DEMO: Subnetting Example C-1

We're ready to hit our first subnetting exercise. We're going to do a class C, because that's one of the easier ones to do. Although, class B and A are really just the same thing, it's just on a bigger scale. So let's tackle a class C and see what a subnetting exercise looks like.

Everything that we've been learning is leading up to these exercises. This is really where most of the bulk of what we're going to be doing is at. So you've learned most of the tools. We're going to learn a couple new things here, but for the most part, you've learned all the tools to be successful at this. You just need a little bit of practice.

Setting Up the Problem

Let's go through this first practice here. We're given some information, and we're going to use that to solve this. We need to fill in some information here, so we'll do some calculations to figure that out. And then we've got an area for a scratchpad here where we're going to actually fill in some of this data and figure things out.

So what are we given? We're given the number of subnets. We're given the number of hosts that we're going to need. We're given an IP address range, and they've already filled this in right here for us. And this favor thing, we don't need to worry about that. That's not applicable on this go-round, and I'll show you why here in a second.

Class, Mask, and Borrowed Bits

The first question asks us what class it is. Well, it's a C. What's the default mask for a class C? Well, that's 255.255.255.0. How many borrowed bits? We still need to figure that out. So we need four subnets. So this one's not going to work, but this one will. We need 60 hosts, so that won't work, that won't work, that won't work, that won't work — this would. So there's only one place that we would draw the line. If there were multiple places, that's where we would need to know if we're going to favor the network side of it, favor the host side of it, or find something in between.

So number of borrowed bits — this tells us there's two borrowed bits. What is the custom subnet mask with that? Well, two borrowed bits gives us 192. So 255.255.255.192 is the answer for that. Total number of subnets, we have four there, so just enough. Total number of hosts, we've got 62 — well, 64, but there's 62 usable hosts. So 62 usable hosts.

Finding the Network Ranges

Now let's start figuring this out. What are our actual network ranges? If we've borrowed two bits here, then we have two to the second power, which is four possibilities. There's four networks that we've created. What are those networks? Well, here's where your binary counting will come in handy, because we have 0 0, we've got 0 1, we've got 1 0, and we've got 1 1.

Now, for each of these, to figure out what the starting range is going to be — that is the network address — the network address would have all zeros here. So that's not too difficult to figure out. It's going to be whatever these first two bits actually make it up. So the starting range for this first one is all zeros for this last octet. So we're going to have a zero network. So that's zero. The next one, all zeros except for the 64, so that's a 64. We've got a 128, and 128 plus 64 is 192. So there are our network addresses.

First Usable Address

The first usable address, since we can't use the network address, is just going to be one up from that. So we've got a dot one. Let me space it out a little more here. So we've got a dot one, we've got a dot 65, we've got a dot 129, and a dot 193. So that's the first usable host in the range, and we are going to figure out the whole range here.

Broadcast Addresses

Let's figure out what the broadcast address is. In this first example, I'm going to just put all ones across here for the broadcast address. So if the first two are 0 0 and then with all ones, we'd add it up: 32 plus 16 plus 8 plus 4 plus 2 plus 1 would be 63. I'm going to put that at the end here. So 63 is our broadcast address.

If I were to do the same thing, 63 plus 64 — because this bit right here, we're using these two — so 63 plus 64 is .127. So .127 is our broadcast address for this second network right here. I have a 128 plus the 63 to find the end of that. So 63, because all of these are ones right there. So to figure out, 128 plus 63 is going to be 191. And then the last one is double ones here, with all ones here. So the last one is 255.

To double check our work, what we could do is compare the end of this range should be the beginning of the next range. So 63, 64; 127, 128; 191, 192; 255 is the last one. So that makes sense. So this is the broadcast, this is our network, and these are your usable hosts.

Last Usable Host

The last usable host in this range would be just one less than the 63, so 62. In this range would be 126, so 126. And this range right here would be 190, so 190. And in this range would be 254, so 254.

There you go. So now we have four different networks: 1, 2, 3, 4. We have the network addresses for each of those ranges. We've got the usable host range for each one of those. And we've got the broadcast address for each one of those. So there's the answer. That is how you solve these subnetting problems. And now we've got a solution for each of our networks and the network range, and how many usable hosts are in each range. We have figured it all out.