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QUESTION 161
Refer to the exhibit. Which two statements are true about the loopback address that is configured on RouterB? (Choose two.)

 

A.    It ensures that data will be forwarded by RouterB.
B.    It provides stability for the OSPF process on RouterB.
C.    It specifies that the router ID for RouterB should be 10.0.0.1.
D.    It decreases the metric for routes that are advertised from RouterB.
E.    It indicates that RouterB should be elected the DR for the LAN.

Answer: BC
Explanation:
A loopback interface never comes down even if the link is broken so it provides stability for the OSPF process (for example we use that loopback interface as the router-id) – The router-ID is chosen in the order below:
+ The highest IP address assigned to a loopback (logical) interface. + If a loopback interface is not defined, the highest IP address of all active router’s physical interfaces will be chosen.
-> The loopback interface will be chosen as the router ID of RouterB –

QUESTION 162
Which characteristics are representative of a link-state routing protocol? (Choose three.)

A.    provides common view of entire topology
B.    exchanges routing tables with neighbors
C.    calculates shortest path
D.    utilizes event-triggered updates
E.    utilizes frequent periodic updates

Answer: ACD
Explanation:
Each of routers running link-state routing protocol learns paths to all the destinations in its “area” so we can say although it is a bit unclear.
Link-state routing protocols generate routing updates only (not the whole routing table) when a change occurs in the network topology so
Link-state routing protocol like OSPF uses Dijkstra algorithm to calculate the shortest path -> . Unlike Distance vector routing protocol (which utilizes frequent periodic updates), link-state routing protocol utilizes event-triggered updates (only sends update when a change occurs) ->

QUESTION 163
Refer to the exhibit. The network is converged.After link-state advertisements are received from Router_A, what information will Router_E contain in its routing table for the subnets 208.149.23.64 and 208.149.23.96?

 

A.    208.149.23.64[110/13] via 190.173.23.10, 00:00:07, FastEthemet0/0 208.149.23.96[110/13] via
190.173.23.10, 00:00:16, FastEthemet0/0
B.    208.149.23.64[110/1] via 190.172.23.10, 00:00:07, Serial1/0 208.149.23.96[110/3] via 190.173.23.10,
00:00:16, FastEthemet0/0
C.    208.149.23.64[110/13] via 190.173.23.10, 00:00:07, Serial1/0 208.149.23.96[110/13] via 190.173.23.10,
00:00:16, Serial1/0 208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
D.    208.149.23.64[110/3] via 190.172.23.10, 00:00:07, Serial1/0 208.149.23.96[110/3] via 190.173.23.10,
00:00:16, Serial1/0

Answer: A
Explanation:
Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A through FastEthernet interface. The interface cost is calculated with the formula 108 / Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 / 100,000,000 = 1.
Therefore the cost is 12 (learned from Router_A) + 1 = 13 for both subnets ->
The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps = 64; T3 cost = 108 / 45 Mbps = 2) so surely OSPF will choose the path through T3 link -> Router_E will choose the path from Router_A through FastEthernet0/0, not Serial1/0.
In fact, we can quickly eliminate answers B, C and D because they contain at least one subnet learned from Serial1/0 -> they are surely incorrect.

QUESTION 164
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: AC

QUESTION 165
Lab Simulation – CLI Configuration

  
Attention:
In practical examinations, please note the following, the actual information will prevail.
1. Name of the router is xxx
2. Enable. secret password is xxx
3. Password In access user EXEC mode using the console is xxx
4. The password to allow telnet access to the router is xxx
5. IP information

Answer:

Router>enable
Router#config terminal
Router(config)#hostname Gotha
Gotha(config)#enable secret mi222ke
Gotha(config)#line console 0
Gotha(config-line)#password G8tors1
Gotha(config-line)#exit
Gotha(config)#line vty 0 4
Gotha(config-line)#password dun63lap
Gotha(config-line)#login
Gotha(config-line)#exit
Gotha(config)#interface fa0/0
Gotha(config-if)#no shutdown
Gotha(config-if)#ip address 209.165.201.4 255.255.255.224
Gotha(config)#interface s0/0/0
Gotha(config-if)#ip address 192.0.2.190 255.255.255.240
Gotha(config-if)#no shutdown
Gotha(config-if)#exit
Gotha(config)#router rip
Gotha(config-router)#version 2
Gotha(config-router)#network 209.165.201.0
Gotha(config-router)#network 192.0.2.176
Gotha(config-router)#end
Gotha#copy run start

QUESTION 166
Lab Simulation – OSPF

 

Answer:
The question mentioned Boston router was not configured correctly or incomplete so we should check this router first. Click on PC-B to access the command line interface (CLI) of Boston router.

Boston>enable (type cisco as its password here)
Boston#show running-config
 
First, remember that the current OSPF Process ID is 2 because we will need it for later configuration. Next notice that in the second “network” command the network and wildcard mask are 192.168.155.0 and 0.0.0.3 which is equivalent to 192.168.155.0 255.255.255.252 in term of subnet mask. Therefore this subnetwork’s range is from 192.168.155.0 to 192.168.155.3 but the ip address of s0/0 interface of Boston router is192.168.155.5 which don’t belong to this range -> this is the reason why OSPF did not recognize s0 interface of Boston router as a part of area 0. So we need to find a subnetwork that s0 interface belongs to.
IP address of S0 interface: 192.168.155.5/30
Subnet mask: /30 = 1111 1111.1111 1111.1111 1111.1111 1100
Increment: 4
Network address (which IP address of s0 interface belongs to):
192.168.155.4 (because 4 * 1 = 4 < 5)
Therefore we must use this network instead of 192.168.155.0 network

Boston#configure terminal
Boston(config)#router ospf 2
Boston(config-router)#no network 192.168.155.0 0.0.0.3 area 0
Boston(config-router)#network 192.168.155.4 0.0.0.3 area 0
Boston(config-router)#end
Boston#copy running-config startup-config

Finally, you should issue a ping command from Boston router to Lancaster router to make sure it works well.

Boston#ping 192.168.43.1

QUESTION 167
This topology contains 3 routers and 1 switch. Complete the topology.

Drag the appropriate device icons to the labeled Device
Drag the appropriate connections to the locations labeled Connections.
Drag the appropriate IP addresses to the locations labeled IP address

(Hint: use the given host addresses and Main router information)
To remove a device or connection, drag it away from the topology.
Use information gathered from the Main router to complete the configuration of any
additional routers. No passwords are required to access the Main router. The config
terminal command has been disabled for the HQ router. The router does not require
any configuration.
Configure each additional router with the following:

Configure the interfaces with the correct IP address and enable the interfaces.
Set the password to allow console access to consolepw
Set the password to allow telnet access to telnetpw
Set the password to allow privilege mode access to privpw

Note: Because routes are not being added to the configurations, you will not be able
to ping through the internetwork.
All devices have cable autosensing capabilities disabled.
All hosts are PC’s
 

Answer:

 

Explanation:
(1) Configure two routers on the left and right with these commands:

Router1 = router on the left

(2) Assign appropriate IP addresses to Fa0/0 & Fa0/1 interfaces:

Router1>enable
Router1#configure terminal
Router1(config)#interface fa0/0
Router1(config-if)#ip address 192.168.152.190 255.255.255.240
Router1(config-if)#no shutdown

Router1(config-if)#interface fa0/1
Router1(config-if)#ip address 192.168.152.142 255.255.255.240
Router1(config-if)#no shutdown

(3) Set passwords (configure on two routers)

+ Console password:

Router1(config-if)#exit
Router1(config)#line console 0
Router1(config-line)#password consolepw
Router1(config-line)#login
Router1(config-line)#exit

+ Telnet password:

Router1(config)#line vty 0 4
Router1(config-line)#password telnetpw
Router1(config-line)#login
Router1(config-line)#exit

+ Privilege mode password:

Router1(config)#enable password privpw

Save the configuration:

Router1(config)#exit
Router1#copy running-config startup-config

(4) Configure IP addresses of Router2 (router on the right)

Router2>enable
Router2#configure terminal
Router2(config)#interface fa0/0
Router2(config-if)#ip address 192.168.152.238 255.255.255.240
Router2(config-if)#no shutdown

Router2(config-if)#interface serial0/0
Router2(config-if)#ip address 192.168.152.174 255.255.255.240
Router2(config-if)#no shutdown

QUESTION 168
To what type of port would a cable with a DB-60 connector attach?

A.    Serial port
B.    Console port
C.    Ethernet port
D.    Fibre optic port

Answer: A

QUESTION 169
A network administrator is connecting PC hosts A and B directly through their Ethernet interfaces as shown in the graphic. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts? (Choose two.)

 

A.    A crossover cable should be used in place of the straight-through cable.
B.    A rollover cable should be used in place of the straight-through cable.
C.    The subnet masks should be set to 255.255.255.192
D.    A default gateway needs to be set on each host.
E.    The hosts must be reconfigured to use private IP addresses for direct connections of this type.
F.    The subnet masks should be set to 255.255.255.0

Answer: AF

QUESTION 170
Refer to the exhibit. What kind of cable should be used to make each connection that is identified by the numbers shown?

 

A.    1 – Ethernet Crossover cable
2 – Ethernet straight-through cable
3 – Fiber Optic cable
4 – Rollover cable
B.    1 – Ethernet straight-through cable
2 – Ethernet straight-through cable
3 – Serial cable
4 – Rollover cable
C.    1 – Ethernet rollover cable
2 – Ethernet crossover cable
3 – Serial cable
4 – Null-modem cable
D.    1 – Ethernet straight-through cable
2 – Ethernet Crossover cable
3 – Serial cable
4 – Rollover cable
E.    1 – Ethernet straight-through cable
2 – Ethernet Crossover cable
3 – Serial cable
4 – Ethernet Straight-through cable

Answer: B

QUESTION 171
Which of the following are types of flow control? (Choose three.)

A.    buffering
B.    cut-through
C.    windowing
D.    congestion avoidance
E.    load balancing

Answer: ACD
Explanation:
http://www.info-it.net/cisco/ccna/exam-tips/flow-control.php
During Transfer of data, a high speed computer is generating data traffic a lot faster than the network device can handle in transferring to destination, so single gateway or destination device cannot handle much amount of traffic that is called “Congestion”.
Buffering
The Technie is used to control the data transfer when we have congestion, when a network device receive a data it stores in memory section and then transfer to next destination this process called “Buffering”.
Windowing Whereas Windowing is used for flow control by the Transport layer. Say the sender device is sending segments and the receiver device can accommodate only a fixed number of segments before it can accept more, the two devices negotiate the window size during the connection setup.
This is done so that the sending device doesn’t overflow the receiving device’s buffer. Also the receiving device can send a single acknowledgement for the segments it has received instead of sending an acknowledgement after every segment received. Also, this window size is dynamic meaning, the devices can negotiate and change the window size in the middle of a session. So if initially the window size is three and the receiving device thinks that it can accept more number of segments in its buffer it can negotiate with the sending device and it increase it to say 5 for example. Windowing is used only by TCP since UDP doesn’t use or allow flow control.

QUESTION 172
How many simultaneous Telnet sessions does a Cisco router support by default?

A.    1
B.    2
C.    3
D.    4
E.    5
F.    6

Answer: E

QUESTION 173
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:
http://whatis.techtarget.com/definition/flow-control 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.

QUESTION 174
Refer to the exhibit. How many collision domains are shown?

 

A.    one
B.    two
C.    three
D.    four
E.    six
F.    twelve

Answer: B
Explanation:
Hubs create single collision and broadcast domains.

QUESTION 175
Which IP address is a private address?

A.    12.0.0.1
B.    168.172.19.39
C.    172.20.14.36
D.    172.33.194.30
E.    192.169.42.34

Answer: C
Explanation:

 

QUESTION 176
Which statement is true?

A.    An IPv6 address is 64 bits long and is represented as hexadecimal characters.
B.    An IPv6 address is 32 bits long and is represented as decimal digits.
C.    An IPv6 address is 128 bits long and is represented as decimal digits.
D.    An IPv6 address is 128 bits long and is represented as hexadecimal characters.

Answer: D
Explanation:
http://www.cisco.com/en/US/technologies/tk648/tk872/technologies_white_paper0900aecd 8026003d.pdf
One of the key advantages IPv6 brings is the exponentially larger address space. The following will outline the basic address architecture of IPv6.
128-bit-long addresses
Represented in hexadecimal format:
Uses CIDR principles: prefix/prefix length
x:x:x:x:x:x:x:x, where x is a 16-bit hex field
The last 64 bits are used for the interface ID

QUESTION 177
If an Ethernet port on a router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?

A.    1024
B.    2046
C.    4094
D.    4096
E.    8190

Answer: C
Explanation:
Each octet represents eight bits. The bits, in turn, represent (from left to right): 128, 64, 32 , 16 , 8, 4, 2, 1
Add them up and you get 255. Add one for the all zeros option, and the total is 256. Now take away one of these for the network address (all zeros) and another for the broadcast address (all ones). Each octet represents 254 possible hosts. Or 254 possible networks. Unless you have subnet zero set on your network gear, in which case you could conceivably have 255.
The CIDR addressing format (/20) tells us that 20 bits are used for the network portion, so the maximum number of networks are 2^20 minus one if you have subnet zero enabled, or minus 2 if not.
You asked about the number of hosts. That will be 32 minus the number of network bits, minus two. So calculate it as (2^(32-20))-2, or (2^12)-2 = 4094

QUESTION 178
Which statements are TRUE regarding Internet Protocol version 6 (IPv6) addresses? (Choose three.)

A.    An IPv6 address is divided into eight 16-bit groups.
B.    A double colon (::) can only be used once in a single IPv6 address.
C.    IPv6 addresses are 196 bits in length.
D.    Leading zeros cannot be omitted in an IPv6 address.
E.    Groups with a value of 0 can be represented with a single 0 in IPv6 address.

Answer: ABE
Explanation:
IPv6 addresses are divided into eight 16-bit groups, a double colon (::) can only be used once in an IPv6 address, and groups with a value of 0 can be represented with a single 0 in an IPv6 address.
The following statements are also true regarding IPv6 address:
IPv6 addresses are 128 bits in length.
Eight 16-bit groups are divided by a colon (:).
Multiple groups of 16-bit 0s can be represented with double colon (::).
Double colons (::) represent only 0s.
Leading zeros can be omitted in an IPv6 address.
The option stating that IPv6 addresses are 196 bits in length is incorrect. IPv6 addresses are 128 bits in length.
The option stating that leading zeros cannot be omitted in an IPv6 address is incorrect.
Leading zeros can be omitted in an IPv6 address.

QUESTION 179
Which of the following IP addresses are valid Class B host addresses if a default Class B mask is in use? (Choose two.)

A.    10.6.8.35
B.    133.6.5.4
C.    192.168.5.9
D.    127.0.0.1
E.    190.6.5.4

Answer: BE
Explanation:
The IP addresses 133.6.5.4 and 190.6.5.4 are both valid Class B addresses when a default mask is in use.
The Class B default mask is 255.255.0.0 and the range of valid addresses is 128.0.0.0- 191.255.255.255.
The IP address 10.6.8.35 is a Class A address. The Class A default mask is 255.0.0.0 and the range of valid addresses is 1.0.0.0 – 127.255.255.255, with the exception of the range 127.0.0.1 – 127.255.255.255, which is reserved and cannot be assigned. The IP address 192.168.5.9 is a Class C address. The Class C default mask is 255.255.255.0 and the range of valid addresses is 192.0.0.0 – 223.255.255.255. The IP address 127.0.0.1 is a Class A address, but it comes from a reserved portion that cannot be assigned.
The range 127.0.0.1 – 127.255.255.255 is used for diagnostics, and although any address in the range will work as a diagnostic address, 127.0.0.1 is known as the loopback address. If you can ping this address, or any address in the 127.0.0.1 – 127.255.255.255 range, then the NIC is working and TCP/IP is installed. The Class A default mask is 255.0.0.0 and the range of valid addresses is 1.0.0.0 – 127.255.255.255, with the exception of the range 127.0.0.1 – 127.255.255.255, which is reserved and cannot be assigned.

 

QUESTION 180
How many addresses will be available for dynamic NAT translation when a router is configured with the following commands?

Router(config)#ip nat pool TAME 209.165.201.23 209.165.201.30 netmask 255.255.255.224
Router(config)#ip nat inside source list 9 pool TAME

A.    7
B.    8
C.    9
D.    10
E.    24
F.    32

Answer: B
Explanation:
209.165.201.23 to 209.165.201.30 provides for 8 addresses.

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