Examcollection offers free demo for 100 105 icnd1 pdf exam. "Cisco Interconnecting Cisco Networking Devices Part 1 (ICND1 v3.0)", also known as icnd1 100 105 exam, is a Cisco Certification. This set of posts, Passing the Cisco ccent 100 105 exam, will help you answer those questions. The cisco 100 105 pdf Questions & Answers covers all the knowledge points of the real exam. 100% real Cisco cisco 100 105 pdf exams and revised by experts!
Q121. - (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.
Q122. - (Topic 5)
A receiving host has failed to receive all of the segments that it should acknowledge. What can the host do to improve the reliability of this communication session?
A. decrease the window size
B. use a different source port for the session
C. decrease the sequence number
D. obtain a new IP address from the DHCP server
E. start a new session using UDP
Answer: A Explanation:
The Window bit in the header determines the number of segments that can be sent at a time. This is done to avoid overwhelming the destination. At the start of the session the window in small but it increases over time. The destination host can also decrease the window to slow down the flow. Hence the window is called the sliding window. When the source has sent the number of segments allowed by the window, it cannot send any further segments till an acknowledgement is received from the destination. On networks with high error rates or issues, decreasing the window size can result in more reliable transmission, as the receiver will need to acknowledge fewer segments. With a large window size, the sender will need to resend all the frames if a single one is not received by the receiver.
Q123. - (Topic 1)
Refer to the exhibit.
Which option describes how SwitchA will handle the frame just received?
A. It will drop the frame.
B. It will forward the frame out of port Fa0/3 only.
C. It will flood the frame out all ports.
D. It will flood the frame out of all the ports except Fa0/3.
Answer: D
Q124. - (Topic 3)
Refer to the exhibit.
The internetwork is using subnets of the address 192.168.1.0 with a subnet mask of
255.255.255.224. The routing protocol in use is RIP version 1. Which address could be assigned to the FastEthernet interface on RouterA?
A. 192.168.1.31
B. 192.168.1.64
C. 192.168.1.127
D. 192.168.1.190
E. 192.168.1.192
Answer: D
Explanation:
Subnet mask 255.255.255.224 with CIDR of /27 which results in 32 hosts per.
192.168.1.31 is the broadcast address for sunbet '0'
192.168.1.64 is the network address for subnet '2'
192.168.1.127 is the broadcast address for subnet '3'
192.168.1.192 is the network address for subnet '6'
Q125. - (Topic 3)
What is the default administrative distance of the OSPF routing protocol?
A. 90
B. 100
C. 110
D. 120
E. 130
F. 170
Answer: C
Explanation:
Default Distance Value Table This table lists the administrative distance default values of the protocols that Cisco supports: If the administrative distance is 255, the router does not believe the source of that route and does not install the route in the routing table.
Q126. - (Topic 3)
Which parameter or parameters are used to calculate OSPF cost in Cisco routers?
A. Bandwidth
B. Bandwidth and Delay
C. Bandwidth, Delay, and MTU
D. Bandwidth, MTU, Reliability, Delay, and Load
Answer: A
Explanation:
The well-known formula to calculate OSPF cost is Cost = 108/ Bandwidth
Q127. - (Topic 1)
Which statements are true regarding ICMP packets? (Choose two.)
A. They acknowledge receipt of TCP segments.
B. They guarantee datagram delivery.
C. TRACERT uses ICMP packets.
D. They are encapsulated within IP datagrams.
E. They are encapsulated within UDP datagrams.
Answer: C,D
Explanation:
Ping may be used to find out whether the local machines are connected to the network or whether a remote site is reachable. This tool is a common network tool for determining the network connectivity, which uses ICMP protocol instead of TCP/IP and UDP/IP. This protocol is usually associated with the network management tools, which provide network information to network administrators, such as ping and traceroute (the later also uses the UDP/IP protocol). ICMP is quite different from the TCP/IP and UDP/IP protocols. No source and destination ports are included in its packets. Therefore, usual packet-filtering rules for TCP/IP and UDP/IP are not applicable. Fortunately, a special "signature" known as the packet’s Message type is included for denoting the purposes of the ICMP packet. Most commonly used message types are namely, 0, 3, 4, 5, 8, 11, and 12 which represent echo reply, destination unreachable, source quench, redirect, echo request, time exceeded, and parameter problem respectively. In the ping service, after receiving the ICMP "echo request" packet from the source location, the destination
Q128. - (Topic 5)
Refer to the exhibit.
A problem with network connectivity has been observed. It is suspected that the cable connected to switch port Fa0/9 on Switch1 is disconnected. What would be an effect of this cable being disconnected?
A. Host B would not be able to access the server in VLAN9 until the cable is reconnected.
B. Communication between VLAN3 and the other VLANs would be disabled.
C. The transfer of files from Host B to the server in VLAN9 would be significantly slower.
D. For less than a minute, Host B would not be able to access the server in VLAN9. Then normal network function would resume.
Answer: D
Explanation:
Because Switch1 has multiple redundant links in this network, traffic would not work for less than a minute, and then it would get rerouted along the longer path to the host. The 1 minute outage would be the length of time it takes STP to converge.
Q129. - (Topic 3)
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: A,C,D
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)
->
Q130. - (Topic 5)
An administrator has connected devices to a switch and, for security reasons, wants the dynamically learned MAC addresses from the address table added to the running configuration.
What must be done to accomplish this?
A. Enable port security and use the keyword sticky.
B. Set the switchport mode to trunk and save the running configuration.
C. Use the switchport protected command to have the MAC addresses added to the configuration.
D. Use the no switchport port-security command to allow MAC addresses to be added to the configuration.
Answer: A
Explanation:
http://www.cisco.com/en/US/docs/switches/lan/catalyst6500/ios/12.2SX/configuration/guide
/port_sec.pdf
One can configure MAC addresses to be sticky. These can be dynamically learned or manually configured, stored in the address table, and added to the running configuration. If these addresses are saved in the configuration file, the interface does not need to dynamically relearn them when the switch restarts, hence enabling security as desired.
Q131. - (Topic 7)
What is the default lease time for a DHCP binding?
A. 24 hours
B. 12 hours
C. 48 hours
D. 36 hours
Answer: A
Explanation: By default, each IP address assigned by a DHCP Server comes with a one-day lease, which is the amount of time that the address is valid. To change the lease value for an IP address, use the following command in DHCP pool configuration mode:
Q132. - (Topic 3)
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.
Q133. - (Topic 7)
Configuration of which option is required on a Cisco switch for the Cisco IP phone to work?
A. PortFast on the interface
B. the interface as an access port to allow the voice VLAN ID
C. a voice VLAN ID in interface and global configuration mode
D. Cisco Discovery Protocol in global configuration mode
Answer: B
Explanation:
Configure the Switch Port to Carry Both Voice and Data TrafficWhen you connect an IP phone to a switch using a trunk link, it can cause high CPU utilization in the switches. As all the VLANs for a particular interface are trunked to the phone, it increases the number of STP instances the switch has to manage. This increases the CPU utilization. Trunking also causes unnecessary broadcast / multicast / unknown unicast traffic to hit the phone link. In order to avoid this, remove the trunk configuration and keep the voice and access VLAN configured along with Quality of Service (QoS). Technically, it is still a trunk, but it is called a Multi-VLAN Access Port (MVAP). Because voice and data traffic can travel through the same port, you should specify a different VLAN for each type of traffic. You can configure a switch port to forward voice and data traffic on different VLANs. Configure IP phone ports with a voice VLAN configuration. This configuration creates a pseudo trunk, but does not require you to manually prune the unnecessary VLANs. The voice VLAN feature enables access ports to carry IP voice traffic from an IP phone. The voice VLAN feature is disabled by default. The Port Fast feature is automatically enabled when voice VLAN is configured. When you disable voice VLAN, the Port Fast feature is not automatically disabled.
Q134. - (Topic 5)
Refer to the exhibit.
After configuring two interfaces on the HQ router, the network administrator notices an error message. What must be done to fix this error?
A. The serial interface must be configured first.
B. The serial interface must use the address 192.168.1.2
C. The subnet mask of the serial interface should be changed to 255.255.255.0
D. The subnet mask of the FastEthernet interface should be changed to 255.255.255.240
E. The address of the FastEthernet interface should be changed to 192.168.1.66
Answer: D
Explanation:
The IP address 192.168.1.17 255.255.255.0 specifies that the address is part of the 192.168.1.0/24 subnet
24 mask bits = 255.255.255.0 28 mask bits = 255.255.255.240 192.168.1.0/24 subnet has a host range of 192.168.1.1 to 192.168.1.254 (0 being network and 255 being broadcoast) 192.168.1.17/28 subnet has a host range of 192.168.1.17 to 192.168.1.30 (16 being network and 31 being broadcast) 192.168.1.65/28 subnet has a host range of 192.168.1.65 - 192.168.1.78 (64 being network and 79 being broadcast)
if fa0/0 was left as /24, you can see that the host range includes the host range of 192.168.1.64/28 which conflicts. Simply speaking, you can't overlap the subnets. By changing the subnet mask of fa0/0 to 255.255.255.240, these networks would no longer overlap.
Q135. - (Topic 3)
An administrator is working with the 192.168.4.0 network, which has been subnetted with a /26 mask. Which two addresses can be assigned to hosts within the same subnet? (Choose two.)
A. 192.168.4.61
B. 192.168.4.63
C. 192.168.4.67
D. 192.168.4.125
E. 192.168.4.128
F. 192.168.4.132
Answer: C,D
Explanation:
Increment: 64 (/26 = 11111111.11111111.11111111.11000000)
The IP 192.168.4.0 belongs to class C. The default subnet mask of class C is /24 and it has
been subnetted with a /26 mask so we have 2(26-24).= 22.= 4 sub-networks:
1st subnet: 192.168.4.0 (to 192.168.4.63)
2nd subnet: 192.168.4.64 (to 192.168.4.127)
3rd subnet: 192.168.4.128 (to 192.168.4.191)
4th subnet: 192.168.4.192 (to 192.168.4.225)
In all the answers above, only answer C and D are in the same subnet.
Therefore only IPs in this range can be assigned to hosts.