Cisco Certified Network Associate 200-301 CCNA Exam Questions and Answers – Page 2

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 171

Which of the following are Wide Area Network (WAN) protocols? (Choose three.)
A. PPP
B. AAA
C. WEP
D. STP
E. HDLC
F. Frame Relay

Correct Answer:
A. PPP
E. HDLC
F. Frame Relay
Answer Description:
Point-to-Point Protocol (PPP), High-Level Data Link Control (HDLC), and Frame Relay are WAN protocols.

PPP is a WAN protocol is defined in Request for Comments (RFCs) 1332, 1661, and 2153. PPP works with asynchronous and synchronous serial interfaces as well as High-Speed Serial Interfaces (HSSI) and Integrated Services Digital Network (ISDN) interfaces (BRI and PRI). Some of the characteristics of PPP are:

  • Can be used over analog circuits
  • Can encapsulate several routed protocols, such as TCP/IP
  • Provides error correction
  • Should be used rather than HDLC when non-Cisco routers are involved, as it is implemented consistently among vendors
  • PPP authentication can be used between the routers to prevent unauthorized callers from establishing an ISDN circuit

To change the encapsulation from the default of HDLC to PPP when connecting to a non-Cisco router, such as a Juniper, you would use the following command:

router(config)#interface serial S0
router(config-if)#encapsulation ppp

HDLC is a WAN protocol used with synchronous and asynchronous connections. It defines the frame type and interaction between two devices at the Data Link layer.

Frame Relay is a group of WAN protocols, including those from International Telecommunication Union (ITU-T) and American National Standards Institute (ANSI). Frame Relay defines interaction between the Frame Relay customer premises equipment (CPE) and the Frame Relay carrier switch. The connection across the carrier’s network is not defined by the Frame Relay standards. Most carriers, however, use Asynchronous Transfer Mode (ATM) as a transport to move Frame Relay frames between different sites.

Authentication, Authorization, and Accounting (AAA) is incorrect because this is a scheme to monitor access control and activities on networked devices.

Wired Equivalent Privacy (WEP) is a security scheme for wireless networks and therefore it is incorrect.

Spanning Tree Protocol (STP) is for loop avoidance in redundant topologies. This option is incorrect because this protocol is used on Local Area Network (LAN).

Objective: WAN Technologies
Sub-Objective: Describe WAN access connectivity options

Exam Question 172

Which statement is supported by the following output?
router# show ip protocols
Routing Protocol is "eigrp 3"
Sending updates every 90 seconds, next due in 24 seconds
<Some output omitted>
EIGRP metric weight K1=1, K2=0, K3=1, K4=0, K5=0
EIGRP maximum hopcount 100
EIGRP maximum metric variance 1
Redistributing: eigrp 3
Automatic network summarization is not in effect
Maximum path: 4
Routing for Networks:
172.160.72.0
192.168.14.0
<Output omitted>

A. EIGRP supports load-balancing over three equal-cost paths
B. EIGRP supports load-balancing over three unequal-cost paths
C. EIGRP supports load-balancing over four equal-cost paths
D. EIGRP supports load-balancing over four unequal-cost paths

Correct Answer:
C. EIGRP supports load-balancing over four equal-cost paths
Answer Description:
The Maximum path: 4 output indicates that Enhanced Interior Gateway Routing Protocol (EIGRP) will support round-robin load-balancing over four equal-cost paths. This is a default setting, and is a true statement for most routing protocols (including RIP, OSPF and IS-IS). Equal-cost paths are different routes to the same destination network with identical metrics, as determined by the routing protocol. Most routing protocols allow this maximum to be raised up to 16 with the maximum-paths command.

EIGRP has the additional benefit of allowing unequal cost load-balancing. With unequal cost loadbalancing, the router can be configured to include less desirable (higher-metric) paths in the routing table. The router will then send a balanced percentage of traffic over both the best route and the less desirable paths, such as sending two packets over the best path plus one over a less desirable path. EIGRP will never perform unequal-cost load-balancing by default; it must be configured with a variance command. Therefore, you cannot state that EIGRP supports load-balancing over unequal-cost paths in this example.

You cannot state that EIGRP will support load-balancing over three paths because the output displays the Maximum path: 4 value.

Objective: Routing Fundamentals
Sub-Objective: Configure, verify, and troubleshoot EIGRP for IPv4 (excluding authentication, filtering, manual summarization, redistribution, stub)

Exam Question 173

Which type of switching process requires a switch to wait for the entire frame to be received before forwarding it to a destination port?
A. store and forward
B. cut-through
C. fragment free
D. frame-forward

Correct Answer:
A. store and forward
Answer Description:
The store and forward switching process requires a switch to wait until the entire frame is received before forwarding it to a destination port. The store and forward method increases latency as it buffers the entire frame and runs a Frame Check Sequence (FCS) before forwarding it to destination port. However, it ensures error-free frame forwarding because its filters all frame errors.

The cut-through switching process does NOT require a switch to verify the FCS in a frame before forwarding it to the destination port. This type of internal switching method is faster than the store and forward process, but may forward error frames.

The fragment-free switching process only waits to receive the first 64 bytes of the frame before forwarding it the destination port. Fragment-free internal switching assumes that if there is no error in the first 64 bytes of the data, the frame is error free. The assumption is based on the fact that if a frame suffers a collision, it occurs within the first 64 bytes of data. Fragment-free forwarding speed lies between that of store and forward and cut-through.

The term frame-forward is not a valid internal switching process for Cisco switches.

Objective: LAN Switching Fundamentals
Sub-Objective: Describe and verify switching concepts

Exam Question 174

Which type of Dynamic Host Configuration Protocol (DHCP) transmission is used by a host to forward a DHCPDISCOVER packet to locate a DHCP server on the network?
A. unicast
B. broadcast
C. multicast
D. anycast

Correct Answer:
B. broadcast
Answer Description:
Hosts broadcast DHCPDISCOVER messages to locate a DHCP server. The following steps are followed during the allocation of the IP address dynamically using a DHCP server:

  • The client device broadcasts a DHCPDISCOVER message to locate a DHCP server.
  • The DHCP server replies with a DHCPOFFER unicast message with configuration parameters, such as an IP address, a MAC address, a domain name, and a lease for the IP address for the client device.
  • The client returns a DHCPREQUEST broadcast, which is a formal request for the offered IP address to the DHCP server.
  • The DHCP server replies to client device with DHCPACK unicast message, acknowledging the allocation of the IP address to this client device.

Dynamic Host Configuration Protocol (DHCP) is an enhancement over Bootstrap Protocol (BOOTP) and is used to automate the distribution of IP address to clients from a central server. BOOTP protocol was also used to distribute IP addresses, but was inflexible to changes in the network.

DHCP offers the following three advantages that also addressed the inflexibility of the BOOTP protocol:

  • Automatic allocation of permanent IP addresses
  • Automatic allocation of time bound (leased) IP addresses
  • Provision of assigning static IP address or defining a pool of reserved IP address

DHCP does not use multicast messages.

Anycast is a concept of IPv6 protocol and is not valid type used by DHCP.

Objective: Infrastructure Services
Sub-Objective: Configure and verify DHCP on a router (excluding static reservations)

Exam Question 175

Examine the partial output of the show ip interface command below.

Examine the partial output of the show ip interface command below.
Examine the partial output of the show ip interface command below.

What is the subnet broadcast address of the LAN connected to the router from which the command was executed?
A. 192.168.93.15
B. 192.168.93.255
C. 1.1.1.255
D. 1.1.1.127

Correct Answer:
A. 192.168.93.15
Answer Description:
In the output we can see there are two interfaces, a serial interface (which goes to another router) and a GigabitEthernet interface (the LAN interface). The LAN interface has an address of 192.168.93.1/28, which is a mask of 255.255.255.240. When this mask is used against the 192.168.93.0 classful network, it yields the following subnets:

192.168.93.0
192.168.93.16
192.168.93.32
192.168.93.48

and so on, incrementing in intervals of 16 in the last octet.

Since the LAN interface has an address of 192.168.93.1, the interface is in the 192.168.93.0/28 network. That networks broadcast address is the last address before the next subnet address of 192.168.93.16.

Therefore, the broadcast address of the LAN connected to the router from which the command was executed is 192.168.93.15.

The address 192.168.93.255 is not the broadcast address. If a standard 24-bit mask were used instead of the /28, this would be the broadcast address.

The address 1.1.1.255 is the broadcast address of the network in which the Serial interface resides. The question asked for the LAN interface.

The address 1.1.1.127 would be the broadcast address of the network in which the Serial interface resides if the mask used on the interface were 255.255.255.128. However, that is not the mask, and the question asked for the LAN interface.

Objective: Network Fundamentals
Sub-Objective: Configure, verify, and troubleshoot IPv4 addressing and subnetting

Exam Question 176

Which Cisco command will display the version and configuration data for Secure Shell (SSH)?
A. show ssh
B. show ip ssh
C. debug ssh
D. debug ip ssh

Correct Answer:
B. show ip ssh
Answer Description:
The show ip ssh command is used to display the version and configuration data for SSH on a Cisco router. The following is sample output of the show ip ssh command:

router#show ip ssh
SSH Enabled - version 1.5
Authentication timeout: 120 secs; Authentication retries: 2

This show ip ssh command output displays the enabled status of the SSH protocol, the retries parameter (configured at two attempts), and the timeout of 120 seconds.

The following message will appear when the show ip ssh command is issued and SSH has been disabled:

router# show ip ssh
%SSH has not been enabled

To enable SSH include the transport input SSH command when configuring authentication on a line. For example, the configuration of a Cisco network device to use SSH on incoming communications via the virtual terminal ports, with a specified password as shown from the partial output of the show run command is shown below:

line vty 0 4
password 7 030752180500
login
transport input ssh

It is important to note the login command on the third line of the above ouput is critical for security. This command instructs the device to prompt for a username and password using SSH. If this line reads no login, SSH might be otherwise be correctly configured, but the device will never prompt for the username and password.

The show ssh command will display the status of the SSH connections on the router. The following is the sample output of the show ssh command:

The show ssh command will display the status of the SSH connections on the router.
The show ssh command will display the status of the SSH connections on the router.

The debug ip ssh command is used to display debug messages for SSH.

The debug ssh command is not a valid Cisco command.

Objective: Infrastructure Management
Sub-Objective: Use Cisco IOS tools to troubleshoot and resolve problems

Exam Question 177

You are the network administrator for your company. You want to use both IPv6 and IPv4 applications in the network. You also want to ensure that routers can route both IPv6 and IPv4 packets.
Which deployment model should be implemented to accomplish the task?
A. IPv6 over IPv4 tunnels
B. IPv6 over dedicated Wide Area Network (WAN) links
C. Dual-Stack Backbones
D. Protocol translation

Correct Answer:
C. Dual-Stack Backbones
Answer Description:
A dual-stack backbone deployment model should be used to accomplish the task in this scenario. When routers route both IPv6 and IPv4 packets, it is called dual stack routing or a dual-stack backbone.

The following deployment models are available for IPv4 to IPv6 migration:

  • IPv6 over IPv4 tunnels: IPv6 traffic is encapsulated into IPv4 packets. Then these packets are transferred over an IPv4 WAN. This model eliminates the need to create separate circuits to connect to the IPv6 networks. This model increases protocol overhead because of the IPv6 headers and requires one end to be capable of both protocols
  • Protocol translation: A translation method of allowing an IPv6 host to communicate with an IPv4 host. This is accomplished with the help of Network Address Translation – Protocol Translation (NAT-PT) used to configure translation between IPv6 and IPv4 hosts. NAT-PT allows communication between IPv6 hosts and applications, and native IPv4 hosts and applications.
  • IPv6 over dedicated WAN links: A new deployment of IPv6 is created. In this model, IPv6 hierarchy, addressing, and protocols are used by all nodes. However, this model involves cost for creating IPv6 WAN circuits. This solution is not designed for LAN translation but rather translation over WAN links.
  • Dual-Stack Backbones: A hybrid model in which backbone routers have dual-stack functionality, which enables them to route both IPv4 and IPv6 packets. It is suitable for an enterprise that uses both IPv4 and IPv6 applications. Running IPv6 and IPv4 together in a network is known as dual-stack routing.

Objective: Network Fundamentals
Sub-Objective: Identify the appropriate IPv6 addressing scheme to satisfy addressing requirements in a LAN/WAN environment

Exam Question 178

Your assistant has been assigned the task of configuring one end of a WAN link between two offices. The link is a serial connection and the router on the other end is a non-Cisco router. The router in the other office has an IP address of 192.168.8.6/24. The connection will not come up, so you ask your assistant to show you the commands he configured on the Cisco router. The commands he executed are shown below.

Ciscorouter(config)# interface serial0/0
Ciscorouter(config-if)# ip address 192.168.8.5 255.255.255.0
Ciscorouter(config-if)# no shut

What command(s) should he run to correct the configuration?
A. Ciscorouter(config-if)# no ip address 192.168.8.5
Ciscorouter(config-if)# ip address 192.168.8.10
B. Ciscorouter(config-if)# encapsulation ppp
C. Ciscorouter(config-if)# encapsulation ansi
D. Ciscorouter(config-if)# authentication chap

Correct Answer:
B. Ciscorouter(config-if)# encapsulation ppp
Answer Description:
There are three encapsulation types available for a serial connection: High-Level Data Link Control (HDLC), Point-To-Point (PPP), and Frame Relay. HDLC is the default on Cisco routers and the form of HDLC used on a Cisco router is incompatible with routers from other vendors. Since the encapsulation command was not run, the router is set for HDLC. To correct this, you should execute the encapsulation ppp command. Frame Relay could also be used if the other router were running Frame Relay, since it also is an industry standard.

The IP address does not need to be changed. It is currently set for 192.168.8.5/24. This is correct since it is in the same subnet as the IP address of the other end,192.168.8.6/24.

The command authentication chap should not be run because the scenario does not indicate that authentication is configured on the other end. If it is set on one end, it must be set on the other as well.

The command encapsulation ansi should not be run because ANSI is not an encapsulation type. It is an LMI type used in Frame Relay. The three LMI options available are Cisco, ANSI, and ITU.

Objective: WAN Technologies
Sub-Objective: Configure and verify PPP and MLPPP on WAN interfaces using local authentication

Exam Question 179

Which is the valid IP address range that can be assigned to hosts on the subnet that includes the address 172.16.4.6/23?
A. 172.16.2.1 – 172.16.4.254
B. 172.16.3.1 – 172.16.5.254
C. 172.16.4.1 – 172.16.5.254
D. 172.16.4.1 – 172.16.4.254

Correct Answer:
C. 172.16.4.1 – 172.16.5.254
Answer Description:
172.16.4.1 – 172.16.5.254 is the valid IP address range that can be assigned to hosts on the subnet that includes the address 172.16.4.6/23.

To determine the range of addresses that can be assigned in a subnet, you must first determine the network ID and broadcast address of the subnetwork. All addresses that can be assigned to hosts will lie between these two endpoints. The network ID can be obtained by determining the interval between subnet IDs. With a 23-bit mask, the decimal equivalent of the mask will be 255.255.254.0. The interval between subnets can be derived by subtracting the value of the last octet of the mask from 256. In this case that operation would be 256 – 254. Therefore, the interval is 2, and it is applied in the third octet where the subnet mask ends.

The first network ID will always be the classful network you started with (in this case 172.16.0.0). Then each subnetwork ID will fall at 16-bit intervals as follows:
172.16.0.0
172.16.2.0
172.16.4.0
172.16.6.0

At 172.16.6.0 we can stop because the address that we are given in the scenario, 172.16.4.6, is in the network with a subnet ID of 172.16.4.0. Therefore, since the broadcast address for this network will be 1 less than the next subnet ID, or 172.16.5.255, the valid range is 172.16.4.1 – 172.16.5.254.

All the other options are incorrect because these are not valid IP address ranges for this scenario.

Objective: Network Fundamentals
Sub-Objective: Apply troubleshooting methodologies to resolve problems

Exam Question 180

You are working with an Internet Service Provider (ISP) as network manager. A corporate client approaches you to lease a public IP subnet that can accommodate 250 users. You have assigned him the 192.25.27.0 subnet.

What subnet mask should be assigned to this IP address so that it can accommodate the number of users required by the corporate client?
A. 255.255.255.0
B. 255.255.255.128
C. 255.255.255.224
D. 255.255.255.252

Correct Answer:
A. 255.255.255.0
Answer Description:
The 192.25.27.0 subnet should be assigned the subnet mask of 255.255.255.0 to accommodate 250 users. This subnet mask can accommodate a maximum of 254 hosts. The number of hosts that can reside on a subnet can be calculated using the formula 2n – 2 = x, where n is equal to the number of hosts bits in the mask and x is the resulting number of hosts. 2 is subtracted from the results to represent the two address, the network ID and the broadcast address, that cannot be assigned to computers in the subnet. Since the 255.255.255.0 mask leaves 8 bits at the end of the mask, the formula will be 28 – 2, which is 256 – 2, which equals 254.

In situations where the same subnet mask must be used for multiple interfaces on a router, the subnet mask that is chosen must provide capacity sufficient for the largest number of hosts on any single interface while also providing the required number of subnets. For example, in the diagram below, the three interfaces on the router R2 have 16, 32 and 58 users respectively on each interface:

For example, in the diagram below, the three interfaces on the router R2 have 16, 32 and 58 users respectively on each interface.
For example, in the diagram below, the three interfaces on the router R2 have 16, 32 and 58 users respectively on each interface.

If each interface must have the same subnet mask, the subnet mask would need to be one that yields at least 58 addresses to support the interface with the highest host count and yields at least 3 subnets as well.

If the chosen classful networks were 128.107.4.0/24, the correct mask would be 255.255.255.192. Since the mask is currently 255.255.255.0 (/24), by borrowing 2 bits to /26 or 255.255.255.192, we will get 4 subnets (22 = 4) and each subnet will yield 62 hosts (26 – 2 = 62).

With a subnet mask of 255.255.255.128, the 192.25.27.0 subnet can accommodate only 126 hosts. The mask 255.255.255.128 leaves 7 host bits in the mask and when we plug that into the formula we get 27 – 2, which equals 126.

With a subnet mask of 255.255.255.224, the 192.25.27.0 subnet can accommodate only 30 hosts. The mask 255.255.255.224 leaves 5 host bits in the mask and when we plug that into the formula we get 25 – 2, which equals 30.

With a subnet mask of 255.255.255.252, the IP address 192.25.27.24 can accommodate only two hosts. The mask 255.255.255.252 leaves 2 host bits in the mask and when we plug that into the formula we get 22 – 2, which equals 2.

Objective: Network Fundamentals
Sub-Objective: Apply troubleshooting methodologies to resolve problems