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Cisco Certified Network Associate 200-301 CCNA Exam Questions and Answers – Page 6

The latest Cisco Certified Network Associate 200-301 CCNA certification actual real practice exam question and answer (Q&A) dumps are available free, which are helpful for you to pass the Cisco Certified Network Associate 200-301 CCNA exam and earn Cisco Certified Network Associate 200-301 CCNA certification.

Exam Question 551

Which of the following technologies allows a switch port to immediately transition to a forwarding state?

A. Rapid STP
B. PortFast
C. VTP
D. CDP
Correct Answer:
B. PortFast
Answer Description:
PortFast is a technology that allows a switch port connected to an end node such as a workstation, server, or printer to bypass the normal Spanning Tree Protocol (STP) convergence process. When a new device is powered up on a switch port, it will immediately transition to a forwarding state.

NOTE: PortFast should only be used on access ports. It should not be used on trunk ports or on ports that connect to hubs, routers and other switches.

Rapid STP (RSTP) is a new STP standard that provides faster convergence than the original 802.1d STP. RSTP supports PortFast, but it must be configured explicitly.

The VLAN Trunking Protocol (VTP) does not allow for immediate transition to a forwarding state. VTP is used to synchronize VLAN databases between switches, and has no effect on STP.

The Cisco Discovery Protocol (CDP) does not allow for immediate transition to a forwarding state. CDP is used to verify connectivity and document directly connected Cisco devices. CDP is not related to STP.

Exam Question 552

Which command enables HSRP on an interface?

A. hsrp
B. standby ip
C. standby mode hsrp
D. switchport mode hsrp
Correct Answer:
B. standby ip
Answer Description:
The standby ip interface configuration command enables Hot Standby Router Protocol (HSRP). The syntax for this command is as follows: switch(config-if)# standby group-number ip ip-address

The group-number argument specifies the HSRP group number on the interface. You do not need to enter a group number if there is only one HSRP group.

At least one interface on one of the routers in the group must be configured with the virtual IP address of the group. It is optional on all other interfaces on the other routers, which can learn the address through the hellos sent among the group.

A complete HSRP configuration is shown below with an explanation of each command.

RouterA (config) #interface Fa0/1
RouterA (config-if) # ip address 192.168.5.6 255.255.255.0
RouterA (config-if) # standby 2 ip 192.168.5.10
RourerA(config-if) # standby 2 priority 150
RouterA (config-if) #standby 2 Preempt
RouterA(config-if) #standby 2 track interface fa0/2

  • Line 1 specifies the interface
  • Line 2 addresses the interface
  • Line 3 specifies the HSRP group number and the virtual IP address
  • Line 4 sets the HSRP priority
  • Line 5 allows the router to take the active role if its priority becomes higher than that of the active router

In the above, the router is tracking its own Fa0/2 interface. If that interface goes down it will reduce its priority by 10 (this is the default decrement when not specified). The new value would be 140 if that happened. To specify a decrement value, add it to the track command, as in this example: track interface Fa0/2 20.

When you configure routers to be part of an HSRP group, they listen for the HSRP MAC address for that group as well as their own burned-in MAC addresses.

HSRP uses the following MAC address: 0000.0c07.ac** (where ** is the HSRP group number)

The switchport mode interface configuration command will configure the VLAN membership mode of a port. It is not used to enable HSRP.

The options standby mode hsrp and hsrp are not valid commands.

Exam Question 553

Which Cisco IOS command configures the clock rate to 64,000 bits per second on an interface?

A. clock-rate 64000
B. clock rate 64k
C. clock rate 64000
D. clockrate 64000
Correct Answer:
C. clock rate 64000
Answer Description:
The clock rate 64000 command would configure the clock rate to 64,000 bits per second on an interface. The clock rate command is used to configure the clock rate for hardware connections on serial interfaces. These interfaces can be network interface modules (NIMs) and interface processors. The syntax of this command is clock rate bps.

A serial connection between two routers that are connected with a v.35 serial cable requires a clock rate on the Data Communications Equipment (DCE) end of the cable, but not on the Data Terminal Equipment (DTE) end. When the router is connected to a CSU/DSU for connection to the outside world, the DCE end will be the CSU/DSU. In a lab environment or any situation where you have two routers connected with this type of serial cable, a clock rate must be set on the DCE end of the cable.

When troubleshooting a connection of this type between routers, the state of the clock rate (set or unset) can be determined by running the show controllers command on the DCE end. The output will display as follows if the clock rate is NOT set:

Router#show controllers s0/0
Interface Serial0/0
Hardware is PowerQUICC MPC860
DTE V.35 clocks stopped
More output omitted

Notice the DTE V.35 clocks stopped line, which indicates no clock rate is set. Another clue that there is a Layer 2 problem is the output of the show ip interface S0/0 command, executed on the same interface below:

Router# show ip interface s0/0
Serial0/0 is up, line protocol is down
Internet address is 192.168.1.2/24
Broadcast address is 255.255.255.255

Notice the Serial0/0 is up, line protocol is down line. Serial0/0 is up indicates that the physical connection is good, but line protocol is down indicates a problem with Layer 2 . If you were troubleshooting from the bottom layer to the top, you would now check Layer 2, which would be the clock rate.

If you want to change a DCE interface to a DTE device, you should use the no clock rate command.

All the other options are incorrect because these commands are syntactically incorrect.

Exam Question 554

Which of the following commands sets the local router to serve as an authoritative time source?

A. ntp server
B. ntp master
C. ntp authenticate
D. ntp peer
Correct Answer:
B. ntp master
Answer Description:
The ntp master command sets the local router to serve as an authoritative time source.

The ntp server command is used to specify an external time source that the local router should use as its time source.

The ntp authenticate command is used to enable the authentication of time source to which the local router has been configured to use. It is the first step in a process that must also include the specification of a hashing algorithm and a key, both of which must match on the time source.

The ntp peer command is used to configure the local router to synchronize a peer or to be synchronized by a peer. It does not make the local router authoritative as a time source like the ntp master command.

Exam Question 555

A packet is received with a destination IP address of 10.2.16.10. What would the next hop IP address be for this packet?
A packet is received with a destination IP address of 10.2.16.10. What would the next hop IP address be for this packet?

A. 192.168.1.10
B. 192.168.4.2
C. 192.168.10.254
D. None; the packet will be dropped.
Correct Answer:
B. 192.168.4.2
Answer Description:
The packet will be routed to the next hop IP address of 192.168.4.2, since this routing table entry is the most specific match for the remote network. Packets are routed according to the most specific, or “longest,” match in the routing table.

The packet in the scenario has a destination IP address of 10.2.16.10, which matches two entries in the routing table.

10.0.0.0 /8: this matches based on the /8 mask, where only the first byte has to match. The destination IP address of 10.2.16.10 has a first byte matching 10. If this were the only matching route table entry, it would be selected.

10.2.16.0 /24: The first 24 bits of this entry match the first 24 bits of the destination IP address of 10.2.16.10.

Therefore, the 10.2.16.0 /24 entry is selected for routing this packet because it most specifically matches the destination IP address, or has the longest number of matching bits.

The next hops of 192.168.1.10 and 192.168.10.254 will not be used, as these routes are not the most specific matches for the destination IP address of the packet.

It is interesting to note that packets that are destined for the 10.2.32.0 network will be load balanced across both serial 0/0 and serial 0/1 because the cost (2172425) is the same for both paths.

The packet will not be dropped because there is at least one routing table entry that matches the destination IP address of the packet.

To ensure that no packets are dropped, even if there is no matching route in the routing table, a default route could be configured as follows (next hop picked at random for illustration): Router(config)# ip route 0.0.0.0 0.0.0.0 192.168.1.1

This configuration would instruct the router to send any packets that do match the existing routes to 192.168.1.1. For example, a packet destined for 201.50.6.8/24 would not match any routes in the table, and would thus be forwarded to 192.168.1.1.

If you understand how routing tables and routing advertisements work, it is relatively simple to describe the contents of a router’s routing table without seeing the table directly. To do so, you would view the router’s configuration and the configuration of its neighbors using show run, along with a diagram of its network connections. For example, examine the diagram of the two routers shown below along with their respective configurations:
For example, examine the diagram of the two routers shown below along with their respective configurations.

It will contain S*0.0.0.0/0 [1/0] via 192.35.87.5 because of the static default route indicated in line 4 of its configuration output.

It will contain R 192.168.110.128/26 [120/1] via 192.35.87.5 00:00:22, Serial 0/0 because Router 2 has a network 192.168.110.128 statement indicating that it will advertise this network to its neighbors.

It will contain the two routes C 192.35.87.4/30 is directly connected, S0/0 and C 192.168.54.64/26 is directly connected, Fa0/0 because all directly connected routes are automatically placed in the table.

Exam Question 556

Which of the following statements are true when discussing link state and distance vector routing protocols? (Choose all that apply.)

A. After convergence, routing advertisements are only triggered by changes in the network with distance vector protocols
B. Packets are routed based upon the shortest path calculated by an algorithm with link state protocols
C. Only one router in an OSPF area can represent the entire topology of the network
D. Distance vector protocols send the entire routing table to a neighbor
E. Distance vector protocols send updates regarding the status of their own links to all routers in the network
F. Link-state protocols place a high demand on router resources running the link-state algorithm
G. Distance vector protocols require a hierarchical IP addressing scheme for optimal functionality
H. Link-state protocols use hello packets and LSAs from other routers to build and maintain the topological database
I. Link-state protocols require a hierarchical IP addressing scheme for optimal functionality.
Correct Answer:
B. Packets are routed based upon the shortest path calculated by an algorithm with link state protocols
D. Distance vector protocols send the entire routing table to a neighbor
F. Link-state protocols place a high demand on router resources running the link-state algorithm
H. Link-state protocols use hello packets and LSAs from other routers to build and maintain the topological database
I. Link-state protocols require a hierarchical IP addressing scheme for optimal functionality.
Answer Description:
The following statements are true of link-state and distance vector routing protocols:

  • Packets are routed based upon the shortest path calculated by an algorithm with link state protocols.
  • Distance vector protocols send the entire routing table to a neighbor.
  • Link-state protocols place a high demand on router resources running the link-state algorithm.
  • Link-state protocols use hello packets and LSAs from other routers to build and maintain the topological database.
  • Link-state protocols require a hierarchical IP addressing scheme for optimal functionality.

Link state protocols like OSPF use the Shortest Path First algorithm to calculate the shortest path based on a metric called cost, while distance vector protocols like RIP consider only hop count when determining the best route. Running the algorithm places a high demand on router resources. Distance vector protocols are required to send the entire routing table with each update, while link state protocols only send updates when required by changes in the network. Therefore, less traffic is created with link state protocols.

Sending routing advertisements after convergence only when changes occur in the network is a characteristic of link state protocol’s not distance vector protocols. With distance vector protocols, updates occur regularly and include the entire routing table.

All routers in an OSPF area can represent the entire topology of the network, not just one.

Distance vector protocols do not send updates regarding the status of their own links to all routers in the network. Updating link status is a characteristic of link state protocols. Distance vector protocols send the entire routing table.

Distance vector protocols do NOT require a hierarchical IP addressing scheme for optimal functionality. Link-state protocols do require this for optimal functionality, as it supports more efficient route aggregation or summarization. This reduces the number of routes in the table and the number of calculations required by the SPF algorithm, thereby lowering router resource demand.

Exam Question 557

Which two are NOT features of Cisco NAT implementation? (Choose two.)

A. overload
B. override
C. overrule
D. static NAT
E. dynamic NAT
Correct Answer:
B. override
C. overrule
Answer Description:
Override and overrule are NOT features of Cisco’s Network Address Translation (NAT) implementation. NAT translates internal IP address to external IP address and vice versa. NAT is typically used by firewalls or routers.

The following are some of the characteristics of NAT:

  • It can act as an address translator between Internet and the local network.
  • It conserves IP addresses and simplifies the process of IP address allocation.
  • It allows the local network to connect to Internet using unregistered IP addresses.
  • It can present only one address for the entire network to the outside world when using dynamic NAT.
  • It enhances network security, as it does not disclose internal network addresses to the outside world.

All of the other options are incorrect because they are valid NAT features.

With static NAT, translation mappings are created statically and are placed in the translation tables whether or not there is traffic flowing. In this case, no registered addresses are saved because a registered address is still required for each mapping.

With dynamic NAT, the translation table is populated as the required traffic flows through NAT-enabled devices. In this case, a single address or multiple public addresses can be used multiple times to represent multiple private addresses.

The overload keyword allows the ip nat inside command to translate multiple devices in the internal network to the single address in the IP address pool. This process is also called overloading in that the same public IP address is mapped to all private addresses from inside the network. Since the router performing the NAT overload function will use the unique TCP source port from each host for identification, while mapping all of them to the same public IP address, it is sometimes referred to as Port Address Translation or PAT.

For example: ip nat pool test 172.28.15.1 172.28.15.1 prefix 24
In this example, the NAT pool named “test” only has a range of one address.

Another variant of this command is given below, which configures NAT to overload on the address assigned to the serial 0 interface: ip nat inside source list 3 interface serial 0 overload
When this variation is used, the command uses a list named 3 to determine the addresses in the pool.

Exam Question 558

Which classful protocols perform an automatic summarization of routes when routers send updates across major classful network boundaries? (Choose two.)

A. RIPv1
B. RIPv2
C. IGRP
D. OSPF
E. EIGRP
F. BGPv4
Correct Answer:
A. RIPv1
C. IGRP
Answer Description:
The classful routing protocols Routing Information Protocol version1 (RIPv1) and Interior Gateway Routing Protocol (IGRP) summarize routes at classful network boundaries. RIPv1 is a standard distance vector protocol that uses hop count as a metric. IGRP is a Cisco Systems proprietary distance vector routing protocol that has a composite metric based on bandwidth, delay, load, reliability, and maximum transmission unit (MTU).

In classless routing protocols RIPv2, Open Shortest Path First (OSPF), Enhanced IGRP (EIGRP) and Border Gateway Protocol version 4 (BGPv4), route summarization can be controlled manually at any bit position in the IP address. Classless routing protocols transmit subnet mask along with the routes, and therefore manual summarization may be required at times to keep the routing table size in control.

It should be noted that RIPv2 and EIGRP, although classless protocols, will perform automatic summarization by default unless the no auto-summary command is configured. Once no auto-summary is configured, you can manually configure summarization on any bit position in the IP address. Since you can override auto-summarization in both RIPv2 and EIGRP, RIPv1 and IGRP are better answers to this question.

Exam Question 559

You have configured a router as shown in the following output:
You have configured a router as shown in the following output.
Hosts on the LAN cannot receive an IP address. What is wrong?

A. The IP address on the serial interface is incorrect.
B. The default-router command in the DHCP pool is incorrect.
C. An IP address needs to be configured on the FastEthernet interface.
D. The NAT pool is not large enough.
Correct Answer:
C. An IP address needs to be configured on the FastEthernet interface.
Answer Description:
An IP address needs to be configured on the FastEthernet interface. Dynamic Host Control Protocol (DHCP) is used to dynamically provide IP network configurations to workstations as they are booted up. DHCP minimizes network administration overload, allowing devices to be added to the network with little or no manual configuration.

The router configuration in the scenario has created a DHCP address pool called POOLNAME. The network statement in the exhibit, network 10.2.10.0 255.255.255.0, identifies the range of IP addresses that the pool will provide to host systems (10.2.10.0 /24). However, a DHCP pool can only provide IP addresses over a subnet to which it is directly connected. Because neither of the interfaces in the exhibit has an IP address on the 10.2.10.0 /24 subnet, the solution is to assign the FastEthernet0/0 interface the IP address specified in the default-router statement, 10.2.10.254 /24.

The IP address on the serial interface has no impact on the DHCP pool.

The default-router statement is correctly providing the IP address that DHCP hosts will use as their default gateway. The problem is not with the default-router statement, but with the lack of a correct IP address assigned to the FastEthernet0/0 interface.

The NAT configuration in the exhibit has no impact on the DHCP pool. If the NAT pool were not large enough, the result would be that some of the hosts would be able to get to the Internet and others would not. For example, the output from the diagram shown below indicates that there are fourteen addresses in the pool (205.2.1.1 to 205.2.1.14). If the network contained 30 computers, only fourteen would be able to use the Internet at the same time because of the number of public addresses in the pool:

ip nat pool NATPOOL 205.2.1.1 205.2.1.14 netmask 255.255.255.240
ip nat inside source list 1 pool NATPOOL

Exam Question 560

Which of the following is NOT a dynamic table maintained by a router running the EIGRP routing protocol?

A. topology table
B. CAM table
C. routing table
D. neighbor table
Correct Answer:
B. CAM table
Answer Description:
All are tables maintained by a router running the EIGRP routing protocol except a Content Addressable Memory (CAM) table. This table is only present on a switch. It is used to maintain the two MAC addresses involved in a conversation between computers so that the conversation can be routed once and then switched thereafter which is a much faster process.

EIGRP maintains three dynamic tables in RAM:

  • Neighbor table, which is a list of all neighboring EIGRP routers on shared subnets
  • Topology table, which contains all discovered network paths in the internetwork
  • Routing table, which contains the best path (based on lowest metric) to each destination