# Cisco CCNA: Subnetting class C address made easy, from start to finish.

Statistics have shown that apart from access control list and OSPF, the major area that CCNA students encounter challenges while studying for their certification is subnetting. While there are many applications to you help subnet, knowing how to get by manually is very important, after all, you must do it manually to pass the exam. So in this article, I will be sharing how to subnet the class C address with a detailed example.

Subnetting is simply the process of taking a block of IP and breaking it down into smaller blocks called subnets. See it like set and sub-sets in mathematics.  Subnetting helps segment a network, bringing about network security while helping engineers to avoid broadcast storm.

Before we go into sunetting proper, let’s talk about the range of class addresses.
Range of a class C address
If the values in the first octet of an IP address fall between 192 and 223, then it is a class C address. Example, 192.x.x.x to 223.x.x.x.
Now, assuming we have 192.168.0.0/24, with this network address and subnetmask, no subnetting is done because /24 which equals 255.255.255.0, is the default subnetmask for class C addresses. Subnetting starts when you have subnetmasks from /25 to /30. /31 and /32 cannot be subnetted because they will leave you with no IP addresses for hosts.
Example:lets subnet 192.168.0.0/27.
Step one: write out the subnet mask from the given slash notation. /27=255.255.255.224 (/27 means 27 bits on. Each 8 bit is 255 while the last octet of 3 bits equals 224 in decimal)
Step two: determine the number subnets that the subnetmask above will give. Here you use this formula: 2^X, where x is equal to the number of bits borrowed. Since the IP given is a class C IP and the default subnetmask for Class is /24 but we are given /27 in this example, it therefore means we have borrowed 3 bits. Substituting 3 into the formula, we have 2^3=8. This means we well have 8 subnets.
Step 3: How many hosts will the given network address and subnetmask produce per subnet? To answer this question, you will use this formula; 2^y-2, where y equals the number of bits off. If in the question we were given /27 out of a total of 32 (IP version 4 is a 32-bit address), it means we have 32-27=5. We have 5 bits off. This gives us 2^5-2. This will be 32-2 which gives us 30. (The 2 we subtracted is for the broadcast and network addresses. You cannot assign those addresses to hosts on your network). From here we know that we will have 30 valid hosts for each of the 8 subnets we will get from 192.168.0.0/27. Next is to determine the range of these subenets as well as the valid IPs in each subnet.
Step 4: Block size. We will use block size to get the range of the subnets as well as the valid IPs in each subnet. To get the block size you subtract the first non-255 value in the given subnetmask from 256. 256 is a constant value and in this example, 224 is the first non-255 value in our subnetmask; 255.255.255.224. Block size=256-224=32. With a block size of 32, our subnets will be:

You may also like: subnetting class B, from start to finish.

 Subnet 192.168.0.0 192.168.0.32 192.168.0.64 192.168.0.96 192.168.0.128 192.168.0.160 192.168.0.192 192.168.0.224 First IP 192.168.0.1 192.168.0.33 192.168.0.65 192.168.0.97 192.168.0.129 192.168.0.161 192.168.0.193 192.168.0.225 Last IP 192.168.0.30 192.168.0.62 192.168.0.94 192.168.0.126 192.168.0.158 192.168.0.190 192.168.0.222 192.168.0.254 Broadcast 192.168.0.31 192.168.0.63 192.168.0.95 192.168.0.127 192.168.0.159 192.168.0.191 192.168.0.223 192.168.0.255

You can see a video on this here.

That is all you need to answer any question regarding class C subnetting. A post on class B subnetting will be coming soon.