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Question No: 1
Refer to the exhibit.
From R1, a network administrator is able to ping the serial interface of R2 but, unable to ping any of the subnets attached to RouterB. Based on the partial outputs in the exhibit, what could be the problem?
A. EIGRP does not support VLSM.
B. The EIGRP network statements are incorrectly configured.
C. The IP addressing on the serial interface of RouterA is incorrect.
D. The routing protocol has summarized on the classful boundary.
E. EIGRP has been configured with an invalid autonomous system number.
Answer: D
Explanation:
CCNA - EIGRP Common Question Reference:
http://www.orbitco-ccna-pastquestions.com/CCNA---EIGRP-Common-Question.php
If you look carefully at the R2 ip route, you will discover that the R2 does not learn any network from R1; this is because the routing protocol used here (EIGRP) performs auto summary when advertising routes to peers across a network. So in this case the address 172.17.0.0/16 is a summarized address. If the router was configured with no auto summary command, R2 LAN addresses would have been advertised and reached.
Question No: 2
The command frame-relay map ip 10.121.16.8 102 broadcast was entered on the router. Which of the following statements is true concerning this command?
A. This command should be executed from the global configuration mode.
B. The IP address 10.121.16.8 is the local router port used to forward data.
C. 102 is the remote DLCI that will receive the information.
D. This command is required for all Frame Relay configurations.
E. The broadcast option allows packets, such as RIP updates, to be forwarded across the PVC.
Answer: E
Explanation:
The command frame-relay map ip 10.121.16.8 102 broadcast means to map the remote IP 10.121.16.8 to the local DLCI 102. When the u201cbroadcastu201d keyword is included, it turns Frame Relay network as a broadcast network, which can forward broadcasts.
Question No: 3
What are two benefits of using NAT? (choose two)
A. NAT protects network security because private networks are not advertised.
B. NAT accelerates the routing process because no modifications are made on the packets.
C. Dynamic NAT facilitates connections from the outside of the network.
D. NAT facilitates end-to-end communication when IPsec is enable.
E. NAT eliminates the need to re-address all host that require external access.
F. NAT conserves addresses through host MAC-level multiplexing.
Answer: A,E
Question No: 4
At which layer of the OSI model does PPP perform?
A. Layer 2
B. Layer 3
C. Layer 4
D. Layer 5
Answer: A
Explanation:
Point-to-Point Protocol (PPP) is a data link protocol commonly used in establishing a direct connection between two networking nodes. It can provide connection authentication, transmission encryption (using ECP, RFC 1968), and compression.
Question No: 5
R1# show running-config interface Loopback0
description ***Loopback***
ip address 192.168.1.1 255.255.255.255
ip ospf 1 area 0
!
interface Ethernet0/0
description **Connected to R1-LAN** ip address 10.10.110.1 255.255.255.0
ip ospf 1 area 0
!
interface Ethernet0/1
description **Connected to L2SW**
ip address 10.10.230.1 255.255.255.0
ip ospf hello-interval 25 ip ospf 1 area 0
!
router ospf 1
log-adjacency-changes
R2# show running-config
R2
!
interface Loopback0 description **Loopback**
ip address 192.168.2.2 255.255.255.255
ip ospf 2 area 0
!
interface Ethernet0/0
description **Connected to R2-LAN** ip address 10.10.120.1 255.255.255.0
ip ospf 2 area 0
!
interface Ethernet0/1
description **Connected to L2SW**
ip address 10.10.230.2 255.255.255.0
ip ospf 2 area 0
!
router ospf 2
log-adjacency-changes
R3# show running-config R3
username R6 password CISCO36
!
interface Loopback0 description **Loopback**
ip address 192.168.3.3 255.255.255.255
ip ospf 3 area 0
!
interface Ethernet0/0
description **Connected to L2SW**
ip address 10.10.230.3 255.255.255.0
ip ospf 3 area 0
!
interface Serial1/0
description **Connected to R4-Branch1 office** ip address 10.10.240.1 255.255.255.252
encapsulation ppp ip ospf 3 area 0
!
interface Serial1/1
description **Connected to R5-Branch2 office** ip address 10.10.240.5 255.255.255.252
encapsulation ppp
ip ospf hello-interval 50 ip ospf 3 area 0
!
interface Serial1/2
description **Connected to R6-Branch3 office** ip address 10.10.240.9 255.255.255.252
encapsulation ppp ip ospf 3 area 0
ppp authentication chap
!
router ospf 3
router-id 192.168.3.3
!
R4# show running-config R4
!
interface Loopback0 description **Loopback**
ip address 192.168.4.4 255.255.255.255
ip ospf 4 area 2
!
interface Ethernet0/0
ip address 172.16.113.1 255.255.255.0
ip ospf 4 area 2
!
interface Serial1/0
description **Connected to R3-Main Branch office** ip address 10.10.240.2 255.255.255.252
encapsulation ppp ip ospf 4 area 2
!
router ospf 4
log-adjacency-changes
R5# show running-config R5
!
interface Loopback0 description **Loopback**
ip address 192.168.5.5 255.255.255.255
ip ospf 5 area 0
!
interface Ethernet0/0
ip address 172.16.114.1 255.255.255.0
ip ospf 5 area 0
!
interface Serial1/0
description **Connected to R3-Main Branch office** ip address 10.10.240.6 255.255.255.252
encapsulation ppp ip ospf 5 area 0
!
router ospf 5
log-adjacency-changes
R6# show running-config R6
username R3 password CISCO36
!
interface Loopback0 description **Loopback**
ip address 192.168.6.6 255.255.255.255
ip ospf 6 area 0
!
interface Ethernet0/0
ip address 172.16.115.1 255.255.255.0
ip ospf 6 area 0
!
interface Serial1/0
description **Connected to R3-Main Branch office** ip address 10.10.240.10 255.255.255.252
encapsulation ppp ip ospf 6 area 0
ppp authentication chap
!
router ospf 6
router-id 192.168.3.3
!
An OSPF neighbor adjacency is not formed between R3 in the main office and R4 in the Branch1 office. What is causing the problem?
A. There is an area ID mismatch.
B. There is a Layer 2 issue; an encapsulation mismatch on serial links.
C. There is an OSPF hello and dead interval mismatch.
D. The R3 router ID is configured on R4.
Answer: A
Question No: 6
Refer to the exhibit.
Assume that all of the router interfaces are operational and configured correctly. How will router R2 be affected by the configuration of R1 that is shown in the exhibit?
A. Router R2 will not form a neighbor relationship with R1.
B. Router R2 will obtain a full routing table, including a default route, from R1.
C. R2 will obtain OSPF updates from R1, but will not obtain a default route from R1.
D. R2 will not have a route for the directly connected serial network, but all other directly connected networks will be present, as well as the two Ethernet networks connected to R1.
Answer: A
Explanation:
Open Shortest Path First http://en.wikipedia.org/wiki/Open_Shortest_Path_First
The configuration of R1 shows "router ospf 1" however, the diagram also shows that both routers should be in the backbone OSPF Area of "0". When routers are in different OSPF areas they will not form a neighbor relationship.
Neighbor relationships
As a link state routing protocol, OSPF establishes and maintains neighbor relationships in order to exchange routing updates with other routers. The neighbor relationship table is
called an adjacency database in OSPF.
Provided that OSPF is configured correctly, OSPF forms neighbor relationships only with the routers directly connected to it. In order to form a neighbor relationship between two routers, the interfaces used to form the relationship must be in the same area. Generally an interface is only configured in a single area, however you can configure an interface to belong to multiple areas. In the second area, such an interface must be configured as a secondary interface. (A neighbor state simulation shows how neighbor state changes from Down to Full Adjacency progressively with exchanging Hello, DD, Request, Update, and Ack packets).
Question No: 7
A network administrator needs to configure a serial link between the main office and a remote location. The router at the remote office is a non-Cisco router. How should the network administrator configure the serial interface of the main office router to make the connection?
A. Main(config)# interface serial 0/0
Main(config-if)# ip address 172.16.1.1 255.255.255.252 Main(config-if)# no shut
B. Main(config)# interface serial 0/0
Main(config-if)# ip address 172.16.1.1 255.255.255.252 Main(config-if)# encapsulation ppp
Main(config-if)# no shut
C. Main(config)# interface serial 0/0
Main(config-if)# ip address 172.16.1.1 255.255.255.252 Main(config-if)# encapsulation frame-relay
Main(config-if)# authentication chap Main(config-if)# no shut
D. Main(config)# interface serial 0/0
Main(config-if)#ip address 172.16.1.1 255.255.255.252 Main(config-if)#encapsulation ietf
Main(config-if)# no shut
Answer: B
Explanation:
Cisco High-Level Data Link Controller (HDLC) is the Cisco proprietary protocol for sending data over synchronous serial links using HDLC. So HDLC runs only in Cisco router. PPP is not proprietary protocol it's a open source every cisco router and non-cisco router understand the PPP protocol.So we need to configure the PPP protocol if connection is between cisco and non-cisco router.
Question No: 8
Which command do you enter to enable an interface to support PPPoE on a client?
A. Dev1(config)# bba-group pppoe bba 1.
B. Dev1(config-if)# pppoe-client dial-pool-number 1.
C. Dev1(config-if)# pppoe enable group bba1.
D. Dev1(config-if)# pppoe enable.
Answer: B
Question No: 9
What are the two default metrics used by EIGRP for route selection? (Choose two.)
A. Bandwidth
B. Delay
C. Reliability
D. Load
E. MTU
Answer: A,B
Explanation:
EIGRP adds together weighted values of different network link characteristics in order to calculate a metric for evaluating path selection.
These characteristics include:
u2022 Delay (measured in 10s of microseconds)
u2022 Bandwidth (measured in kilobytes per second)
u2022 Reliability (in numbers ranging from 1 to 255; 255 being the most reliable)
u2022 Load (in numbers ranging from 1 to 255; 255 being saturated)
Various constants (K 1 through K 5) are able to be set by a user to produce varying routing behaviors. However by default, only delay and bandwidth are used in the weighted formula to produce a single 32bit metric:
Note: Default K values are: K1 =K3 = 1 andK2 =K4 =K5 = 0When K5 is equal to 0 then [
K5/( K4 + reliability)] is defined to be 1
Use of the default constants effectively reduces the formula above to:
Reference: http://www.cisco.com/c/en/us/products/collateral/ios-nx-os-software/enhanced- interior-gateway-routing-protocol-eigrp/whitepaper_C11-720525.html
Question No: 10
Refer to the exhibit.
Which two statements are true about interVLAN routing in the topology that is shown in the exhibit? (Choose two.)
A. Host E and host F use the same IP gateway address.
B. Router1 and Switch2 should be connected via a crossover cable.
C. Router1 will not play a role in communications between host A and host D.
D. The FastEthernet 0/0 interface on Router1 must be configured with subinterfaces.
E. Router1 needs more LAN interfaces to accommodate the VLANs that are shown in the exhibit.
F. The FastEthernet 0/0 interface on Router1 and the FastEthernet 0/1 interface on Switch2 trunk ports must be configured using the same encapsulation type.
Answer: D,F
Explanation: http://www.cisco.com/en/US/tech/tk389/tk815/technologies_configuration_example09186a 00800949fd.shtml
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