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Q51. - (Topic 3) 

Refer to the exhibit. 

What is the simplest way to configure routing between the regional office network 10.89.0.0/20 and the corporate network? 

A. router1(config)#ip route 10.89.0.0 255.255.240.0 10.89.16.2 

B. router2(config)#ip route 10.89.3.0 255.255.0.0 10.89.16.2 

C. router1(config)#ip route 10.89.0.0 255.255.240.0 10.89.16.1 

D. router2(config)#ip route 0.0.0.0 0.0.0.0 10.89.16.1 

Answer:

Explanation: 

The next hop of the static route should be 10.89.16.1, which is the IP address for router R1 in this example. Since this router is a stub router with only a single connection to the WAN, the simplest thing to do is to configure a single static default route back to the HQ network. 


Q52. - (Topic 5) 

Select three options which are security issues with the current configuration of SwitchA. (Choose three.) 

A. Privilege mode is protected with an unencrypted password 

B. Inappropriate wording in banner message 

C. Virtual terminal lines are protected only by a password requirement 

D. Both the username and password are weak 

E. Telnet connections can be used to remotely manage the switch 

F. Cisco user will be granted privilege level 15 by default 

Answer: A,B,D 


Q53. - (Topic 3) 

Which two statements describe the IP address 10.16.3.65/23? (Choose two.) 

A. The subnet address is 10.16.3.0 255.255.254.0. 

B. The lowest host address in the subnet is 10.16.2.1 255.255.254.0. 

C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0 

D. The broadcast address of the subnet is 10.16.3.255 255.255.254.0. 

E. The network is not subnetted. 

Answer: B,D 

Explanation: 

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 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 


Q54. - (Topic 7) 

Under which circumstance should a network administrator implement one-way NAT? 

A. when the network must route UDP traffic 

B. when traffic that originates outside the network must be routed to internal hosts 

C. when traffic that originates inside the network must be routed to internal hosts 

D. when the network has few public IP addresses and many private IP addresses require outside access 

Answer:

Explanation: NAT operation is typically transparent to both the internal and external hosts. Typically the internal host is aware of the true IP address and TCP or UDP port of the external host. Typically the NAT device may function as the default gateway for the internal host. However the external host is only aware of the public IP address for the NAT device and the particular port being used to communicate on behalf of a specific internal host. 

NAT and TCP/UDP 

"Pure NAT", operating on IP alone, may or may not correctly parse protocols that are totally concerned with IP information, such as ICMP, depending on whether the payload is interpreted by a host on the "inside" or "outside" of translation. As soon as the protocol stack is traversed, even with such basic protocols as TCP and UDP, the protocols will break unless NAT takes action beyond the network layer. IP packets have a checksum in each packet header, which provides error detection only for the header. IP datagrams may become fragmented and it is necessary for a NAT to reassemble these fragments to allow correct recalculation of higher-level checksums and correct tracking of which packets belong to which connection. The major transport layer protocols, TCP and UDP, have a checksum that covers all the data they carry, as well as the TCP/UDP header, plus a "pseudo-header" that contains the source and destination IP addresses of the packet carrying the TCP/UDP header. For an originating NAT to pass TCP or UDP successfully, it must recompute the TCP/UDP header checksum based on the translated IP addresses, not the original ones, and put that checksum into the TCP/UDP header of the first packet of the fragmented set of packets. The receiving NAT must recompute the IP checksum on every packet it passes to the destination host, and also recognize and recompute the TCP/UDP header using the retranslated addresses and pseudo-header. This is not a completely solved problem. One solution is for the receiving NAT to reassemble the entire segment and then recompute a checksum calculated across all packets. The originating host may perform Maximum transmission unit (MTU) path discovery to determine the packet size that can be transmitted without fragmentation, and then set the don't fragment (DF) bit in the appropriate packet header field. Of course, this is only a one-way solution, because the responding host can send packets of any size, which may be fragmented before reaching the NAT. 


Q55. - (Topic 3) 

Refer to the exhibit. 

When running OSPF, what would cause router A not to form an adjacency with router B? 

A. The loopback addresses are on different subnets. 

B. The values of the dead timers on the routers are different. 

C. Route summarization is enabled on both routers. 

D. The process identifier on router A is different than the process identifier on router B. 

Answer:

Explanation: 

To form an adjacency (become neighbor), router A & B must have the same Hello interval, Dead interval and AREA numbers 


Q56. - (Topic 3) 

What information can be used by a router running a link-state protocol to build and maintain its topological database? (Choose two.) 

A. hello packets 

B. SAP messages sent by other routers 

C. LSAs from other routers 

D. beacons received on point-to-point links 

E. routing tables received from other link-state routers 

F. TTL packets from designated routers 

Answer: A,C 

Explanation: 

Reference 1: http://www.ciscopress.com/articles/article.asp?p=24090&seqNum=4 

Link state protocols, sometimes called shortest path first or distributed database protocols, are built around a well-known algorithm from graph theory, E. W. Dijkstra'a shortest path algorithm. Examples of link state routing protocols are: Open Shortest Path First (OSPF) for IP The ISO's Intermediate System to Intermediate System (IS-IS) for CLNS and IP DEC's DNA Phase V Novell's NetWare Link Services Protocol (NLSP) Although link state protocols are rightly considered more complex than distance vector protocols, the basic functionality is not complex at all: 

1.

 Each router establishes a relationship—an adjacency—with each of its neighbors. 

2.

 Each router sends link state advertisements (LSAs), some 

3.

 Each router stores a copy of all the LSAs it has seen in a database. If all works well, the databases in all routers should be identical. 

4.

 The completed topological database, also called the link state database, describes a graph of the internetwork. Using the Dijkstra algorithm, each router calculates the shortest path to each network and enters this information into the route table. OSPF Tutorial 


Q57. - (Topic 3) 

Refer to the exhibit. 

Which two statements are correct? (Choose two.) 

A. This is a default route. 

B. Adding the subnet mask is optional for the ip route command. 

C. This will allow any host on the 172.16.1.0 network to reach all known destinations beyond RouterA. 

D. This command is incorrect, it needs to specify the interface, such as s0/0/0 rather than an IP address. 

E. The same command needs to be entered on RouterA so that hosts on the 172.16.1.0 network can reach network 10.0.0.0. 

Answer: A,C 

Explanation: 

This is obviously the default route which is set between the routers and since it is entered in such a manner that it ensures connectivity between the stub network and any host lying beyond RouterA. 


Q58. - (Topic 3) 

Which statement describes the process ID that is used to run OSPF on a router? 

A. It is globally significant and is used to represent the AS number. 

B. It is locally significant and is used to identify an instance of the OSPF database. 

C. It is globally significant and is used to identify OSPF stub areas. 

D. It is locally significant and must be the same throughout an area. 

Answer:

Explanation: 

The Process ID for OSPF on a router is only locally significant and you can use the same number on each router, or each router can have a different number-it just doesn't matter. The numbers you can use are from 1 to 65,535. Don't get this confused with area numbers, which can be from 0 to 4.2 billion. 


Q59. - (Topic 3) 

ROUTER# show ip route 

192.168.12.0/24 is variably subnetted, 9 subnets, 3 masks C 192.168.12.64 /28 is directly connected, Loopback1 C 192.168.12.32 /28 is directly connected, Ethernet0 C 192.168.12.48 /28 is directly connected, Loopback0 O 192.168.12.236 /30 [110/128] via 192.168.12.233, 00:35:36, Serial0 C 192.168.12.232 /30 is directly connected, Serial0 O 192.168.12.245 /30 [110/782] via 192.168.12.233, 00:35:36, Serial0 O 192.168.12.240 /30 [110/128] via 192.168.12.233, 00:35:36, Serial0 O 192.168.12.253 /30 [110/782] via 192.168.12.233, 00:35:37, Serial0 O 192.168.12.249 /30 [110/782] via 192.168.12.233, 00:35:37, Serial0 O 192.168.12.240/30 [110/128] via 192.168.12.233, 00:35:36, Serial 0 

To what does the 128 refer to in the router output above? 

A. OSPF cost 

B. OSPF priority 

C. OSPF hop count 

D. OSPF ID number 

E. OSPF administrative distance 

Answer:

Explanation: 

The first parameter is the Administrative Distance of OSPF (110) while the second parameter is the cost of OSPF. 


Q60. - (Topic 1) 

Refer to the topology and switching table shown in the graphic. 

Host B sends a frame to Host C. What will the switch do with the frame? 

A. Drop the frame 

B. Send the frame out all ports except port 0/2 

C. Return the frame to Host B 

D. Send an ARP request for Host C 

E. Send an ICMP Host Unreachable message to Host B 

F. Record the destination MAC address in the switching table and send the frame directly to Host C 

Answer: