Free CCNA | Subnetting (Part 3 - VLSM) | Day 15 | CCNA 200-301 Complete Course — Transcript

Full transcript

1. 0:01Welcome to Jeremy’s IT Lab.
2. 0:03This is a free, complete course for the CCNA.
3. 0:06If you like these videos, please subscribe to follow along with the series.
4. 0:10Also, please like and leave a comment, and share the video to help spread this free series
5. 0:15of videos.
6. 0:16Thanks for your help.
7. 0:18This video will be the last of the subnetting series.
8. 0:22Subnetting is very important to understand so I wanted to take it slowly to make sure
9. 0:26you don’t get confused.
10. 0:28In this video we’ll cover the last few important things you should know.
11. 0:33So, what exactly will we cover in this video?
12. 0:37First of all, I’ll go over the answers last video’s quiz questions.
13. 0:41Then, we’ll do a few practice questions using class A addresses.
14. 0:46Then, we’ll look at another topic in subnetting, called ‘VLSM’, which stands for ‘variable-length
15. 0:52subnet masks’.
16. 0:53I’ll explain exactly what that is when we get to it.
17. 0:57Finally, I’ll give you some other resources where you can get more practice subnetting.
18. 1:02Okay here’s question 1 from the last video.
19. 1:07You have been given the 172.30.0.0/16 network.
20. 1:12Your company requires 100 subnets with at least 500 hosts per subnet.
21. 1:17What prefix length should you use?
22. 1:20This is a simple one.
23. 1:22How many bits do we have to borrow to make 100 subnets?
24. 1:26You can just count on your fingers.
25. 1:28Borrowing one bit allows for 2 subnets, borrowing 2 bits allows 4 subnets, 3 borrowed bits allows
26. 1:35for 8 subnets, 4 borrowed bits allows for 16 subnets, 5 borrowed bits allows for 32
27. 1:41subnets, 6 borrowed bits allows for 64 subnets, and 7 borrowed bits allows for 128 subnets.
28. 1:49So, we need to borrow 7 bits to create enough subnets.
29. 1:53Does that leave enough host bits for at least 500 hosts?
30. 1:57Let’s check.
31. 1:59So, borrowing 7 bits means we will use a /23 prefix length, which leaves 9 host bits.
32. 2:06Once more, the red bits are the host bits.
33. 2:099 host bits allows 510 usable addresses, which is enough for the 500 we need.
34. 2:15So, that’s the answer to question 1, the prefix length will be /23.
35. 2:21/23 is written as 255.255.254.0 in dotted decimal by the way.
36. 2:29As I mentioned before, make sure you know how to write different prefix lengths in dotted
37. 2:33decimal, too.
38. 2:35Let’s go on to question 2.
39. 2:38What subnet does host 172.21.111.201/20 belong to?
40. 2:46These questions are easy if you remember the steps I taught you.
41. 2:50First write out the address in binary, like this.
42. 2:53Then, change all of the host bits to 0.
43. 2:58Convert the address back to dotted decimal, and that’s the answer.
44. 3:02So, host 172.21.111.201/20 belongs to the subnet 172.21.96.0/20.
45. 3:13Let’s check the answer for question 3 next.
46. 3:17What is the broadcast address of the network 192.168.91.78/26 belongs to?
47. 3:25So, to solve this problem you follow the same steps as for the previous question, except
48. 3:30instead of changing all host bits to 0, you change them to 1.
49. 3:34Let’s try it out.
50. 3:37Write out the address in binary.
51. 3:38Then, since we’re looking for the BROADCAST address, not the network address, change all
52. 3:44of the host bits to 1.
53. 3:46Convert the address back to dotted decimal, and that’s the broadcast address for the
54. 3:51subnet.
55. 3:52So the answer for question 3 is 192.168.91.127, that’s the broadcast address for this network.
56. 4:01Let’s check the answer for question 4.
57. 4:05You divide the network 172.16.0.0/16 into 4 subnets of equal size.
58. 4:13Identify the network and broadcast addresses of the second subnet.
59. 4:17So, the first step to solving this question is to determine the prefix length we must
60. 4:21use.
61. 4:23To divide the network into 4 subnets of equal size, we have to borrow 2 bits, because 2
62. 4:29to the power of 2 is 4.
63. 4:31So, when we borrow 2 bits, that makes a /18 prefix length.
64. 4:37Change this last bit to 1, and write it in dotted decimal again.
65. 4:42172.16.64.0.
66. 4:45This is the network address of the second subnet.
67. 4:49Now to find the broadcast address, simply change all of the host bits to 1.
68. 4:53So, the broadcast address is 172.16.127.255.
69. 4:59So, these are the answers to question 4.
70. 5:03Finally, question 5.
71. 5:06You divide the 172.30.0.0/16 network into subnets of 1000 hosts each.
72. 5:13How many subnets are you able to make?
73. 5:16Well, to know how many subnets we are able to make, we have to know how many bits we
74. 5:21can borrow.
75. 5:22But, to know how many bits we can borrow, we need to find how many host bits we need
76. 5:26for 1000 hosts.
77. 5:29We will need 10 host bits for 1000 hosts, because 2 to the power of 10, minus 2, allows
78. 5:35for 1022 hosts.
79. 5:38That will leave us with 6 borrowed bits.
80. 5:416 borrowed bits means we can make 64 subnets, so that’s the answer to question 5.
81. 5:47Okay, let’s get to subnetting class A networks.
82. 5:53Look at the SIZE OF REST BIT FIELD, that is the host portion.
83. 5:57There are 24 host bits that we can borrow from, meaning lots of room to make subnets.
84. 6:03However, let me remind you that the process of subnetting class A, class B, and class
85. 6:08C networks is exactly the same!
86. 6:10So, let’s just do 2 practice questions subnetting class A networks, and then let’s move on
87. 6:16to VLSM.
88. 6:19You have been given the 10.0.0.0/8 network.
89. 6:23You must create 2000 subnets which will be distributed to various enterprises.
90. 6:28What prefix length must you use?
91. 6:31How many host addresses, or usable addresses, will be in each subnet?
92. 6:35Let’s solve this question just like the other ones.
93. 6:39Here is 10.0.0.0/8 in binary and dotted decimal.
94. 6:44It’s /8, so only the first octet is the network portion, and we have 3 whole octets
95. 6:50to borrow from and make subnets.
96. 6:53Notice that I wrote the /8 subnet mask down here, 255.0.0.0.
97. 6:58Currently, we are borrowing 0 bits, so we can’t make any subnets.
98. 7:04Instead of going through the whole process of borrowing 1 bit, 2 bits, 3 bits etc. let’s
99. 7:09see if you can do it in your head.
100. 7:11So, 2 to the power of WHAT?
101. 7:14Equals at least 2000?
102. 7:16Remember, each bit you borrow doubles the number of subnets you can make.
103. 7:212, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048.
104. 7:35Thats 11 borrowed bits, so the answer is 2 to the power of 11.
105. 7:41So, borrowing 11 bits allows us to create the required 2000 subnets.
106. 7:48This means we will use a /19 prefix length.
107. 7:51Now, how many hosts will be in each subnet?
108. 7:54Well, there are 13 host bits remaining, so, that means 8190 hosts per subnet.
109. 8:02So, those are the answers.
110. 8:05We will use a /19 prefix length, and there will be 8190 hosts addresses, AKA usable addresses,
111. 8:12in each subnet.
112. 8:14This is the same process we used for Class B and Class C networks, the numbers are just
113. 8:19larger.
114. 8:20Let’s do one more practice question for a class A network.
115. 8:24PC1 has an IP address of 10.217.182.223/11.
116. 8:32Identify the following for PC1’s subnet: 1) Network address:
117. 8:372) Broadcast address: 3) First usable address:
118. 8:424) Last usable address: 5) Number of host (or usable) addresses:
119. 8:47So, feel free to pause this video and figure out the answer for yourself.
120. 8:53But let’s walk through the answers together now.
121. 8:56Here’s the address written out as a normal /8, so 8 network bits and 24 host bits.
122. 9:05But it’s not /8, it’s /11, so there are 3 borrowed bits.
123. 9:10To find the network address, change all of the host bits to 0.
124. 9:14Then change it back to dotted decimal.
125. 9:16So, there’s the network address, 10.192.0.0.
126. 9:22Add 1 to the network address, and you get the first usable address, which is 10.192.0.1.
127. 9:30Change all of the host bits to 1, and you get the broadcast address, 10.223.255.255.
128. 9:39Subtract one from the broadcast address and you get the last usable address.
129. 9:43Finally, to determine the number of host addresses, count the number of host bits.
130. 9:50There are 21, so that means 2 million, 97 thousand, 1 hundred 50 hosts per subnet.
131. 9:58So, here are the answers to each part of the question.
132. 10:02As I said before, the process for subnetting Class A networks is the same as for class
133. 10:06B and class C networks, you just have to get used to the larger numbers.
134. 10:12Also I hope you’re getting used to converting between binary and dotted decimal by now,
135. 10:17as you have probably noticed it’s absolutely essential for subnetting.
136. 10:22So let’s move on to a very important topic, VLSM, which stands for variable-length subnet
137. 10:28masks.
138. 10:30Until now, we have practiced subnetting used FLSM (Fixed-Length Subnet Masks).
139. 10:36This means that all of the subnets use the same prefix length for example, subnetting
140. 10:40a class C network into 4 subnets using /26.
141. 10:44However, VLSM (Variable-Length Subnet Masks) is the process of creating subnets of different
142. 10:51sizes, to make your use of network addresses more efficient.
143. 10:55Now, VLSM IS more complicated than FLSM, but it’s easy if you follow the steps correctly.
144. 11:02So, that’s what I want to teach you now, the steps to subnet a network using VLSM.
145. 11:09So here’s an example small enterprise network.
146. 11:12There are two LANs in Tokyo, and two LANs in Toronto.
147. 11:17Tokyo LAN A has has 110 hosts, and Tokyo LAN B has 8 hosts.
148. 11:23Toronto LAN A has 29 hosts and Toronto LAN B has 45 hosts.
149. 11:30Also, there is a point-to-point connection between the two routers that we must assign
150. 11:34IP addresses for.
151. 11:37We are assigned the 192.168.1.0/24 network, and must divide it into 5 subnets to provide
152. 11:44IP addresses for all hosts in the enterprise network.
153. 11:48If we were to try this with fixed-length subnet masks, we would need to borrow 3 bits to make
154. 11:53enough subnets.
155. 11:55That would leave 5 host bits left.
156. 11:575 host bits only allows 30 host addresses, so that’s not enough addresses for Tokyo
157. 12:03LAN A or Toronto LAN B. However, if we use VLSM we can assign different subnet sizes
158. 12:11to each LAN, which will allow us to make sure each LAN has enough addresses available.
159. 12:16So, what are the steps to subnet using VLSM?
160. 12:21First, assign the largest subnet at the start of the address space.
161. 12:25Then, assign the second-largest after it.
162. 12:29And then repeat the process until all subnets have been assigned, from largest to smallest.
163. 12:35If you look at our network here, that means we will assign subnets in this order.
164. 12:40First, Tokyo LAN A, which requires 110 hosts.
165. 12:45Then Toronto LAN B. Then, Toronto LAN A, Tokyo LAN B, and finally the point-to-point connection
166. 12:53between the two routers.
167. 12:55So, let’s do Tokyo LAN A. I want you to pause the video and figure out these five
168. 13:01values by yourself.
169. 13:04Network address, broadcast address, first usable address, last usable address, and total
170. 13:09number of usable host addresses.
171. 13:11We’ve done lots of practice for this, so I know you can do it by yourself.
172. 13:16Pause the video now to find the answers….Okay, let’s check out the answers.
173. 13:23So, I decided to use a /25 prefix length.
174. 13:27Why is that?
175. 13:29Well, a /25 prefix length leaves 7 host bits, which means a total of 128 address, so 126
176. 13:37usable host adressess.
177. 13:40We need 110, so /25 is the correct prefix length.
178. 13:45That means the network address is 192.168.1.0/25.
179. 13:51Convert all of the host bits to 1, and this is the broadcast address, 192.168.1.127.
180. 13:58So, now we have our answers for Tokyo LAN A. The first usable address is the network
181. 14:05address plus one, and the last usable address is the broadcast address minus one.
182. 14:11And as I just mentioned, 7 host bits allows for 126 usable host addresses, which is 2 to the
183. 14:18power of 7, minus 2.
184. 14:20So, we now have our first subnet, 192.168.1.0/25 for Tokyo LAN A. That /25 subnet uses up half
185. 14:31of the address space of the 192.168.1.0/24 network, but that’s no problem.
186. 14:39Using VLSM we can assign smaller subnets to these other LANs, and you’ll see that there
187. 14:44is enough address space left.
188. 14:46So, next we should assign Toronto LAN B. So, 192.168.1.127 is the broadcast address of
189. 14:56Tokyo LAN A. If we add one to that, we will get the network address of the next subnet,
190. 15:02which will be used for Toronto LAN B. Therefore, 192.168.1.128 is the network address of Toronto
191. 15:10LAN B. But, we’re missing one important point.
192. 15:14What prefix length should we use for Toronto LAN B?
193. 15:18So, with this information I want you to do the same thing for Toronto LAN B. Pause the
194. 15:24video, and find the prefix length we must use, then the broadcast address, first and
195. 15:29last usable addresses, and total number of usable host addresses for this subnet.
196. 15:35Pause the video now….okay, let’s check the answer.
197. 15:41So, to accommodate the 45 hosts we will use a /26 prefix length.
198. 15:47That leaves 6 host bits, which allows for 62 host addresses.
199. 15:52That’s more than we need, but if we make a smaller subnet with a /27 prefix length,
200. 15:58we can only have 30 hosts, which is not enough.
201. 16:01So, the complete network address is 192.168.1.128/26.
202. 16:10Change all of the host bits to 1, and you get the broadcast address, 192.168.1.191.
203. 16:18So here are the answers for Toronto LAN B.
204. 16:21For Tokyo LAN A and Toronto LAN B, we’ve used address space from 192.168.1.0 through 191.
205. 16:31These two subnets take up three quarters of the address space, but that’s no problem.
206. 16:37There is still room for more, smaller subnets.
207. 16:40192.168.1.191 is the broadcast address of Toronto LAN B, so 192.168.1.192 is the network
208. 16:50address of Toronto LAN A. Once again though, we have to figure out what prefix length to
209. 16:56use for Toronto LAN A.
210. 16:59Let’s do the same process as with the other LANs.
211. 17:03Please pause the video now to find the remaining information….okay, hopefully you found the
212. 17:11answers, let’s check.
213. 17:14Toronto LAN A requires 29 hosts, so we should use a /27 prefix length, which leaves 5 host
214. 17:21bits, and therefore 30 host addresses.
215. 17:25So, 192.168.1.192/27 is the network address, and 192.168.1.223 is the broadcast address.
216. 17:37With this information, we can determine the first and last usable addresses.
217. 17:42So, here are the answers for Toronto LAN A. Let’s check out Tokyo LAN B next.
218. 17:50So, we’ve used all the way to 192.168.1.223 , which is the broadcast address for Toronto
219. 17:58LAN A. There’s not a lot of address space left, but with VLSM we can fit the last two
220. 18:05subnets into the address space.
221. 18:08The next address after Toronto LAN A’s broadcast address is Tokyo LAN B’s network address.
222. 18:15Once again, we need to find the prefix length to use for this subnet.
223. 18:20So, please pause the video here to find these different addresses for Tokyo LAN B.
224. 18:28Okay, let’s check the answers.
225. 18:31So, because Tokyo LAN B requires 8 hosts, we must use a /28 prefix length.
226. 18:39One possible mistake here is using a /29 prefix length.
227. 18:43Although /29 allows 8 addresses, remember we must subtract two for the network and broadcast
228. 18:50addresses, so really /29 only allows 6 usable addresses.
229. 18:56Therefore, we must use /26, which allows 14 host addresses.
230. 19:00So, the network address for Tokyo LAN B is 192.168.1.224/28.
231. 19:09Change the host bits to 1, and the broadcast address is 192.168.1.239.
232. 19:17So, here are the answers for Tokyo LAN B. Now there is only one subnet left that we must
233. 19:24assign, and that is the point-to-point connection between these two routers.
234. 19:29So, we’ve used address space all the way up to 192.168.1.239.
235. 19:35There’s not a lot of space left, but that’s no problem.
236. 19:39Point-to-point connections only require 2 addresses.
237. 19:43192.168.1.239 is the network address of Tokyo LAN B, therefore 192.168.1.240 is the network
238. 19:53address of the point-to-point connection.
239. 19:56Now, what prefix length should we use?
240. 19:59So, as I mentioned in the first subnetting video, it IS possible to use a /31 prefix length for
241. 20:06a subnet requiring only two hosts.
242. 20:09However, for the CCNA test, if you are asked what prefix length to use for a subnet that
243. 20:14requires two hosts, I recommend NOT using a /31.
244. 20:19Instead, what other prefix length allows for 2 hosts?
245. 20:24Pause the video here and find your answers...okay, let’s check.
246. 20:31So, a /30 prefix length allows for 2 hosts.
247. 20:35192.168.1.240 is the network address, and 192.168.1.243 is the broadcast address.
248. 20:46So here are the answers for the point-to-point connections.
249. 20:49There are only 2 usable host addresses, 1 for the Tokyo router and one for the Toronto
250. 20:55router.
251. 20:56Okay, so we successfully subnetted this network using VLSM, and there is still a little bit
252. 21:02of address space left.
253. 21:05Notice that each subnet uses a different prefix length.
254. 21:09If we tried to use the same prefix length for each subnet, there wouldn’t be enough
255. 21:13address space, but with VLSM we were able to do it and leave some extra address space
256. 21:18at the end.
257. 21:20So, here’s a reminder of the steps of VLSM.
258. 21:24Start with the largest subnet and assign it at the beginning of the address space, and
259. 21:29then proceed to the second largest subnet, etc. and repeat until you have assigned all
260. 21:34necessary subnets.
261. 21:36VLSM is a great way to use network address space more efficiently.
262. 21:41So, over these three videos we have done lots of subnetting practice questions.
263. 21:46However, I recommend that you continue practicing subnetting using other resources.
264. 21:52Here are three websites that provide excellent practice.
265. 21:56subnettingquestions.com, subnetting.org, and subnettingpractice.com.
266. 22:03Check them out and try out questions from each site.
267. 22:06Here’s an example question from subnettingquestions.com.
268. 22:11It’s similar to the ones we have been doing in these videos.
269. 22:16And here’s an example question from subnetting.org, once again the same kind of question as we have
270. 22:22done in these videos.
271. 22:25Those other two websites are good, however my personal favorite is this last one, subnettingpractice.com.
272. 22:32It has lots of challenging subnetting questions like this, which go beyond the sort of questions
273. 22:37we have been doing in these videos.
274. 22:40As for today’s quiz, There is no quiz for this video.
275. 22:44Instead, here is some homework: do at least ONE practice question from EACH of those practice
276. 22:50websites every day for at least one week.
277. 22:55There is also no flashcard deck for this video.
278. 22:59Those three subnetting practice websites should give you all of the subnetting practice you
279. 23:03need.
280. 23:04However, there will be a packet tracer practice lab, where you will practice subnetting a
281. 23:08network and assigning the appropriate IP addresses to different devices.
282. 23:15Before I wrap up this video, I want to give a shoutout to my JCNP-level channel members,
283. 23:20Vance, Mike, Yonatan, and Lito.
284. 23:24Thank you again for your support.
285. 23:26Thank you to my JCNA-level channel members too.
286. 23:31Thank you for watching.
287. 23:32Please subscribe to the channel, like the video, leave a comment, and share the video
288. 23:36with anyone else studying for the CCNA.
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292. 23:50That's all for now.