WEBVTT 0:00:09.720000 --> 0:00:13.100000 Hello and welcome to this video which is part of the CCNA exam assessment 0:00:13.100000 --> 0:00:16.740000 series. In this video I'm going to be assessing you on your knowledge 0:00:16.740000 --> 0:00:22.220000 and memory of wireless principles. 0:00:22.220000 --> 0:00:25.220000 Question number one. 0:00:25.220000 --> 0:00:32.600000 What IEEE standard, which has been amended many times over the years, 0:00:32.600000 --> 0:00:36.820000 defines wireless LAN operations and protocols. 0:00:36.820000 --> 0:00:46.860000 And that would be answer D, 802.11. 0:00:46.860000 --> 0:00:57.780000 Question two. A single wireless access point and it's a set of associated 0:00:57.780000 --> 0:01:04.020000 wireless clients all comprise a single what? 0:01:04.020000 --> 0:01:10.080000 That would be a single BSS. 0:01:10.080000 --> 0:01:13.260000 Now I'm not going to tell you what a BSS is because there might be another 0:01:13.260000 --> 0:01:17.320000 question coming up asking you to define what that acronym is but hopefully 0:01:17.320000 --> 0:01:23.300000 you know it. Question three. 0:01:23.300000 --> 0:01:37.840000 Which of the following statements about a BSSID are true? 0:01:37.840000 --> 0:01:45.340000 Answer A. The BSSID is the MAC address of the access point. 0:01:45.340000 --> 0:01:56.880000 Question four. When clicking on your Wi-Fi adapter you see the following 0:01:56.880000 --> 0:01:58.960000 available networks. 0:01:58.960000 --> 0:02:04.360000 Each of the names of these networks represents a what that has been advertised 0:02:04.360000 --> 0:02:06.180000 by an access point. 0:02:06.180000 --> 0:02:08.220000 So what do those names represent? 0:02:08.220000 --> 0:02:11.880000 What's the technical term for those names? 0:02:11.880000 --> 0:02:25.120000 Those names would be called an SSID, a service set identifier. 0:02:25.120000 --> 0:02:36.660000 Question five. In 802.11 terminology what name is given to the upstream 0:02:36.660000 --> 0:02:41.680000 wired connection or the wired network that's connected to a wireless access 0:02:41.680000 --> 0:02:51.040000 point? That is considered the distribution system. 0:02:51.040000 --> 0:02:54.900000 So within 802.11 documents that's what they would call it the distribution 0:02:54.900000 --> 0:03:04.620000 system. Question six. 0:03:04.620000 --> 0:03:10.600000 Which of the following is a benefit of advertising multiple SSIDs each 0:03:10.600000 --> 0:03:15.700000 assigned to its own VLAN from a single wireless access point? 0:03:15.700000 --> 0:03:23.940000 None of those answers are correct. 0:03:23.940000 --> 0:03:26.500000 So let's go through each of these answers here. 0:03:26.500000 --> 0:03:30.560000 Answer A. It allows more clients to be associated to that access point 0:03:30.560000 --> 0:03:33.820000 than if it were advertising on a single SSID. 0:03:33.820000 --> 0:03:35.580000 That is incorrect. 0:03:35.580000 --> 0:03:41.940000 An access point only has a single maximum quantity of clients that can 0:03:41.940000 --> 0:03:43.300000 be associated to it. 0:03:43.300000 --> 0:03:48.740000 Whether all those clients are associated to one SSID or six SSIDs it doesn't 0:03:48.740000 --> 0:03:53.960000 matter. So adding more SSIDs to your access point does not increase that 0:03:53.960000 --> 0:03:57.620000 maximum quantity of clients it can support. 0:03:57.620000 --> 0:04:02.320000 B. It allows more clients to communicate to the access point at the same 0:04:02.320000 --> 0:04:05.420000 time. That is also incorrect. 0:04:05.420000 --> 0:04:10.300000 Through most of the 802.11 standards only one client can communicate to 0:04:10.300000 --> 0:04:15.060000 the access point at any given time because a radio is only listening to 0:04:15.060000 --> 0:04:19.840000 one receiver. In some of the more advanced standards it's true that if 0:04:19.840000 --> 0:04:22.240000 clients are using some of the more advanced standards they might be able 0:04:22.240000 --> 0:04:26.460000 to communicate at the same time but that has nothing to do with how many 0:04:26.460000 --> 0:04:28.880000 SSIDs you happen to have. 0:04:28.880000 --> 0:04:33.500000 C. It places clients into separate collision domains. 0:04:33.500000 --> 0:04:34.900000 That's incorrect. 0:04:34.900000 --> 0:04:40.340000 So whether you have one SSID or ten SSIDs they're all being advertised 0:04:40.340000 --> 0:04:42.080000 by the same radio. 0:04:42.080000 --> 0:04:45.100000 Let's just say they're all I'm not going to say what that radio is because 0:04:45.100000 --> 0:04:49.220000 it might be another question but one radio on a particular frequency is 0:04:49.220000 --> 0:04:51.160000 advertising all those SSIDs. 0:04:51.160000 --> 0:04:55.200000 So everybody is in one big collision domain whether they're all in one 0:04:55.200000 --> 0:05:00.300000 SSID or they divide between six SSIDs they're all contending for the same 0:05:00.300000 --> 0:05:03.680000 airspace. And D. 0:05:03.680000 --> 0:05:07.220000 It allows clients from different SSIDs to be placed onto different wireless 0:05:07.220000 --> 0:05:09.340000 channels. That's also wrong. 0:05:09.340000 --> 0:05:14.300000 Once again only a single radio is advertising those SSIDs and that radio 0:05:14.300000 --> 0:05:17.140000 is on one particular channel. 0:05:17.140000 --> 0:05:21.880000 A single radio cannot operate simultaneously across different channels 0:05:21.880000 --> 0:05:23.360000 it can't do that. 0:05:23.360000 --> 0:05:28.540000 So none of these answers were correct. 0:05:28.540000 --> 0:05:38.040000 Question seven. Which of the following statements is true about an 802 0:05:38.040000 --> 0:05:50.240000 .11 ESS? So the correct answers here the two answers were it requires at 0:05:50.240000 --> 0:05:54.060000 least two access points connected to the same distribution system. 0:05:54.060000 --> 0:05:59.220000 In other words connected to the same wired network and if using two or 0:05:59.220000 --> 0:06:04.420000 more access points they should all advertise the same SSIDs that way clients 0:06:04.420000 --> 0:06:08.560000 can move back and forth between the access points seamlessly without even 0:06:08.560000 --> 0:06:11.560000 really knowing that they're changing access points. 0:06:11.560000 --> 0:06:14.640000 That is the definition of an ESS. 0:06:14.640000 --> 0:06:22.340000 Question number eight. 0:06:22.340000 --> 0:06:29.040000 Which of the following elements would you find in an 802.11 IBSS? 0:06:29.040000 --> 0:06:39.200000 You would find two wireless stations and an RF channel. 0:06:39.200000 --> 0:06:45.280000 Remember that in an IBSS that is the situation where two wireless clients 0:06:45.280000 --> 0:06:50.620000 like two laptops for example are wirelessly connecting to each other. 0:06:50.620000 --> 0:06:53.500000 They're not using an access point they're connecting directly to each 0:06:53.500000 --> 0:06:57.840000 other and speaking wirelessly in like a point to point type of scenario. 0:06:57.840000 --> 0:07:01.640000 So it's two stations and they're still using a radio frequency channel 0:07:01.640000 --> 0:07:05.380000 to communicate with each other. 0:07:05.380000 --> 0:07:17.460000 Question nine. What is the basic unit of measurement for measuring the 0:07:17.460000 --> 0:07:21.660000 frequency of an electromagnetic wave? 0:07:21.660000 --> 0:07:28.860000 The correct answer is B, Hertz. 0:07:28.860000 --> 0:07:35.260000 Hertz is used to measure the frequency of something. 0:07:35.260000 --> 0:07:46.020000 Question ten. Which of the following answers is equivalent to a wave that 0:07:46.020000 --> 0:07:54.880000 oscillates or changes 5 ,000 times per second? 0:07:54.880000 --> 0:07:59.920000 The correct answer there is C, 5 kilohertz. 0:07:59.920000 --> 0:08:07.720000 The first answer A, 5 Hertz is changing only 5 times per second. 0:08:07.720000 --> 0:08:20.280000 The second answer is changing 500 times per second. 0:08:20.280000 --> 0:08:25.320000 So that's why C is correct. 0:08:25.320000 --> 0:08:28.220000 Question eleven. 0:08:28.220000 --> 0:08:40.480000 Which two radio frequency bands are most commonly used for wireless LANs? 0:08:40.480000 --> 0:08:46.060000 Select two answers. 0:08:46.060000 --> 0:08:53.520000 The correct answers are the 2.4 gigahertz range and the 5 gigahertz range. 0:08:53.520000 --> 0:08:58.260000 Question twelve. 0:08:58.260000 --> 0:09:07.200000 When using the 2.4 gigahertz band for Wi-Fi, what channels, if designed 0:09:07.200000 --> 0:09:14.540000 properly, can be three answers? 0:09:14.540000 --> 0:09:26.720000 The correct answer is our channels one, six and eleven. 0:09:26.720000 --> 0:09:30.120000 Question thirteen. 0:09:30.120000 --> 0:09:39.900000 Which of the following IEEE 802.11 amendments do not operate within the 0:09:39.900000 --> 0:09:41.320000 5 gigahertz band? 0:09:41.320000 --> 0:09:45.460000 Select two answers. 0:09:45.460000 --> 0:09:54.860000 That would be the 802.11B and the 802.11G amendments. 0:09:54.860000 --> 0:10:03.080000 Both of those specify only using the 2.4 gigahertz bands. 0:10:03.080000 --> 0:10:09.720000 Question fourteen. 0:10:09.720000 --> 0:10:15.780000 Which of the statements about using encryption in Wi-Fi networks is true? 0:10:15.780000 --> 0:10:17.820000 Select two answers. 0:10:17.820000 --> 0:10:27.180000 The correct answers are A, all clients in the same BSS must use the same 0:10:27.180000 --> 0:10:33.680000 encryption method and D, each client is given a unique encryption key. 0:10:33.680000 --> 0:10:36.900000 That's how you and I can be connected to the same access point. 0:10:36.900000 --> 0:10:40.540000 We're both doing encryption, but I can't read your stuff and you can't 0:10:40.540000 --> 0:10:44.320000 read mine because we both have unique encryption keys. 0:10:44.320000 --> 0:10:53.640000 Question fifteen and our final question for this section. 0:10:53.640000 --> 0:10:58.840000 When using Wi-Fi encryption, what is the purpose of the group encryption 0:10:58.840000 --> 0:11:09.680000 key? And the answer is A, to encrypt packets sent from the access point 0:11:09.680000 --> 0:11:13.640000 meant for every station to receive. 0:11:13.640000 --> 0:11:17.840000 So for example, if from the distribution system from the wired network 0:11:17.840000 --> 0:11:22.700000 comes some sort of broadcast or multicast packet that's meant for all 0:11:22.700000 --> 0:11:27.760000 of the clients in the Wi-Fi to get that, the access point will encrypt 0:11:27.760000 --> 0:11:33.360000 that using the group encryption key which everybody can decrypt. 0:11:33.360000 --> 0:11:38.460000 That way that packet can be sent and heard by all of the clients. 0:11:38.460000 --> 0:11:42.800000 So that concludes this section of our assessment. 0:11:42.800000 --> 0:11:45.180000 Thank you for watching and I hope it was helpful for you.