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200-301: Cisco Certified Network Associate (CCNA)

200-301: Cisco Certified Network Associate (CCNA) Certification Video Training Course

200-301: Cisco Certified Network Associate (CCNA) Certification Video Training Course includes 271 Lectures which proven in-depth knowledge on all key concepts of the exam. Pass your exam easily and learn everything you need with our 200-301: Cisco Certified Network Associate (CCNA) Certification Training Video Course.

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Curriculum for Cisco CCNA 200-301 Certification Video Training Course

200-301: Cisco Certified Network Associate (CCNA) Certification Video Training Course Info:

The Complete Course from ExamCollection industry leading experts to help you prepare and provides the full 360 solution for self prep including 200-301: Cisco Certified Network Associate (CCNA) Certification Video Training Course, Practice Test Questions and Answers, Study Guide & Exam Dumps.

Subnetting

4. Subnetting Class C Networks and VLSM

Okay, moving on. Assume we've been assigned to class C 215, dot 10, and dot 24. The same as in the previous example. And here we're going to move the line all the way to the right. We're going to go with a slash of 31. Obviously, we can't do a slash 32 you if we want to have more than one available host. Now you can use a slash 32 subnetmask, but there's only one host there that's mostly used for loopback addresses. Don't worry about it yet. We're going to talk about that later on. So if you do need to have multiple hosts, then the furthest right you can go is a 31. If we wrote that subnet mask in dotted-decimal notation, it would be two 5525-525-5254. We get two five four because one two eight plus 64, 30, 216, and eight four plus two come up to two five four. So that leaves one bit for the host address, which obviously has a possible value of a one or a zero. So, as a class C 31, it borrows seven bits for the network address from the host address. That gives us 128 subnets, which is to the power of seven, getting what's counted up: two, 4816, 32, 64, 128. and that's going to accommodate two hosts each. So that's what we bring with So thaThe valid host addresses would be 200, 15100 to dot 1215, ten, two to three, then 215 to dot ten, dot four to five, etc. Up to and including 215, ten, 254, and two fives. You can check this by we're using 215, dot ten. So the first three octets are never going to change. The last octet, you just start with all zeros, and then it would be one and then a zero out of one and then 0010, and then a zero with a one, and so on. But wait, there are only two possible values to assign to hosts when using a slash 31. What about the network and broadcast address? Yeah, a slash 31 subnet breaks the standard rules of IP addressing where we have to have the network address and the broadcast address at the top and bottom of the range. Slash 31 subnets are, however, supported onCisco routers for point to point links. If you think about it, a point-to-point link between just one side and the other side means that any traffic that comes from here needs to go to the other side, and vice versa. There's nowhere else for it to go. So there isn't really any point in having a network and a broadcast address. So that's why Cisco allowed that exception. You can have a slash 31 on a point-to-point link. Okay, next up, moving on, we're going to move the line back a space. We're going to create a slash 30. Now the subnet mask and dots additional255, 255, 5255, dot two five, two, that, and 30—they both mean the same thing. That leaves two bits for the host address because we have two bits to the power of two for the host address, which is four minus two for the network and the broadcast address. That gives us two possible hosts. It borrows six bits for the network address. We can't miss up to two, 48, 16, 32, 64, and our 64 bits, which can accommodate two hosts each. The valid addresses are on the first subnet. The network address is 200, dot 15, dot 10, dot O, and the broadcast address is 215, dot ten, dot three. The valid addresses are dot one and dot two. The next subnet is 215 ten, five to six, and the network for the broadcast is seven. The next subnet, the network, would be eight. The valid addresses would be nine and ten, and the broadcast would be eleven and so on. As you look at this, the way that you can calculate what the different valid subnets are going to be, notice where the line is. It's after the number four. So the network address is going to go up in values by four; the first network address is zero. The next network address is four; the next network addresses are 812, 16, and so on. And because we know the network addresses, we can work out what the broadcast address and the range of valid IP addresses are going to be. So here it's going up in value by four. The first subnet is going to be a zero. So zero is the network address. And because the next range starts at four, the broadcast address must be three. So if there is a network address and three is the broadcast address, the valid host addresses must be one and two. The next subnet starts with four. The next subnet after that is eight. So the broadcast address must be "dot seven." So we have a network address of four and a broadcast address of seven. That leaves the addresses in between five and six available to be assigned to our hosts. Let's look at a comparison betweenthe 31 and the slash 30. Now, hopefully, you notice that the bow supported two hosts per subnet, but Aslash 31 supported 128 subnetworks. 30 only supported 64. So you may be thinking of how we're always going to use 31 then.And a slash 31 is useful if you need to maximise the use of your address space. But a slash of 30 is more commonly used. It's more standardised for the CC and E exams. If you get a question where you got a choice where youcould use either a 31 or a 30, always use a 30. So if the question requires you to come up with a subnet that supports two hosts, use a 30 unless the question explicitly tells you to use a 31. Okay, so we covered lines 30 and 31. Next up is a slash of 29. So we're going to move the line back into space again. a 29 and dot. A decimal is two 5525-525-5248, and we get the 248 because it's 128 plus 64 plus 32 plus 16 plus eight up to where the line is. That leaves three bits for the host address; the power of three is 2488 minus two because of the network address. And the broadcast address gives us six available host addresses. And in class C, the default subnet mask is 24. A slash 29 is used here. So we're borrowing five bits for the network address; the power of five is two, 4816, 32. So we've got 32 available subnets. Here are the valid host addresses. Notice that the line is after the 8th, so the network addresses are going to go up in values of it.The first one will be zero, then eight, then 16, and so on. So the first subnet, the network address, is zero. The next subnet, the network address, is eight. So the broadcast address for that first subnet must be one less. It's going to be 215, 10, and 7 for our example. So the network address is zero and the broadcast address is seven. That leaves one to six available to be assigned for our hosts. The next subnet is going to be, as we said already, eight. The next one is at 16. So our broadcast address now must be 15 less, which leaves the available host addresses at nine to fourteen. And we can go on like this all the way up to 215 ten, 248 as the network address, 255, the broadcast address with host addresses, and 24925 four. And we can continue to move the line back. So we did 31 39.We could also do a 28 (two 5525-525-5248). That is going to give us 16 networks of 14 hosts each on a class C network. Again, the default is a slash (24).If we're using slash 28, we've borrowed four bits. Four bits are two 4816 available subnets, and after 28, it goes up to 32. So we've got four beds available for our host. That is two 4816 minus two for the host, giving us 14 available hosts. We can move the line back again to 27. That's going to be borrowing three bits. So we have two, four, eight networks available, leaving five bits for our hosts, which will be two 4816. 32 minus two gives us 30 available hosts. You can see how I'm getting that now. Four networks and 62 hosts are represented by a slash of 26. Two networks and 126 hosts are represented by a slash of 25. And the default network is slash 24, which has 254 hosts. Okay, so that's how we carry out subnetting and figure out how many networks and how many hosts we're going to have. The next thing to tell you about is variable-length subnet masks with early routing protocols like Rip version 1. They only supported fixed lamp subnet masks, which meant you could subnet, but all subnets in a particular network had to be the same size. So all of your different subnets had to accommodate 14 hosts, for example, or 30 hosts, for example. And if you had one subnet that needed up to 30 hosts, you had to have all your subnets at the size of 30 hosts, even if one of them only had three hosts in it. Then, with later routing protocols, we came out with support for variable-length subnet masks. That means that within the same range, you can have different sizes of subnets in there. So say I've got one part of my network that has ten hosts in it; I could give them 14 available hosts, and in another part of my network that's got 28 hosts, I could give them a subnet mask that gives them 30 available hosts available on subnet masking. It means we can use different lampsubnet masks within the same network. This is a good thing because it lets us be much more precise with the size of our networks, and we're going to be wasting a lot fewer addresses. We will cover how to do videobleweb subnet masking in the next lecture.

5. Subnetting Practice Questions

You've got a good handle on all of the subnetting information that we've covered so far, and you'd be able to work your way through a subnetting problem, but let's actually verify that now by doing practise questions. So the question is, what are the network address, the broadcast mask address, and the valid host addresses for the IP address of one 9822, 45, 173, 26? And the second part of the question: what is the subnet mask in dotted decimal notation? So stop the video now, please, and figure that out when you come back. I'll walk you through how we arrive at the answer. Okay? So let's look at how we are going to figure out the answer to the question. We're actually going to do the second part of the question first because it's more logical and easier to do it that way. So the second part of the question was: What is the subnet mask? 26 dotted decimals. And the way to do this whenever you're working out a subnet problem, particularly at the start, is to write out the bit pattern at the top of a piece of paper. Now later on, you'll be able to do a lot of these in your head, but when you're first learning, it's way easier to write it all out like this. So take a piece of paper, start off at the top, start off at the right, and write one, two, 4816, 326-4128; then you can put a dot in; then 1248; and so on until you've drawn out the bit pattern for the four octets in your subnet mask. Then the question we had was how to work out the dotted decimal notation for a slash of 26. So just under your bit pattern at the top of the paper, write out 26 ones underneath that, and then you can put in six o's for the host portion of the address. Also draw a line in where the line is for the subnet mask, and then we can just add these up. So obviously the first octet is all ones. It's going to be two five five. The second octet is also going to be two five five.The third octet is also two five five.And then in the fourth octet, where we're doing the subnetting here, the ones are in the 128 and the 64. So we add one, two, eight, and 64 together to give us one nine two. So 5525-525-5192 is a slash 26 and dotaddress monetization. Let's look at what the second part of the question was. What is the network address, the broadcast address, and a valid host address for the IP address? 109, 822-451-7326. So again we use our piece of paper, and we're going to write out the IP address now as well. You can see that I've done it on the top part here. That is the bit pattern for one, nine, eight, dot 22, dot 45, dot 173. That's how we would write it in decimal. And the network portion of the address is the first 26 bits. So I can see that that is one, nine, eight, dot, 22, dot, 45. And then I just count whatever is in the first two bits here. And I've got a one for the one-two-eight and a zero for the 64. So that means it's one, two, eight. So the network address is 198-224-5128. Don't worry about the text yet, by the way. This is the second way that we can figure out the answer, which I'll get to in a second. I'm going to explain how to do it the long way first. So the long way is the logical way to do it. First, can we write out the IP address and the subnet mask in binary notation, and can we draw the line where the subnet mask is? From that, we can figure out what the network address is, like I just said, 198-224-5128. I can see it here. Then we figure out what the broadcast address is. So that would be if we had one host in all of the host bids. That is a one on the 32 up to the one. And 32 plus 16 plus eight plus four plus two plus 132 plus 16 equals 48, eight equals 56, four equals 60, and 262 plus one equals 63. So the network portion of the address was 12812. Eight plus 63 equals one nine one. So again, the network address is 192-224-5128. The broadcast address is 198-224-5191. So we've got the network address and the broadcast address. The possible host addresses are going to be the addresses that lie between those two numbers. So that would be 198-224-5129 going up to 12819, 822, 45, one, 90. Okay, so that's how we can figure it out, very manually and very logically. We can take a little bit of a shortcut, and what we do is obviously, in the last lecture, look at where the line is. So the line is here after the 64. So I know that my subnets go up in multiples of 64. So the first one is going to be 198-22450, then 45, 64, then one, two, eight, and so on. I can see that this address that I'm interested in is 198-224-5173. So the first subnet is zero, then 64, then 128. The next one would be one two eight plus 64, which is going to be one nine two. So the answer is 173, which is less than one and a half. As a result, it must be in the one-two-eight network. So I know 198-224-5128 is in the network. And then add 64. I knew to add on 64 because that's where the line is. And the following network address is one nine two. So the next network address is one nine two.It must mean that the broadcast address for this subnet is one less than that. It's going to be one, nine, one. So I deduced that my network address is 1, 2, 8. I figured out that my broadcast address is one nine one. The available host must be everything in between. That would be ones 9822-45, 12921, 9822-45, and 190. Okay, so that's how we can figure it out. And a final important point to mention is that everything in this example was done on the 26th subnet line. So that is in the last octet. So we don't actually need to worry about the first three octets. They're always going to be unchanged. is 198, 22, and 45 in our example. So you can save some time by only writing out the octet that the subnetting is happening on. You don't have to write the whole thing out in full every time. OK, so that was a quick example of how we will solve a subnet in question, the kind of thing that you could see on the CCS exam.

6. Variable Length Subnet Masking Example Part 1

In this lecture, we're going to carry on with our subnetting theme, and by the end of the lecture, you will be able to carry out a variable length subnetmask design for a real-world network topology. Now, it's only a small network in the example that we're going to work through, but you can apply the same things that you're going to learn here to larger networks in the real world. The slide that you see here is the same slide that you saw at the end of the last lecture, where we explained that, back in the day with early routing protocols, whenever you did a subnet, each of the subnets had to be exactly the same size. So if you had one subnet that was a slash 28, for example, they all had to be a slash 28. You couldn't mix and match 27, 28, and 29 within the same larger network, but that ended up wasting a lot of addresses. So with later routing protocols, they did have support for variable-length subnet masks. And now we can mix and match the size of our subnets within that greater network. and we're going to work through an example here. The things that we need to consider when we're going to do the design are how many different locations that we have in the network that are going to need subnets, and how many hosts are going to be in each of those subnets. What are the IP addressing requirements for each of the different locations? So should different departments or types of hosts be on different subnets? For example, you might have one office, but within that office maybe you've got a sales department, a research and development department, and an accounting department, and you want the accounting department to be secured from different departments. In that case, you'd definitely put it in its own subnet because different subnets have different routes. At a router, it's easy to secure them at the Layer 3 level based on their IP address. And the last question is, what size is appropriate for each subnet? We don't want to waste addresses, but we want to leave some room for growth. So we need to have a bit of a balance here. Don't make it too big; we're wasting addresses, but don't make it so small that we're going to run out of addresses and I'm going to have to do a redesign later. That would just be giving ourselves a big headache. So for an example, this is our network topology diagram. We've got an organization. They've got an office in New York and a branch office in Boston. New York is their headquarters, and they've got 28 horses in the Engineering Department and 14 hosts in Sales in Boston. They've also got engineering and sales. They've got 28 in Engineering, the same as in New York, and they've only got seven hosts in the Sales department. So those are the different parts of the network that we need to apply IP addresses to. And we've been allocated the Class C network of 215, Dot Ten, Dot O 24, from our Internet service provider. Another thing that we need to do is remember our point-to-point links between the routers. They need to have connectivity with each other, so they're going to need IP addresses too. It's a point-to-point link. So we've just got two host IP addresses there. The outside interface on the New York router is connected to the outside interface on the Boston router. So that was our requirement. Now we need to think about how we are actually going to design this. As well as the steps you take. First off, find the largest segment and allocate a suitable subnet size for it. You then create that subnet at the start of your available address space, and then that's it. You just keep working your way down. So you start off with the largest subnet. You then move on to the next one, the next one, and the next one until you've allocated addresses for all of your different subnets. Now, in a real-world deployment, you want to have a scalable design, so you want to have room for growth in your networking design. So let's say that I've got a subnet that's got 14 hosts on it. You know that one of our subnets has exactly 14 hosts. Don't give the subnet that size because maybe in a few weeks, a couple of extra people are going to join the department. And now how are you going to fix that problem? So make the subnet size big enough that it's going to be able to grow. Another thing is that we're going to go sequentially from largest to smallest. Maybe we're going to have a subnet with 30 hosts there, another subnet with 30 hosts, and then a subnet with 14 hosts. Well, leave a spare subnet with 30 horses in it. So don't just do two subnets of 30, do three or four subnets of 30. Because then if you do have a new department that requires up to 30 hosts later, you've got a spare subnet there waiting for it. And everything is still sequential, contiguous, and logical going from largest to smallest. So, hopefully, you got the point. Don't design for what's right now. Design for what is going to be the best fit in the future. Leave some scalability there. Leave some room for growth. So that's what you do in the real world. Don't do that in the CCNA exam. Do not think about, Oh, this is what I would do in the real world. This would be best practice. Do exactly what the exam question tells you to do, even if you think that would be a stupid thing to do. Okay, don't overthink things. Don't think, Oh, well, it would be better to do it this way. Do exactly what the question asks you to do, and then you're going to get the question right. Okay? So for our example, we had the engineering departments in New York and Boston, and they have 28 hosts each. and that was the largest subnet that we required. For example, let's say the exam question says that the departments will not grow and we need to use the smallest subnet possible to maximise our available address space. Do not leave any room for future growth here because the question says to make it optimal and maximize the use of the address base. Okay? So that's a question. What I want you to do is pause the video here and calculate the optimal subnet mask for the engineering departments. So make it no bigger than is necessary, but make sure it's at least big enough to support 28 hosts. Once you've done that, you'll be able to determine the network and broadcast addresses. Do that for both engineering departments, both in New York and Boston, and figure out what the range of host addresses will be. Okay? So go ahead and do that. Please pause. You know, when you come back in a second, I'll walk you through how to get the answer. OK, so let's look at how to get the answer. So we've been allocated numbers 200, 1510, and 24. We've been allocated a Class C, and we've got two different departments that we want two different subnets for. And they had 28 hosts each. Aslash 27 supports 30 hosts. A slash of 28 is 14. So it's not big enough. We can't use that. The smallest character we can use, as instructed, is a slash 27. Real world. You might argue that you'd prefer to use a slash 26.This is not the real world. It's an exam question. We'll use as few as possible, which was a slash 27. We wrote that in a decimal notation. It's 2552-552-5224. OK, so that was the first part of the question, figuring out the subnet mask. The next thing we have to do is actually divide up our address space. The headquarters was in New York; let's give them the first subnet. So the network address is going to be 215, 10: 00:27. And if we look at the line, we can see it's after 32. So 27 is going to go up in increments of 32. So the next network address is going to be 215 1032, which means that the first subnet broadcast address is going to be one less than that; it's going to be 215 1031. And that leaves addresses available for our hosts of 215—from one up to 30. Okay, so that's the engineering subnet. Next is the Boston engineering subnet. Well, the last broadcast address used was 31. So the network address we're going to be using is 32. As a result, the network address is 215 1032 27. Again, we're still using that 27 where the line is after the 32. So the next network address would be 64. So our broadcast address must be 215, 1063, which is one less. And our hosts are going to be... Between the network and broadcast addresses, that's 215 dot ten, dot 33, up to 62. Okay, so that's the engineering department taken care of. All right, we're coming up to the ten-minute mark in this lecture. So if you want to take a break, now is a good time to go and stretch your legs. When I see you back here for part two, we'll go through the subnetting for the rest of the department.

7. Variable Length Subnet Masking Example Part 2

The next largest subnet is New York Sales, which requires 14 hosts. So I want you to pause the video again. Here, calculate the optimal subnet mask for New Yorksales and again determine the network and broadcast addresses and the range of available host addresses. Okay, so stop the video now. I'll see you in a second once you've worked that out. Okay, so let's look at the answer. We need to support 14 hosts. The smallest possible subnet we can use is 28. 28 means we've got four bits available for host addresses. That's two 4816 minus two, which gives us 14 hosts, from 215 100 to 215 1063. We're already being used by the engineering department, so the network address we'll start with here is 215 1064.If we look at the line, we can see it's after the 16. So 64 plus 16 is dot EA; take one away. Our broadcast address is going to be 2151079, and the available host addresses are 215 1065 to 215 1078, the addresses that are between the network address and the broadcast address. OK, that's our first three departments. Done. The last department was BostonSales, which requires seven horses. So let's do the same thing again. Calculate the optimal subnet mask and then the network address, broadcast address, and host addresses that we're going to use. So stop the video now, and I'll see you back with the answer. Okay, so we need to support 14 hosts. We're going to use a slash 28 again here, the same as what we did for the last department. You can use a slash 29, which is a possible mistake that some people would make because we need to support seven hosts. So they go 2480. Okay, that's three bits, and they'll make a slash 29. They forgot to take away the two for the network address and the broadcast address, so we get 29 supports and eighteen hosts We require seven here, so that's enough. So we're going to use a slash28 again, which supports 14 hosts. The last broadcast address was 215 1079, so our network address will be 215 1080. Again, the line is after the 16. So the next network address would be 96, which means our broadcast address is going to be 215 1095. The valid host addresses are 81 to 94. Okay, so that's it. That was each of our four departments. So we're down right now. Remember, we have to allocate addresses for the point-to-point link between the routers in Boston and New York. Another thing you would do in the real world is to also allocate address space for your loopback addresses. Loopback addresses are used for management, but with a logical address, there's not anything physical on the other end. As a result, we usually assign a slash32 to our loopback addresses. Again, we'll talk about loopback addresses more in later lectures. Don't worry about them for now. I just mentioned it here for completeness. So that last subnet, the link between the New York and Boston routers, So let's do the same thing again. Pause the video, determine the optimal subnet mask, the network broadcast addresses, and the host addresses that we're going to use. So pause the video and I'll see you back with the answer. OK, so we want to support two hosts. Remember: a 31 and a slash 30 support two hosts. And hopefully you remember what I told you before: Unless the exam explicitly tells you if you need to support two hosts, go with a slash 30 because that's the standard that we use. It complies with all of the Internet standards. So we're going to use a slash 30 here for our two hosts. We are already using up to 215 1095 in our departments. So our network address will be one up. From there, we're going to use 215, 1096. If you look at the line, it's after before.So the network address is going to go up in increments before.So the network address would be 215 ten, 101 less than that, which gives us our broadcast of 215 1099, which leaves the host addresses to be dot 97 and 98. OK, there's one more thing that I want you to do. I want you to have a look again at the network topology diagram that you saw at the start of this lecture. You don't need to scroll back. I'm going to put it up on the next slide. Then what I want you to do is get a piece of paper and a pencil, and I want you to draw the network diagram. But this time I want you to include the networking information—the different subnets that we just figured out. Let me go onto the next slide. So make it look exactly like this with the routers and the switches. But rather than saying, Sales 14 hosts," I want you to see Sales. And then I want you to tell me the subnet that is going to be in use here. So the subnet and the subnet mask and slash notation do that for the four departments. also do it for the point-to-point link. And another thing I want you to do is also put in the IP addresses that will be used on our router interfaces. For the router interfaces, use the first available host address in that particular subnet. Okay? So get your paper and pencil out. Go ahead and do that. Now on the next slide, I'll show you the answer. Okay, so here we are with the answer. Remember when we did our design? We started off with the largest subnet, which was the engineering departments. So, in my network diagrams, I would show engineering in New York at 215 ten 00:27 and engineering in Boston at 215, 1032, 27. I also have a sales department in New York (215) 1064-28. And sales in Boston were 215, 1080, and 28. And the subnet I used for my point-to-point link is 215 1096 30. Then we're going to use the first available address (the IP address) and our router interfaces. So that was one on the interface for the New York Engineering Department, 33 for engineering, 65 for New York sales, and 81 for Boston sales, and then the point-to-point link, which I'm musing is 97 on the left and 98 on the right. OK. And this is actually what would be an acceptable network diagram in a real-world environment. This is typical of how we would write it. Okay, that's it. So you now know how to do a variable-amp subnet mask design for a network. I'll see you in the.

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