WEBVTT 0:00:09.380000 --> 0:00:14.520000 Hello and welcome to this video which is testing you on your Cisco CCNA 0:00:14.520000 --> 0:00:20.180000 IPB6 addressing and subnetting concepts. 0:00:20.180000 --> 0:00:22.980000 Question number one. 0:00:22.980000 --> 0:00:29.120000 How many bits does an IPB6 address consume? 0:00:29.120000 --> 0:00:34.800000 The correct answer is 128 bits. 0:00:34.800000 --> 0:00:41.520000 An IPB6 address is four times as large as an IPB4 address. 0:00:41.520000 --> 0:00:51.920000 Question two. What is the full expanded version of the IPB6 address shown 0:00:51.920000 --> 0:00:56.940000 here? So what you see here is an abbreviated address. 0:00:56.940000 --> 0:01:00.120000 What would that address look like if it was not abbreviated? 0:01:00.120000 --> 0:01:13.280000 And the correct answer is C. 0:01:13.280000 --> 0:01:21.740000 Question three. Which of the following answers represents the shortest 0:01:21.740000 --> 0:01:25.720000 way one could abbreviate the IPB6 address that follows? 0:01:25.720000 --> 0:01:29.020000 So here you see a fully expanded IPB6. 0:01:29.020000 --> 0:01:33.620000 Now you're asked to abbreviate this to represent it as short with as few 0:01:33.620000 --> 0:01:35.580000 characters as possible. 0:01:35.580000 --> 0:01:43.160000 The correct answer is D. 0:01:43.160000 --> 0:01:48.520000 Remember that double colons can only show once in an IPB6 address. 0:01:48.520000 --> 0:01:53.100000 That's why answers B and C are incorrect because they have the double 0:01:53.100000 --> 0:01:55.400000 colon more than once. 0:01:55.400000 --> 0:02:03.200000 Answer A technically is correct as far as one way that you can abbreviate 0:02:03.200000 --> 0:02:07.040000 it but it is not the shortest way you could abbreviate it and this question 0:02:07.040000 --> 0:02:13.360000 was asking for the shortest way which is answer D. 0:02:13.360000 --> 0:02:23.820000 Question four. Maps the following IPB6 addresses to their respective type 0:02:23.820000 --> 0:02:33.340000 identification. And here are the answers. 0:02:33.340000 --> 0:02:38.100000 Every IPB6 address that begins with FE80 is a link local address. 0:02:38.100000 --> 0:02:40.840000 Regardless of what follows that pattern. 0:02:40.840000 --> 0:02:45.000000 Multicast addresses always begin with FF. 0:02:45.000000 --> 0:02:49.440000 If an address begins with a two or three that is a global unicast address 0:02:49.440000 --> 0:02:57.760000 and every address that begins with an FC or an FD is a unique local address. 0:02:57.760000 --> 0:03:09.160000 Question five. What subnet mask is strongly recommended to be applied 0:03:09.160000 --> 0:03:19.040000 to IPB6 subnets where end user devices and VLANs reside? 0:03:19.040000 --> 0:03:24.400000 That would be a slash 64. 0:03:24.400000 --> 0:03:36.100000 Question six. Your company is given the IPB6 network of 2006 colon 111 0:03:36.100000 --> 0:03:42.420000 colon 2222 colon colon slash 48 from your service provider. 0:03:42.420000 --> 0:03:46.820000 You wish to divide this into subnets with each subnet containing a slash 0:03:46.820000 --> 0:03:52.880000 64 net mask. Given the above information what will be the address of your 0:03:52.880000 --> 0:04:04.620000 22nd subnet? In this case the answer is A and just to help you work that 0:04:04.620000 --> 0:04:12.740000 out. So we know here that in this particular case that all the IPB6 addresses 0:04:12.740000 --> 0:04:20.360000 will begin with 1222. 0:04:20.360000 --> 0:04:23.060000 Those are the first 48 bits. 0:04:23.060000 --> 0:04:28.480000 Now the next word before you get to the colon is going to bring it out 0:04:28.480000 --> 0:04:33.960000 to 64 bits. So that next word the very first network the very first subnet 0:04:33.960000 --> 0:04:38.340000 would be 000 colon colon slash 64. 0:04:38.340000 --> 0:04:40.500000 That's the first subnet. 0:04:40.500000 --> 0:04:50.600000 The second subnet will be 0001. 0:04:50.600000 --> 0:04:54.200000 That's the second subnet colon colon slash 64. 0:04:54.200000 --> 0:04:57.160000 Then the third subnet will be 0002. 0:04:57.160000 --> 0:04:59.720000 So hopefully you see at this point a pattern. 0:04:59.720000 --> 0:05:02.320000 The subnet is one less. 0:05:02.320000 --> 0:05:06.700000 So for example if I was to say what's the fourth subnet it would be 0003. 0:05:06.700000 --> 0:05:08.740000 What's the ninth subnet? 0:05:08.740000 --> 0:05:13.820000 0008. But we're asked for the 22nd subnet. 0:05:13.820000 --> 0:05:18.360000 So you might think oh well that would be 00021. 0:05:18.360000 --> 0:05:21.840000 But remember IPB6 addresses are in hex. 0:05:21.840000 --> 0:05:26.600000 So you have to ask yourself what is 21 in hexadecimal? 0:05:26.600000 --> 0:05:31.500000 Well in hexadecimal 21 is 1. 0:05:31.500000 --> 0:05:36.660000 That would be 16 plus 5. 0:05:36.660000 --> 0:05:41.220000 So remember these octets here the first one is the ones position. 0:05:41.220000 --> 0:05:45.500000 Each one of these can have 0 through f in them. 0:05:45.500000 --> 0:05:54.420000 So 000f that is the number 15 which is our 16th subnet. 0:05:54.420000 --> 0:05:56.160000 Remember it started at 000. 0:05:56.160000 --> 0:05:59.200000 So 000f which is 15 that's our 16th subnet. 0:05:59.200000 --> 0:06:02.740000 So our 17th subnet now has to move into the next position. 0:06:02.740000 --> 0:06:04.580000 That's the 17th subnet. 0:06:04.580000 --> 0:06:08.860000 That's actually 0010 that's 16 in hex. 0:06:08.860000 --> 0:06:12.840000 0010 in hex is 16 in decimal. 0:06:12.840000 --> 0:06:19.020000 So 16 plus 5 gives us 22. 0:06:19.020000 --> 0:06:35.840000 Actually that gives us 21. 0:06:35.840000 --> 0:06:47.680000 Question 7. Given the IPB6 address of 2001 colon 2222 colon colon 1A33 0:06:47.680000 --> 0:06:50.180000 colon 555 slash 64. 0:06:50.180000 --> 0:06:58.540000 What portion of this address is denoted by colon colon 1A33 colon 555. 0:06:58.540000 --> 0:07:07.400000 Well that last part of the address would be the interface identifier. 0:07:07.400000 --> 0:07:11.360000 In the world of IPB4 we would call those the host bits. 0:07:11.360000 --> 0:07:16.480000 But this is IPB6 and the name is the interface ID. 0:07:16.480000 --> 0:07:26.840000 Question 8. Which of the following is the correct command to add the IPB6 0:07:26.840000 --> 0:07:32.500000 address that you see here to interface gig 00 along with a slash 64 net 0:07:32.500000 --> 0:07:40.460000 mask? The correct answer is C. 0:07:40.460000 --> 0:07:46.060000 This is an interface level command that starts with IPB6 address. 0:07:46.060000 --> 0:07:59.060000 Question 9. Interface gig 00 on router 2. 0:07:59.060000 --> 0:08:03.840000 R2 contains the MAC address that you see right there. 0:08:03.840000 --> 0:08:06.920000 This interface has been configured with the following command. 0:08:06.920000 --> 0:08:14.380000 IPB6 address 2001 colon 2 colon colon slash 64 EUI dash 64. 0:08:14.380000 --> 0:08:19.560000 Which of the following answers display the full IPB6 address that will 0:08:19.560000 --> 0:08:22.220000 be applied to this interface? 0:08:22.220000 --> 0:08:30.240000 And the correct answer is E. 0:08:30.240000 --> 0:08:35.240000 So this question was testing whether or not you could remember how the 0:08:35.240000 --> 0:08:40.060000 EUI 64 interface ID works when creating a dynamic interface ID. 0:08:40.060000 --> 0:08:46.500000 And remember EUI 64 takes as a basis the MAC address of an interface. 0:08:46.500000 --> 0:08:51.140000 Takes that MAC address splits it right in half. 0:08:51.140000 --> 0:08:58.100000 So it would be 2022 colon FA puts F.F. 0:08:58.100000 --> 0:09:06.560000 colon F.E. So that expands it from a 48 bit to a 64 bit. 0:09:06.560000 --> 0:09:07.600000 But that's not all. 0:09:07.600000 --> 0:09:10.840000 So if you would just stop there you probably would have thought that one 0:09:10.840000 --> 0:09:14.920000 of these other answers for example D you probably would have picked D 0:09:14.920000 --> 0:09:19.460000 as the correct answer. 0:09:19.460000 --> 0:09:22.540000 But actually you probably would have picked C as the correct answer. 0:09:22.540000 --> 0:09:27.300000 But the problem with this is the other thing it does is it goes into the 0:09:27.300000 --> 0:09:29.500000 very first octet. 0:09:29.500000 --> 0:09:37.620000 So 2 0 is our first octet and it changes the seventh bit of that octet. 0:09:37.620000 --> 0:09:40.700000 So if we have the octet right here. 0:09:40.700000 --> 0:09:52.060000 So 0 0 1 0 that's 2 0 0 0 0 so that's 2 0. 0:09:52.060000 --> 0:09:55.440000 It changes this bit right here it flips it. 0:09:55.440000 --> 0:10:01.600000 So you would end up with 0 0 1 0 0 1 0. 0:10:01.600000 --> 0:10:11.100000 So now instead of being 2 0 2 2 you have 2 2 2 2 colon F.F. 0:10:11.100000 --> 0:10:19.340000 colon F.E. F.B. which gets you to this correct answer right here. 0:10:19.340000 --> 0:10:32.720000 Question 10. What command must exist on a Cisco router? 0:10:32.720000 --> 0:10:36.920000 So that router can be enabled with a dynamic IPv6 routing protocol such 0:10:36.920000 --> 0:10:40.060000 as OSPF or EIGRP. 0:10:40.060000 --> 0:10:52.460000 And that would be the global configuration command IPv6 unicast-routing. 0:10:52.460000 --> 0:11:02.520000 Question 11. Which of the following is a stateful method for providing 0:11:02.520000 --> 0:11:11.100000 dynamic IPv6 addresses to host? 0:11:11.100000 --> 0:11:13.980000 And that would be DHCP version 6. 0:11:13.980000 --> 0:11:21.340000 Now Slack can also provide dynamic IPv6 addresses to host but that is 0:11:21.340000 --> 0:11:24.480000 stateless that is not stateful. 0:11:24.480000 --> 0:11:29.020000 Only DHCP v6 is stateful. 0:11:29.020000 --> 0:11:38.460000 Question 12. Which of the following statements about packets addressed 0:11:38.460000 --> 0:11:42.520000 with IPv6 link local addresses are true? 0:11:42.520000 --> 0:11:44.640000 Select two answers. 0:11:44.640000 --> 0:11:52.980000 Well packets addressed with link local addresses will start with the well 0:11:52.980000 --> 0:11:54.740000 -known pattern of F.E.80. 0:11:54.740000 --> 0:11:58.720000 So that's how you recognize a link local address by that pattern. 0:11:58.720000 --> 0:12:03.280000 And the link local address of a node is automatically generated. 0:12:03.280000 --> 0:12:05.580000 You don't have to type it in or configure it. 0:12:05.580000 --> 0:12:07.840000 It's automatically generated. 0:12:07.840000 --> 0:12:12.420000 The rest of these answers are false. 0:12:12.420000 --> 0:12:25.500000 Question 13. What Cisco iOS command enables IPv6 processing as well as 0:12:25.500000 --> 0:12:32.060000 a link local address on an interface, but adds no other unicast IPv6 address 0:12:32.060000 --> 0:12:34.080000 to that interface. 0:12:34.080000 --> 0:12:38.500000 In other words, if you want to enable IPv6 processing on an interface 0:12:38.500000 --> 0:12:43.920000 so that it could send and receive IPv6 packets, but you did not want to 0:12:43.920000 --> 0:12:48.700000 configure an IPv6 address on that interface, which of these commands would 0:12:48.700000 --> 0:12:54.060000 do that for you? 0:12:54.060000 --> 0:12:57.760000 And that is simply the command IPv6 enable. 0:12:57.760000 --> 0:13:01.740000 As soon as you type that, a link local address will be dynamically generated 0:13:01.740000 --> 0:13:07.600000 and IPv6 processing will be enabled on that interface. 0:13:07.600000 --> 0:13:21.100000 Question 14. Match the following IPv6 multicast addresses with their usage. 0:13:21.100000 --> 0:13:32.320000 And here are the correct answers for those. 0:13:32.320000 --> 0:13:42.980000 Question 15. Which of the following is an IPv6 multicast address that 0:13:42.980000 --> 0:13:47.020000 is bounded by a site local scope? 0:13:47.020000 --> 0:13:53.580000 And the answer is C. 0:13:53.580000 --> 0:14:00.700000 Remember, after the FF0, the very next number, that fourth number denotes 0:14:00.700000 --> 0:14:04.440000 the scope of an IPv6 multicast address. 0:14:04.440000 --> 0:14:08.540000 And if that number is a five, that means that multicast address can travel 0:14:08.540000 --> 0:14:14.640000 within your site, but no further. 0:14:14.640000 --> 0:14:27.840000 Question 16. Which of the following represents the IPv6 loopback address? 0:14:27.840000 --> 0:14:39.040000 And that would be answer B colon colon one. 0:14:39.040000 --> 0:14:44.060000 And that brings us to the end of this self-assessment for IPv6 within 0:14:44.060000 --> 0:14:47.500000 the category of network fundamentals for CCNA. 0:14:47.500000 --> 0:14:48.200000 Thank you for watching.