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Welcome to Jeremy's IT Lab. This is a free, complete course for the CCNA. If you like

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In this video, we will be talking about subnetting. This is a very big topic for the CCNA, but

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not just for the test--it’s an essential skill for a network engineer. Many people

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have trouble understanding subnetting, but let me assure you, it is not difficult. Subnetting

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is very simple if you take it step by step. So, I’m going to split subnetting into two,

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or maybe even three, videos so we can take our time to really understand subnetting without

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getting lost. Now, because subnetting is such an important topic, and many people have trouble

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with it, there are already plenty of subnetting videos on YouTube. Of course, feel free to

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check out those videos too--there are some different tricks and techniques people teach

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that can speed up the subnetting process. I, however, will simply outline the basic

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steps involved in subnetting. I will avoid overcomplicating the topic. My end

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goal for these videos is that you understand and can do subnetting. So, let’s get started.

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So, what will we cover in this video? Just a couple of things. First is CIDR, pronounced

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“CIDR,” which stands for Classless Inter-Domain Routing. What exactly is that? Well, remember

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I introduced the IPv4 address classes, such as Class A, B, and C? Well, CIDR throws all

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that away and lets us be more flexible with our IPv4 networks. Then, of course, we’ll

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cover the process of subnetting, taking it step by step so you don’t get lost.

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Now, before I get into CIDR, let’s review these IPv4 address classes so we can then

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understand the need for classless IPv4 addressing. There are five classes of IPv4 addresses:

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A, B, C, D, and E. Class A addresses have a first octet beginning with zero, and the rest

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of the bits can either be zero or one. This leads to a decimal range for the first octet of

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0 to 127. Remember, an IPv4 address is 32 bits, so there are 4 octets--4 groups of 8

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bits--in an IPv4 address. This makes the Class A address range from 0.0.0.0 through 127.255.255.255.

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Now, remember, there are some special and reserved addresses in these ranges that can’t be

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used for normal IP addresses on a device, but for this video, we’ll just include all

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of them in Class A. Class B addresses have a first octet beginning with 10, and the

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other 6 bits can be either 0 or 1. This gives a range for the first octet of 128 through

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191. The address range for Class B is 128.0.0.0 through 191.255.255.255. Class C addresses

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have the first three bits set to 110, and the others can be either zero or one. If you write

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that range in decimal, it is 192 through 223. The address range is therefore 192.0.0.0

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through 223.255.255.255. Class D addresses begin with 1110 in binary, which gives

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a range of 224 through 239 for the first octet of the address. This means that the address range

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for Class D is 224.0.0.0 through 239.255.255.255. Finally, Class E addresses begin with 1111

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in binary, so the first octet range is 240 through 255, and therefore the address range is 240.0.0.0

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through 255.255.255.255.

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However, only the Class A, B, and C addresses can be assigned to a device as an IP address,

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as Classes D and E have special purposes, as I mentioned in the IPv4 addressing videos. Class

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A addresses have an 8-bit prefix length, meaning the first octet identifies the network and

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the other three octets are used for individual hosts within the network. Class B addresses

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have a 16-bit prefix length, so the first two octets identify the network, and the last

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two octets identify individual hosts within that network. Class C addresses have a prefix

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length of 24, so the first three octets are used to identify the network, and only the

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last octet is used to identify individual hosts within that network.

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The different prefix lengths give different characteristics to these classes. As you can

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see, there are few Class A networks available--only 128, actually less than that because

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some are reserved, like the 127.0.0.0 range, which you may remember is used for loopback

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addresses. Because only the first octet of a Class A address is used for the network ID,

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there are three whole octets available for addresses within each Class A network,

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so there are 16,777,216 addresses in each Class A network. That is

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2 to the power of 24, because there are three octets (3 times 8 = 24 bits). Class B

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addresses are different. There are more Class B networks--16,384--but fewer addresses per

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network, 65,536, which is still many addresses, of course. Finally, there are very

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many Class C networks--2,097,152 networks--but only 256 addresses per network.

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So, how does a company get their own network address range to use? Well, IP addresses are assigned to

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companies or organizations by a nonprofit American corporation called the IANA, the

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Internet Assigned Numbers Authority. The IANA assigns IPv4 addresses and networks to companies

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based on their size. For example, a very large company might receive a Class A or Class B

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network. Remember, there are lots of available addresses to use for hosts in each Class A

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and Class B network. While a small company might receive a Class C network, because there

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are fewer addresses in each Class C network--only 256. However, this system led to many

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wasted IP addresses, so multiple methods of improving this system have been created. Let

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me give you an example of how this strict system of addresses can waste IP addresses.

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So, here are two routers. As you can see, R1 has three networks connected to it here.

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Remember that routers are used to connect different networks, so each of these links is a separate

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Layer 3 network, different IP networks. R2 also has three networks connected here. Perhaps

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each of these networks will have a few switches, with many end hosts such as PCs and servers

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connected to these switches. However, there is one more network here. That’s this network

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connecting these two routers. This is known as a point-to-point network, meaning

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that it’s a network connecting two points, in this case, R1 and R2. For example, this

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might be a connection between offices in different cities, let’s say San Francisco and New York.

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So, because this is a point-to-point network, we don’t need a large address block, so

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let’s use a Class C network, 203.0.113.4. Because this is a Class C network, there are

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256 addresses in the network, minus one for the network address (203.0.113.0), minus one

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for the broadcast address (203.0.113.255), minus one for R1’s address, which I’ll

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assign as 203.0.113.1, and minus one for R2’s address, which I’ll assign as 203.0.113.2.

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That’s a total of four addresses used and 252 addresses wasted.

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Clearly, this is not an ideal system.

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Before introducing CIDR, here’s another quick example of address waste. A company,

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Company X, needs IP addressing for 5,000 end hosts. This is a problem, why? A Class C network

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does not provide enough addresses, so a Class B network must be assigned. Because a Class

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B network allows for about 65,000 addresses, this results in about 60,000 addresses being wasted.

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When the Internet was first created, the creators did not predict that the Internet would become

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as large as it is today. This resulted in wasted address space, like the examples I showed

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you, and there are many more examples that I could show you. The total IPv4 address space

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includes over 4 billion addresses, and that seemed like a huge number of addresses when

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IPv4 was created, but now address space exhaustion is a big problem. There's not enough addresses. One way to solve, or remedy, this problem is

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CIDR. The IETF (Internet Engineering Task Force) introduced CIDR in 1993 to replace

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the classful addressing system.

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With CIDR, the requirements of Class A addresses to use an 8-bit network mask, Class

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B to use 16, and Class C to use 24 were removed. This allowed larger networks

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to be split into smaller networks, allowing greater efficiency. These smaller networks

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are called subnetworks, or subnets. Let’s look at an example of splitting a

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larger network into a smaller network so you can see how it works.

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Here’s the same point-to-point network we looked at before. Previously, it was assigned

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the 203.0.113.0/24 network space, but that resulted in lots of wasted addresses. Let’s

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write this out in binary. Here’s the binary, with the dotted decimal underneath. Now, the

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prefix length is 24, so here’s the network mask, also known as the subnet mask: 255.255.255.0.

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Remember, all 1s in the subnet mask indicate that the same bit in the address

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is the network portion. In this case, I’ve made the network portion blue, and the host portion

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is red. Well, how many host bits are there? 8, because it’s one octet. So, how many potential hosts, or how

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many usable addresses, are there? Well, the formula is this: 2 to the power of 8 minus

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2 equals 254 usable addresses. What is the 8? Well, it’s the number of host bits, which is

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8 in this case. And why minus 2? Those are the network address and the broadcast address.

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We can’t assign them to a device, so we have to remove them from the number of usable addresses.

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So, we have 254 usable addresses, but we only need two--one for R1 and one for R2.

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However, CIDR allows us to use different prefix lengths, so it doesn’t have to be 24.

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Let’s get some practice calculating the number of hosts within different prefix lengths.

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203.0.113.0/25, 203.0.113.0/26, 203.0.113.0/27, /28, /29, /30, /31, and finally /32. I’ve

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put /31 and /32 in red because they’re a little bit special, as you’ll see when you

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try to calculate it. So, pause the video here and try to calculate how many usable addresses

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are on each network. Okay, let’s check out the answers.

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So, here is 203.0.113.0/25, but this time with a /25 mask. Notice that the network portion

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of the address has extended into the first bit of the last octet, and the mask

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in dotted decimal is now written as 255.255.255.128. I changed the color of the extra bit to purple,

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but it is part of the network portion, which is the blue part. If you don’t remember how to convert

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from binary to dotted decimal, make sure you review that; it’s very important for subnetting.

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Now, there are 7 bits in the host portion of the address, so the number of usable addresses

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is 2 to the power of 7 minus 2, which equals 126. Once again, we only need two addresses--

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one for R1 and one for R2--so we will be wasting 124 addresses. That’s better than wasting

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252 addresses with a /24 prefix length, but it’s still wasteful.

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How about a /26 prefix length? Notice that it’s now written as 255.255.255.192 in dotted

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decimal, because two bits of the last octet are now part of the network portion. Since

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there are six host bits, there are now 62 usable addresses in this network. If we were to use

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a /26 network mask for the 203.0.113.0 network, we would be wasting 60 addresses. Getting

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better, but we can make this network even smaller.

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Now that you get the idea, let’s speed it up. For a /27 prefix length, the mask is written

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as 255.255.255.224 in dotted decimal. There are now five host bits, so that means there are

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30 usable addresses. As you can see, the address space is getting smaller and smaller as we extend the network mask.

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For a /28 prefix length, the mask is written as 255.255.255.240 in dotted decimal. There

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are now only four host bits, so that means there are 14 usable addresses. After assigning addresses

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to R1 and R2, this would mean only 12 wasted addresses, but we can make this address space

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even smaller to make our addressing even more efficient.

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If we use a /29 prefix length, the mask is written as 255.255.255.248 in dotted decimal.

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Now we have only three host bits, so that means there are just six usable addresses. Again,

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after we give R1 and R2 addresses, there would be only four wasted addresses.

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If we use a /30 prefix length, the mask is written as 255.255.255.252 in dotted decimal.

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There are now only two host bits, so that means two usable addresses. So, this is perfect. There

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are four total addresses: the network address, the broadcast address, R1’s address, and

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R2’s address. That means zero wasted addresses.

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Before moving on to look at the /31 and /32 prefix lengths, let me clarify a little bit. So, instead of 203.0.113.0/24,

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we will use 203.0.113.0, which is a subnet of that larger Class C network. 203.0.113.0

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includes the address range of 203.0.113.0 through 203.0.113.3. Let me show you that

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in binary. Here is 203.0.113.0 in binary, the host portion all zeroes. Here is 203.0.113.1,

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203.0.113.2, and 203.0.113.3. These are the four addresses in the network, with these two being

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the two usable addresses, which are assigned to R1 and R2. So, we took up four addresses with

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this subnet. What about the other addresses in the 203.0.113.0/24 range? The remaining

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addresses in the address block, which are 203.0.113.4 through 203.0.113.255, are now available

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to be used in other subnets. That’s the magic of subnetting. Instead of using 203.0.113.0/24

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and wasting 252 addresses, we can use /30 and waste no addresses. Or, perhaps there is another

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way to make this even more efficient. Let’s look into it.

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If we use a /31 prefix length, the mask is written as 255.255.255.254 in dotted decimal.

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There is now only one host bit, so that means zero usable addresses. Two to the power of one is two,

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minus two for the network and broadcast addresses, means zero addresses that we can assign to devices.

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So, you used to not be able to use /31 network prefixes because of this. However, for a point-to-point

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connection like this, it actually is possible to use a /31 mask. Let's check it out.

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So, here’s the 203.0.113.0/31 network. R1 is 203.0.113.0, and R2 is 203.0.113.1. The

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203.0.113.0/31 network consists of addresses from 203.0.113.0 through 203.0.113.1, which

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is actually only two addresses. Here they are in binary. There’s 203.0.113.0, and

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there’s 203.0.113.1. Normally, this would be a problem because it leaves no usable

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addresses after subtracting the network and broadcast addresses, but for point-to-point

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networks like this, a dedicated connection like this between two routers, there is actually

0:19:15.530,0:19:20.900
no need for a network address or a broadcast address. So, we can break the rules in this

0:19:20.900,0:19:27.290
case and assign the only two addresses in this network to our routers. Note that if

0:19:27.290,0:19:32.020
you try this configuration on a Cisco router, you’ll get a warning like this, reminding

0:19:32.020,0:19:37.490
you to make sure that this is a point-to-point link, but it is a totally valid configuration.

0:19:37.490,0:19:49.560
So, once again, the remaining addresses in the 203.0.113.0/24 address block, which are 203.0.113.2 through 203.0.113.255,

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are now available to be used in other networks. But this time, we've

0:19:54.730,0:20:02.110
saved even more addresses, using only two addresses instead of four for this point-to-point connection.

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People still do use 30 for point-to-point connections at times, but 31 masks are totally

0:20:08.130,0:20:12.380
valid and more efficient than 30, so I recommend this method.

0:20:14.130,0:20:23.030
But, we still haven't looked at the 32 mask. A 32 mask is written as 255.255.255.255 in

0:20:23.030,0:20:29.620
dotted decimal, making the entire address the network portion. There are no host bits.

0:20:29.620,0:20:35.280
If you calculate this using our formula, you will get one usable address. Clearly, the

0:20:35.280,0:20:41.510
formula doesn't work in this case. You won't be able to use a 32 mask in this case, and

0:20:41.510,0:20:47.461
you will probably never use a 32 mask to configure an actual interface. However, there

0:20:47.461,0:20:53.180
are some uses for a 32 mask. For example, when you want to create a static route not

0:20:53.180,0:21:00.410
to a network, but to just one specific host, you can use a 32 mask to specify that exact host.

0:21:00.410,0:21:05.710
Anyway, I'll talk about that later in the course. Just know that 32 masks are

0:21:05.710,0:21:10.090
used at some points, but you don't have to worry about them for now.

0:21:10.090,0:21:15.550
Here's a simple chart showing the dotted decimal subnet masks and their equivalent

0:21:15.550,0:21:21.170
in CIDR notation. That's right, the way of writing a prefix with a slash followed

0:21:21.170,0:21:29.780
by the prefix length, like 25, 26, etc., is called CIDR notation because it was introduced

0:21:29.780,0:21:36.880
with the CIDR system. Previously, only the dotted decimal method was used. Note that

0:21:36.880,0:21:41.630
I've shown you only how to subnet a class C network so far, but we will look at

0:21:41.630,0:21:49.790
class B and class A networks as well, with prefix lengths like 17, 11, 9, etc.

0:21:49.790,0:21:55.500
I spent a lot of time on just that one example, but I hope you can see the use of

0:21:55.500,0:21:59.891
subnetting--dividing a larger network into smaller networks called subnets.

0:21:59.891,0:22:07.130
Instead of using the whole 203.0.113.0/24 network for the point-to-point connection, we can

0:22:07.130,0:22:14.070
use a 30 subnet and use only four addresses, or even better, use a 31 subnet and use only

0:22:14.070,0:22:20.290
two addresses. I'll give one more example of subnetting before finishing up this video.

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In the next video, I'll give you some practice problems and walk you through them so you

0:22:23.870,0:22:26.710
can get some hands-on practice with subnetting.

0:22:26.710,0:22:33.620
So, here's a scenario: There are four networks connected to R1, with many hosts connected

0:22:33.620,0:22:40.760
to each switch. There are 45 hosts per network. R1 needs an IP address in each network, so

0:22:40.760,0:22:49.710
its address is included in that 45-host number. You have received the 192.168.0.14 network,

0:22:49.710,0:22:54.080
and you must divide the network into four subnets that can accommodate the number of

0:22:54.080,0:23:01.950
hosts required. First off, are there enough addresses in the 192.168.0.14 network in

0:23:01.950,0:23:09.010
the first place? We need 45 hosts per network, including R1, but also remember that each

0:23:09.010,0:23:15.800
network has a network and broadcast address, so that's plus two. So, we need 47 addresses per subnet.

0:23:15.800,0:23:24.240
47 times 4 equals 188, so there's no problem in terms of the number of hosts.

0:23:24.240,0:23:32.560
192.168.0.0/24 is a class C network, so there are 256 addresses. Therefore, we will be able to assign

0:23:32.560,0:23:36.460
four subnets to accommodate all hosts, no problem.

0:23:36.460,0:23:43.160
Okay, let's see how we can calculate the subnets we need to make. We need four equal-sized subnets

0:23:43.160,0:23:50.450
with enough room for at least 45 hosts. Here, I've written out 192.168.0.10

0:23:50.450,0:24:00.480
with a 30 mask, 255.255.255.252. I skipped 32 and 31 since these aren't point-to-point links.

0:24:00.480,0:24:08.190
We can't use 31 and definitely can't use 32. Since there are two host bits,

0:24:08.190,0:24:14.290
the formula to determine the number of usable addresses is

0:24:14.290,0:24:20.970
2^2 - 2. 2^2 is 2 times 2, which is 4, so that means there are two usable addresses

0:24:20.970,0:24:28.220
in a 30 network. Clearly, not enough room to accommodate the 45 hosts we have.

0:24:28.220,0:24:36.330
How about if we use a 29 mask to make these subnets? Can we fit the 45 hosts we need? There are three host bits,

0:24:36.330,0:24:43.710
so the formula is 2^3 - 2. 2^3 is 2 times 2 times

0:24:43.710,0:24:51.910
2, which is 8. Therefore, there are six usable addresses, not enough for 45 hosts.

0:24:51.910,0:25:00.250
How about if we use 28? There are four host bits, so the formula is 2^4 - 2.

0:25:00.250,0:25:07.710
2^4 is 2 times 2 times 2 times 2, which is 16. So, that means there are

0:25:07.710,0:25:13.410
14 usable addresses--once again, not enough for 45 hosts.

0:25:13.410,0:25:23.400
How about 27? There are five host bits, so the formula is 2^5 - 2. And 2^5

0:25:23.400,0:25:30.169
is 2 times 2 times 2 times 2 times 2, which equals 32. So that means

0:25:30.169,0:25:35.370
30 usable addresses. Again, not enough for 45 hosts.

0:25:35.370,0:25:43.740
How about a 26 subnet mask? There are now six host bits, so the formula is 2^6 - 2.

0:25:43.740,0:25:51.480
2^6 is 2 times 2 times 2 times 2 times 2 times 2, which equals 64.

0:25:51.480,0:25:59.120
That means there are 62 usable addresses. So, it looks like we've found our number. 27

0:25:59.120,0:26:04.960
doesn't provide enough address space, but 26 provides more than we need, so we have to

0:26:04.960,0:26:10.940
go with 26. Unfortunately, you can't always make subnets have exactly the number of addresses

0:26:10.940,0:26:16.570
you want. There might be some unused address space. That's actually fine, since it's good

0:26:16.570,0:26:20.500
to have some room for growth anyway.

0:26:20.500,0:26:26.571
So, I think this video has gone on long enough. Instead of finishing this task in this video, I'll make

0:26:26.571,0:26:36.660
it this week's quiz. The first subnet, subnet one, is 192.168.0.16. What are the remaining

0:26:36.660,0:26:42.660
subnets? To help you out, here's a hint: Find the broadcast address of subnet one.

0:26:42.660,0:26:50.120
The next address after that is the network address of subnet two. And then just repeat the process for subnets

0:26:50.120,0:26:56.780
three and four. Post your answers in the comment section, and I'll also go over the answer in the next video.

0:26:57.970,0:27:05.300
So, what did we cover in this video? We covered CIDR (Classless Inter-Domain Routing), which

0:27:05.300,0:27:10.680
removes the rules of class A, B, and C networks and lets us be more flexible with network

0:27:10.680,0:27:17.030
addressing, according to the size of the network. We also covered the process of subnetting,

0:27:17.030,0:27:22.040
but mostly just the basics. Hopefully, you understand the purpose of subnetting and

0:27:22.040,0:27:27.120
know a little bit about how to do it. I will clarify and expand upon many things in the

0:27:27.120,0:27:33.960
next video, but also feel free to ask any questions you have in the comments section.

0:27:33.960,0:27:38.760
For today's video, there won't be a practice lab; that will be after I've finished explaining everything about

0:27:38.760,0:27:43.760
subnetting. There will be flashcards, however, to help you review some of the things learned

0:27:43.760,0:27:48.120
in this video. You can download them from the link in the description.

0:27:48.120,0:27:53.280
I've also recently enabled the membership feature for my channel. If you want to leave

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a monthly tip to support me, this is another great way to do so. Click "Join" here under

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0:28:01.710,0:28:08.230
For those who become a JCNP (Jeremy Certified Network Professional) level supporter, I'll

0:28:08.230,0:28:14.700
give you a shoutout at the end of my videos. So, first of all, thank you so much to Vance Simmons. I just

0:28:14.700,0:28:19.080
enabled the membership feature and haven't said anything about it yet, and he became my first

0:28:19.080,0:28:24.670
JCNP level supporter. Thank you so much for supporting the channel. I hope the videos are helping

0:28:24.670,0:28:30.107
you out. And for my JCNA level supporters, thanks to you too.

0:28:33.080,0:28:37.950
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0:28:37.950,0:28:43.030
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