnet mask is: 11111111.11111111.11111111.11100000 or 255.255.255.224

Step 3 – Determine how many hosts you will have available on each network
Use the formula 2n -2 where n = the number of 0’s in your subnet mask
25-2 = 30
Step 4 – Determine the new network addresses
Determine the decimal value of the last bit (the last 1) in the subnet mask. In our example the last 1 is in the 32 position so our networks will be in multiples of 32

 128 64 32 16 8 4 2 1 1 1 1 0 0 0 0 0

Network:

1 – 192.168.10.0 5 – 192.168.10.128

2 – 192.168.10.32 6 – 192.168.10.160

3 – 192.168.10.64 7 – 192.168.10.192

4 – 192.168.10.96 8 – 192.168.10.224

The broadcast address will always be the last address in the range or 1 bit less than the next network.

 IP Address First Host Address Last Host Address Broadcast Address 192.168.10.0 192.168.10.1 192.168.10.30 192.168.10.31 192.168.10.32 192.168.10.33 192.168.10.62 192.168.10.63 192.168.10.64 192.168.10.65 192.168.10.94 192.168.10.95 192.168.10.96 192.168.10.97 192.168.10.126 192.168.10.127 192.168.10.128 192.168.10.129 192.168.10.158 192.168.10.159 192.168.10.160 192.168.10.161 192.168.10.190 192.168.10.191 192.168.10.192 192.168.10.193 192.168.10.222 192.168.10.223 192.168.10.224 192.168.10.225 192.168.10.254 192.168.10.255