Subnetting on the 3rd and 4th Octet – examples

Cisco CCNA Subnetting Examples

In this Cisco CCNA tutorial I’ll work through a couple of subnetting examples. One where we’re subnetting on the 3rd octet, and one where we’re subnetting on the 4th octet.

Vytas Banionis

Vytas Banionis

Awesome course! Everything was explained clearly. The lab demos helped very much in seeing how configurations were done followed by lab exercises. During the course, often I’d have a question pop up and sure enough, Neil knew this would be asked and already was ahead of the game knowing you’d be asking the question. I found Neil to be VERY thorough in everything he explained. I’d reach out to him personally and always got a prompt reply with a clear answer.
I have completed the course and come back to many of the sections I needed more review on. This course is great to have as a reference tool.
I’m glad I chose this course and HIGHLY recommend taking it. You will not be disappointed. After taking this course, you will come out confident of knowing everything you need to know for the CCNA exam.

Vytas Banionis

Subnetting on the 3rd Octet – Example

“For the IP address 172.19.216.50/23, what is the network address, broadcast address and range of valid hosts?”

 

The first thing to figure out is which octet we are subnetting on.

/1 to /8 = 1st octet

/9 to /16 = 2nd octet

/17 to /24 = 3rd octet

/25 to /31 = 4th octet

The example uses /23 so we’re subnetting on the 3rd octet.

 

I like to underline the octet I’m working on at this point so that I don’t accidentally confuse the octets later:

172.19.216.50/23

The 172.19. part of the address (the first 2 octets) is never going to change in our example. Values in the 3rd and 4th octet can change, but the values in the first 2 octets will always remain the same.

 

Next up let’s figure out the network address.

When the subnet mask is /23, the first 23 bits are used for the network portion of the address and the remaining 9 bits are used for the host portion.

Write the octets down in binary and then count along 23 bits.

Subnetting on the 3rd Octet

We can see that the 23rd bit falls on a ‘2’ (the number in the top row), so the subnets will go up in multiples (a block size) of 2 on the 3rd octet:

172.19.0.0/23

172.19.2.0/23

172.19.4.0/23

172.19.6.0/23

Etc…

172.19.214.0/23

172.19.216.0/23

172.19.218.0/23

172.19.220.0/23

Etc...

 

172.19.216.0/23 is the network address we are working on.

172.19.218.0/23 is the next network address.

 

Next we figure out the broadcast address. This is one less than the next network address:

172.19.217.255 is the broadcast address.

172.19.217.255 is one less than the next network address of 172.19.218.0

When subnetting on the 3rd octet, remember the broadcast address will be 255 in the 4th octet – NOT 0 in the 4th octet (this is a common mistake to make). The broadcast address is 172.19.217.255, not 172.19.217.0.

(172.19.217.0 is actually a valid address in the range which could be assigned to a PC or other host.)

 

The last thing to figure out is the range of valid host addresses. This is everything between the network address and broadcast address.

172.19.216.1 to 172.19.217.254 is the range of valid host addresses.

172.19.216.1 is one more than the network address of 172.19.216.0

172.19.217.254 is one less than the broadcast address of 172.19.217.255 (don’t write 172.19.216.254 by mistake.)

 

Subnetting on the 4th Octet – Written Example

“For the IP address 172.19.216.50 255.255.255.240, what is the network address, broadcast address and range of valid hosts?”

 

The first thing to figure out is which octet we are subnetting on.

The subnet mask is 255.255.255.240

We can see we are not subnetting on the first 3 octets as the value is 255.

We are subnetting on the 4th octet with a value of 240.

 

I like to underline the octet I’m working on at this point so that I don’t accidentally confuse the octets later:

172.19.216.50/28

The 172.19.216 part of the address (the first 3 octets) is never going to change in our example.

 

Next up let’s figure out the network address.

Write down the 4th octet in binary then add the digits together from the left until they add up to the ‘240’ used in the subnet mask.

Subnetting on the 4th Octet

128 + 64 = 192

192 + 32 = 224

224 + 16 = 240

The first 4 bits in the 4th octet are being borrowed for the network portion of the address and the remaining 4 bits are used for the host portion.

Subnetting on the 4th Octet 1

We can see that the network portion of the address falls on a ‘16’, so the subnets will go up in multiples (a block size) of 16 on the 4th octet:

172.19.216.0/28

172.19.216.16/28

172.19.216.32/28

172.19.216.48/28

172.19.216.64/28

Etc…

 

At this point we can see which subnet the IP address in our example is in. 172.19.216.50 falls in the range between 172.19.216.48 and 172.19.216.64, so it is in the 172.19.216.48 subnet.

(‘50’ is between ‘48’ and ‘64’.)

172.19.216.48/28 is the network address we are working on.

172.19.216.64/28 is the next network address.

 

Next we figure out the broadcast address. This is one less than the next network address:

172.19.216.63 is the broadcast address.

(‘63’ is one less than ‘64’.)

 

The last thing to figure out is the range of valid host addresses. This is everything between the network address and broadcast address.

172.19.216.49 to 172.19.216.62 is the range of valid host addresses.

(‘49’ is one more than the network address of ‘48’, ‘62’ is one less than the broadcast address of ‘63’.)

 

Additional Resources

 

Subnetting Overview

Subnetting Class A and Class B Networks Tutorial

VLSM Variable Length Subnet Mask Example

Subnetting Practice Questions

 

Want to practice Cisco CCNA technologies on your laptop? Download my complete 350-page Cisco CCNA Lab Guide for free.

 

Click Here to get my Cisco CCNA Gold Bootcamp, the highest rated CCNA course online with a 4.8 star rating from over 20,000 public reviews.