Get Latest 100-105 Exam Dumps Questions - 100-105 Exam Dumps - Dumps4Download.us

Question No : 1

How does TCP differ from UDP? (Choose two.)

A. TCP provides best effort delivery.
B. TCP provides synchronized communication.
C. TCP segments are essentially datagrams.
D. TCP provides sequence numbering of packets.
E. TCP uses broadcast delivery.

Answer: B,D
Explanation:
Because TCP is a connection-oriented protocol responsible for ensuring the transfer of a datagram from the source to destination machine (end-to-end communications), TCP must receive communications messages from the destination machine to acknowledge receipt of the datagram. The term virtual circuit is usually used to refer to the handshaking that goes on between the two end machines, most of which are simple acknowledgment messages (either confirmation of receipt or a failure code) and datagram sequence numbers. Rather than impose a state within the network to support the connection, TCP uses synchronized state between the two endpoints. This synchronized state is set up as part of an initial connection process, so TCP can be regarded as a connection-oriented protocol. Much of the protocol design is intended to ensure that each local state transition is communicated to, and acknowledged by, the remote party.

Reference:
http://en.wikibooks.org/wiki/Communication_Networks/TCP_and_UDP_Protocols

Question No : 2

Which two characteristics apply to Layer 2 switches? 
(Choose two.)

A. Increases the number of collision domains
B. Decreases the number of collision domains
C. Implements VLAN
D. Decreases the number of broadcast domains
E. Uses the IP address to make decisions for forwarding data packets

Answer: A,C
Explanation:
Layer 2 switches offer a number of benefits to hubs, such as the use of VLANs and each switch port is in its own separate collision domain, thus eliminating collisions on the segment.

Valid And Updated 100-105 Exam Certifications Dumps Questions - Dumps4Download.us

Question No : 3

Which option is a valid IPv6 address?

A. 2001:0000:130F::099a::12a
B. 2002:7654:A1AD:61:81AF:CCC1
C. FEC0:ABCD:WXYZ:0067::2A4
D. 2004:1:25A4:886F::1

Answer: D
Explanation:
IPv6 Address Notation
IPv6 addresses are denoted by eight groups of hexadecimal quartets separated by colons
in between them.
Following is an example of a valid IPv6 address:
2001:cdba:0000:0000:0000:0000:3257:9652
Any four-digit group of zeroes within an IPv6 address may be reduced to a single zero or
altogether omitted.
Therefore, the following IPv6 addresses are similar and equally valid:
2001:cdba:0000:0000:0000:0000:3257:9652
2001:cdba:0:0:0:0:3257:9652
2001:cdba::3257:9652
Reference: http://www.ipv6.com/articles/general/IPv6-Addressing.htm

Question No : 4

To allow or prevent load balancing to network 172.16.3.0/24, which of the following commands could be used in R2? 
(Choose two.)

A. R2(config-if)#clock rate

B. R2(config-if)#bandwidth

C. R2(config-if)#ip ospf cost

D. R2(config-if)#ip ospf priority

E. R2(config-router)#distance ospf

Answer: B,C

Explanation: 

http://www.cisco.com/en/US/tech/tk365/technologies_white_paper09186a0080094e9e.shtml#t6

The cost (also called metric) of an interface in OSPF is an indication of the overhead required to send packets across a certain interface. The cost of an interface is inversely proportional to the bandwidth of that interface. A higher bandwidth indicates a lower cost. There is more overhead (higher cost) and time delays involved in crossing a 56k serial line than crossing a 10M Ethernet line. The formula used to calculate the cost is:

Cost = 10000 0000/bandwidth in bps

For example, it will cost 10 EXP8/10 EXP7 = 10 to cross a 10M Ethernet line and will cost 10 EXP8/1544000 =64 to cross a T1 line.By default, the cost of an interface is calculated based on the bandwidth; you can force the cost of an interface with the ip ospf cost <value> interface subconfiguration mode command.

Get 2018 Latest 100-105 Exam Dumps Questions - 100-105 Exam Dumps - Dumps4Download.us

Question No : 5

What is the subnet address for the IP address 172.19.20.23/28?

A. 172.19.20.0

B. 172.19.20.15

C. 172.19.20.16

D. 172.19.20.20

E. 172.19.20.32

Answer: C

Explanation:

From the /28 we can get the following:

Increment: 16 (/28 = 11111111.11111111.11111111.11110000)

Network address: 172.19.20.16 (because 16 < 23)

Broadcast address: 172.16.20.31 (because 31 = 16 + 16 – 1)

Question No : 6

After the network has converged, what type of messaging, if any, occurs between R3 and R4?

A. No messages are exchanged

B. Hellos are sent every 10 seconds.

C. The full database from each router is sent every 30 seconds.

D. The routing table from each router is sent every 60 seconds.

Answer: B

Explanation:

HELLO messages are used to maintain adjacent neighbors so even when the network is converged, hellos are still exchanged. On broadcast and point-to-point links, the default is 10 seconds, on NBMA the default is 30 seconds. Although OSPF is a link-state protocol the full database from each router is sent every 30

minutes (not seconds) therefore, C and D are not correct.

100-105 Exam Preparation Tips - Pass In 24 Hours - Dumps4Download.us

Question No : 7

Refer to the exhibit.

The junior network support staff provided the diagram as a recommended configuration for the first phase of a four-phase network expansion project. The entire network expansion will have over 1000 users on 14 network segments and has been allocated this IP address space.

192.168.1.1 through 192.168.5.255

192.168.100.1 through 192.168.100.255

What are three problems with this design? (Choose three.)

A. The AREA 1 IP address space is inadequate for the number of users.

B. The AREA 3 IP address space is inadequate for the number of users.

C. AREA 2 could use a mask of /25 to conserve IP address space.

D. The network address space that is provided requires a single network-wide mask.

E. The router-to-router connection is wasting address space.

F. The broadcast domain in AREA 1 is too large for IP to function.

Answer: A,C,E

Explanation:

The given IP addresses of areas 1 and 3 along with network masks of /24 cannot accommodate 500 users so are inadequate, while the area 2 is having over capacity so its network mask can be reduced to /25 to accommodate the only 60 users it has.

Question No : 8

Refer to the exhibit.

 Which two statements are true of the interface configuration? (Choose two)

A. The encapsulation in use on this interface is PPP.

B. The default serial line encapsulation is in use on this interface.

C. The address mask of this interface is 255.255.255.0.

D. This interface is connected to a LAN.

E. The interface is not ready to forward packets.

Answer: A,C

Download Verified 100-105 Exam Certifications Questions - Dumps4Download.us

Question No : 9

Which command displays the number of times that an individual router translated an inside address to an outside address?

A. show ip protocol 0

B. show ip nat translation

C. show counters

D. show ip route

E. show ip nat statistics

Answer: D

Question No : 10

Which OSI layer header contains the address of a destination host that is on another network?

A. application

B. session

C. transport

D. network

E. data link

F. physical

Answer: D

Explanation:

 Only network address contains this information. To transmit the packets the sender uses network address and datalink address. But the layer 2 address represents just the address of the next hop device on the way to the sender. It is changed on each hop. Network address remains the same.

Get 2018 Latest 100-105 Exam Dumps Questions - 100-105 Exam Dumps - Dumps4Download.us