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Alright, welcome back, everybody, tothis lecture on subnetting examples.I've been waiting to record this lecturefor a very long time now.I'm so excited, I'm so eager to talk to youabout subnetting and to teach you how to subnet.So without wasting any further time,let's start with some examples.I hope you have a pen and a paper ready.What you see on the screen right nowis the four steps for performing subnetting.There are multiple methods for subnetting,I've tried a few of them.But what I'm going to show you right now iswhat I feel is the simplest of all methods.So if you have a pen and apaper, make a note of these four steps.Step number one, we'll start by convertingthe subnet mask into binary format.So we've already completed decimalto binary conversion, right?That's exactly what we're going to do.We'll take the subnet mask that has been givento us and we'll convert that into binary format.Once we complete that, we will then determine thenumber of host bits that need to be borrowed.Number three, we need to determine the increment.And number four, we need to addthe increment to get the new subnet.And that's about it.Let me show you how simple that is with an example.Example number one, divide 192,168, 100:24 into five networks.Now, 192, 168, single network.But we've been given the task of dividingthis one large network into five smaller subnetworks.Step number one, very simple, convertthe subnet mask into binary format.So in decimal format, 24 equals 255-255-2550.When we convert that over to binary, we have thefirst three octets made up of all ones, and thelast octet is made up of all zeros.That's step number one.Step number two says determine the numberof host bits to be borrowed.Now, initially, you might findthis a little bit confusing.Why do I need to borrow host bits?So the answer is 109, 216-8102.Four is one network.But I need to break this down into fivenetworks, which means the existing network bids the first24 bits are giving me only one network.I need additional network bitsto have additional networks.So we need to borrow some of thehost bits and convert them into network bits.To determine the number of host bits tobe borrowed, we have a simple formula.The formula says two power, x is greater thanor equal to the required number of subnets.In this case, we've been told todivide the network into five subnets.So the formula now becomes two power, xis greater than or equal to five.Now we need to start trying the values ofx from zero and try to find that valueof X that will satisfy this equation.Let's start with X is equal to zero.If you put x is equal to zero, weget two power zero, which is one and thatis not greater than or equal to five.If we put X is equal to one, weget two power one, which is two, and thatis not greater than or equal to five.If we put X is equal to two, wehave two power two, which is four, and thatis also not greater than or equal to five.But if we put X is equal to three,we have two power three, which is eight, andthat is definitely greater than or equal to five.So we now know X is equal to three.And that is the number of hostbits that I need to borrow.So we have the original subnet mask, which isslash 24, which has the first 24 bits asones and the last eight bits as zeros.Now, when I borrow three more bits, I havethe new subnet mask, which is slash 27.24 plus three borrowed bits makes a total of slash 27.And when we write that in binary format, you havethese three zeros that have been converted over to once.So we have 24, which we know is two five, fivedot, two five, five dot, two five, five dot zero.We now need to convert slash 27 into decimal format.So I've just written it again in binary format.You have the first 27 bits as ones andyou have the last five bits as zeros.We need to convert that last octet over to decimal.We're going to follow the exact same method that welearned in the lecture on decimal to binary conversion.So we first write all the eight bits in a single row.We then write the number two against every bit.We then assign the powers of two against every bit.So you have two power zero on the right hand sideand it increments as we move to the left hand side.And the last one on the lefthand side is two power seven.The last step is to add the powers oftwo, but then we are only going to addthose powers of two that correspond to the ones.In this case, we only have thefirst three bits that are once.So we're going to add two power seven, twopower six and two power five, which is 128plus 64 plus 32, and the total is 224.So that gives me my new subnet mask.27 equals two five, five dot, two five,five dot, two five, five dot 224.Step number three says determine the increment.Now, increment is the power of two corresponding tothe least significant bit or the right most bit.So we have this new subnet mask, which is27, which has the first 27 bits as onesand the last five bits as zeros.When we align the powers of two corresponding tothe last octane, this is how it looks like.You have the last octet over here andyou have the powers of two over here.This is exactly what we saw in theprevious slide when we were trying to findout the decimal format of the subdued mask.This is the previous slide that we looked at.This is the right most network bit.The power of two corresponding to that is 32.And that's exactly what I've written.Again, over here, you have these eight bits overhere, and you have the powers of two.Over here, the power of two correspondingto the right most bit is 32.That is going to be my increment.Step number four, add the incrementto get the new subnet.My original network address is 192-1681, dot zero.We need to start adding theincrement to get our desired subnet. Very important.Please remember, the increment should only be addedto the octet from which bits were borrowed.In this case, we borrowed bits from the fourth octet.We're going to start adding 32 to the fourth octet.So we have the first networkaddress, which is 109 216810.We add 32 to the last octet, we get 109 216-8132.That's my second network.My third network is 109 216-8164.And you keep adding all the way until youreach the last subnet, which is 192, 168, 1224.If you're wondering why we stop at 224, the answer isif you add 32 over to 224, that becomes 256.And once you reach 256, that becomes another network.Because the largest possible value for an octet is 255.If we reach 256, that's going to go overT019-21-6820, but we have to restrict ourselves T019-21-6810.So those are my eight new networks, or subnetworks.Now, you may have a question.We were told to divide it into five networks.Why do we have eight networks?The answer is the number of networks into which youcan subnet will always be a power of two.It is not possible to divide intonetworks that are not powers of two.For example, we cannot divide it exactlyinto seven networks or six networks.We can divide it into two networks, fournetworks, eight networks, 16 networks and so on,which are basically powers of two.So that's the first example.Did you find it slightly difficult?I guess so, because when I learned subnetting for thefirst time, my first example actually spun my head.So before I take you to the second example, weare just going to do a 1 minute quick revisionof what we saw in the first example.So I'm going to take you back to the first slide.This is where we started.We were told to divide 192,116, 100:24 into five networks.We had these four steps to perform subnetting.Step one says convert thesubnet mask into binary format.So the decimal format for slash 24 is 255-255-2550.We converted that over to binary.We had the first 24 bits as onesand the last eight bits as zeros.Step number two was to determine thenumber of host bits to be borrowed.And for that, we had a formula whichsays two power x is greater than orequal to the required number of subnets.The required number of subnets is five.When we put X is equal to three, we haveeight, which is definitely greater than or equal to five.So the required number of bits that need to beborrowed is the value of X, which is three.So we wrote the subnet mask one moretime in binary format, and then we borrowedthree bits from the host portion.So we got our new subnet mask, which is slash 27.Now, we had to convert thisover to decimal one more time.So we aligned the last octet against the powers of two,and we added only the powers of two corresponding to theones, which was 128 plus 64 plus 32 equals 224.So our new subnet mask wasslash 27, which equals 255-25-5255, 224.Step number three was to determine the increment.The increment is the power of two corresponding tothe least significant bit or the rightmost bit.In this case, the rightmost bitcorresponds to the value of 32.That's our increment.Step number four is very simple.We just need to start adding theincrements to the original network address.But we add the increments to theoctate from which we borrowed the bits.So the original network address is 192-16810.We start adding 32 to the fourthoctet, and we get eight new subnets.Very important.All these networks have a new subnet mask of slash 27.Let's try one more example.This one is a class B address.So we have to divide 172 1016 into ten subnets.Step number one is to convertthe subnet mask into binary format.You have the decimal format of slash 16, which is two,five, five dot, two five five dot zero dot zero.When we convert that to binary, we have the first16 bits that is made up of ones and thelast eight bits that is made up of zeros.Step number two says, determine the numberof host bits to be borrowed.The formula for that is two power.X is greater than or equalto the required number of subnets.That means two power, x is greaterthan or equal to the number ten.We need to find the valueof X that satisfies this equation.Two power zero is one.Two power one is two.Two power, two is four.Two power, three is eight.None of these actually satisfy the equation.When we put X is equal to four, wehave two power four, which is 16, that isdefinitely greater than or equal to ten.So the number of bits that we need toborrow is the value of X, which is four.So we have the original subnet mask, which is 16,which has the first 16 bits as the network bits.When we borrow four additional bits, the new subnet mask isnow 20, which has the first 20 bits as the networkbits and the last twelve bits as the host bits.Notice here that we borrowed bits from the third octetand not from the fourth octet, because we always borrowbits from the octave that is right next to it.So now we need to convert 20 into decimal format.So we write the third octet, which is 111000,and we write the number two against every bit.We then assign the powers of two, starting with twopower zero on the right hand side and incrementing thepower as we move to the left hand side.So we have two power zero on the right handside and two power seven on the left hand side.Now we need to add the powers of two correspondingonly to the ones we have first four ones.So we're going to add the powers of two corresponding tothe first four ones, which is 128 plus 64 plus 32plus 16, which gives you a total of 240.So my new subnet mask is 20, which is 25525.Five dot, 240 dot, zero.Step number three is to determine the increment.We now know the increment is the power of twocorresponding to the rightmost bit or the least significant bit.So when we align the third octet againstthe powers of two, we have the number16, which corresponds to the rightmost bit.That is going to be my increment.Step number four is to add theincrement to the original network address.So we have 172, 1016.We're going to start adding the increment tothe octet from which we borrowed the bits.This is where it's a little bit trickyand we have to be very careful.Naturally, it is possible that we may start addingthe increment to the last octet, but we actuallyborrowed the bits from the third octet.That's where we are going to add the increment.So we start with 172-1002, because20 is my new subnet mask.We then have 172, ten, 116, 00:22, 00:20 and so on,and we go all the way up to 172, ten, 2400.We stop here because if you add 16 one moretime to 240, that will result in 256, which isa totally new network, and that will become one 72110.We have to stay inside 172, 100, zero.So the question was to divide it into ten subnets.We've actually divided into 16 subnets because we just discussedthat the number of subnets that we can actually divideinto will always be a power of two.In this case, it was two power four.So we have a total of 16 new subnets.We need ten subnets.We use only ten subnets and we reserve therest of the six subnets for future purposes.Next.It's homework time, guys.Before I see you in the next lecture, Iwant you to practice two more examples of subnetting.Number one, break 2230 00:24 into two, subnetsnumber two, break one hundred and thirty eight,zero, zero, sixteen into seven, subnets.We'll discuss the answers to thesequestions in the next lecture.Some important tips.Remember the four steps of subnetting.If you have not made a note ofit already, make sure you write it down.And make sure you memorize the four steps.The second tip is to practice, practice, and practice.When I started practicing subnetting, it did take mea while to get my head around it.So make sure you practice and practice.I promise you, you'll find subnetting very interesting and veryeasy as you continue to practice and play around within the next lecture, we're going to look at somemore and slightly trickier examples of submitting.So that's it for this lecture, guys.Let me know if you have any questions.If not, I'd like to thank you for watching, andI'm going to catch you in the next lecture.

Alright, welcome everyone back to the second lecture on subnetting. In this lecture, we're going to practise some more examples of subnetting, but they are not going to be exactly the same as we practised earlier. These examples are going to be slightly different. But before we do that, if you guys remember, we had some homework from the last lecture, right? So let's discuss the answers to those questions, and then we'll begin with the content of this lecture. We had a couple of exercises to complete in the last lecture. Number one, we had to break 223 00:24 into two subnets. The answer to that is 223-0025 and 223-001-2825. The second exercise was to break one hundred and thirty-eight zero-zero-sixteen into seven subnets. And you have the answers on your screen. We have 100 and 380-032-0640 all the way up to 224 arezero, all with a slash 19 subnet mask. I hope you got the answers right. If you did not get it correct, please look back at the last lecture, follow the exact same steps, and try to identify where you may have gone wrong. If you do have a question about any of these exercises, please feel free to let me know. So now let's begin with the examples from this lecture. Before we actually get to the examples, let's revise the steps once again. Step number one is to convert the subnet mask into binary format. Step number two is to determine the number of host bits that need to be borrowed. Step number three is to determine the increment. And step number four is to add the increment to get the new subnet. So I hope you have a pen and paper ready. Let's begin with the first example. So the first example says subnet one hundred and twenty-four zero zero eight, such that each network has 500 hosts. The earlier example that we looked at required us to subnet based on the number of networks. So we had an example that says "subnet 223 slashes 24 into two subnets." So we were actually told the number of subnets that we need to break it into here. We need to subnet based on the number of hosts. The steps for subnetting remain almost the same. Step number one: convert the subnet mass into binary format. So in decimal form, eight equals 2550. If you convert that into binary, the first octet is made up of only ones, and the last three octets are made up of all zeros. Step number two says to determine the number of host bits to be fixed. Originally, step number two was to determine the number of host bits to be borrowed. Here. Instead of borrowing host bits and then converting them to network bits, we'll fix the host bits that will remain unchanged. Because here we are required to subnet based on the number of hosts instead of the number of networks. So when we fix the hostbits, that will remain unchanged. We are sure to have the required number of hosts per subnet. To determine the number of host bits to be fixed, we have a formula. And that formula looks very similar to what we already know. Two Power X is greater than the required number of hosts plus two. The original formula was "two powers of X" are greater than or equal to the required number of subnets. Here, the formula is almost the same. It says, "Two Power X is greater than or equal to the required number of hosts plus the number two." I'm assuming you already know why we add the number two at the end. Because we already discussed, the first address in a network is the network address, and the last address is the broadcast address. Now, those two addresses cannot be assigned to hosts, so we need to make sure that we supplement that by adding the number two. So two Power X is greater than 500, which is the required number of hosts plus the number two. This implies that Two Power X is greater than or equal to 502. Like always, we start with X is equal to zero, X is equal to one, and so on. And when we reach X being equal to nine, we will get the value of 512, which is greater than or equal to 50 two.So the total number of bits that we need to fix equals nine. So now let's look at the subnet mask. The original subnet mask is eight. After fixing nine host bits, we are left with 23 network bits. So the new subnet mask becomes 23, where you have the first 23 bits as the network bits and the last nine bits as the host bits. So let's quickly convert that into decimals. We have the first octet, which is allones, which we know is two five five. The second octet is also all ones, which is also two five five. The last octet is all zeros, which is zero. We need to figure out the decimal value for the third octet. So you guys know what we need to do now. So we'll align all the bits and then write the number two against every bit. The powers of two will then be assigned, beginning with two power zero on the right and progressing to two power seven on the left. And then we'll only add those powers of two, which correspond to the ones. So that means we'll add 128 plus 64 plus 32 plus 16 plus eight plus four plus two. And that gives us a value of 254. So our new subnet mask is now 23, which in decimal equals 255-255-2540. Step number three says, "Determine the increment." We now know it. The increment is the power of two corresponding to the least significant bit or the rightmost bit. In this case, the least significant bit, or rightmostmost bit, is in the third octet. And the power of two corresponding to the least significant bit is the number two. So that's going to be our increment. Step number four: add the increment to get the new subnets. The first network address is 124. Looking at all those zeros, we may get confused as to where we are going to add the increment. Important: The increment should always be added to the octet that contains the least significant bit. In this case, the least significant bit is in the third octet. So let's start adding the increments. You have the first network address, which is 124. When we add the number two to the third octet, we get 1240 20. We have a total of 124 zeros, four zeros, all the way up to 102 402540. When we add two one more time, that becomes 256. So that means we have to carry over to the next octet on the left-hand side. So now we have 124100, 124, one 20, and it goes on up to 102, 412540. And you keep doing this, on and on and on. You'll arrive at 124 dot 254, dot zero, dot zero. We again keep adding the number two to the third action. We get 124, 254, 204060, and eventually 102, 425-42540. The total number of subnets equals two powers of n, where n is the total number of network bits. So the total number of subnets equals two powers of 23, which is 838-8608. The total number of hosts per subnet equals two power H, which is equal to two power nine, because we fix the nine host bits, which is equal to 512. The total number of usable host bits per subnet equals two power H minus two, which is two power nine minus two, which is 512 minus two, which is 510. So the original question was to subnet that network in such a way that we have 500 hosts per subnet. We now have 510 hosts per subnet. Let's now look at the second example. The second example says, "To which subnet does 201 one 100:29 belong?" Now here, instead of finding the subnets, we've been given an IP address that is already subnetted. And we've been asked to find which subnet the IP address belongs to. Interestingly, the steps remain the same. Step number one: convert the subnet mask into binary format. The decimal format of 29 equals 255-25-5255 248. And by now, you already know how we got the number dot 248 in binary format. We have the first 29 bits as the network bids and the last three bits as the host bits. Step number two: determining the number of host bits to be borrowed. In this case, the host bits have already been borrowed, which is why we have a subnet mask of slash 29. Otherwise, we would have had the default subnetmask of slash 24 because IP addresses beginning with 200 belong to the class C network. So we are not going to perform Step Number Two. Step number three says, "Determine the increment." The decimal subnet mask is 29, which I've written in binary format. We already know now that the increment is the power of two, corresponding to the least significant bit. So what I've done is what you already know. I've written the bits of the last octet and then aligned the powers of two corresponding to those bits. The power of two corresponding to the least significant bit is the number eight. That is going to be our increment. Step number four: Add the increment to get the new subnets. The first network address is 201, 10. We'll add the increment to the last octet because that is what contains the least significant bit. So adding the increment to 201, we get 201 1, 8201, 1, 6, and so on all the way up to 201 1248. Looking at the networks that we have derived, we now know that 201 10 belongs to the network 201 1829. I've added one more exercise, which is to identify the network address and the broadcast address of 201. We already know that 201 10 belongs to 201 1829. We also know the first address is the network address and the last address is the broadcast address. The first address of this subnet is 201 one eight, which is the network address. The next subnet begins at 201 one six.So one address before that is going to be the last address in this network. So the broadcast address, which is the last address of the network, is 201 1 5 Once again, it is homework time. So I have a couple of exercises for you. Number one is subnet 110 zero eight, such that each network has 1010 hosts per subnet. And exercise number two, which subnet does222 one two, oneone 30 belong to? My tips for this lecture remain exactly the same as the previous one. Just remember the four steps of subnetting and, most important, practice, practise, practice, and you will find that subnetting is really easy. So this is the last lecture for IP version four. In the next lecture, we'll get into a very interesting topic, which is IP version six. That's it for this lecture, guys. Please let me know if you have any questions. If not, I'd like to thank you for watching, and I'm going to catch you in the next lecture.

Hello and welcome back to this lecture, which is IP version six, part two. In the last lecture, we started by talking about IP version six. We understood how an IP version 6 address looks, so we looked at the address format. We also looked at the rules for writing IP Version 6 addresses. In this lecture, we'll continue where we left off. So we'll start by talking about the six IP address types. We'll talk about IP version six addresses, and then finally we'll touch a little bit on IP version six subnetting. If you're ready, let's begin. First up, let's start by talking about IP version six address types. And essentially, there are three address types. You have unicast, multicast, and Anycast. Unicast is used for one-to-one communication, while multicast is used for one-to-one communication. We had broadcast in IP version 4, which was one-to-all communication. Broadcast has now been replaced with any cast in IP version six. Anycast is one-to-many communication. However, the packet only reaches the node that is closest, and IP version six does not support broadcast addresses or broadcast communication. This diagram is a representation of unicast communication. You have one source and you have one destination, so it's one-to-one communication. This diagram represents multicast communication. So you have one source, and you have multiple destinations. It's too much communication. And this diagram shows you any cast communication. You have one source; you have multiple destinations. However, the communication is only between the source and the closest destination. Now let's talk about IP version six addresses. There are many different types of IP version six addresses, but we're only going to look at the most important ones. So on the list we've got unspecified addresses, loopback addresses, link local addresses, unique local addresses, global unicast addresses, and multicast addresses. Let's start with the first one, which is unspecified. So when a computer boots up and has no address assigned, it uses the unspecified address. If the computer is on a network that has DHCP enabled, this address is used even before it gets an address assigned via DHCP. The unspecified address can be represented by 128, which essentially means it is all zeros. And this can be compared with zero, zero, zero. In IP Version 4, this address should not be assigned manually to a node. Even if you did, the router would not forward the packet. The next one is addressing loopback devices. And we spoke a little bit about this in IP version four, right? It's the exact same thing over here. In IP version six, it is denoted by colon oneforward slash 128, which means it is all zeros. And the last bit is set as a one. This is the same as one hundred and twenty-seven, zero zero one, in IP version four. And like you all know by now, a loopback address is used by a host to communicate with itself. for example, when one application on the host is trying to communicate with another application on the same host. We then have links to local addresses. This is an address that the computer assigns to itself from a specific range, which is Fe Ay. So the first ten bits of that address are fixed, which is why it is represented as FE 800:10. This address is unique on the subnet, and this address is used for communicating over the land or a single link. It cannot be used to send packets to other countries. The link local address is analogous to a piper, or automatic private IP address configuration. In IP.Version four, I believe we did not discussa Pi in IP version four section. So APA is a range of addresses in IP Version 4, which a computer assigns to itself if it does not have a statically configured address or if it does not get an address via DHCP. In IP version 4, the range is 16925 400:16.The range in IP version 6 is Fe 80slash ten links. Local addresses are used for automatic address configuration when no static address is assigned by the user and when no DHCP server is present. We then have unique local addresses. These are unique and local addresses used for communicating inside the LAN. These can be compared to the RFC 1918 addresses. In IP version four, we had 00817, 2160, 00:16, and 192. We can compare these ranges with the unique local addresses in IP version six. These addresses cannot be routed over the Internet. They are just for local communication. The range for these addresses is FC seven. Next. We have global Unicast addresses. These are public addresses that can be routed over the Internet, and these addresses range from 2003 to 3 FFF 3. We then have multicast addresses. These addresses are used to send data to multiple hosts at the same time, especially for routing updates. The range would be FF eight.Multicast addresses should only be used as destination addresses. They should never be used as source addresses. All right, that's about the addresses. Now let's talk a little bit about IP version six subnetting. In IP version four, we spent a considerable amount of time with subnetting; in IP version six, we hardly touched it. It's going to be much more simple. We don't have to know everything about subnetting. We just need to know at a very high level: how does subnetting look in IP version six? All right, so let's begin. So, IP version six, subnetting, is similar to IP version four, but on a larger scale. Out of the 128 bits of an IP version 6 address, the first 48 bits represent the network address, also known as the routing prefix. This is assigned by the Internet Service Provider. The next 16 bits are known as subnetbits, and the last 64 bits represent the host bits, also known as the interface ID. So this is how it looks like: You have the 128-bit address, out of which the first 48 bits represent the routing precision. The 16 bits represent the subnet, and the 64 bits in the end represent the interface ID. The 16 bits in the subnet field can be used to break down the networks into subnets. The total number of possible subnets for each routing policy would be two powers of 16 because you have 16 bits and the two possible values are zero and one. So for every routing prefix, you have two power-16 possible subnets. For every subnet that we have, the total number of possible host addresses would be two powers of 64. Well, that's all we need to know about subnetting. So what are my important tips from this lecture? Remember the address types, which are unicast, multicast, and any cast. And also remember that broadcast is not supported in IP version six. Remember the different IP Version 6 addresses And remember how subnetting works in IP version six. This lecture was the last one for section one. So in the next lecture, we're going to summarise what we learned in Section One. That's it for this lecture. Please let me know if you have any questions. If not, I'd like to thank you for watching, and I'll catch you in the next lecture. Thank you.

Alright, welcome back, everybody. In this lecture, we are going to talk about IP version six. In the last two or three lectures, we looked at IP version four. Now it's time to talk about the next version of Internet Protocol, which is IP version six. But before we get there, first, let's discuss the end answers to the homework from the previous lecture. So the first question was to subnet 110 0/8, such that each network has 1010 hosts, and you have the answers on the screen. So the first subnet starts at eleven 00:22.The increment is four, which is added into the third octet. So we have eleven dot zero 40 slash 20 2110 80and so on, all the way up to 110 2520. When you add four to 252, you get 256, which means you have to move into the next octet. So you have eleven, dot one, dot 00:20 211 dot one,dot four, dot zero, dot eight, dot zero and so on. And if you keep continuing this way, you will end up with eleven, 252-25-2022. Which subnet does 222 dot one, dot two, dot eleven, slash 30 belong to? When we look at the subnets, we have 222 one.So doteleven on 222 dot one, dot two is part of the network 222 dot one, 2830. and eleven happens to be the broadcast address for that network. I hope you got the answers right. If you did not, please look back at the last lecture and try to find out what may have gone wrong. If you have a question, please feel free to ask me. Alright, now let's talk about IP version six. IP version six or Internet Protocol version six isintended to be the successor of IP version four. And if you're wondering what the reason is, the reason is very simple. We do not have enough IP Version 4 addresses left anymore. We are actually running out of public IPv4 addresses, so we had to move to another version that had the capacity to give us more IP addresses to play with. IP version six addresses are 128 bits in length. If you remember from IP version four, it was 32 bits in length, while IP version six is 128 bits in length, which means we have a significantly larger address space to play with. The total number of addresses that are available in IP version 6 is two powers of 128. Now, IP version four has three types of communication. You had unicast, you had multicast and broadcast. IP version six introduces a new type of communication known as any cast.When you send a communication to an anycast address, it is similar to a multicast address, which means you're trying to send a packet to multiple destinations. However, it reaches the nearest node of the group of destination nodes. One significant change is that IP Version 6 does not support broadcast at all. IP version six provides something known stateless address auto configuration." This means an IP Version 6 device has the ability to generate a unique address that can be used for communication over the network. Very important, guys, IP version six is not compatible with IP version four, which means you need to use some mechanisms such as dual stack and tunnelling to facilitate the transition. These two mechanisms are not the only ones. We actually have a few more, but I've just given you two examples, which are dual stack and tunneling. All right, now let's talk about the address format. An IP version six address is written in hexadecimal format. I'm sure you guys remember that an IP Version 4 address is written in decimal format. When you convert the IP version six address into binary format, you get 128 bits. That's why we call it a 128-bit address. The hexadecimal digits range from 0 to 9 and 8 F. An IP version six address is writtenas eight groups separated by colons. If you remember from IPV 4, it is written in four octets separated by dots. Here we have eight groups. We don't call it Octets anymore. They are known as groups, and they are separated by colons. And we have an example IP version six address as well. So you have 20010 DB eight four C8522, A eight, and B five A B 378-9177. If this is the first time you're looking at an IPV 6 address, this might spin your head. It is significantly longer than an IP Version 4 address. The good news is that we have some rules for writing IP version 4 addresses that can make these addresses appear as simple as possible, if not simpler. But one thing that we need to remember is that IP version six is the way forward. So we have to learn this, and we have to get comfortable with it because this is what we are going to be using in the future. Every hexadecimal character, which means zero to nine and eight to F, can be represented by four binary bits. In the IP version of six addresses, you have eight groups. Each group has four characters, which means you have a total of 32 hexadecimal characters. And every hexadecimal character can be represented by four binary bits. So 32 times four, you have a total of 128 bits in the IP version 6 address. Now let's talk about the rules for writing IP version six addresses. There are only two rules. Rule number one says a leader who leads zeros in a group can be discarded. So let's consider this IP version six address. You have 20010 DB 848 b five one. Now, rule number one says you can actually discard the leading zeros. Here you have leading zeros in the 2nd, 3rd, 4th, 6th, and 7th groups. and these leading zeros can be discarded. So that means we can rewrite this IP version six address so that 2001 remains the same because we do not have any leading zeros. Zero DB eight can be represented as DB eight. After dropping the leading zero, the third one has all zeros, which means you can drop the first three zeros and write it as zero. The next one has one leading zero. So we'll write it as C three A, and thenext one, four eight B five remains the same. This one again has four leading zeros. We'll drop the first three, so you have the zero over here. The 7th group has three leading zeros. So we'll drop that and just write the one, and then you have 9177. Rule two says any one group of two or more zeros can be replaced with two colons. However, this can only be done once. So let's look at an IPV 6 address. You have 2001, and you'll notice the next three groups are all zeros. Next, you have four eighteen five, and then you have two more groups of zeros and 9177. Let's start by applying rule one, which says leading zeros in a group can be discarded. So we can rewrite this address as "2001." You drop the first three zeros, and you write your zero. The same applies for the next. You have zero, zero, zero, zero, zero, zero, zero, zero, zero, zero, zero, zero, zero, zero, zero, zero, zero, zero As is, for these two groups, we can drop the first three leading zeros. So we only write it as zero and zero. Finally, there is 9177, which is as such. Let's now apply Rule Two. Rule two says any one group of two or more zeros can be replaced with a colon. Notice carefully that we have two groups of two or more zeros. This one has three groups of zeros, and this one has two groups of zeros. Because rule two states that you can only replace one of these two groups with a colon, we can replace either of these two groups with a colon. So there are two ways in which you can write this answer. Number one, if we replace this one with "colon," you'll have this answer, which is 2000 and 148 b 509-1770r. Option number two is to leave this as is and replace it with a colon. So you've got 2001 0048-B-5. You have a colon, which is what we get by replacing these two zeros. And then you have 9177. Both of these are valid representations. However, you'll always notice it is written like the first one, which is this one here, where we placed three consecutive groups with a colon. If you're wondering why we would choose that, look at the number of bits. The number of bits in this representation is less than the number of bits in this representation. So even though both of them are valid, you'll always find this one being used. Most importantly, please remember that we cannot use a colon more than once in an address. If you're wondering why we can't use the colon more than once, here is the answer. Consider the same IP address again. You've got 2001. You have three groups of zeros. Four, eight, and five You have two groups of zeros, 9177. Now, we can replace any one of these two groups of zeros with a colon. Let's imagine that we actually replace both groups of zeros with colon colon.This is what it would look like. You have 2001, and then you replace this with colon colon four, eight, or five, and you also replace this one with colon colon. And then you have 9177. If we did it this way, imagine the confusion that would happen when we had to put the zeros back in place. We would have more than one correct answer. For example, we have one group of three zeros over here and one group of two zeroes here. When we had to put the zeros back, we could actually do it in two ways. Number one, you put three groups of zeros over here and two groups of zeros over here. You could also do it in one more way where you put two groups of zeros over here and three groups of zeros over here. Because you have the two colons over here, you have no idea where you put the three groups of zeros and the two groups of zeros. In this situation, the operating system would not know which one was the correct answer. So to avoid this kind of confusion, it is allowed to use a colon only once in an IP version 6 address. Some important tips from this video Remember that IP version six is a 128-bit address written in hexadecimal SMIL format. And remember the rules for writing IP version six addresses. Rule number one says leading zeros can be discarded. And rule number two says any one group of two or more zeros can be replaced with a colon. This can be used only once in an IP version 6 address. In the next lecture, we'll discuss some remaining topics from IP version six. Initially, I planned to put all these topics in one lecture, but I felt we might be just cramming in too much into a single video. So let's do it in a separate lecture. In the next lecture, we'll discuss IP version six addresses, IP version six address types, and a little bit about IP version six subnetting. That's all for this lecture, guys. Let me know if you have any questions. If not, I'd like to thank you for watching, and I'll catch you in the next lecture.

Alright, welcome back. In this lecture, we're going to do a quick summary of what we learned in Section One. We'll take a high-level overview of the different topics that we learned in Section One. So in the first lecture of section one, we understood the OSI model, also known as the Open Systems Interconnection Model. We learned that the OSI model is a logical model used to divide network communication into seven layers. The different layers are named Application Layer, Presentation Layer, Session Layer, Transport Layer, Network Layer, Data Link Layer, and the Physical Layer. We also discussed the mnemonics to remember the OSI model. We saw a couple of them. The first one says Please donot throw sauces pizza away. and the second one is what I personally like. It says people don't need these stupid packets anyway. So those are the two mnemonics that you can remember to remember the layers of the OSI model. From an examination standpoint, remember the functions of every layer. And guys, also remember that the layers of the OSI model do not necessarily correspond to the physical layers or the physical components of network communication. We then discussed the TCP/IP model, which is another way of representing network communication. The TCP IP model divides network communication into four layers. They're called the Application Layer, Transport Layer, Internet Layer, and the Network Access Layer. And we also saw a diagram that compares the OSI model with the TCP/IP model. We then spoke about Ethernet. And we said that Ethernet is a suite of technologies and protocols that define communication at layer two. It is popularly known as IEEE 802 3. We then discussed CS MACD. It stands for Carrier-Sensitive Multiple Access with Collision Detection. When two devices on a shared medium try to send data at the same time, it results in what is known as a collision. And to avoid this, every device that wants to transmit must listen to the traffic on the network or the medium. A device is allowed to transmit only when the medium is determined to be free. If two or more devices try to send data at the same time, it results in collisions. If a collision is detected, all sending devices must back off and wait for a random amount of time before they attempt to retransmit the data. We then spoke about Mac addresses. The Mac address, also known as the Media Access Control Address, is a 48-bit address that is burned onto the chip of your device. Hence it is also known as real address of your device." Very important. Remember the representation of a Mac address? It is represented as six groups of two hexadecimal characters. and we have an example on the screen. Every hexadecimal character can be represented by four binary bits. We have a total of twelve hexadecimal characters. So twelve times four is a 48-bit address. The first three groups of characters are known as "Oui" or the organisational unit identifier. It identifies the manufacturer of the network equipment. We then spoke about collision domains and broadcast domains. A collision domain is a part of the network where packets can collide with one another. I also give you a simpler way to understand. A collision domain refers to how many devices can send data at the same time. When we talk about a hub, it's only one collision domain, because only one device can send data at the same time. When we talk about a switch or a router, every port is its own collision domain because every device can send data at the same time. A broadcast domain is a logical division of a computer network in which all nodes can reach each other by broadcasting at the data link layer, or in other words, broadcasting at layer two. The simple way to remember this is that the abroadcast domain indicates how far a broadcast can reach. When we talk about AHUB, the broadcast reaches everywhere. It reaches every device that is connected to the hub, which is why it is one broadcast domain. When we talk about a switch without any additional VLANs configured, which means only the default VLAN, it is just one broadcast domain. However, you can actually configure additional VLANs on the switch, and every VLAN would be its own broadcast domain. When we talk about a router, every port on a router is a broadcast domain. Because routers do not forward broadcast traffic, abroadcast will end on the port. So every port is its own broadcast domain. Next, we spoke about VLANs, or virtual LANs, to deal with broadcast traffic and logically separate devices on the same LAN into different groups, VLANs can be used.A VLAN is a logical separation of devices on the same LAN. It allows you to divide a land segment into multiple logical lands, also known as virtual lands. Now, each VLAN is a different network with separate layers we're addressing, and each VLAN is a different broadcast domain. Having different VLANs allows you to separate the traffic, and it also allows you to apply different quality of service to different traffic from different VLANs. We also discussed that by default, VLANs do not talk to each other. You need a layer-three device to route the traffic between VLANs. And when a switch is divided into VLANs, each VLAN is identified by a unique two-word QID, also known as a VLAN tag. We then discussed the different network devices. We discussed that a hub is a layer one device and a bridge is a layer two device. A switch is also a layer-two device. However, we do have layer-three switches as well, and a router is a layer-three device. We also discussed the different switch interface modes. There are three interface modes. You have access mode, you have trunk mode, and then you have tagged access mode. And we also discussed the differences between all thesemodes It is important from an examination standpoint as well to know the different modes on a switch. Next, we discussed a very important topic known as "spanning tree protocol." We started by discussing what a layer-two loop is, and we understood that layer-two loops are good and bad. A layer 2 loop provides a backup, but it can also cause the packet to loop around in the network forever and keep consuming valuable network resources. The spanning tree protocol solves the problem by disabling the redundant links. It does this by electing what is known as a "root bridge." The route bridge is a switch from whose perspective the entire network is visualized, and we use something known as bridge ID to elect the root bridge. Every bridge has a priority number, which can be manually configured from zero to 65536. The default priority number is 32768. The bridge priority number, along with the MAC address of the bridge, forms what is known as the bridge ID. The bridge with the lowest span is elected as the route bridge. Once the route bridge has been elected, all the ports of the route bridge will be in a "forwarding state," which means they are ready to forward traffic. All other switches compute what is known as the "best path" or the shortest path to reach the root bridge. The ports other than the best to reach the root bridge are placed into blocking mode, ready to be activated when required. We then discussed IP version four. We discussed what an IP address looks like. It's a 32-bit logical address that is assigned to your computer, and it is written in decimal format. It has four octets, and each of those octets is separated by a dot. For example, 192 168 1 Every bit in that IP address can have two possible values: zero and one. Hence, the total number of IP addresses possible is two powers of 32. We then discuss the different classes of IP addresses. We have classes ABCD and E, and we also discuss the range of IP addresses in every class. Class A numbers range from zero zero zero to 102 725-525-5255. Class B starts at and goes all the way up to 109, 125, 525, and 5255. Class C starts all the way up at 202 325-525-5255.Class D starts at 2240 zero and goes all the way up to 239 255-25-5255, and class E starts at 2400 zero and goes all the way up to 2 5525-525-5255. Class D is reserved for multicast purposes, and Class E is reserved for experimental purposes. Also, in each class A, B, and C, we have a block of addresses that is reserved for internal use, also known as private addresses, and this is defined in RFC 1918. In class A, we have the range from 100 all the way up to ten, 255, 25, 5, 255. This can also be represented as 100 zero eight. For class B, the range is 170 2160, all theway up to 170 216, 31 two five five. This can also be represented as 170 2160 00:12, and for class C the range is 192 all the way up to 192 106 825-5255, which can be represented as 192 1680 00:16. We also discussed loopback addresses, which are also a reserved range of IP addresses. The range for loopback addresses is one hundred and twenty-seven zero-odd eight. And then we also discussed APAaddresses, also known as automatic private IP addresses. The address range for that is 169–25, 40 00:16. We then discussed what a subnet mask is. We understood that every IP address has two parts. It has a network portion, and it has a host portion. And it is the separate mask that allows you to separate the network portion from the host portion. The IP address can be denoted in two ways. You can denote that as ten-eight or ten 25500.These are two ways to represent the exact same IP address. When the subnet mask is converted into binary, the ones denote the network portion and the zeros denote the host portion. for example, 100 zero eight.When the subnet mask is written in binary format, we have the first eight bits made up of ones, and we have the last 24 bits made up of zeros. The first eight bits, which are all ones, are the network portion. Classes A, B, and C have a default subnet mask. The default mask for class A is eight, for class B it is 16, and for class C it is 24. The total number of host addresses in any network equals two powers of H, where H is the number of host bits. And then we also discussed that in any network, the first address belongs to the network, also known as the network address, and the last address is the broadcast address. Both these IP addresses cannot be assigned to devices. Hence, the total number of usable IP addresses would be two powers of H minus two. We then talked about subnetting and what its importance is, and we discussed the four steps for subnetting. Step number one is to start by converting the subnet mask into binary format. Once we've done that, we need to determine the number of host bits that need to be borrowed. Following that, we will determine the increment, and finally we'll add the increment to get the new subnet. We also looked at different examples of subnetting across different classes. Classes A, B, and C. We then briefly discussed IP version six. IP version six addresses are 128 bits in length, and the total number of addresses equals two powers of 128. It is written in hexadecimal format, and we have an example of that on the screen as well. Broadcast traffic has been removed from IP version six. Instead, it has been replaced with any other type. We then discussed the rules for writing IP version six addresses. There's only two rules. Rule number one is that leading zeros in a group can be discarded. And the second one says any group of two or more zeros can be replaced with a colon. Colon. However, this can only be done once. We also discussed the different IP version six addresses. We talked about unspecified addresses, loopback addresses, link local addresses, unique local addresses, global, unicast, and multicast addresses. So those are the topics that we covered in Section One. I'm very happy with the way we proceeded. We've actually covered a lot of ground, and we've actually laid a very strong foundation for the next section, which is going to be an introduction to Juno's. So in section two, we'll start by talking about Juniper's device portfolio, which means we'll understand the different types of devices available from Juniper. We'll understand the architecture of the Juniper operating system, and then we'll also discuss concepts related to traffic processing in Juno's operating system. That's all for this lecture, guys. Let me know if you have any questions. If not, I'd like to thank you for watching. I'll catch you in the next lecture of Section Two. Thank you.

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