CompTIA Network+ N10-008 – IP Addressing Part 2

  1. Computer Mathematics Practice

Now that we’ve covered the basics of computer mathematics, let’s go through a couple of practice problems. Now as we go through this lesson, I want you to pause the video and try to answer the questions yourself, and then restart the video to see if you got the right answer. Now we want to make sure that you can convert from binary to decimal and from decimal to binary, because this is crucial as we move into subletting. So remember, if you’re going from binary to decimal, we want to add things together. So for our first practice, let’s convert from binary to decimal. If you have the number 0110-1011, what is that in decimal? Now for your first step, you’re going to fill in the number from right to left into your chart, and then we’re going to add it together.

So did you get the right answer? Well, if we fill in our chart from right to left, we’re going to add up the numbers 64, 32, eight, two, and one, which gives us 107. If you got 107, you’re doing great on binary to decimal conversions. Now let’s try one more. Let’s convert the number 1001-0100.Pause the video, and we’ll return to see if you got the correct answer. All right, we take our first step of filling in the chart from right to left, and then we look everywhere.

There’s one. So we have 128 plus 16 plus four. What is that going to equal? That’s right, 148. You’re on your way to understanding binary to decimal conversions if you got the correct answer of 148. Next, let’s convert decimal to binary. So I’m going to give you the number 49, and I want you to pause the video and convert it to binary. Alright, thank you for coming back. So if we have 49, we’re going to fill that in by subtracting from left to right using our table. So can 128 go into 49? Well, no, can you? No, but 32 can. So we’re going to have one. And then we’ll subtract 32 from 49, leaving us with 1716can go into 17, which then leaves us with one.

And then we have eight. Four and two can go into one. So we get zeros there. And finally, one minus one is zero. So we have three in place: one in the 32 column, one in the 16 column, and one in the one column. Now to double-check our math, we’re going to add those back together. So if we add 32 plus 16 plus one, we do get 49. And you can see that 49 does convert to the binary number 001–1001, and we have solved that problem. Did you get it right?

If so, you’re doing great. If not, you may want to find some more practice problems. How do you find practice problems? Well, you can make any number you want go from binary to decimal and from decimal to binary. Just go to Google and in the search bar type “convert the number to binary.” So, if you convert eight to binary, you’ll get something that says 1000, whereas if you convert 49 to binary, you’ll get the answer 001-001. As a result, simply converting X numbers to binary allows you to create an infinite number of problems. Or if you have a binary number, convert that binary number to decimal, and you can go back and forth and practice as much as you want until you solidify these concepts. But just remember the basics: when going from binary to decimal, you add; when going from decimal to binary, you subtract.

  1. Subletting

Subletting. So now that we’ve gotten the basics under our belt, let’s dive into subletting. Subletting is where we can take a large network and split it up into smaller networks. And when we do this, we are logically addressing the default group. Subnet masses are rarely the optimal choice for subnet size. Subnets can be modified using subnet mass to create networks that are much better scoped. For example, if I gave you a ten-something network as a class A address, that means there are 16.7 million IPS in that range. Now, you may have a large network, but your network has probably not reached 16.7 million clients. So instead, we want to create subnets that will borrow bits from the original host portion and add them to the network portion to make smaller networks. Now, if I had one like 100 0/8, that is a full class A subnet with 16 million hosts. But really, I probably only need 256 hosts or less. As a result, I can use a 24 instead, which is the Apache default subnet mask.

So 100 00:24 gives me all the IPS in that subnet, and then I can have a second subnet with the 100 subnet of 100. I still have millions of IPS that I can subnet out later. Now, why do I even need subnets? Well, it’s an efficient use of IP addresses instead of using the full default, because even in your home network, if you’re using a class C address of, that gives you 256 possibilities. But you probably only have five or ten devices on your network.

This will allow us to create additional VLANs and subnets and allow separation of our networks for better security and better bandwidth control. Now, here on the screen, you’re going to see a chart that has the addresses for A, B, and C subnets and their default masks, which should be a review for you. On the right side, you’ll see the assignable number of IP addresses, which is, for class A, 16.7 million addresses. For class B, there are 65,000 plus IP addresses, and for class C, there are 254 usable IP addresses. That is a lot of IPS, so we need to be able to start necking those down into usable chunks. And that’s what we use subletting for. Now, in the private IPS, this isn’t as big of a deal, but with the public IPS, every public IP costs you money. So you only want the amount that you need, and subletting allows you to do that. How do we do that? Well, we use what’s called the subnet mask. And we’ve talked about classes A,  B, and C default masks before.

As you can see here on the screen, they are 816 and slash 24. When converted to binary, you can see that each octave is filled with either all ones or all zeros, making them class full. Now, the ones on the bottom of the screen, from 25 down to slash 30, are actually smaller subnets. And they are referred to as class less because they will not use all ones or all zeros. And you can see how we’re borrowing bits from the host portion, which is all zeros, and taking some of that by adding ones there and putting it more into the network portion. Now we’ll dive a little bit deeper into that as we go in the next couple of slides. So let’s take a look at some formulas that we need to understand for calculating the number of subnets and the number of assignable hosts. So if I’m calculating the number of subnets, the formula is two to the S, where S is the number of borrowed bits from the host space. So, for instance, if I’m using a slash 25, I am borrowing one bit from the host space and putting it into the network space.

So it would be two to one or two. Now, if I’m looking at the number of assignable IP addresses, this is two to the number of host bits minus two. So in the case of my 25 networks, there are 32 bits total. I have 25 of them that are being used for the network portion, leaving me seven host bits. So two to the seventh minus two equals 128 minus two equals 126 available. Now, why do we have a minus two? Well, every network has a network ID and a broadcast ID. The network ID is the first IP. The broadcast ID is going to be the last IP. As a result, no matter which network you choose, you will always sacrifice the first and last as the network name and broadcast. So, whenever you calculate this number of usable, it will always be the number minus two. Let’s take a look at classful versus subnetted networks. If you recall, a classy network is one that begins with slash 816 or ends with slash 24. So if I use the example of 192, 168, dot one, dot zero, and slash 24, this is a classful class C network. Because it is two to the zero, there is only one network. because we have borrowed zero bits. Because 24 is a whole number, that means there are two to the eighth minus two assignable IPS. So 256 IPS minus two would give us 254 usable IPS.

Now, as you can see here on the screen, we have all of the network bits, all 24 of them, and we have eight of the host bits. Now, let’s say I wanted to make smaller subnets. How could I do that? Well, I can borrow two bits from the host and create a subnet. So if I want to make this a 26, for instance, I’m going to borrow two host bits. So it now becomes two to the S or two to the second, which gives me four networks. Now, how many IPS can I have on each of those four networks? Well, originally I had 256 total. Now I’m going to have two borrowed host bits, so I only have two of the six host bits left. So that is going to give me 64 IPS per subnet. And remember, of those, I have to give away two. One for the broadcast and one for the network name. So I only have 62 usable IPS for each of those four created subnets. Now, what is that going to look like? Let me show you a diagram here in the blue At the bottom, you can see my original 256 possible IPS, which does include the broadcast and the network name.

If I subnet that down, I am going to have four subnets, or two for the second. And I am going to have 62 possible IPS used in each one because the first and the last are always going to be the name of the network. In our case, 101-64-1128 or 1192 And we have the broadcast, which is the lastIP of the subnet, the dot 63, the dot127, the dot 191, and the dot 255.So are you starting to see how these subnets start working together? Now, if we wanted to calculate those IPS, we were going to do that based on the mask. Again, we take 32, which is the total number of bits minus whatever our site or slash notation is. In this case, 26. So 32 minus 26 equals six host bits remaining; that would be two to the sixth minus two, which is 64 minus two or 62 usable assignable IPS. I’m hoping this is all starting to come together. Subletting is a very complex topic that a lot of students struggle with. So keep practising it and keep working through these problems because I guarantee you’re going to get a number of subletting questions on test day. So let’s talk about how we can list out these subnets out. Let’s keep going with our example of 192. Dot 168, dot one, dot zero, slash 26, which means we’ve created four subnets of 62 usable IPS each.

Where does each network begin and end? Well, we’ve already kind of gone over this by saying that it was the zero, the 64, the 128, and the 192. But where did I come up with those? Well, I started with what the original IP was. 192, 168, 10, So zero is my first one. And each subnet was 64 IPS in length. So there are zero, 64, 128, and 192. What is each one’s most recent IP address? That’s the broadcast. So all I have to do is add 63 to each of those numbers. So I become 6312-7191 and 255. And that will be the start and end of each subnet, as well as everything in between. For example, 1–62, 65–126, 129–190, and 193–254 are all usable hosts for servers, clients, tablets, laptops, and other devices. So when we do all of this, we can write all of these out and do 192, 168, one dot zero, 192, 168, one dot 1127. But if we did that for each of these four subnets, it would get rather long. So instead we have a shorthand notation called “Cider classless inner domain routing.” So instead of advertising multiple individual routes, we can summarise them all as a single route. And we do this through cider notation. Now, in the case of this slash, 26, we would want to be able to consolidate all of those under one.

And we can do that by summing up these continuous networks using route aggregation. So when I do this slash 26, notice that the first 26 bits are all equivalent. They’re all the same. And so by consolidating those, that allows me to put all those networks together in one notation and just write them simply. Next, we have variable-length subnet masking, or VLSM. This allows subnets of various sizes to be used. So far, everything I’ve done has been equal. I took a big chunk of 256 IPS and broke them into 464-bit chunks. So far, everything I have done has been equal. I took one big group of 256 IPS and broke it equally into four chunks of 64. But we don’t have to do that. We can actually break them up into 16 and 32 and 64, as long as they all add back up to the total amount. Now, this is going to require a routing protocol that supports it, but all modern protocols do, including Rip, OSPF, ISIS, EIGRP, and BGP.

Now, we haven’t talked much about routing protocols yet, but we will when we get to the routing section of this course. Basically, variable length subnet masking is the subletting of subnets without VLSM. All subnets would have to be the same size, like you see here on the screen. But by using VLSM, I can break them up into whatever chunks I want. Now, I’m going to give you the key to subletting here. Look at this chart. It is the subletting exam tip. Now the one I want you to memorise is the small chart off to the right. You can do all the math on test day, and you can add two to the S and add two to the H minus two. Or you can memorise this small chart on the right. If you start with a slash 24, you have one subnet because it’s classful, and you have 256 IPS. As I add one to the Cider notation, it becomes 25. I take my one network and multiply it by two, which gets me to two. And then I divide the 256 IPS by two. and that gives me 128.

Every time my Cider notation goes up, my number of networks goes up, but my number of IPS per network goes down. So if you can remember the chart on the right or calculate it on test day, you’re going to do great because you’re going to be able to answer any question they give you on subletting. The nice thing about Network Plus is that they’re not going to give you difficult subnets. They’re going to ask you something like, “If you have a slash 28, how many subnets are created?” Well, 16, because if I go from 24 to 28, I have to keep doubling it. Double it to get to 25, double it again to get to 26, double it again to get to 27, and double it again to get to 16, which gets you to 16. Or they might ask you how many IPS you would have in each subnet if you had a slash of 30. Well, we use slash 30s for point-to-point connections. There are four IPS, only two of which are actually usable because the first one is used by the name and the last one is used by the broadcast. And so you have four IPS, two of which are usable. So if you can remember this chart, you’re going to do great come Subletting Day. Now, before test day, do lots of practise subletting problems, especially in the class C range; everything from 24 through slash 30 is fair game for Network Plus. Get those down pat, memorise those slash notations with Cider, and you are going to answer any question they throw at you.

  1. Subletting Practice

The first thing I want to point out is that I have this chart on the screen that slashes 25 and 30. Now, why do I do this? Because this is something you should commit to memory and put it on your dump sheet as soon as you sit down for the exam.

So, taking our first problem, you are the network administrator for Dion Training. We’ve decided we’re going to open a small branch office in another city. Support that. I need a subnet for a private IP address given to me for several different networks that I need in this new location. The new office has been given the range of ten, 101000:24, which means you have 256 total IPS to work with. When you set up your new network, I want you to configure separate subnets for each of the departments. I’m going to tell you, and you’re going to tell me with Cidernotation which ones are going to be in which place. Now, you have to resist the urge to go big. For each subnet, I want you to tell me the minimum subnet that will support my requirements. So in my new company, we are going to have this many users. We have 54 new people in it: 32 new instructors, five people in sales, three in administration, and then there’s going to be an unused portion that is whatever is left over. When you’re done with each of those five areas, I need you to tell me what the site or notation is for it. That’s how we’re going to answer this question.

So let’s take it step by step. First, you’ve memorized the table. So this is going to be a pretty simple problem for you. We don’t have to use the two to the S or the two to the H minus two calculations because we have our table here. So we’re going to round up our department numbers to multiples of two. So either one or two, 4816, 32, 64, or 128. Remember that for each of these, I told you how many clients needed to be on the network. But you still need to account for the host and the broadcast for those networks, which are the first and last IPS. So if I have 54, I really need 56. And moving up to the next closest range of two would give me 64. If I look at instructors, we had 32. When we had two, that got us to 34, which meant the next multiple of two was 64. same as it. The next one we have is sales. There were five. You add two, one for the broadcast and one for the network name, and that gets you to seven.

Rounding up, that gets me to eight. Next, we’re going to look at administrative. Administrative had three people; two more brings me to five. You round that up; the next one is eight because we go from one to two to four to eight. And lastly, we have unused to calculate unused is going to be a little different. We’re going to start with the total number of IPS we had. Since you were given a slash-24 subnet, that means you had 256 possible IPS. I’m then going to subtract it, which was 64; human resources, which is 64; sales, which is eight; and administrative, which is eight. So that gives me 112 left. But I can’t round up because I’m discussing unused. So I need to round down, which would get me to 64 again. So now this really tough problem only required calculating two things. What is the definition of a subnet for 64,IPS for instructors, and unused? And what is the subnet range, or the “Cider,” for the sales and the administrative, which both have an eight? So we’re going to have to figure that one out. Well, let’s look at our chart.

That becomes a slash 26 for 64. If we have instructors, that’s a 64, which is a slash 26. For sales, that’s eight. That means we’re going to go to 29. Administrative is again eight, which is slash 29, and unused is a 64, which goes back to slash 26. See how easy that problem was when you break it down and take it step by step? All right, for our next question, we’re going to make it a little bit simpler. The last one may have been something you might see in more of a simulation where you have to calculate multiple things, but during the test itself, it’s mostly multiple choice. So what will that problem look like? Well, subletting practice problem number two is a great example of that. How many assignable IP addresses exist in the network? 172 dot 16, one, zero, slash 27. Well, first, before we answer this question, let me ask you this: Does it matter what the IP address is? Well, no, it doesn’t, really. It’s only the Cider notation that slashes 27 that matters.

The IP address itself is there just to mess with you, so you can ignore it in this case. So the question becomes, how many assignable IPS use a 27 in the network? Now, I’m going to have you think about it here for a second. Is it 30, 30, 214, or 64? Well, looking at our chart, we know it’s either going to be 30 or 32 because a 27 shows that there are 32 IPS. But which one is it? Well, the keyword here is assignable. Assignable IPS means I can give it to a client; I can’t give the broadcast to a client, and I can’t give the host ID or the network name to the client. So the first and last of those 32 have been used up. So this becomes a matter of knowing that it isactually 32 minus two or 30 is the answer. Now, how could you calculate this if you remember your subnet formulas?

Well, it would be two to the H. Your available hosts are reduced by two. Since we have a 27 As our cider notation shows, it becomes two to the 32nd (27), which is going to give us two to the fifth, which gives us 32. And again, minus two gets us back to 30. So whether you wanted to do it the formula way, which I think is a lot more work, or by memorizing my table, either way is going to get you the right answer. Just remember, are they asking about assignable or usable IPS, or are they asking about total IPS? The answer would have been 32 if they had asked how many total IPS exist in the network. But because they said assignable IPS, you have to take out the network ID and the broadcast ID. Now, let’s go on to our third practice problem. How many assignable IPS exist in the network? 192. 168?100:28. Well, this is just like the last problem, right? So if it’s 28, we look at our chart, and we come up with 16. But they said assignable, so we have to take off the broadcast and the network name. And so that is going to give us 14 as our answer. 16 minus one minus one, which gives us 14.

  1. IPv6 Addresses

IPV-6 addressing, or Internet Protocol version six So far, we’ve covered everything there is to know about IPV4, including how to subnet it. But one of the big problems we had with IPV4 was the limited address space. Because there are only 32 bits, there are only 4.2 billion addresses. Now, I know that sounds like a lot, but when we took out entire portions of things like IPA addresses, local host addresses, and private IPS addresses, and then just a large waste of them before we used subletting, this became a big issue and we started running out of network addresses. So they started looking to the future, and they came up with IPV 6. IPV-6 is actually a 128-bit address instead of a 32-bit address.

This gives us the possibility of 340 undecillionth IP addresses, and that is enough IP addresses for every man, woman, and child on the planet. In fact, many, many IP addresses for each of them. Now, you may be asking, “What about version five?” Well, version five never went into production. A lot of the concepts from version five, because it was an experimental protocol, were brought into IP version six, and we’ll talk about those as we go through and learn about IP version six in this lesson. Now, what are some benefits of IPV Six? Well, one of the benefits is a much bigger address space because there are 128 bits of addresses instead of 32, giving us so much more space. But some of the other benefits we have are that there’s no broadcast anymore, which increases the efficiency of our networks. There’s no fragmentation, and so there’s no maximum transmission units or discovery for each session. With IPV4, there was a certain size for each packet, and that’s called an MTU.

If you went over the maximum transmission unit, you had to fragment those packets and send them, then put them back together on the other side, and that actually slowed down your networking, making it an inefficient way to do things. So IPV6 did away with fragmentation. It also had a really big benefit, which helped its adoption, in that it can coexist with IPV Four. Now, one of the reasons they had to do this was because people have been dragging their feet and moving to IPV 6. It’s now 2018, and I still run across networks that are not using IPV6, which has been in production for at least eight to ten years. Now, the way that they’ve done it is that they allow you to run either dual stack, where you can run both IPV 4 and IPV 6 simultaneously because your router can do both, or you can use IPV 6 tunneled over IPV 4, which is called tunneling. Now, the other good benefit of IPsec is that it has a simplified header.

So instead of the twelve fields that we had in IPVFour, we only have five fields, making it a slimmer header that is a lot more efficient. Now, what does that look like? Well, here it is for you on the screen. Now, again, you don’t need to memorise this. This is just to show you the different fields that were in IPV 4, which is on the top, versus IPV 6, which is on the bottom. And you can see just how much simpler IPV-6 is. What does an IPV6 address look like now? Well, because it’s 128 bits, that would be a lot of octets to write in decimal. So instead we use hexadecimal, which is base 16. Every single hexadecimal digit is actually four bits. And the way we write an IPV6 address is by combining four hexadecimal digits together, which means you get 16 bits between each colon. There are 128 bits in an IPV6 address and no more than 32 hexadecimal digits.

Why don’t they say anything else? Why wouldn’t it just be 32 hexadecimal digits? because 32 times four is 128. Well, IPV6 allows you to shorten things. So if you had four zeros, for instance, you could put 10. All leading zeroes are dropped. So in the example on the screen, I have a really long IPV6 address of 2018 and then a zero colon, 0004-8154 Ae. Now I can actually summaries that even further because whenever I have all these sets of zeros, I can put them as a double colon one time inside IPsec and summaries the whole thing down. As a result, the big, long address, which was already shorter than 32 characters, is now 2018-4-8, 1554 AE. How do you recognise an IPV-6 address versus an IPV-4 address? Well, IPV Four is always going to use dotted octet notation or dotted decimal notation in those four octaves. IPV6 is going to use colons in between its numbers, and it’s written in hexadecimal. Now, the only place this is going to cause confusion is with Mac addresses, because Mac addresses are written with colons as well and are hexadecimal.

On test day, a Mac address is something that looks like an IPsec and has twelve, exactly twelve hexadecimal digits separated by colons. Otherwise, if it looks like this, it’s an IPV6 address. And for the exam, you just need to be able to recognise what the IPV6 address looks like, and you should be able to summaries it by taking those zeros and consolidating them down to a double colon. You can do those two things. You’ll be fine for IPV6 addressing on test day. Now, IPV6 has three different address types. There are globally routed unicast addresses, just as there were with IPV4. And they all start with the numbers 2000 to 3 9 9. So if your address starts somewhere between 20, 39, and 99, it’s unicast and globally routable. A link local address is now only used on the local area network, and it will begin with Fe 80. So any IPV6 address that starts with FE80 is a link local address, and then there’s a multicast address, and those all start with FF.

So if you have an FF that’s multicast, do you need DHCP for IPV six? Well, that’s a great question. IPV Six uses auto configuration to discover the current network it’s on and select its own host ID based on its Mac address using the UEI 64 process. So you can use DHCP and automatically assign things through a DHCP server. But built into the IPV Six protocol, you really don’t need to use DHCP if you do want to assign what address it’s going to get. You can do that using DHCP version six, which is compatible with IPV Six. IPV Six, by default, uses the Neighbor Discovery Protocol, or NDP, to learn about the layer-2 addresses on the network based on their Mac addresses and pick its own host ID. Now let’s talk a little bit more about NDP, or the Neighbor Discovery Protocol. It’s used, like I said, to learn layer two addresses on the network. It performs router solicitation, router advertisement, neighbor solicitation, neighbor advertisement, and redirection. Router solicitation is when your client is going to send a message to locate the routers on its network because it has to figure out what the default gateway is so it can get out. Routers can now advertise via NDP, saying, “Hey, I’m a router; I’m over here.” You guys can solicit me.

Neighbor Solicitation is when your IPV6 starts going around and saying what other nodes are on the network, and if I can determine what link layer addresses are out there or layer 2 addresses, I can then learn how to talk to them directly. Neighbor advertisement is similar to radio advertising, but it is done with neighbors. And so those clients start saying, “Hey, I’m here; here are the services I offer.” And finally, redirection: this is where routers can inform hosts of better first-hop routers. So, for example, if I’m sitting in Washington, DC, And I think my best first-hop router is in New York; to get over to California, there might be one in West Virginia that says, “Hey, I’m closer; you should talk to me instead.” So you should talk to me, and I’ll talk to California for you. Right? That’s exactly what it’ll do, Redirection. Now, you don’t need to know NDP in a large amount of depth for the exam, but you should understand that NDP is used in IPV 6, and it takes a lot of the functions from layers 2 and 3 of the OSI model for router advertisement and neighbor discovery.

  1. IPv6 Data Flows

IPV has six data flows. So we talked about data flows back in IPVFour, but they changed just a little bit. In IPV 6, there are still three dataflow, but one of them has changed. We still have unicast and multicast, but now we’ve done away with broadcast and introduced what’s called Any Cast. Unicast works the same way it did in IPV4. But instead of using IPV 4-addressing, we use IPV Six Addressing. So in this case, my server wants to send a message to PC 1 and PC 2. It has to send out two copies of the same message, one intended for each host, because it’s going from one server to one host and then from one server to the other host. Notice that the only difference here is their IP addresses.

We’re using ABCD 1, and all of that space in between would be filled up with zeros to get it back to a 32-hexadecimal digit address. Next, we have multicast. And in multicast, just like IPv4, we use multicast groups. And so the server addresses the message to the multicast group, in this case, FF Zero A. Because remember, anything that starts with “FF” is a multicast. And then, once it gets to the switch, it distributes an additional copy to each of the recipients. In multicast group number one, data travels from the single source to multiple but specific destination devices. And we now have a cast. And this is the unique one. This is the one that is unique to IPV 6. It’s designed to let one host initiate an efficient update of router tables for a group of other hosts. IPV Six is able to determine which gateway the host is closest to and sends the packets to that gateway as though it were a unicast communication.

And then that host, that gateway, is then going to broadcast that through any cast to any other host in the group until all of the routing tables are updated. Now, what does that really look like? Let’s take a look at the diagram here. We have a server, and the router there on the left is going to send out its Any Cast. So it sends it out to the Internet, and it’s going to go to whatever is the closest place that has a router. In this case, it’s the router on top. Now, once it gets that router on top, it can then get to the DNS server and then reroute down to the routers on the bottom and keep pushing that around. But the original router on the left side of the screen only had to send out that message once. And then from there, all the other routers will figure out the best way to update everybody else’s tables.