1 00:00:00,000 --> 00:00:06,000 welcome back, my name is David Bombal CCIE 1123 2 00:00:06,000 --> 00:00:10,000 and in this section, we're going to look at Variable Length Subnet Mask or VLSM 3 00:00:10,000 --> 00:00:12,000 and route summarization. 4 00:00:12,000 --> 00:00:15,000 VLSM allows us to better use IP addresses 5 00:00:15,000 --> 00:00:21,000 by allowing for Variable Length Subnet Mask on a single network. 6 00:00:21,000 --> 00:00:25,000 so what we're going to cover in this section is firstly VLSM 7 00:00:25,000 --> 00:00:28,000 once again Variable Length Subnet Mask, we're going to look at CIDR 8 00:00:28,000 --> 00:00:31,000 or classless Inter-Domain Routing 9 00:00:31,000 --> 00:00:33,000 I want to explain summarization and show you 10 00:00:33,000 --> 00:00:38,000 how you can summarize multiple routes into fewer or single a route. 11 00:00:38,000 --> 00:00:40,000 we're going to look at routing choices 12 00:00:40,000 --> 00:00:43,000 and how routers choose one route over another 13 00:00:43,000 --> 00:00:45,000 not just based on administrative distance 14 00:00:45,000 --> 00:00:49,000 but also based on the legnth of the prefix match 15 00:00:49,000 --> 00:00:53,000 and then lastly we're going to look at issues regarding discontiguous networks. 16 00:00:53,000 --> 00:00:55,000 so Variable Length Subnet Mask 17 00:00:55,000 --> 00:01:00,000 allow us to have varying or variable mask throughout our network 18 00:01:00,000 --> 00:01:03,000 so for example, a classful network of 10 19 00:01:03,000 --> 00:01:06,000 will have a mask of /8 with the first octet 20 00:01:06,000 --> 00:01:08,000 10 is the network portion of the address 21 00:01:08,000 --> 00:01:11,000 and the last 3 octets, in other words, the last 24 bits 22 00:01:11,000 --> 00:01:13,000 is the host portion of the address. 23 00:01:13,000 --> 00:01:16,000 that classful network was subnetted 24 00:01:16,000 --> 00:01:21,000 we could, for instance, have a subnet of 10.1.1.0/24 25 00:01:21,000 --> 00:01:24,000 where 10.1.1 is the network portion of the address 26 00:01:24,000 --> 00:01:27,000 in other words, the most siginificant 24 bits 27 00:01:27,000 --> 00:01:32,000 or most significant 3 octets is the netwok portion and the last octet 28 00:01:32,000 --> 00:01:36,000 or least significant portion of the address is the host portion. 29 00:01:36,000 --> 00:01:43,000 we could also subnet a 10 network down to 10.1.0.0/16 30 00:01:43,000 --> 00:01:46,000 where half of the address is the network portion 31 00:01:46,000 --> 00:01:49,000 and half of the address is the host portion. 32 00:01:49,000 --> 00:01:52,000 please refer to the ICND 1 part of this course 33 00:01:52,000 --> 00:01:57,000 or my subnetting explained e-book, for lots of detail on subnetting. 34 00:01:57,000 --> 00:02:03,000 CIDR or Classless Inter-Domain Routing was introduced in 1993 35 00:02:03,000 --> 00:02:07,000 to replace the prior addressing architecture of classful networks. 36 00:02:07,000 --> 00:02:10,000 classful networks were not scalable 37 00:02:10,000 --> 00:02:14,000 classful networking introduce 3 classes of addresses 38 00:02:14,000 --> 00:02:18,000 for allocation of IP addresses to hosts and networking devices 39 00:02:18,000 --> 00:02:22,000 we had A B and C networks. 40 00:02:22,000 --> 00:02:29,000 so for instance 10.0.0.0/8 is an example of a classful A network. 41 00:02:29,000 --> 00:02:35,000 172.16.0.0/16 is an example of a class B network 42 00:02:35,000 --> 00:02:41,000 and 192.168.1.0/24 is an example of a class C network. 43 00:02:41,000 --> 00:02:44,000 the problem with this method is that you were forced to use 44 00:02:44,000 --> 00:02:50,000 for instances a /16 mask which gave you approximately 65000 host addresses 45 00:02:50,000 --> 00:02:53,000 or you could use a class C address like this 46 00:02:53,000 --> 00:02:56,000 which only gave you 254 host addresses. 47 00:02:56,000 --> 00:02:58,000 this scheme would have resulted 48 00:02:58,000 --> 00:03:02,000 in the quick extortion of IP addresses and inflexibility. 49 00:03:02,000 --> 00:03:06,000 so CIDR replaced this which is based on Variable Length Subnet Mask 50 00:03:06,000 --> 00:03:10,000 were masks can vary on arbitrary lengths 51 00:03:10,000 --> 00:03:12,000 so they allow for different size subnets 52 00:03:12,000 --> 00:03:17,000 so rather than being forced to use for instance 10.0.0.0/8 53 00:03:17,000 --> 00:03:25,000 you can also use 10.0.0.0/16 or 24 or 30 depending on the requirements 54 00:03:25,000 --> 00:03:29,000 for the number of subnets or hosts in your network 55 00:03:29,000 --> 00:03:37,000 or you can use for instance 10.1.1.0/27 or 10.1.2.0/26 and so forth and so on. 56 00:03:37,000 --> 00:03:40,000 so you could take a single class A address 57 00:03:40,000 --> 00:03:44,000 and subdivide it into multiple subnets based on your requirements. 58 00:03:44,000 --> 00:03:47,000 it's also important to note that in CIDR 59 00:03:47,000 --> 00:03:52,000 we don't just advertise a network, we advertise a routing prefix 60 00:03:52,000 --> 00:04:00,000 in other words, we would advertise 192.168.1.0/24 not just 192.168.1.0 61 00:04:00,000 --> 00:04:03,000 CIDR also allows for the summarization of addresses 62 00:04:03,000 --> 00:04:06,000 which I'm going to explain in more detail in a moment. 63 00:04:06,000 --> 00:04:09,000 but this is were we take multiple subnets 64 00:04:09,000 --> 00:04:12,000 and put them into a super net or summary network 65 00:04:12,000 --> 00:04:14,000 which will reduce routing table sizes. 66 00:04:14,000 --> 00:04:19,000 Variable Length Subnet Mask have many advantages 67 00:04:19,000 --> 00:04:23,000 including the better utilization of IP addresses and subnets. 68 00:04:23,000 --> 00:04:28,000 on a WAN link as an example only 2 IP addresses are required. 69 00:04:28,000 --> 00:04:30,000 1 IP address for 1 end of the link 70 00:04:30,000 --> 00:04:32,000 and another IP address for the other end of the link 71 00:04:32,000 --> 00:04:39,000 thus a /30 mask would be better in say a /16 or /24 mask. 72 00:04:39,000 --> 00:04:43,000 a /30 mask only support 2 hosts on that subnet 73 00:04:43,000 --> 00:04:48,000 and thats all that's required in the point-to-point scenario like this. 74 00:04:48,000 --> 00:04:52,000 a /24 mask for instance would support 254 hosts 75 00:04:52,000 --> 00:04:55,000 and that would be over kill on this specific link. 76 00:04:55,000 --> 00:04:58,000 so VLSM gives assist the administrator 77 00:04:58,000 --> 00:05:02,000 the ability to better utilize IP addresses and subnets. 78 00:05:02,000 --> 00:05:04,000 where we subnet down subnets 79 00:05:04,000 --> 00:05:09,000 to support the required number of IP addresses on that specific segment. 80 00:05:09,000 --> 00:05:12,000 so as an example, here we have subnets 81 00:05:12,000 --> 00:05:17,000 that are using a /27 mask, on the WAN links we are using /30 mask 82 00:05:17,000 --> 00:05:22,000 and for the instance at the headquarters we are using a /16 mask. 83 00:05:22,000 --> 00:05:25,000 it also allows us to implement better summarization 84 00:05:25,000 --> 00:05:30,000 and scalability of networks as rather than the entire world 85 00:05:30,000 --> 00:05:32,000 being forced to use a specific subnet mask 86 00:05:32,000 --> 00:05:37,000 we are able to use varying mask in our own networks 87 00:05:37,000 --> 00:05:41,000 and advertise summaries to other companies or to the Internet. 88 00:05:41,000 --> 00:05:44,000 so heres a simple network to explain the problem 89 00:05:44,000 --> 00:05:46,000 when summarization is not used. 90 00:05:46,000 --> 00:05:51,000 on the left hand side we have networks 10.1.1.0/24 91 00:05:51,000 --> 00:05:54,000 up to 10.1.2.200.0/24 92 00:05:54,000 --> 00:05:57,000 on the right hand side we have networks 10.2.1.0/24 93 00:05:57,000 --> 00:06:05,000 up to 10.2.200.0/24 now summarization is now implemented 94 00:06:05,000 --> 00:06:08,000 and let's say you're running a routeing protocol like RIP 95 00:06:08,000 --> 00:06:11,000 which advertises the entire routing table every 30 seconds 96 00:06:11,000 --> 00:06:14,000 you're going to have scenario like this 97 00:06:14,000 --> 00:06:16,000 with the router on the left-hand side 98 00:06:16,000 --> 00:06:21,000 advertises 200 routes every 30 seconds and the router on the router side 99 00:06:21,000 --> 00:06:25,000 also advertises 200 routes every 30 seconds. 100 00:06:25,000 --> 00:06:29,000 RIP will continually sends out its entire routing table 101 00:06:29,000 --> 00:06:32,000 every 30 seconds even though there are no changes. 102 00:06:32,000 --> 00:06:36,000 so 400 routes are advertised across this WAN link 103 00:06:36,000 --> 00:06:39,000 which is very inefficient, those routing updates 104 00:06:39,000 --> 00:06:42,000 consume a lot of bandwidth and don’t accomplish much 105 00:06:42,000 --> 00:06:44,000 because they will continuously be advertise 106 00:06:44,000 --> 00:06:46,000 even though there are no changes 107 00:06:46,000 --> 00:06:49,000 so it makes more sense to summarize the routes 108 00:06:49,000 --> 00:06:53,000 so on the left-hand side, we have 10.1.1.0 109 00:06:53,000 --> 00:06:56,000 all the way up to 10.1.200.0 110 00:06:56,000 --> 00:07:01,000 So these are subnets of the 10.1.0.0 network 111 00:07:01,000 --> 00:07:05,000 so we could summarize those 200 routes into a single route 112 00:07:05,000 --> 00:07:11,000 and on the left hand side 10.2.1.0 up to 10.2.200.0 113 00:07:11,000 --> 00:07:14,000 could be summarize as 10.2.0.0/16 114 00:07:14,000 --> 00:07:20,000 so rather than advertising 400 routes a single route from the left-hand side 115 00:07:20,000 --> 00:07:22,000 is advertised to the left-hand side. 116 00:07:22,000 --> 00:07:26,000 and a single route from right-hand side is advertised to the left-hand side 117 00:07:26,000 --> 00:07:29,000 With summarization what we are have looking for 118 00:07:29,000 --> 00:07:33,000 is most significant bits that are equal, in other words 119 00:07:33,000 --> 00:07:37,000 starting from the left-and side and working towards the right-hand side 120 00:07:37,000 --> 00:07:40,000 we look for bits that are the same 121 00:07:40,000 --> 00:07:43,000 so in the first octet all the subnets contain 10 122 00:07:43,000 --> 00:07:47,000 so that’s common, in the second octet all the subnets contain 1 123 00:07:47,000 --> 00:07:52,000 so that's common but in the 3rd octet the values vary from 1 - 200 124 00:07:52,000 --> 00:07:56,000 so it’s not as easy to see what’s in common 125 00:07:56,000 --> 00:07:59,000 so we could just summarize it as 10.1.0.00/16 126 00:07:59,000 --> 00:08:03,000 knowing that 10.1 is common throughout 127 00:08:03,000 --> 00:08:06,000 and the router over here will summarize all of those routes 128 00:08:06,000 --> 00:08:09,000 and only send out 1 advertisement 129 00:08:09,000 --> 00:08:13,000 the router on the left-hand side can still get to all of this subnets 130 00:08:13,000 --> 00:08:17,000 because they are subnets of this network 131 00:08:17,000 --> 00:08:20,000 so even though we've summarized routes 132 00:08:20,000 --> 00:08:23,000 we still have full connectivity throughout the network 133 00:08:23,000 --> 00:08:25,000 here’s some more complicated example 134 00:08:25,000 --> 00:08:30,000 assume that we have networks 172.16.32.0/24 135 00:08:30,000 --> 00:08:36,000 all the way up to 172.16.63.0/24 136 00:08:36,000 --> 00:08:40,000 now rather than this router advertising all of those networks 137 00:08:40,000 --> 00:08:47,000 can we summarize this network into 1 or at least a few summaries 138 00:08:47,000 --> 00:08:51,000 now to work this out you start from the most significant bits 139 00:08:51,000 --> 00:08:53,000 in other words, you start from the left-hand side 140 00:08:53,000 --> 00:08:58,000 and you look for what is common throughout all of this networks 141 00:08:58,000 --> 00:09:04,000 so in the first octet 172 is common, so we know that’s common 142 00:09:04,000 --> 00:09:10,000 16 is also common, so we know does far that the 1st 2 octets are common 143 00:09:10,000 --> 00:09:15,000 in the third octet, however, the numbers are changing 144 00:09:15,000 --> 00:09:20,000 so we have 32, 33, 34, 35, all the way up to 63 145 00:09:20,000 --> 00:09:24,000 so it's not as easy to visualize or see what’s common here 146 00:09:24,000 --> 00:09:27,000 so what we're going to do is we can convert the third octet 147 00:09:27,000 --> 00:09:37,000 into binary to look for common bits, so for example 32 = 0010 0000 148 00:09:37,000 --> 00:09:41,000 now obviously there’s no gap halfway through an octet 149 00:09:41,000 --> 00:09:45,000 I’ve just put the space here to try and make it easier to read 150 00:09:45,000 --> 00:09:50,000 so an octet consists of 8 binary values, 32 would look at follows 151 00:09:50,000 --> 00:09:53,000 now as soon as you convert into binary 152 00:09:53,000 --> 00:09:56,000 you should convert the remaining bits into binary as well 153 00:09:56,000 --> 00:09:59,000 now in this example, you don’t need to do that 154 00:09:59,000 --> 00:10:04,000 because you can see that the last octet contains just a 0 155 00:10:04,000 --> 00:10:11,000 but I’ve done it here for completeness, the second network is 172.16.33.0 156 00:10:11,000 --> 00:10:18,000 and 33 in binary looks as follows, 34 looks as follows, 35 looks as follows 157 00:10:18,000 --> 00:10:22,000 and I wouldn’t recommend that you convert all the addresses into binary 158 00:10:22,000 --> 00:10:24,000 I would just do say the first 3 or 4 159 00:10:24,000 --> 00:10:26,000 and then the last one to see what’s in common. 160 00:10:26,000 --> 00:10:31,000 in the real world, however, you may come across far more complicated examples 161 00:10:31,000 --> 00:10:34,000 and you may need to convert all the addresses into binary 162 00:10:34,000 --> 00:10:38,000 but at the CCNA level that may not be necessary. 163 00:10:38,000 --> 00:10:41,000 So once we’ve done that we can look for bits that are in common 164 00:10:41,000 --> 00:10:45,000 so the first binary 0 in the 3rd octet is common throughout 165 00:10:45,000 --> 00:10:51,000 the second binary 0 is also common throughout, the 3rd binary bit is a 1 166 00:10:51,000 --> 00:10:55,000 and that is common throughout, however, in the 4th binary bit position 167 00:10:55,000 --> 00:11:02,000 the values change as you can see here the first few networks have 0's 168 00:11:02,000 --> 00:11:05,000 but the last network has a 1 in the fourth binary bit position 169 00:11:05,000 --> 00:11:11,000 so we can draw a line to the right of all the bits that are common 170 00:11:11,000 --> 00:11:15,000 so as you can see here to the left of the line in the 3rd octet 171 00:11:15,000 --> 00:11:18,000 we have 001 in binary 172 00:11:18,000 --> 00:11:23,000 the remaining bits in binary are not common in other words bits vary. 173 00:11:23,000 --> 00:11:27,000 Now Cisco does not support discontiguous subnet masks 174 00:11:27,000 --> 00:11:33,000 so you can see these bits are not in common and these this bits are in common 175 00:11:33,000 --> 00:11:36,000 there has to be a contiguous grouping of common bits 176 00:11:36,000 --> 00:11:40,000 starting from the left-hand side moving from the right-hand side 177 00:11:40,000 --> 00:11:43,000 and as soon as there are bits that are not common 178 00:11:43,000 --> 00:11:47,000 you have to draw a line to make a differentiation between common bits 179 00:11:47,000 --> 00:11:49,000 and bits that are not common 180 00:11:49,000 --> 00:11:54,000 so we know now that the first octet is common, the second octet is common 181 00:11:54,000 --> 00:12:00,000 so 172.16 in the third octet, the first 3 binary bits are common 182 00:12:00,000 --> 00:12:05,000 so 001 now any bits that are not in common are just set to 0 183 00:12:05,000 --> 00:12:11,000 so notice, we’ve populate the remaining portion of the address with binary 0's 184 00:12:11,000 --> 00:12:14,000 and then we convert the binary back to decimal 185 00:12:14,000 --> 00:12:17,000 so hopefully remember this from the ICND 1 course. 186 00:12:17,000 --> 00:12:23,000 So the first octet is 172, the second octet is 16 187 00:12:23,000 --> 00:12:28,000 the third octet if you convert this to decimal is equal to 32 188 00:12:28,000 --> 00:12:32,000 and the last octet converted to decimal is 0. 189 00:12:32,000 --> 00:12:37,000 So the summary address is 172.16.32.0 190 00:12:37,000 --> 00:12:41,000 The last step is to work out the bits that are in common 191 00:12:41,000 --> 00:12:45,000 now the first octet is in common and an octet is 8 bits 192 00:12:45,000 --> 00:12:49,000 the second octet is in common and that’s an additional 8 bits 193 00:12:49,000 --> 00:12:54,000 so thus far we have 8 bits + 8 bits in other words, 16 bits that are in common 194 00:12:54,000 --> 00:13:02,000 3 bits in the third octet are in common, so 8+8+3 will give you 19 bits in common 195 00:13:02,000 --> 00:13:10,000 thus the summary address for this subnets will be 172.16.32.0/19 196 00:13:10,000 --> 00:13:14,000 it’s a simple as that to work out summarization 197 00:13:14,000 --> 00:13:16,000 and I’m going to show you trick in a moment 198 00:13:16,000 --> 00:13:19,000 that allows you to work this out in the few seconds 199 00:13:19,000 --> 00:13:29,000 here’s another example, we have subnets 172.16.64.0/24 up to 172.16.127.0/24 200 00:13:29,000 --> 00:13:36,000 can these subnets be summarized into a single subnet or fewer subnets? 201 00:13:36,000 --> 00:13:40,000 so using the same process we start from the left-hand side 202 00:13:40,000 --> 00:13:42,000 and we look for bits that are in common 203 00:13:42,000 --> 00:13:47,000 in the first octet, we have 172 throughout all of those subnets 204 00:13:47,000 --> 00:13:51,000 so that’s common, in the second octet we have 16 205 00:13:51,000 --> 00:13:53,000 so that’s common throughout 206 00:13:53,000 --> 00:13:58,000 So we know that the first 2 octets 172.16 are common throughout 207 00:13:58,000 --> 00:14:01,000 but in the third octet the values are changing 208 00:14:01,000 --> 00:14:06,000 we’ve got 64, 65, 66, 67 all the way up to 127 209 00:14:06,000 --> 00:14:10,000 so that third octet we're going to convert to binary 210 00:14:10,000 --> 00:14:13,000 to be able to better see what’s in common 211 00:14:13,000 --> 00:14:20,000 so converting 64 into binary will give you 00100 0000 212 00:14:20,000 --> 00:14:26,000 65 will look as follows, 66 as this, 67 as that 213 00:14:26,000 --> 00:14:30,000 and you could convert a third octet of all of the subnets until you get a 127 214 00:14:30,000 --> 00:14:32,000 which looks as follows 215 00:14:32,000 --> 00:14:38,000 the last octet is a 0 in decimal which looks as follows in binary 216 00:14:38,000 --> 00:14:40,000 so now what we need to do is we need to look for the common bits 217 00:14:40,000 --> 00:14:44,000 so the common bits are once again 172.16 218 00:14:44,000 --> 00:14:50,000 and then the first binary 0 is in common throughout the subnets 219 00:14:50,000 --> 00:14:55,000 the second binary bit which is set to 1 is common throughout 220 00:14:55,000 --> 00:15:00,000 but the third binary bit is not common throughout notice there’s 0 221 00:15:00,000 --> 00:15:02,000 and then here there’s a binary 1 222 00:15:02,000 --> 00:15:05,000 so we can draw a line after the second binary bit 223 00:15:05,000 --> 00:15:09,000 to denote that everything to the left of the line is in common 224 00:15:09,000 --> 00:15:12,000 and everything to the right of the line is not in common 225 00:15:12,000 --> 00:15:16,000 so all of this is in common and all of this is not in common. 226 00:15:16,000 --> 00:15:23,000 therefore 172.16 in decimal, the first 2 octets are in common 227 00:15:23,000 --> 00:15:26,000 and the first 2 binary bits of the third octet are in common 228 00:15:26,000 --> 00:15:30,000 the remaining binary bits need to be set to 0's 229 00:15:30,000 --> 00:15:34,000 so filling the remaining binary bits is to 0 will look as follows 230 00:15:34,000 --> 00:15:36,000 and then converting the binary back to decimal 231 00:15:36,000 --> 00:15:41,000 will get 172 in the first octet, 16 in the second octet 232 00:15:41,000 --> 00:15:47,000 this is 64 in decimal, so the third octet is 64 233 00:15:47,000 --> 00:15:52,000 and the last octet binary 0's is equal to 0 in decimal 234 00:15:52,000 --> 00:16:01,000 so the address is 172.16.64.0 that is the summary network for all of these subnets. 235 00:16:01,000 --> 00:16:04,000 the last step is to count the number of bits in common 236 00:16:04,000 --> 00:16:08,000 the first octet is 8 bits, the second octet is 8 bits 237 00:16:08,000 --> 00:16:12,000 so that’s 16 bits in total followed by another 2 binary bits 238 00:16:12,000 --> 00:16:15,000 which gives you 18 bits in common 239 00:16:15,000 --> 00:16:19,000 so the first 18 bits are in common throughout all of this subnets 240 00:16:19,000 --> 00:16:25,000 so our final answer will be 172.16.64.0/18 241 00:16:25,000 --> 00:16:30,000 that is the summary network for all of the listed subnets.