1 00:00:01,140 --> 00:00:07,230 Welcome to Jeremy's IT Lab. This is a free, complete course for the CCNA. If you like 2 00:00:07,230 --> 00:00:12,670 these videos, please subscribe to follow along with the series. Also, please like, leave 3 00:00:12,670 --> 00:00:15,670 a comment, and share the video to help spread this free series of videos. 4 00:00:15,670 --> 00:00:17,648 Thanks for your help. 5 00:00:18,770 --> 00:00:24,749 In this video, we will be talking about subnetting. This is a very big topic for the CCNA, but 6 00:00:24,749 --> 00:00:29,900 not just for the test--it’s an essential skill for a network engineer. Many people 7 00:00:29,900 --> 00:00:35,591 have trouble understanding subnetting, but let me assure you, it is not difficult. Subnetting 8 00:00:35,591 --> 00:00:41,809 is very simple if you take it step by step. So, I’m going to split subnetting into two, 9 00:00:41,809 --> 00:00:46,879 or maybe even three, videos so we can take our time to really understand subnetting without 10 00:00:46,879 --> 00:00:52,610 getting lost. Now, because subnetting is such an important topic, and many people have trouble 11 00:00:52,610 --> 00:00:58,260 with it, there are already plenty of subnetting videos on YouTube. Of course, feel free to 12 00:00:58,260 --> 00:01:02,920 check out those videos too--there are some different tricks and techniques people teach 13 00:01:02,920 --> 00:01:08,210 that can speed up the subnetting process. I, however, will simply outline the basic 14 00:01:08,210 --> 00:01:14,310 steps involved in subnetting. I will avoid overcomplicating the topic. My end 15 00:01:14,310 --> 00:01:20,430 goal for these videos is that you understand and can do subnetting. So, let’s get started. 16 00:01:20,430 --> 00:01:28,020 So, what will we cover in this video? Just a couple of things. First is CIDR, pronounced 17 00:01:28,020 --> 00:01:35,420 “CIDR,” which stands for Classless Inter-Domain Routing. What exactly is that? Well, remember 18 00:01:35,420 --> 00:01:42,610 I introduced the IPv4 address classes, such as Class A, B, and C? Well, CIDR throws all 19 00:01:42,610 --> 00:01:48,750 that away and lets us be more flexible with our IPv4 networks. Then, of course, we’ll 20 00:01:48,750 --> 00:01:54,369 cover the process of subnetting, taking it step by step so you don’t get lost. 21 00:01:54,369 --> 00:02:00,009 Now, before I get into CIDR, let’s review these IPv4 address classes so we can then 22 00:02:00,009 --> 00:02:06,909 understand the need for classless IPv4 addressing. There are five classes of IPv4 addresses: 23 00:02:06,909 --> 00:02:14,800 A, B, C, D, and E. Class A addresses have a first octet beginning with zero, and the rest 24 00:02:14,800 --> 00:02:20,480 of the bits can either be zero or one. This leads to a decimal range for the first octet of 25 00:02:20,480 --> 00:02:29,250 0 to 127. Remember, an IPv4 address is 32 bits, so there are 4 octets--4 groups of 8 26 00:02:29,250 --> 00:02:42,180 bits--in an IPv4 address. This makes the Class A address range from 0.0.0.0 through 127.255.255.255. 27 00:02:42,180 --> 00:02:46,960 Now, remember, there are some special and reserved addresses in these ranges that can’t be 28 00:02:46,960 --> 00:02:52,060 used for normal IP addresses on a device, but for this video, we’ll just include all 29 00:02:52,060 --> 00:02:59,510 of them in Class A. Class B addresses have a first octet beginning with 10, and the 30 00:02:59,510 --> 00:03:06,560 other 6 bits can be either 0 or 1. This gives a range for the first octet of 128 through 31 00:03:06,560 --> 00:03:18,750 191. The address range for Class B is 128.0.0.0 through 191.255.255.255. Class C addresses 32 00:03:18,750 --> 00:03:25,019 have the first three bits set to 110, and the others can be either zero or one. If you write 33 00:03:25,019 --> 00:03:33,530 that range in decimal, it is 192 through 223. The address range is therefore 192.0.0.0 34 00:03:33,530 --> 00:03:42,750 through 223.255.255.255. Class D addresses begin with 1110 in binary, which gives 35 00:03:42,750 --> 00:03:49,709 a range of 224 through 239 for the first octet of the address. This means that the address range 36 00:03:49,709 --> 00:04:02,000 for Class D is 224.0.0.0 through 239.255.255.255. Finally, Class E addresses begin with 1111 37 00:04:02,000 --> 00:04:11,470 in binary, so the first octet range is 240 through 255, and therefore the address range is 240.0.0.0 38 00:04:11,470 --> 00:04:14,090 through 255.255.255.255. 39 00:04:14,090 --> 00:04:21,970 However, only the Class A, B, and C addresses can be assigned to a device as an IP address, 40 00:04:21,970 --> 00:04:28,680 as Classes D and E have special purposes, as I mentioned in the IPv4 addressing videos. Class 41 00:04:28,680 --> 00:04:34,550 A addresses have an 8-bit prefix length, meaning the first octet identifies the network and 42 00:04:34,550 --> 00:04:40,200 the other three octets are used for individual hosts within the network. Class B addresses 43 00:04:40,200 --> 00:04:45,530 have a 16-bit prefix length, so the first two octets identify the network, and the last 44 00:04:45,530 --> 00:04:51,610 two octets identify individual hosts within that network. Class C addresses have a prefix 45 00:04:51,610 --> 00:04:57,600 length of 24, so the first three octets are used to identify the network, and only the 46 00:04:57,600 --> 00:05:03,340 last octet is used to identify individual hosts within that network. 47 00:05:03,340 --> 00:05:08,280 The different prefix lengths give different characteristics to these classes. As you can 48 00:05:08,280 --> 00:05:14,130 see, there are few Class A networks available--only 128, actually less than that because 49 00:05:14,130 --> 00:05:21,280 some are reserved, like the 127.0.0.0 range, which you may remember is used for loopback 50 00:05:21,280 --> 00:05:26,870 addresses. Because only the first octet of a Class A address is used for the network ID, 51 00:05:26,870 --> 00:05:32,040 there are three whole octets available for addresses within each Class A network, 52 00:05:32,040 --> 00:05:40,410 so there are 16,777,216 addresses in each Class A network. That is 53 00:05:40,410 --> 00:05:48,220 2 to the power of 24, because there are three octets (3 times 8 = 24 bits). Class B 54 00:05:48,220 --> 00:05:55,270 addresses are different. There are more Class B networks--16,384--but fewer addresses per 55 00:05:55,270 --> 00:06:03,250 network, 65,536, which is still many addresses, of course. Finally, there are very 56 00:06:03,250 --> 00:06:12,850 many Class C networks--2,097,152 networks--but only 256 addresses per network. 57 00:06:12,850 --> 00:06:19,850 So, how does a company get their own network address range to use? Well, IP addresses are assigned to 58 00:06:19,850 --> 00:06:26,250 companies or organizations by a nonprofit American corporation called the IANA, the 59 00:06:26,250 --> 00:06:32,850 Internet Assigned Numbers Authority. The IANA assigns IPv4 addresses and networks to companies 60 00:06:32,850 --> 00:06:39,160 based on their size. For example, a very large company might receive a Class A or Class B 61 00:06:39,160 --> 00:06:44,370 network. Remember, there are lots of available addresses to use for hosts in each Class A 62 00:06:44,370 --> 00:06:49,910 and Class B network. While a small company might receive a Class C network, because there 63 00:06:49,910 --> 00:06:56,810 are fewer addresses in each Class C network--only 256. However, this system led to many 64 00:06:56,810 --> 00:07:02,881 wasted IP addresses, so multiple methods of improving this system have been created. Let 65 00:07:02,881 --> 00:07:08,680 me give you an example of how this strict system of addresses can waste IP addresses. 66 00:07:08,680 --> 00:07:15,820 So, here are two routers. As you can see, R1 has three networks connected to it here. 67 00:07:15,820 --> 00:07:20,910 Remember that routers are used to connect different networks, so each of these links is a separate 68 00:07:20,910 --> 00:07:28,280 Layer 3 network, different IP networks. R2 also has three networks connected here. Perhaps 69 00:07:28,280 --> 00:07:33,300 each of these networks will have a few switches, with many end hosts such as PCs and servers 70 00:07:33,300 --> 00:07:39,650 connected to these switches. However, there is one more network here. That’s this network 71 00:07:39,650 --> 00:07:45,290 connecting these two routers. This is known as a point-to-point network, meaning 72 00:07:45,290 --> 00:07:51,591 that it’s a network connecting two points, in this case, R1 and R2. For example, this 73 00:07:51,591 --> 00:07:57,570 might be a connection between offices in different cities, let’s say San Francisco and New York. 74 00:07:58,540 --> 00:08:04,900 So, because this is a point-to-point network, we don’t need a large address block, so 75 00:08:04,900 --> 00:08:14,020 let’s use a Class C network, 203.0.113.4. Because this is a Class C network, there are 76 00:08:14,020 --> 00:08:22,620 256 addresses in the network, minus one for the network address (203.0.113.0), minus one 77 00:08:22,620 --> 00:08:30,740 for the broadcast address (203.0.113.255), minus one for R1’s address, which I’ll 78 00:08:30,740 --> 00:08:40,300 assign as 203.0.113.1, and minus one for R2’s address, which I’ll assign as 203.0.113.2. 79 00:08:40,300 --> 00:08:45,326 That’s a total of four addresses used and 252 addresses wasted. 80 00:08:45,326 --> 00:08:48,730 Clearly, this is not an ideal system. 81 00:08:50,840 --> 00:08:56,980 Before introducing CIDR, here’s another quick example of address waste. A company, 82 00:08:56,980 --> 00:09:05,310 Company X, needs IP addressing for 5,000 end hosts. This is a problem, why? A Class C network 83 00:09:05,310 --> 00:09:11,140 does not provide enough addresses, so a Class B network must be assigned. Because a Class 84 00:09:11,140 --> 00:09:18,020 B network allows for about 65,000 addresses, this results in about 60,000 addresses being wasted. 85 00:09:19,610 --> 00:09:23,970 When the Internet was first created, the creators did not predict that the Internet would become 86 00:09:23,970 --> 00:09:30,150 as large as it is today. This resulted in wasted address space, like the examples I showed 87 00:09:30,150 --> 00:09:35,890 you, and there are many more examples that I could show you. The total IPv4 address space 88 00:09:35,890 --> 00:09:40,970 includes over 4 billion addresses, and that seemed like a huge number of addresses when 89 00:09:40,970 --> 00:09:52,060 IPv4 was created, but now address space exhaustion is a big problem. There's not enough addresses. One way to solve, or remedy, this problem is 90 00:09:52,060 --> 00:10:00,190 CIDR. The IETF (Internet Engineering Task Force) introduced CIDR in 1993 to replace 91 00:10:00,190 --> 00:10:02,110 the classful addressing system. 92 00:10:03,760 --> 00:10:10,311 With CIDR, the requirements of Class A addresses to use an 8-bit network mask, Class 93 00:10:10,311 --> 00:10:18,190 B to use 16, and Class C to use 24 were removed. This allowed larger networks 94 00:10:18,190 --> 00:10:24,270 to be split into smaller networks, allowing greater efficiency. These smaller networks 95 00:10:24,270 --> 00:10:29,340 are called subnetworks, or subnets. Let’s look at an example of splitting a 96 00:10:29,340 --> 00:10:33,990 larger network into a smaller network so you can see how it works. 97 00:10:33,990 --> 00:10:40,780 Here’s the same point-to-point network we looked at before. Previously, it was assigned 98 00:10:40,780 --> 00:10:48,640 the 203.0.113.0/24 network space, but that resulted in lots of wasted addresses. Let’s 99 00:10:48,640 --> 00:10:55,830 write this out in binary. Here’s the binary, with the dotted decimal underneath. Now, the 100 00:10:55,830 --> 00:11:04,680 prefix length is 24, so here’s the network mask, also known as the subnet mask: 255.255.255.0. 101 00:11:04,680 --> 00:11:11,210 Remember, all 1s in the subnet mask indicate that the same bit in the address 102 00:11:11,210 --> 00:11:16,850 is the network portion. In this case, I’ve made the network portion blue, and the host portion 103 00:11:16,850 --> 00:11:26,381 is red. Well, how many host bits are there? 8, because it’s one octet. So, how many potential hosts, or how 104 00:11:26,381 --> 00:11:33,520 many usable addresses, are there? Well, the formula is this: 2 to the power of 8 minus 105 00:11:33,520 --> 00:11:41,830 2 equals 254 usable addresses. What is the 8? Well, it’s the number of host bits, which is 106 00:11:41,830 --> 00:11:48,840 8 in this case. And why minus 2? Those are the network address and the broadcast address. 107 00:11:48,840 --> 00:11:53,760 We can’t assign them to a device, so we have to remove them from the number of usable addresses. 108 00:11:53,760 --> 00:12:01,560 So, we have 254 usable addresses, but we only need two--one for R1 and one for R2. 109 00:12:01,560 --> 00:12:09,200 However, CIDR allows us to use different prefix lengths, so it doesn’t have to be 24. 110 00:12:09,200 --> 00:12:13,860 Let’s get some practice calculating the number of hosts within different prefix lengths. 111 00:12:13,860 --> 00:12:34,210 203.0.113.0/25, 203.0.113.0/26, 203.0.113.0/27, /28, /29, /30, /31, and finally /32. I’ve 112 00:12:34,210 --> 00:12:39,730 put /31 and /32 in red because they’re a little bit special, as you’ll see when you 113 00:12:39,730 --> 00:12:46,120 try to calculate it. So, pause the video here and try to calculate how many usable addresses 114 00:12:46,120 --> 00:12:53,850 are on each network. Okay, let’s check out the answers. 115 00:12:53,850 --> 00:13:01,720 So, here is 203.0.113.0/25, but this time with a /25 mask. Notice that the network portion 116 00:13:01,720 --> 00:13:06,570 of the address has extended into the first bit of the last octet, and the mask 117 00:13:06,570 --> 00:13:15,910 in dotted decimal is now written as 255.255.255.128. I changed the color of the extra bit to purple, 118 00:13:15,910 --> 00:13:21,140 but it is part of the network portion, which is the blue part. If you don’t remember how to convert 119 00:13:21,140 --> 00:13:27,130 from binary to dotted decimal, make sure you review that; it’s very important for subnetting. 120 00:13:27,130 --> 00:13:31,520 Now, there are 7 bits in the host portion of the address, so the number of usable addresses 121 00:13:31,520 --> 00:13:40,280 is 2 to the power of 7 minus 2, which equals 126. Once again, we only need two addresses-- 122 00:13:40,280 --> 00:13:47,290 one for R1 and one for R2--so we will be wasting 124 addresses. That’s better than wasting 123 00:13:47,290 --> 00:13:53,890 252 addresses with a /24 prefix length, but it’s still wasteful. 124 00:13:53,890 --> 00:14:02,160 How about a /26 prefix length? Notice that it’s now written as 255.255.255.192 in dotted 125 00:14:02,160 --> 00:14:08,351 decimal, because two bits of the last octet are now part of the network portion. Since 126 00:14:08,351 --> 00:14:14,940 there are six host bits, there are now 62 usable addresses in this network. If we were to use 127 00:14:14,940 --> 00:14:23,640 a /26 network mask for the 203.0.113.0 network, we would be wasting 60 addresses. Getting 128 00:14:23,640 --> 00:14:27,730 better, but we can make this network even smaller. 129 00:14:27,730 --> 00:14:33,690 Now that you get the idea, let’s speed it up. For a /27 prefix length, the mask is written 130 00:14:33,690 --> 00:14:42,660 as 255.255.255.224 in dotted decimal. There are now five host bits, so that means there are 131 00:14:42,660 --> 00:14:50,390 30 usable addresses. As you can see, the address space is getting smaller and smaller as we extend the network mask. 132 00:14:50,650 --> 00:15:00,921 For a /28 prefix length, the mask is written as 255.255.255.240 in dotted decimal. There 133 00:15:00,921 --> 00:15:08,180 are now only four host bits, so that means there are 14 usable addresses. After assigning addresses 134 00:15:08,180 --> 00:15:14,500 to R1 and R2, this would mean only 12 wasted addresses, but we can make this address space 135 00:15:14,500 --> 00:15:19,680 even smaller to make our addressing even more efficient. 136 00:15:19,680 --> 00:15:28,330 If we use a /29 prefix length, the mask is written as 255.255.255.248 in dotted decimal. 137 00:15:28,330 --> 00:15:34,920 Now we have only three host bits, so that means there are just six usable addresses. Again, 138 00:15:34,920 --> 00:15:41,510 after we give R1 and R2 addresses, there would be only four wasted addresses. 139 00:15:41,510 --> 00:15:50,140 If we use a /30 prefix length, the mask is written as 255.255.255.252 in dotted decimal. 140 00:15:50,140 --> 00:15:56,921 There are now only two host bits, so that means two usable addresses. So, this is perfect. There 141 00:15:56,921 --> 00:16:03,060 are four total addresses: the network address, the broadcast address, R1’s address, and 142 00:16:03,060 --> 00:16:07,899 R2’s address. That means zero wasted addresses. 143 00:16:07,899 --> 00:16:19,150 Before moving on to look at the /31 and /32 prefix lengths, let me clarify a little bit. So, instead of 203.0.113.0/24, 144 00:16:19,150 --> 00:16:30,170 we will use 203.0.113.0, which is a subnet of that larger Class C network. 203.0.113.0 145 00:16:30,170 --> 00:16:38,089 includes the address range of 203.0.113.0 through 203.0.113.3. Let me show you that 146 00:16:38,089 --> 00:16:50,420 in binary. Here is 203.0.113.0 in binary, the host portion all zeroes. Here is 203.0.113.1, 147 00:16:50,420 --> 00:17:00,380 203.0.113.2, and 203.0.113.3. These are the four addresses in the network, with these two being 148 00:17:00,380 --> 00:17:07,351 the two usable addresses, which are assigned to R1 and R2. So, we took up four addresses with 149 00:17:07,351 --> 00:17:14,880 this subnet. What about the other addresses in the 203.0.113.0/24 range? The remaining 150 00:17:14,880 --> 00:17:24,709 addresses in the address block, which are 203.0.113.4 through 203.0.113.255, are now available 151 00:17:24,709 --> 00:17:33,559 to be used in other subnets. That’s the magic of subnetting. Instead of using 203.0.113.0/24 152 00:17:33,559 --> 00:17:42,480 and wasting 252 addresses, we can use /30 and waste no addresses. Or, perhaps there is another 153 00:17:42,480 --> 00:17:47,340 way to make this even more efficient. Let’s look into it. 154 00:17:47,340 --> 00:17:57,010 If we use a /31 prefix length, the mask is written as 255.255.255.254 in dotted decimal. 155 00:17:57,010 --> 00:18:06,100 There is now only one host bit, so that means zero usable addresses. Two to the power of one is two, 156 00:18:06,100 --> 00:18:12,030 minus two for the network and broadcast addresses, means zero addresses that we can assign to devices. 157 00:18:12,030 --> 00:18:20,350 So, you used to not be able to use /31 network prefixes because of this. However, for a point-to-point 158 00:18:20,350 --> 00:18:26,070 connection like this, it actually is possible to use a /31 mask. Let's check it out. 159 00:18:27,460 --> 00:18:41,179 So, here’s the 203.0.113.0/31 network. R1 is 203.0.113.0, and R2 is 203.0.113.1. The 160 00:18:41,179 --> 00:18:50,740 203.0.113.0/31 network consists of addresses from 203.0.113.0 through 203.0.113.1, which 161 00:18:50,740 --> 00:18:58,580 is actually only two addresses. Here they are in binary. There’s 203.0.113.0, and 162 00:18:58,580 --> 00:19:05,559 there’s 203.0.113.1. Normally, this would be a problem because it leaves no usable 163 00:19:05,559 --> 00:19:10,240 addresses after subtracting the network and broadcast addresses, but for point-to-point 164 00:19:10,240 --> 00:19:15,530 networks like this, a dedicated connection like this between two routers, there is actually 165 00:19:15,530 --> 00:19:20,900 no need for a network address or a broadcast address. So, we can break the rules in this 166 00:19:20,900 --> 00:19:27,290 case and assign the only two addresses in this network to our routers. Note that if 167 00:19:27,290 --> 00:19:32,020 you try this configuration on a Cisco router, you’ll get a warning like this, reminding 168 00:19:32,020 --> 00:19:37,490 you to make sure that this is a point-to-point link, but it is a totally valid configuration. 169 00:19:37,490 --> 00:19:49,560 So, once again, the remaining addresses in the 203.0.113.0/24 address block, which are 203.0.113.2 through 203.0.113.255, 170 00:19:49,560 --> 00:19:54,730 are now available to be used in other networks. But this time, we've 171 00:19:54,730 --> 00:20:02,110 saved even more addresses, using only two addresses instead of four for this point-to-point connection. 172 00:20:02,110 --> 00:20:08,130 People still do use 30 for point-to-point connections at times, but 31 masks are totally 173 00:20:08,130 --> 00:20:12,380 valid and more efficient than 30, so I recommend this method. 174 00:20:14,130 --> 00:20:23,030 But, we still haven't looked at the 32 mask. A 32 mask is written as 255.255.255.255 in 175 00:20:23,030 --> 00:20:29,620 dotted decimal, making the entire address the network portion. There are no host bits. 176 00:20:29,620 --> 00:20:35,280 If you calculate this using our formula, you will get one usable address. Clearly, the 177 00:20:35,280 --> 00:20:41,510 formula doesn't work in this case. You won't be able to use a 32 mask in this case, and 178 00:20:41,510 --> 00:20:47,461 you will probably never use a 32 mask to configure an actual interface. However, there 179 00:20:47,461 --> 00:20:53,180 are some uses for a 32 mask. For example, when you want to create a static route not 180 00:20:53,180 --> 00:21:00,410 to a network, but to just one specific host, you can use a 32 mask to specify that exact host. 181 00:21:00,410 --> 00:21:05,710 Anyway, I'll talk about that later in the course. Just know that 32 masks are 182 00:21:05,710 --> 00:21:10,090 used at some points, but you don't have to worry about them for now. 183 00:21:10,090 --> 00:21:15,550 Here's a simple chart showing the dotted decimal subnet masks and their equivalent 184 00:21:15,550 --> 00:21:21,170 in CIDR notation. That's right, the way of writing a prefix with a slash followed 185 00:21:21,170 --> 00:21:29,780 by the prefix length, like 25, 26, etc., is called CIDR notation because it was introduced 186 00:21:29,780 --> 00:21:36,880 with the CIDR system. Previously, only the dotted decimal method was used. Note that 187 00:21:36,880 --> 00:21:41,630 I've shown you only how to subnet a class C network so far, but we will look at 188 00:21:41,630 --> 00:21:49,790 class B and class A networks as well, with prefix lengths like 17, 11, 9, etc. 189 00:21:49,790 --> 00:21:55,500 I spent a lot of time on just that one example, but I hope you can see the use of 190 00:21:55,500 --> 00:21:59,891 subnetting--dividing a larger network into smaller networks called subnets. 191 00:21:59,891 --> 00:22:07,130 Instead of using the whole 203.0.113.0/24 network for the point-to-point connection, we can 192 00:22:07,130 --> 00:22:14,070 use a 30 subnet and use only four addresses, or even better, use a 31 subnet and use only 193 00:22:14,070 --> 00:22:20,290 two addresses. I'll give one more example of subnetting before finishing up this video. 194 00:22:20,290 --> 00:22:23,870 In the next video, I'll give you some practice problems and walk you through them so you 195 00:22:23,870 --> 00:22:26,710 can get some hands-on practice with subnetting. 196 00:22:26,710 --> 00:22:33,620 So, here's a scenario: There are four networks connected to R1, with many hosts connected 197 00:22:33,620 --> 00:22:40,760 to each switch. There are 45 hosts per network. R1 needs an IP address in each network, so 198 00:22:40,760 --> 00:22:49,710 its address is included in that 45-host number. You have received the 192.168.0.14 network, 199 00:22:49,710 --> 00:22:54,080 and you must divide the network into four subnets that can accommodate the number of 200 00:22:54,080 --> 00:23:01,950 hosts required. First off, are there enough addresses in the 192.168.0.14 network in 201 00:23:01,950 --> 00:23:09,010 the first place? We need 45 hosts per network, including R1, but also remember that each 202 00:23:09,010 --> 00:23:15,800 network has a network and broadcast address, so that's plus two. So, we need 47 addresses per subnet. 203 00:23:15,800 --> 00:23:24,240 47 times 4 equals 188, so there's no problem in terms of the number of hosts. 204 00:23:24,240 --> 00:23:32,560 192.168.0.0/24 is a class C network, so there are 256 addresses. Therefore, we will be able to assign 205 00:23:32,560 --> 00:23:36,460 four subnets to accommodate all hosts, no problem. 206 00:23:36,460 --> 00:23:43,160 Okay, let's see how we can calculate the subnets we need to make. We need four equal-sized subnets 207 00:23:43,160 --> 00:23:50,450 with enough room for at least 45 hosts. Here, I've written out 192.168.0.10 208 00:23:50,450 --> 00:24:00,480 with a 30 mask, 255.255.255.252. I skipped 32 and 31 since these aren't point-to-point links. 209 00:24:00,480 --> 00:24:08,190 We can't use 31 and definitely can't use 32. Since there are two host bits, 210 00:24:08,190 --> 00:24:14,290 the formula to determine the number of usable addresses is 211 00:24:14,290 --> 00:24:20,970 2^2 - 2. 2^2 is 2 times 2, which is 4, so that means there are two usable addresses 212 00:24:20,970 --> 00:24:28,220 in a 30 network. Clearly, not enough room to accommodate the 45 hosts we have. 213 00:24:28,220 --> 00:24:36,330 How about if we use a 29 mask to make these subnets? Can we fit the 45 hosts we need? There are three host bits, 214 00:24:36,330 --> 00:24:43,710 so the formula is 2^3 - 2. 2^3 is 2 times 2 times 215 00:24:43,710 --> 00:24:51,910 2, which is 8. Therefore, there are six usable addresses, not enough for 45 hosts. 216 00:24:51,910 --> 00:25:00,250 How about if we use 28? There are four host bits, so the formula is 2^4 - 2. 217 00:25:00,250 --> 00:25:07,710 2^4 is 2 times 2 times 2 times 2, which is 16. So, that means there are 218 00:25:07,710 --> 00:25:13,410 14 usable addresses--once again, not enough for 45 hosts. 219 00:25:13,410 --> 00:25:23,400 How about 27? There are five host bits, so the formula is 2^5 - 2. And 2^5 220 00:25:23,400 --> 00:25:30,169 is 2 times 2 times 2 times 2 times 2, which equals 32. So that means 221 00:25:30,169 --> 00:25:35,370 30 usable addresses. Again, not enough for 45 hosts. 222 00:25:35,370 --> 00:25:43,740 How about a 26 subnet mask? There are now six host bits, so the formula is 2^6 - 2. 223 00:25:43,740 --> 00:25:51,480 2^6 is 2 times 2 times 2 times 2 times 2 times 2, which equals 64. 224 00:25:51,480 --> 00:25:59,120 That means there are 62 usable addresses. So, it looks like we've found our number. 27 225 00:25:59,120 --> 00:26:04,960 doesn't provide enough address space, but 26 provides more than we need, so we have to 226 00:26:04,960 --> 00:26:10,940 go with 26. Unfortunately, you can't always make subnets have exactly the number of addresses 227 00:26:10,940 --> 00:26:16,570 you want. There might be some unused address space. That's actually fine, since it's good 228 00:26:16,570 --> 00:26:20,500 to have some room for growth anyway. 229 00:26:20,500 --> 00:26:26,571 So, I think this video has gone on long enough. Instead of finishing this task in this video, I'll make 230 00:26:26,571 --> 00:26:36,660 it this week's quiz. The first subnet, subnet one, is 192.168.0.16. What are the remaining 231 00:26:36,660 --> 00:26:42,660 subnets? To help you out, here's a hint: Find the broadcast address of subnet one. 232 00:26:42,660 --> 00:26:50,120 The next address after that is the network address of subnet two. And then just repeat the process for subnets 233 00:26:50,120 --> 00:26:56,780 three and four. Post your answers in the comment section, and I'll also go over the answer in the next video. 234 00:26:57,970 --> 00:27:05,300 So, what did we cover in this video? We covered CIDR (Classless Inter-Domain Routing), which 235 00:27:05,300 --> 00:27:10,680 removes the rules of class A, B, and C networks and lets us be more flexible with network 236 00:27:10,680 --> 00:27:17,030 addressing, according to the size of the network. We also covered the process of subnetting, 237 00:27:17,030 --> 00:27:22,040 but mostly just the basics. Hopefully, you understand the purpose of subnetting and 238 00:27:22,040 --> 00:27:27,120 know a little bit about how to do it. I will clarify and expand upon many things in the 239 00:27:27,120 --> 00:27:33,960 next video, but also feel free to ask any questions you have in the comments section. 240 00:27:33,960 --> 00:27:38,760 For today's video, there won't be a practice lab; that will be after I've finished explaining everything about 241 00:27:38,760 --> 00:27:43,760 subnetting. There will be flashcards, however, to help you review some of the things learned 242 00:27:43,760 --> 00:27:48,120 in this video. You can download them from the link in the description. 243 00:27:48,120 --> 00:27:53,280 I've also recently enabled the membership feature for my channel. If you want to leave 244 00:27:53,289 --> 00:27:59,070 a monthly tip to support me, this is another great way to do so. 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