CCNA Routing and Switching Complete Study Guide. Todd LammleЧитать онлайн книгу.
displays the unique device identifier (UDI) of the router, which comprises the product ID (PID) and serial number of the router. See Chapter 8 for more information.
6. B. The show license feature command allows you to view the technology package licenses and feature licenses that are supported on your router along with several status variables related to software activation and licensing, both licensed and unlicensed features. See Chapter 8 for more information.
7. C, D, F. The SDN architecture slightly differs from the architecture of traditional networks. It comprises three stacked layers: Data, Control and Application. See Chapter 8 for more information.
8. B. There are eight different trap levels. If you choose, for example level 3, level 0 through level 3 messages will be displayed. See Chapter 8 for more information.
9. B, D. Each stack of switches has a single IP address and is managed as a single object. This single IP management applies to activities such as fault detection, VLAN creation and modification, security, and QoS controls. Each stack has only one configuration file, which is distributed to each member in the stack. When you add a new switch to the stack, the master switch automatically configures the unit with the currently running IOS image and the configuration of the stack. You do not have to do anything to bring up the switch before it is ready to operate. See chapter 22 for more information.
10. B. There is no IPv6 default gateway listed in the output, which will be the link-local address of the router interface, sent to the host as a router advertisement. Until this host receives the router address, the host will communicate with IPv6 only on the local subnet. See Chapter 20 for more information.
11. D. The command show ipv6 neighbors provides the ARP cache for on a router. See Chapter 20 for more information.
12. A. If the state is STALE when the interface has not communicated within the neighbor reachable time frame. The next time the neighbor communicates, the state will be REACH. See Chapter 20 for more information.
13. C. There are two successor routes, so by default, EIGRP was load-balancing out s0/0 and s0/1. When s0/1 goes down, EIGRP will just keep forwarding traffic out the second link s0/0. s0/1 will be removed from the routing table. See Chapter 17 for more information.
14. F. There isn't a lot to go on from with the output, but the only commands that provide the FD and AD are show ip eigrp topology and show ipv6 eigrp topology. The addresses in the output are link-local IPv6 addresses, so our answer is the latter. See Chapter 17 for more information.
15. A, B, C, D. Cisco has documented steps, according to the objectives, that you must go through when troubleshooting an adjacency. See Chapter 18 for more information.
16. B, D, E. In order for two OSPF routers to create an adjacency, the Hello and dead timers must match, and they must both be configured into the same area, as well as being in the same subnet. See Chapter 18 for more information.
17. A. The process starts by sending out Hello packets. Every listening router will then add the originating router to the neighbor database. The responding routers will reply with all of their Hello information so that the originating router can add them to its own neighbor table. At this point, we will have reached the 2WAY state – only certain routers will advance beyond to this. See Chapter 19 for more information.
18. C. Referred to as summary link advertisements (SLAs), Type 3 LSAs are generated by area border routers. These ABRs send Type 3 LSAs toward the area external to the one where they were generated. See Chapter 19 for more information.
19. B. Authentication Header (AH) provides authentication of either all or part of the IP packet through the addition of a header that is calculated based on the values in the packet, but it doesn't offer any encryption services. See Chapter 21 for more information.
20. B. Generic Routing Encapsulation (GRE) has no built-in security mechanisms. See Chapter 21 for more information.
21. C. When traffic exceeds the allocated rate, the policer can take one of two actions. It can either drop traffic or re-mark it to another class of service. The new class usually has a higher drop probability. See Chapter 21 for more information.
22. B. This can be a hard question if you don't remember to invert the 7th bit of the first octet in the MAC address! Always look for the 7th bit when studying for the Cisco R/S, and when using eui-64, invert it. The eui-64 autoconfiguration then inserts an FF: FE in the middle of the 48-bit MAC address to create a unique IPv6 address. See Chapter 14 for more information.
23. A. The NDP neighbor advertisement (NA) contains the MAC address. A neighbor solicitation (NS) was initially sent asking for the MAC address. See Chapter 14 for more information.
24. B. Each field in an IPv6 address is 16 bits long. An IPv6 address is a total of 128 bits. See Chapter 14 for more information.
25. A. To enable OSPFv3, you enable the protocol at the interface level, as with RIPng. The command string is area-id. It's important to understand that area 0 and area 0.0.0.0 both describe area 0. See Chapter 19 for more information.
26. C. The command line console 0 places you at a prompt where you can then set your console user-mode password. See Chapter 6 for more information.
27. B, D. The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256–254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254. See Chapter 4 for more information.
28. C. The IP address is configured under a logical interface, called a management domain or VLAN 1, by default. See Chapter 10 for more information.
29. A. 256 – 192 = 64, so 64 is our block size. Just count in increments of 64 to find our subnet: 64 + 64 = 128. 128 + 64 = 192. The subnet is 128, the broadcast address is 191, and the valid host range is the numbers in between, or 129–190. See Chapter 4 for more information.
30. C. An inside global address is considered to be the IP address of the host on the private network after translation. See Chapter 13 for more information.
31. B. Once you create your pool, the command ip nat inside source must be used to say which inside locals are allowed to use the pool. In this question, we need to see if access list 100 is configured correctly, if at all, so show access-list is the best answer. See Chapter 13 for more information.
32. D. Layer 2 switching creates individual collision domains per port. See Chapter 1 for more information.
33. C. The command line vty 0 4 places you in a prompt that will allow you to set or change your Telnet password. See Chapter 6 for more information.
34. B. To see the contents of all access lists, use the show access-lists command. See Chapter 12 for more information.
35. C. VLANs break up broadcast domains at layer 2. See Chapter 11 for more information.
36. A. The command erase startup-config deletes the configuration stored in NVRAM. See Chapter 6 for more information.
37. C. ICMP is the protocol at the Network layer that is used to send messages back to an originating router. See Chapter 3 for more information.
38. A. Class A addressing provides 22 bits for host subnetting. Class B provides 16 bits, but only 14 are available for subnetting. Class C provides only 6 bits for subnetting. See Chapter 3 for more information.
39. B. Only the EIGRP route will be placed in the routing table because EIGRP has the lowest administrative distance (AD), and that is always used before metrics. See Chapter 8 for more information.
40. D. Switches send information about only one VLAN down a link unless it is configured as a trunk link. See Chapter 11 for more information.
41. B, G. To enable OSPF, you must first start OSPF using a process ID. The number is irrelevant; just choose a number from 1 to 65,535 and you're good to go. After you start the