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Q51. CORRECT TEXT - (Topic 6)
Answer: Router>enable
Router#config terminal Router(config)#hostname Apopka 2) Enable-secret password (cisco10): Apopka(config)#enable secret cisco10 3) Set the console password to RouterPass: Apopka(config)#line console 0 Apopka(config-line)#password RouterPass Apopka(config-line)#login Apopka(config-line)#exit 4) Set the Telnet password to scan90: Apopka(config)#line vty 0 4 Apopka(config-line)#password scan90 Apopka(config-line)#login Apopka(config-line)#exit 5) Configure Ethernet interface (on the right) of router Apopka: The subnet mask of the Ethernet network 209.165.201.0 is 27. From this subnet mask, we can find out the increment by converting it into binary form, that is /27 = 1111 1111.1111 1111.1111 1111.1110 0000. Pay more attention to the last bit 1 because it tells us the increment, using the formula: Increment = 2place of the last bit 1 (starts counting from 0,from right to left), in this case increment = 25 = 32. Therefore: Increment: 32 Network address: 209.165.201.0 Broadcast address: 209.165.201.31 (because 209.165.201.32 is the second subnetwork, so the previous IP - 209.165.201.31 - is the broadcast address of the first subnet). -> The second assignable host address of this subnetwork is 209.165.201.2/27 Assign the second assignable host address to Fa0/0 interface of Apopka router: Apopka(config)#interface Fa0/0 Apopka(config-if)#ip address 209.165.201.2 255.255.255.224 Apopka(config-if)#no shutdown Apopka(config-if)#exit 6) Configure Serial interface (on the left) of router Apopka: Using the same method to find out the increment of the Serial network: Serial network 192.0.2.128/28: Increment: 16 (/28 = 1111 1111.1111 1111.1111 1111.1111 0000) Network address: 192.0.2.128 (because 8 * 16 = 128 so 192.0.2.128 is also the network address of this subnet) Broadcast address: 192.0.2.143 -> The last assignable host address in this subnet is 192.0.2.142/28. Assign the last assignable host address to S0/0/0 interface of Apopka router: Apopka(config)#interface S0/0/0 (or use interface S0/0 if not successful) Apopka(config-if)#ip address 192.0.2.142 255.255.255.240 Apopka(config-if)#no shutdown Apopka(config-if)#exit 7) Configure RIP v2 routing protocol: Apopka(config)#router rip Apopka(config-router)#version 2 Apopka(config-router)#network 209.165.201.0 Apopka(config-router)#network 192.0.2.128 Apopka(config-router)#end Save the configuration: Apopka#copy running-config startup-config Finally, you should use the ping command to verify all are working properly!
Topic 7, Mix Questions
Q52. - (Topic 2)
What is the purpose of flow control?
A. To ensure data is retransmitted if an acknowledgement is not received.
B. To reassemble segments in the correct order at the destination device.
C. To provide a means for the receiver to govern the amount of data sent by the sender.
D. To regulate the size of each segment.
Answer: C
Explanation:
Flow control is the management of data flow between computers or devices or between nodes in a network so that the data can be handled at an efficient pace. Too much data arriving before a device can handle it causes data overflow, meaning the data is either lost or must be retransmitted. For serial data transmission locally or in a network, the Xon/Xoff protocol can be used. For modem connections, either Xon/Xoff or CTS/RTS (Clear to Send/Ready to Send) commands can be used to control data flow. In a network, flow control can also be applied by refusing additional device connections until the flow of traffic has subsided.
Reference: http://whatis.techtarget.com/definition/flow-control
Q53. - (Topic 5)
Which two of the following are true regarding the configuration of RouterA? (Choose two.)
A. At least 5 simultaneous remote connections are possible
B. Only telnet protocol connections to RouterA are supported C. Remote connections to RouterA using telnet will succeed
D. Console line connections will nevertime out due to inactivity
E. Since DHCP is not used on Fa0/1 there is not a need to use the NAT protocol
Answer: A,C
Q54. - (Topic 1)
Which two statements describe the operation of the CSMA/CD access method? (Choose two.)
A. In a CSMA/CD collision domain, multiple stations can successfully transmit data simultaneously.
B. In a CSMA/CD collision domain, stations must wait until the media is not in use before transmitting.
C. The use of hubs to enlarge the size of collision domains is one way to improve the operation of the CSMA/CD access method.
D. After a collision, the station that detected the collision has first priority to resend the lost data.
E. After a collision, all stations run a random backoff algorithm. When the backoff delay period has expired, all stations have equal priority to transmit data.
F. After a collision, all stations involved run an identical backoff algorithm and then synchronize with each other prior to transmitting data.
Answer: B,E
Explanation:
Ethernet networking uses Carrier Sense Multiple Access with Collision Detect (CSMA/CD), a protocol that helps devices share the bandwidth evenly without having two devices transmit at the same time on the network medium. CSMA/CD was created to overcome the problem of those collisions that occur when packets are transmitted simultaneously from different nodes. And trust me, good collision management is crucial, because when a node transmits in a CSMA/CD network, all the other nodes on the network receive and examine that transmission. Only bridges and routers can effectively prevent a transmission from propagating throughout the entire network! So, how does the CSMA/CD protocol work? Like this: when a host wants to transmit over the network, it first checks for the presence of a digital signal on the wire. If all is clear (no other host is transmitting), the host will then proceed with its transmission. But it doesn’t stop there. The transmitting host constantly monitors the wire to make sure no other hosts begin transmitting. If the host detects another signal on the wire, it sends out an extended jam signal that causes all nodes on the segment to stop sending data (think, busy signal). The nodes respond to that jam signal by waiting a while before attempting to transmit again. Backoff algorithms determine when the colliding stations can retransmit. If collisions keep occurring after 15 tries, the nodes attempting to transmit will then time out.
Q55. - (Topic 5)
Refer to the exhibit.
A network technician is asked to design a small network with redundancy. The exhibit represents this design, with all hosts configured in the same VLAN. What conclusions can be made about this design?
A. This design will function as intended.
B. Spanning-tree will need to be used.
C. The router will not accept the addressing scheme.
D. The connection between switches should be a trunk.
E. The router interfaces must be encapsulated with the 802.1Q protocol.
Answer: C
Explanation:
The proposed addressing scheme is on the same network. Cisco routers will not allow you to assign two different interfaces to be on the same IP subnet.
Q56. - (Topic 2)
A switch has 48 ports and 4 VLANs. How many collision and broadcast domains exist on the switch (collision, broadcast)?
A. 4, 48
B. 48, 4
C. 48, 1
D. 1, 48
E. 4, 1
Answer: B
Explanation:
A switch uses a separate collision domain for each port, and each VLAN is a separate broadcast domain.
Topic 3, Routing Fundamentals
Q57. - (Topic 5)
Which of the following is true regarding the configuration of SwitchA?
A. only 5 simultaneous remote connections are possible
B. remote connections using ssh will require a username and password
C. only connections from the local network will be possible
D. console access to SwitchA requires a password
Answer: B
Q58. - (Topic 3)
What are two benefits of using a single OSPF area network design? (Choose two.)
A. It is less CPU intensive for routers in the single area.
B. It reduces the types of LSAs that are generated.
C. It removes the need for virtual links.
D. It increases LSA response times.
E. It reduces the number of required OSPF neighbor adjacencies.
Answer: B,C
Explanation:
OSPF uses a LSDB (link state database) and fills this with LSAs (link state advertisement). The link types are as follows:
. LSA Type 1: Router LSA
.......
LSA Type 2: Network LSA
LSA Type 3: Summary LSA
LSA Type 4: Summary ASBR LSA
LSA Type 5: Autonomous system external LSA
LSA Type 6: Multicast OSPF LSA
LSA Type 7: Not-so-stubby area LSA
LSA Type 8: External attribute LSA for BGP
If all routers are in the same area, then many of these LSA types (Summary ASBR LSA, external LSA, etc) will not be used and will not be generated by any router.
All areas in an Open Shortest Path First (OSPF) autonomous system must be physically connected to the backbone area (Area 0). In some cases, where this is not possible, you can use a virtual link to connect to the backbone through a non-backbone area. You can also use virtual links to connect two parts of a partitioned backbone through a non-backbone area. The area through which you configure the virtual link, known as a transit area, must have full routing information. The transit area cannot be a stub area. Virtual links are not ideal and should really only be used for temporary network solutions or migrations. However, if all locations are in a single OSPF area this is not needed.
Q59. - (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
Q60. - (Topic 3)
Which two of these statements are true of IPv6 address representation? (Choose two.)
A. There are four types of IPv6 addresses: unicast, multicast, anycast, and broadcast.
B. A single interface may be assigned multiple IPv6 addresses of any type.
C. Every IPv6 interface contains at least one loopback address.
D. The first 64 bits represent the dynamically created interface ID.
E. Leading zeros in an IPv6 16 bit hexadecimal field are mandatory.
Answer: B,C
Explanation:
A single interface may be assigned multiple addresses of any type (unicast, anycast,
multicast).
Every IPv6-enabled interface must contain at least one loopback and one link-local
address.
Optionally, every interface can have multiple unique local and global addresses.
IPv6 host addresses can be assigned in multiple ways:
Static configuration
Stateless autoconfiguration
DHCPv6
When IPv6 is used over Ethernet networks, the Ethernet MAC address can be used to
generate the 64-bit interface ID for the host. This is called the EUI-64 address.
Since MAC addresses use 48 bits, additional bits must be inserted to fill the 64 bits
required.
Reference:
http://www.cisco.com/en/US/technologies/tk648/tk872/technologies_white_paper0900aecd
8026003d.pdf