WEBVTT 0:00:09.820000 --> 0:00:13.120000 Hello and welcome to this video which is part of the CCNA exam assessment 0:00:13.120000 --> 0:00:17.520000 series where we're going to be testing you on land switching fundamentals. 0:00:17.520000 --> 0:00:22.780000 Now there's one thing I want to make you aware of here in this video which 0:00:22.780000 --> 0:00:29.580000 is that if you look at the Cisco website for the CCNA exam topics they 0:00:29.580000 --> 0:00:34.240000 list each of these things as 1.13 through 1.17. 0:00:34.240000 --> 0:00:39.000000 However if you look in Cisco's official certification guidebooks they 0:00:39.000000 --> 0:00:45.100000 say that all these things are subcategories of 1.13 like 113 A, B, C, 0:00:45.100000 --> 0:00:49.940000 D and so forth. So as it says here at the bottom of the slide this assessment 0:00:49.940000 --> 0:00:54.260000 will reference the values used in the Cisco press official certification 0:00:54.260000 --> 0:00:56.620000 guide for each question. 0:00:56.620000 --> 0:01:02.680000 So I'm going to label my questions as 113 A, 113 C instead of 1.14 or 0:01:02.680000 --> 0:01:07.280000 1.16 but you can see you can match those up here if you need to. 0:01:07.280000 --> 0:01:10.900000 So let's go ahead and get started. 0:01:10.900000 --> 0:01:19.820000 Question 1. Name each section or field of an ethernet frame. 0:01:19.820000 --> 0:01:24.780000 And here we're talking about the original ethernet 2 version. 0:01:24.780000 --> 0:01:28.680000 And no said this frame is going from the left to the right. 0:01:28.680000 --> 0:01:33.900000 And here are your answers. 0:01:33.900000 --> 0:01:36.740000 The very first field that would be received is the preamble. 0:01:36.740000 --> 0:01:40.380000 If you are the receiver the receiver would then receive the start of frame 0:01:40.380000 --> 0:01:45.220000 delimiter followed by the destination Mac, source Mac, a type field otherwise 0:01:45.220000 --> 0:01:47.020000 known as ether type. 0:01:47.020000 --> 0:01:51.740000 The actual data of the frame which is your IPV4, IPV6 packet followed 0:01:51.740000 --> 0:01:53.680000 by the frame checksum. 0:01:53.680000 --> 0:02:03.760000 Question 2. Here we have the same question but this is a different frame. 0:02:03.760000 --> 0:02:07.960000 This is an IEEE 802.3 frame. 0:02:07.960000 --> 0:02:11.100000 Which of the fields are the same as before and which of the fields are 0:02:11.100000 --> 0:02:13.380000 different? See if you can name all the fields here. 0:02:13.380000 --> 0:02:17.420000 Once again being transmitted from left to right. 0:02:17.420000 --> 0:02:23.880000 And here is your answer. 0:02:23.880000 --> 0:02:27.340000 And you can see here that the vast majority of the fields are exactly 0:02:27.340000 --> 0:02:32.660000 the same. What's different is that in 802.3 there is no longer a type 0:02:32.660000 --> 0:02:38.780000 field. Instead, after the source Mac we have a length field and then we 0:02:38.780000 --> 0:02:40.980000 have a logical link control. 0:02:40.980000 --> 0:02:45.880000 And actually within the logical link control there is a couple of fields 0:02:45.880000 --> 0:02:48.260000 which are equivalent to the type field. 0:02:48.260000 --> 0:02:51.980000 They don't go by the name of type but they serve the same purpose. 0:02:51.980000 --> 0:02:56.600000 So that is your 802.3 frame. 0:02:56.600000 --> 0:03:07.340000 Question 3. Which field within an ethernet frame is viewed by an ethernet 0:03:07.340000 --> 0:03:15.700000 switch in order to make a forwarding decision. 0:03:15.700000 --> 0:03:22.960000 And that would be the destination Mac address. 0:03:22.960000 --> 0:03:32.780000 Question 4. What are the three actions that a switch can perform with 0:03:32.780000 --> 0:03:35.160000 a frame that has been received? 0:03:35.160000 --> 0:03:41.880000 Select three answers. 0:03:41.880000 --> 0:03:44.680000 That switch made choose to discard the frame. 0:03:44.680000 --> 0:03:47.200000 In other words, throw it away, destroy it. 0:03:47.200000 --> 0:03:52.240000 It might choose to forward the frame out a single egress interface or 0:03:52.240000 --> 0:03:58.460000 it might choose to flood the frame and send it out multiple egress interfaces. 0:03:58.460000 --> 0:04:08.660000 Question 5. What are the various terms used to describe the layer 2 table 0:04:08.660000 --> 0:04:13.640000 used by a switch when performing a forwarding decision? 0:04:13.640000 --> 0:04:16.480000 That table goes by three different names. 0:04:16.480000 --> 0:04:22.320000 What are those names? 0:04:22.320000 --> 0:04:28.200000 That is called the cam table, the Mac table, or the bridging table. 0:04:28.200000 --> 0:04:32.160000 Most often you will hear it referenced as the Mac table but bridging table 0:04:32.160000 --> 0:04:38.940000 and cam table are also technically correct. 0:04:38.940000 --> 0:04:47.680000 Question 6. Which field within an ethernet frame is viewed by an ethernet 0:04:47.680000 --> 0:04:53.660000 switch in order to make a learning decision and populate the Mac table? 0:04:53.660000 --> 0:05:01.280000 And that would be the source Mac address of the field. 0:05:01.280000 --> 0:05:06.520000 The source Mac address field. 0:05:06.520000 --> 0:05:17.560000 Question 7. Based on the following graphic, when a switch receives a frame 0:05:17.560000 --> 0:05:25.080000 containing a destination Mac address of 0211.222.333, which ports will 0:05:25.080000 --> 0:05:35.160000 be used to transmit or forward this frame. 0:05:35.160000 --> 0:05:38.500000 So hopefully you noticed that that particular destination Mac address 0:05:38.500000 --> 0:05:41.420000 is not in the Mac address table. 0:05:41.420000 --> 0:05:45.640000 So when a switch receives a unicast frame with an unknown destination 0:05:45.640000 --> 0:05:51.060000 Mac address, it will flood that frame. 0:05:51.060000 --> 0:05:55.880000 So we'll actually send it out ports 0202, 0304 and 005. 0:05:55.880000 --> 0:06:03.760000 So answers A through D are correct here. 0:06:03.760000 --> 0:06:14.900000 Question 8. Which of the following destination Mac addresses would result 0:06:14.900000 --> 0:06:18.620000 in a frame being flooded by a switch? 0:06:18.620000 --> 0:06:24.840000 And the correct answer is B. 0:06:24.840000 --> 0:06:30.760000 B. That Mac address of all F's is what we call the broadcast Mac address. 0:06:30.760000 --> 0:06:35.920000 That will result in that frame being flooded. 0:06:35.920000 --> 0:06:47.660000 Question 9. Which of the following answers correctly describes the result 0:06:47.660000 --> 0:06:50.900000 of running STP on a switch? 0:06:50.900000 --> 0:06:53.880000 And I'm not going to tell you what STP is right now because I want to 0:06:53.880000 --> 0:06:55.440000 see if you know it. 0:06:55.440000 --> 0:07:05.500000 So STP stands for the Spanning Tree Protocol. 0:07:05.500000 --> 0:07:10.300000 And the result is that it creates only one active path between any two 0:07:10.300000 --> 0:07:13.220000 pairs of LAN segments. 0:07:13.220000 --> 0:07:17.440000 Recall that the whole intention or purpose of Spanning Tree is to remove 0:07:17.440000 --> 0:07:24.540000 possible bridging and it does that by intentionally removing any redundancy 0:07:24.540000 --> 0:07:28.880000 within your network to where between any two stations there's only one 0:07:28.880000 --> 0:07:31.300000 path those frames could go. 0:07:31.300000 --> 0:07:36.680000 There's not multiple paths that could possibly cause a loop. 0:07:36.680000 --> 0:07:48.380000 Question 10. Which of the following statements are true regarding the 0:07:48.380000 --> 0:07:51.940000 factory default state of Cisco switches? 0:07:51.940000 --> 0:07:54.240000 Select three answers. 0:07:54.240000 --> 0:08:02.280000 The correct answers are interfaces are enabled by default. 0:08:02.280000 --> 0:08:06.360000 By default all of those interfaces are assigned to VLAN 1 and the Spanning 0:08:06.360000 --> 0:08:13.880000 Tree Protocol is enabled and running by default. 0:08:13.880000 --> 0:08:25.120000 Question 11. Once a MAC address is dynamically learned by a switch, what 0:08:25.120000 --> 0:08:30.800000 method is used by the switch or host to keep that entry populated within 0:08:30.800000 --> 0:08:34.320000 the MAC table as time elapses? 0:08:34.320000 --> 0:08:41.000000 The correct answer is C. 0:08:41.000000 --> 0:08:46.820000 The MAC entry is maintained by receiving Ethernet frames from the host. 0:08:46.820000 --> 0:08:51.060000 So as long as that host continues to send Ethernet frames of any type, 0:08:51.060000 --> 0:08:55.480000 its source MAC address of those frames will refresh and repopulate the 0:08:55.480000 --> 0:09:01.500000 timer of the MAC address table within the switch. 0:09:01.500000 --> 0:09:11.440000 Question 12. Host A sends an Ethernet frame to a switch and subsequently 0:09:11.440000 --> 0:09:16.220000 host A's MAC address is learned and placed into the MAC table. 0:09:16.220000 --> 0:09:22.600000 If host A never sends another Ethernet frame, how long will it take until 0:09:22.600000 --> 0:09:31.960000 that MAC address is purged from the table by the switch? 0:09:31.960000 --> 0:09:36.580000 And for Cisco switches, because this is a Cisco exam, that would be 300 0:09:36.580000 --> 0:09:41.940000 seconds equivalent to 5 minutes, that is the default MAC address aging 0:09:41.940000 --> 0:09:46.280000 time. That can be configured to be different, but the default value is 0:09:46.280000 --> 0:09:52.740000 that. Question 13. 0:09:52.740000 --> 0:09:59.520000 A frame is received by a switch and the switch determines that the source 0:09:59.520000 --> 0:10:02.940000 MAC address is new and needs to be learned. 0:10:02.940000 --> 0:10:07.640000 However, the MAC address table is already completely full. 0:10:07.640000 --> 0:10:09.420000 What will happen? 0:10:09.420000 --> 0:10:11.360000 Select two answers. 0:10:11.360000 --> 0:10:16.420000 And by the way, just to be aware, this question is taken off of the Cisco 0:10:16.420000 --> 0:10:19.100000 official certification guidebook. 0:10:19.100000 --> 0:10:23.100000 So the rule of thumb is that what it says in the official certification 0:10:23.100000 --> 0:10:27.240000 guide is what you should know going into the exam. 0:10:27.240000 --> 0:10:32.200000 If you have any practical experience that negates or goes against the 0:10:32.200000 --> 0:10:34.960000 official certification guide, that's good to know for the real world, 0:10:34.960000 --> 0:10:39.740000 but the exam is based on the Cisco Press official certification guide 0:10:39.740000 --> 0:10:45.540000 books. So based on what they say, what will happen in this case, if the 0:10:45.540000 --> 0:10:50.100000 MAC address table is full and a new MAC address needs to be learned, select 0:10:50.100000 --> 0:10:59.260000 two answers. So what will happen is that the oldest MAC address in the 0:10:59.260000 --> 0:11:01.460000 table will be purged. 0:11:01.460000 --> 0:11:09.980000 Therefore, freeing up some room for that new MAC address to be learned. 0:11:09.980000 --> 0:11:18.320000 Question 14. Which of the following interfaces would be most appropriate 0:11:18.320000 --> 0:11:23.600000 to configure an IPv4 address for a switch that requires such an address 0:11:23.600000 --> 0:11:30.700000 for management purposes? 0:11:30.700000 --> 0:11:36.820000 And that will be configured on interface VLAN 1, otherwise known as a 0:11:36.820000 --> 0:11:40.460000 switched virtual interface. 0:11:40.460000 --> 0:11:53.100000 Question 15. Given the following output, which answer could explain why 0:11:53.100000 --> 0:12:03.820000 interface VLAN 2 on the switch is in the down down state? 0:12:03.820000 --> 0:12:10.180000 The correct answer is VLAN 2 does not exist on the switch. 0:12:10.180000 --> 0:12:14.260000 If interface VLAN 2 had the shutdown command, it would show up here as 0:12:14.260000 --> 0:12:16.380000 being administratively down. 0:12:16.380000 --> 0:12:20.060000 That's why A is incorrect. 0:12:20.060000 --> 0:12:22.980000 The interface has not been connected to any networking cable. 0:12:22.980000 --> 0:12:25.580000 Well, interface VLAN 2 is not a physical interface. 0:12:25.580000 --> 0:12:26.960000 There's nothing to plug into it. 0:12:26.960000 --> 0:12:28.080000 It's a logical interface. 0:12:28.080000 --> 0:12:30.380000 So that doesn't make any sense. 0:12:30.380000 --> 0:12:35.140000 The interface does not technically need an IPv4 address and mask to come 0:12:35.140000 --> 0:12:39.580000 up. Now typically you would have an IPv4 address and mask on that interface 0:12:39.580000 --> 0:12:44.140000 so you can manage it, but it's not required to bring the interface up. 0:12:44.140000 --> 0:12:45.720000 And keep a lives. 0:12:45.720000 --> 0:12:47.040000 That's just a distractor. 0:12:47.040000 --> 0:12:49.220000 Keep a lives do not have to be to sit and enable. 0:12:49.220000 --> 0:12:54.800000 They have nothing to do with interface VLAN 2 being up or down. 0:12:54.800000 --> 0:13:01.140000 And that concludes this video within the CCNA exam assessment fundamentals 0:13:01.140000 --> 0:13:03.700000 course. Thank you for watching.