Question 1

In the OSI reference model, acknowledgments, sequencing, and flow control are primary functions of which layer?

Accepted Answer

Introduction / Context: The OSI model is widely used to describe how network protocols and devices interact. For exams, you must know which functions map to which layers. This question specifically targets acknowledgments, sequencing, and flow control, which are closely related to reliable end-to-end transport. Given Data / Assumptions: Acknowledgments confirm receipt of data segments between endpoints. Sequencing ensures data segments can be reordered correctly if they arrive out of order. Flow control helps prevent a fast sender from overwhelming a slower receiver. We want the OSI layer that owns these mechanisms in typical protocol stacks. Concept / Approach: In the OSI model, Layer 4 is the Transport layer. Transport layer protocols such as TCP implement reliable, connection-oriented communication that includes acknowledgments (ACKs), sequence numbers, and flow control mechanisms like sliding windows. These features operate end to end between hosts and are independent of the specific paths taken through the network. Lower layers, such as Data Link, may provide limited reliability on single links, but full end-to-end reliability and flow control are responsibilities of the Transport layer. Step-by-Step Solution: Step 1: Consider acknowledgments: TCP sends ACK segments to confirm successful receipt of data. This happens at the Transport layer.Step 2: Consider sequencing: TCP uses sequence numbers and acknowledgment numbers to ensure that segments are reassembled in order and that lost segments can be retransmitted.Step 3: Consider flow control: TCP adjusts its send window based on feedback from the receiver to match the receiver's capacity and buffer space, which is Transport layer logic.Step 4: Although Layer 2 may include simple reliability features such as frame checks and retransmissions on a single link, it does not provide end-to-end transport semantics. Verification / Alternative check: Protocol analyzers like Wireshark display TCP segments with fields such as Sequence Number, Acknowledgment Number, and Window Size. These fields are clearly part of the Transport layer, confirming that acknowledgments, sequencing, and flow control are implemented there. Why Other Options Are Wrong: Option A is incorrect because the Data Link layer focuses on framing, physical addressing, and link-level error detection or correction, not end-to-end transport control.Option B is wrong because the Network layer is concerned with logical addressing and routing, not with segment sequencing or flow control between endpoints.Option D is incorrect because the Application layer provides services and interfaces to applications but relies on lower layers for transport reliability. Common Pitfalls: A common confusion is between link-layer error handling and end-to-end transport reliability. The key is to remember that TCP and similar protocols provide host-to-host reliability at Layer 4, while Layer 2 protects only individual hops. Final Answer: The OSI layer responsible for these functions is Layer 4, the Transport layer.

Question 2

In the OSI model, which layer is responsible for choosing and managing communication partners, coordinating applications, and agreeing on procedures for controlling data integrity and error recovery between them?

Accepted Answer

Introduction / Context: The OSI model breaks network communication into seven conceptual layers. While Application, Presentation, and Session layers are sometimes blurred in practical TCP or IP stacks, exam questions often ask you to map specific responsibilities to the correct OSI layer. This question focuses on managing communication sessions between applications. Given Data / Assumptions: The layer in question chooses and determines the availability of communicating partners. It coordinates partnering applications and agrees on procedures to control data integrity and error recovery at the session level. We are mapping this description to one of the OSI layers, from Layer 4 to Layer 7. Concept / Approach: The Session layer (Layer 5) is responsible for establishing, managing, and terminating sessions between applications. It handles dialog control (who talks when), session checkpoints, and recovery mechanisms that allow a session to resume after an interruption. While modern implementations often merge Session layer functions into higher level libraries or protocols, in the pure OSI model this is the layer that coordinates communication partners and manages long-lived interactions beyond simple request or response transactions. Step-by-Step Solution: Step 1: The Application layer (Layer 7) provides services directly to end user applications such as email, file transfer, and web browsing, but it does not define general session control mechanisms.Step 2: The Presentation layer (Layer 6) deals with data representation, encryption, and compression, not with partner selection or session checkpoints.Step 3: The Session layer (Layer 5) is explicitly defined as the layer that establishes, maintains, and terminates sessions between applications, coordinates dialog, and can insert checkpoints for data recovery.Step 4: The Transport layer (Layer 4) ensures reliable delivery and flow control between hosts, but it does so at a lower level and is not specifically concerned with which application is talking or how long an application level session lasts. Verification / Alternative check: Classic OSI examples of Session layer functions include Remote Procedure Call (RPC) session management and dialog control in interactive applications. These examples align with the description of choosing partners and coordinating application communication, supporting the mapping to Layer 5. Why Other Options Are Wrong: Option A confuses high level services and user interfaces with the underlying session control that they may rely on.Option B misattributes session management to the Presentation layer, which is more concerned with formats and transformations.Option D confuses transport reliability and flow control with higher level session semantics; Transport handles segments and connections but not application dialog policies. Common Pitfalls: Because many real world protocol stacks collapse the Session, Presentation, and Application layers into a single software library, learners often mix up their OSI roles. Remembering that the Session layer focuses on managing and recovering long lasting interactions between applications helps answer questions like this correctly. Final Answer: The layer described in the question is the Session layer (Layer 5).

Question 3

In Ethernet networking, what is the standard bit length of a MAC address and how is it usually written for human readability?

Accepted Answer

Introduction / Context: Media Access Control (MAC) addresses are used at the Data Link layer to uniquely identify network interface cards on Ethernet and other Layer 2 technologies. For Cisco exams, you must know both the bit length and the typical textual representation of a MAC address. Given Data / Assumptions: The question asks about both the bit length and the expression form of a MAC address. We are talking about the standard Ethernet MAC address format defined by IEEE 802. We want the common way MAC addresses are written for humans, not internal binary representation. Concept / Approach: A standard Ethernet MAC address is 48 bits long. Internally, it is just a 48 bit binary value, but for readability it is shown as 12 hexadecimal digits, grouped into six octets. Each octet is written as two hexadecimal digits, and these octets are typically separated by colons, hyphens, or dots depending on vendor style. Example formats include 00:1A:2B:3C:4D:5E or 001A.2B3C.4D5E. Step-by-Step Solution: Step 1: Recall that each hexadecimal digit represents four bits.Step 2: A six octet MAC address has 6 * 8 = 48 bits. This corresponds to 12 hexadecimal digits because 48 / 4 = 12.Step 3: For human readability, the 12 hex digits are grouped into six pairs; each pair corresponds to one octet, often separated by colons (:) or hyphens (-).Step 4: An example is 00:1A:2B:3C:4D:5E, where each pair like 00 or 1A is a value from 00 to FF in hexadecimal. Verification / Alternative check: On real network devices, commands such as show interfaces or ipconfig /all display MAC addresses in one of these standard formats. You will see 48 bit addresses expressed as six hex pairs, confirming the format described in option A. Why Other Options Are Wrong: Option B describes a 32 bit dotted decimal format, which actually matches IPv4 addresses, not MAC addresses.Option C suggests a 64 bit format like IPv6, but this is not how classic MAC addresses are represented, even though they can be related to IPv6 interface identifiers.Option D gives only 16 bits, which is far too short to uniquely identify interfaces at global scale. Common Pitfalls: Some learners confuse MAC address length with IP address length or with IPv6 interface identifiers derived from MAC addresses. Remember that MAC addresses are 48 bits, while IPv4 addresses are 32 bits and IPv6 addresses are 128 bits. Final Answer: The correct description is that a MAC address is 48 bits, written as six pairs of hexadecimal digits separated by colons or hyphens, for example 00:1A:2B:3C:4D:5E.

Question 4

In an Ethernet or IP network, how does a MAC (Layer 2) address differ from an IP (Layer 3) address, and when is each type of address used?

Accepted Answer

Introduction / Context: MAC addresses and IP addresses are both essential identifiers in modern networks, but they operate at different layers and serve different roles. Cisco certification exams frequently ask you to distinguish between these two types of addresses and to explain how they are used together for data delivery. Given Data / Assumptions: A MAC address is associated with a network interface card on a host or router. An IP address is configured either manually or via DHCP on a host or router interface. We are working within the OSI model, where MAC addresses are at Layer 2 and IP addresses are at Layer 3. Concept / Approach: A MAC address is a Data Link layer identifier that is used for local delivery of frames on a single broadcast domain such as an Ethernet LAN. Switches learn MAC addresses and forward frames based on them. IP addresses are Network layer addresses used for logical addressing and routing across multiple networks. Routers ignore MAC addresses beyond the local link; they forward packets based on IP destination addresses and rewrite MAC headers as packets move from one Layer 2 segment to the next. Step-by-Step Solution: Step 1: When one host sends data to another on the same LAN, it uses the destination IP address to decide where to send and uses ARP to resolve that IP to a MAC address.Step 2: The host constructs an Ethernet frame with the destination MAC address and a payload containing an IP packet whose destination IP address matches the remote host.Step 3: Switches in the LAN forward the frame based on the MAC address table, ensuring the frame reaches the correct port.Step 4: When traffic crosses a router to another network, the router examines the destination IP, chooses a next-hop, and rewrites the Layer 2 header, including source and destination MAC addresses appropriate for the outgoing interface.Step 5: Throughout the path, the IP address remains consistent and is used by all routers for end-to-end forwarding, while the MAC addresses change at each Layer 2 hop. Verification / Alternative check: By capturing packets with a tool such as Wireshark, you can see that as a packet traverses multiple hops, the IP source and destination remain constant, but the Ethernet source and destination addresses change at each router. This demonstrates the different scopes and purposes of MAC and IP addresses. Why Other Options Are Wrong: Option B reverses the roles of MAC and IP addresses and is factually incorrect.Option C claims that both operate only at Layer 2 and are interchangeable, which contradicts the OSI model and real network behavior.Option D incorrectly associates MAC addresses only with wireless and IP addresses only with wired networks, which is not true; both are used in wired and wireless environments. Common Pitfalls: Students sometimes assume that devices use IP addresses at every step, but switches actually forward based on MAC addresses. Others think MAC addresses never change, yet routers deliberately change Layer 2 headers at each hop. Keeping the distinction between local frame delivery and end-to-end packet routing clear helps avoid such confusion. Final Answer: The correct explanation is that A MAC address is a physical Layer 2 identifier burned into or assigned to a network interface and is used for local frame delivery on a single segment, while an IP address is a logical Layer 3 address used by routers to deliver packets end to end across multiple networks.

Question 5

When connecting a PC's serial COM port to the console port of a Cisco router or switch using an RJ45 UTP cable, which type of cable should you use?

Accepted Answer

Introduction / Context: Accessing the console port of a Cisco router or switch is a fundamental task for initial configuration and troubleshooting. The console connection typically uses a special RJ45 UTP cable that is wired differently from standard Ethernet patch cables. Knowing which cable to select is a common exam question. Given Data / Assumptions: We are connecting a PC's serial COM port (often via an RJ45 to serial adapter) to the console port on a router or switch. The physical connector on the network device is RJ45, but the signaling is serial, not Ethernet. We have several cable types to choose from: straight-through, crossover, and rolled. Concept / Approach: A straight-through cable is used to connect devices of different types at Ethernet Layer 2, such as a switch to a router or a switch to a host. A crossover cable historically connected similar Ethernet devices, such as switch to switch or host to host. A rolled, or rollover, cable has one end wired as the reverse of the other and is specifically designed for console connections, not for standard Ethernet data traffic. Step-by-Step Solution: Step 1: Recognize that the console interface is a management port that uses serial signaling and does not act as a normal Ethernet port.Step 2: Cisco traditionally supplies a light blue RJ45 rollover cable that is used to connect the device console port to a PC via a serial adapter.Step 3: The rollover cable has pin 1 on one end wired to pin 8 on the other end, pin 2 to pin 7, and so on, effectively reversing the pinout.Step 4: Straight-through or crossover Ethernet cables would not provide the correct pin mapping for the console serial connection. Verification / Alternative check: In practice, when you purchase a Cisco router or switch, it usually comes with a blue rollover console cable. Cisco documentation and labs explicitly refer to this as a rollover cable for console access, confirming that this is the correct choice. Why Other Options Are Wrong: Option A is used for normal Ethernet data connections between dissimilar devices, not for console connectivity.Option B is historically used for Ethernet connections between similar devices, not for serial console connectivity.Option C introduces a CSU or DSU, which relates to WAN connections, not to local console cables. Common Pitfalls: Because all of these cables may have RJ45 connectors, it is easy to confuse them. Remember that console ports use a special rollover cable, typically color coded, and that Ethernet patch cables, whether straight-through or crossover, are reserved for data ports. Final Answer: You should use a Rolled (rollover) console cable to connect a PC COM port to a router or switch console port.

Question 6

Given the binary number 10110111, what are its decimal and hexadecimal equivalents?

Accepted Answer

Introduction / Context: Number system conversions between binary, decimal, and hexadecimal are basic skills for network engineers. Many configuration registers, addresses, and mask calculations involve understanding how to convert values among these representations. This question asks you to convert a specific binary byte into decimal and hexadecimal. Given Data / Assumptions: The binary number is 10110111. We are asked to find both the decimal and hexadecimal equivalents. We assume standard unsigned interpretation of this 8 bit binary value. Concept / Approach: An 8 bit binary number can be converted to decimal by multiplying each bit by its positional value (128, 64, 32, 16, 8, 4, 2, 1) and summing. To convert to hexadecimal, we group the bits into two four bit nibbles, convert each nibble to a hex digit, and combine the result. The leftmost nibble is the high order hex digit and the rightmost nibble is the low order hex digit. Step-by-Step Solution: Step 1: Write the bit positions for 10110111 from left to right: 1 0 1 1 0 1 1 1.Step 2: Assign weights: the leftmost bit represents 128, then 64, 32, 16, 8, 4, 2, and 1.Step 3: Multiply each 1 bit by its weight and sum: 1*128 + 0*64 + 1*32 + 1*16 + 0*8 + 1*4 + 1*2 + 1*1 = 128 + 32 + 16 + 4 + 2 + 1 = 183.Step 4: For hexadecimal, group the bits as 1011 and 0111. The left nibble 1011 is 8 + 0 + 2 + 1 = 11, which corresponds to hex B. The right nibble 0111 is 0 + 4 + 2 + 1 = 7, which is hex 7.Step 5: Combine the two nibbles to get B7 as the hexadecimal representation. Verification / Alternative check: As a quick consistency check, note that hex B7 is equal to 11*16 + 7 = 176 + 7 = 183, which matches the decimal value we calculated from the binary representation. This confirms the correctness of the conversion. Why Other Options Are Wrong: Option A gives 69 and 0x2102, which do not correspond to the given binary pattern.Option C suggests 173 and A6, which would be 0b10101101, not 0b10110111.Option D states 83 and 0xC5, which represent different binary patterns (01010011 and 11000101 respectively). Common Pitfalls: Common mistakes include misreading bit positions, forgetting which nibble is high and which is low when converting to hex, or mixing up decimal and hexadecimal arithmetic. Carefully writing out the positional weights and grouping bits correctly avoids these errors. Final Answer: The binary number 10110111 equals 183 in decimal and B7 in hexadecimal.

Question 7

On a Cisco Catalyst 2950 switch, which combination of actions correctly erases the old configuration so that the device starts with factory defaults?

Accepted Answer

Introduction / Context: When reusing a Cisco Catalyst 2950 or similar switch, you often need to completely remove the previous configuration, including startup configuration and VLAN information. Cisco exams frequently test your knowledge of the correct sequence of commands to return a switch to a factory default state. Given Data / Assumptions: The device is a Catalyst 2950 switch with nonvolatile memory that stores startup configuration and a separate VLAN database. You want to erase the existing configuration so that the switch boots as if it were new. Typical commands include erasing startup configuration, deleting the VLAN database file, and reloading the device. Concept / Approach: On many Cisco switches, the startup configuration is stored in NVRAM and erased using the erase startup-config or write erase command. VLAN information is often stored in a separate file, usually vlan.dat, in flash memory. To completely clear the configuration, both the startup configuration and VLAN database must be removed, and then the switch must be reloaded so that it starts without any saved configuration. Step-by-Step Solution: Step 1: Use erase startup-config or write erase to remove the NVRAM startup configuration file that contains interface settings, passwords, and other parameters.Step 2: Use delete flash:vlan.dat or a similar command to delete the VLAN database file that stores VLAN and VTP information.Step 3: Confirm the deletion of the vlan.dat file when prompted so that the VLAN configuration is truly removed.Step 4: Reload the switch so that it boots without any startup configuration or VLAN database; on reload, it will behave like a factory default device. Verification / Alternative check: After performing these steps and reloading, the switch should prompt for initial configuration dialog, and show vlan should display only the default VLAN 1 without any custom VLANs. The running configuration should be minimal, confirming that the old configuration has been erased. Why Other Options Are Wrong: Option B incorrectly erases flash, which removes IOS images and other files instead of just configuration, and it does not mention the VLAN database explicitly.Option C suggests that simply restarting the switch clears the configuration, which is false; configuration persists across reboots unless erased.Option D combines actions that are either unnecessary or harmful, such as erasing flash and relying on power-off time, and does not properly address VLAN database handling. Common Pitfalls: Technicians sometimes erase only the startup configuration and forget to delete vlan.dat, leaving old VLANs in place. Others mistakenly erase flash and remove the IOS image. Knowing the proper sequence helps avoid downtime and ensures the switch truly starts clean. Final Answer: The correct set of actions is to Delete the VLAN database file, erase the startup configuration, and then reload the switch.

Question 8

If a host connected to switch Sw-AC3 sends an IP packet to destination 190.0.2.5 on a different subnet, what destination MAC address does the host use in the Ethernet frame?

Accepted Answer

Introduction / Context: This question examines how a host encapsulates traffic when sending data to an IP address on a different subnet. It is a key concept in Cisco studies because it ties together IP routing, ARP, and the role of a default gateway in host communication beyond the local network. Given Data / Assumptions: The host is connected to a switch (Sw-AC3) on some VLAN and has an IP configuration including subnet mask and default gateway. The destination IP address 190.0.2.5 is on a different IP subnet than the host. A router interface on the same VLAN acts as the default gateway for the host. Concept / Approach: When a host needs to send an IP packet to a destination outside its local subnet, it does not try to resolve the remote host's MAC address directly. Instead, it forwards the packet to its configured default gateway. ARP is used to resolve the MAC address of the default gateway interface on the local network. The Ethernet frame is then addressed to that MAC, while the inner IP packet retains the final destination IP of 190.0.2.5. Step-by-Step Solution: Step 1: The host compares its own IP address and subnet mask with the destination IP 190.0.2.5 and determines that the destination is on a different subnet.Step 2: Because the destination is remote, the host decides to send the packet to the default gateway router interface.Step 3: If the host does not already know the MAC address of the default gateway, it sends an ARP request asking "Who has the default gateway IP?"Step 4: The router interface responds with its MAC address. The host stores this mapping in its ARP cache.Step 5: The host builds an Ethernet frame with the destination MAC set to the router's interface MAC, source MAC set to its own interface MAC, and the payload containing an IP packet whose destination IP is still 190.0.2.5. Verification / Alternative check: Packet captures on the host LAN show that frames sent toward off-subnet destinations have the router's MAC as the destination. Only once the packet reaches the router does the router forward it toward the next hop, possibly rewriting the MAC header again on each routed hop. Why Other Options Are Wrong: Option A is incorrect because the host does not know the remote host's MAC address; ARP is limited to the local broadcast domain.Option C is wrong because switches forward frames based on MAC addresses but do not act as IP endpoints; hosts do not use the switch's MAC as a destination.Option D is incorrect because broadcast MAC is used only for discovery or special cases, not for regular unicast communication to remote networks. Common Pitfalls: Many beginners mistakenly think the host must somehow know the final recipient's MAC address. In routed networks, hosts only need to know the MAC of their default gateway for off-subnet destinations; routers handle the rest of the path hop by hop. Final Answer: The host uses The MAC address of the default gateway router interface connected to the host's VLAN as the destination MAC in the Ethernet frame.

Question 9

With a /27 subnet mask and subnet zero usable, which set of three IP addresses are all valid host addresses (not network or broadcast) in their respective /27 subnets?

Accepted Answer

Introduction / Context: Subnetting questions often test your ability to recognize valid host addresses versus network and broadcast addresses. A /27 mask means each subnet contains 32 IP addresses, with the first address being the network and the last being the broadcast. This question asks you to identify a set of three addresses that are all valid host addresses within different /27 subnets. Given Data / Assumptions: The subnet mask is /27, which corresponds to 255.255.255.224. Each /27 subnet contains 32 addresses (0 through 31 in the last five bits for the host part). Subnet zero is usable, so the first subnet block is not reserved. We must avoid any address where the host portion is all zeros (network) or all ones (broadcast) for its /27 block. Concept / Approach: For a /27 mask, the increment in the last octet is 32. Subnets begin at .0, .32, .64, .96, .128, .160, .192, and .224. In each block, the network address is the first (.0, .32, and so on) and the broadcast address is the last (.31, .63, .95, and so on). Any value between these two inclusive boundaries is a valid host address. To decide whether a given IP is a host address, compute the last octet modulo 32 and check whether it is neither 0 nor 31. Step-by-Step Solution: Step 1: For 10.15.32.17, the last octet is 17, which lies between 0 and 31 in the 32.x block and is neither 0 nor 31, so it is a valid host in the 10.15.32.0 /27 subnet.Step 2: For 66.55.128.1, the last octet is 1, which lies between 0 and 31 in the 128.x block and is neither 0 nor 31, so it is a valid host in the 66.55.128.0 /27 subnet.Step 3: For 135.1.64.34, compute 34 modulo 32, which is 2. The 64.x block spans 64 to 95; 34 falls in the next 32 wide block 32 to 63, but using the modulo method, 34 mod 32 = 2, so it is not at the boundaries 0 or 31, making it a valid host in the 135.1.64.32 /27 subnet.Step 4: For 17.15.66.128, the last octet 128 is exactly a multiple of 32, so it is a network address, not a host address.Step 5: For 129.33.192.192, the last octet 192 is also a multiple of 32 and is a network address for the 192.x block.Step 6: For 192.168.5.63, the last octet 63 is equal to 32 + 31, which is the broadcast address for the 192.168.5.32 /27 subnet, not a host address. Verification / Alternative check: Listing a few sample /27 ranges confirms this: for example, 192.168.5.32 to 192.168.5.63 has network 192.168.5.32 and broadcast 192.168.5.63, so anything between .33 and .62 is valid. Applying similar checks to each candidate address yields the same conclusion as the modulo method. Why Other Options Are Wrong: Option B includes 17.15.66.128 and 129.33.192.192, both of which are network addresses for their subnets.Option C consists entirely of non host addresses because 17.15.66.128 and 129.33.192.192 are network addresses and 192.168.5.63 is a broadcast address.Option D includes 192.168.5.63, which again is a broadcast address, making the set invalid. Common Pitfalls: Many candidates focus only on the last bits without computing where the subnet blocks start and end, leading them to mistakenly treat .128 or .192 as host addresses when those octets coincide with network boundaries. Always remember that in a /27, increments of 32 in the last octet mark network addresses, and the last address before the next increment is the broadcast. Final Answer: The only set in which all three addresses are valid hosts is 10.15.32.17, 66.55.128.1, and 135.1.64.34.

Question 10

In Ethernet LANs, which statement best summarizes the correct use of straight-through, crossover, and rolled (rollover) UTP cables?

Accepted Answer

Introduction / Context: Understanding different types of Ethernet cabling and their proper applications is a fundamental networking skill. Before auto MDI or MDIX became common, the distinction between straight-through and crossover cables was critical for making correct physical connections. Rolled cables serve an entirely different purpose, primarily for device management. Given Data / Assumptions: We are dealing with RJ45 UTP copper cables in Ethernet networks and for console access. Device types include hosts, switches, routers, and console ports. We assume some older equipment may not support auto MDI or MDIX, so correct cable choice matters. Concept / Approach: In traditional Ethernet wiring, straight-through cables connect devices that use opposite transmit and receive pairs, such as a host network interface card to a switch port or a switch to a router. Crossover cables swap the transmit and receive pairs, making them appropriate for connecting similar devices such as switch to switch, hub to hub, or host to host. Rolled cables reverse all the pins and are used to carry serial console signals between a PC and a network device's console port, not to carry Ethernet frames. Step-by-Step Solution: Step 1: Identify that straight-through cables keep pin 1 to pin 1, pin 2 to pin 2, and so on, making them suitable for host to switch and switch to router connections.Step 2: Recognize that crossover cables swap the transmit and receive pairs, connecting pin 1 on one end to pin 3 on the other and pin 2 to pin 6, which is needed when similar devices are connected without an intermediate switch.Step 3: Understand that rolled cables map pin 1 to pin 8, pin 2 to pin 7, and so forth, producing a complete reversal. This wiring is designed for RS-232 style serial console connections, not for Ethernet data.Step 4: Conclude that each cable type has a distinct role: straight-through for most access links, crossover for back-to-back connections of similar devices (if auto MDI or MDIX is not present), and rollover for console management. Verification / Alternative check: Network lab manuals and Cisco documentation typically include diagrams that show which cable to use for each device pairing. These diagrams consistently match the summary in option A, differentiating data cables from console rollover cables. Why Other Options Are Wrong: Option B incorrectly ties cable type to duplex mode, which is unrelated to the physical wiring of the pairs.Option C mischaracterizes straight-through cables as WAN only and claims rolled cables are obsolete, which is incorrect.Option D is wrong because the internal wiring of straight-through, crossover, and rolled cables differs, and using the wrong type can prevent devices from linking or communicating. Common Pitfalls: Students often confuse crossover and rolled cables because both look like standard patch leads. Remembering that rollover cables are primarily for console access and that crossover cables are for similar device data links helps avoid miswiring in the lab. Final Answer: The best summary is that Straight-through cables connect different device types such as host to switch or switch to router, crossover cables connect similar device types such as switch to switch or host to host, and rolled cables are used only for console connections, not for Ethernet data links.

Question 11

When you use Telnet or FTP to communicate over a TCP/IP network, which OSI layer are you primarily using to generate the application data?

Accepted Answer

Introduction / Context: Telnet and FTP are classic application protocols used for remote login and file transfer. In OSI model questions, you are often asked to identify which layer a given protocol belongs to. Although real protocol stacks combine several upper layers, logically Telnet and FTP reside at the Application layer. Given Data / Assumptions: Telnet provides remote terminal access. FTP provides file transfer between hosts. Both protocols rely on lower layers such as TCP, IP, and the underlying link technologies for transport and delivery. Concept / Approach: The Application layer (Layer 7) in the OSI model is the interface between the network and end user applications. Protocols at this layer define commands, responses, and data formats that applications use to perform specific tasks. Telnet defines commands for remote terminal emulation, while FTP defines commands for listing directories, transferring files, and managing sessions. These protocols generate the user level data that is then wrapped in Transport layer segments and carried over the network. Step-by-Step Solution: Step 1: Identify Telnet and FTP as user facing services: Telnet for interactive command line access and FTP for file transfers.Step 2: Note that both services use TCP as their transport protocol, which is at the Transport layer. They do not operate directly as Transport protocols themselves.Step 3: Recognize that the OSI Application layer is where specific application protocols define their behavior, commands, and responses.Step 4: Therefore, Telnet and FTP are considered Application layer protocols that use the services of the Transport, Network, Data Link, and Physical layers beneath. Verification / Alternative check: Standard protocol mapping tables from Cisco and other vendors list Telnet and FTP under the OSI Application layer. In protocol analyzers, Telnet and FTP headers appear at the top of the stack above TCP and IP, reinforcing their classification as Application layer protocols. Why Other Options Are Wrong: Option B, the Presentation layer, is concerned with data translation, encryption, and compression, not with defining Telnet or FTP command sets.Option C, the Session layer, manages dialog and session control, but Telnet and FTP are more closely associated with application level functions.Option D, the Transport layer, is where TCP resides, but Telnet and FTP sit above TCP and are not themselves transport protocols. Common Pitfalls: Because Telnet and FTP use TCP, some learners mistakenly place them at the Transport layer. Remember that the presence of TCP ports such as 23 for Telnet or 21 for FTP indicates that these protocols are application level services that use TCP, not part of TCP itself. Final Answer: When you use Telnet or FTP, you are primarily using the Application layer to generate the data.

Question 12

In the four-layer DoD (TCP/IP) model, which layer corresponds functionally to the Network layer of the OSI model?

Accepted Answer

Introduction / Context: The Department of Defense (DoD) model, also called the TCP or IP model, is a four-layer framework that predates but aligns roughly with the seven-layer OSI model. Exam questions often ask you to map specific OSI layers to their approximate counterparts in the DoD model. Given Data / Assumptions: The DoD or TCP or IP model layers are Application, Host-to-Host, Internet, and Network Access. The OSI Network layer (Layer 3) is responsible for logical addressing and routing. We need the DoD layer that performs similar functions. Concept / Approach: In the OSI model, the Network layer handles logical addressing (for example, IP addresses), path selection, and routing between networks. In the DoD model, the Internet layer plays the same role, hosting protocols such as IP, ICMP, and routing protocols. The Host-to-Host layer corresponds more closely to the OSI Transport layer, and the Network Access layer combines OSI Data Link and Physical layer functions. Step-by-Step Solution: Step 1: Recall that the DoD Application layer roughly covers OSI layers 5 to 7, dealing with application services.Step 2: Recognize that the DoD Host-to-Host layer is where TCP and UDP operate, providing end-to-end transport, similar to OSI Layer 4.Step 3: The DoD Internet layer is where IP addressing and routing occur, which is exactly what the OSI Network layer does.Step 4: The DoD Network Access layer handles physical media and link layer functions, equivalent to OSI Layers 1 and 2. Verification / Alternative check: Standard mapping diagrams show IP and routing protocols in the Internet layer and describe it as responsible for addressing and routing. This is the same functionality associated with OSI Layer 3, confirming that the Internet layer is the correct match. Why Other Options Are Wrong: Option A, the Application layer, corresponds to OSI layers 5 through 7 rather than the Network layer.Option B, the Host-to-Host layer, aligns with the OSI Transport layer, not the Network layer.Option D, the Network Access layer, aligns with OSI Data Link and Physical layers, not with logical addressing and routing. Common Pitfalls: Some learners confuse the Internet layer with the Application layer because both sound high level. Remember that the key function of the OSI Network layer is routing based on logical addresses, and that function clearly lives in the Internet layer of the DoD model. Final Answer: The DoD layer equivalent to the OSI Network layer is the Internet layer.

Question 13

Which of the following application protocols uses both TCP and UDP as transport protocols, depending on the type of message being sent?

Accepted Answer

Introduction / Context: Most well-known application protocols use either TCP or UDP as their transport layer protocol, but not both. However, some protocols can use both TCP and UDP depending on the specific operation. Cisco exams often test this nuance, particularly for DNS. Given Data / Assumptions: We are comparing common protocols: FTP, SMTP, Telnet, and DNS. We want to know which protocol can operate over both TCP and UDP. We are focusing on standard, widely implemented behavior. Concept / Approach: DNS primarily uses UDP on port 53 for simple, fast queries and responses because these messages are small and do not require the overhead of TCP. However, DNS also uses TCP on port 53 for tasks such as zone transfers and for responses that are too large to fit in a single UDP packet. FTP, SMTP, and Telnet always use TCP, because they require reliable, connection-oriented sessions for data transfer and interactive communication. Step-by-Step Solution: Step 1: Recall that FTP (ports 20 and 21), SMTP (port 25), and Telnet (port 23) are all designed around TCP session semantics and do not use UDP in normal operation.Step 2: Recall that DNS uses UDP port 53 for standard name resolution queries and replies because it is lightweight and efficient.Step 3: Recognize that DNS also uses TCP port 53 when performing zone transfers between servers or when a UDP response would be truncated due to size limitations.Step 4: Therefore, DNS is the protocol in this list that uses both TCP and UDP. Verification / Alternative check: Checking standard RFCs and network documentation confirms that DNS supports both UDP and TCP on port 53, whereas FTP, SMTP, and Telnet are defined strictly over TCP. Packet captures from real networks also show both UDP and TCP flows for DNS traffic. Why Other Options Are Wrong: Option A (FTP) uses TCP for command and data channels and does not rely on UDP.Option B (SMTP) uses TCP for reliable mail transfer between servers.Option C (Telnet) uses TCP for interactive terminal sessions and does not use UDP. Common Pitfalls: Some learners mistakenly think that because FTP and SMTP move large amounts of data, they might use UDP for performance. In fact, reliability requirements dictate TCP usage. DNS is the notable protocol here that can use both transports, depending on the transaction type and message size. Final Answer: The protocol that uses both TCP and UDP is DNS (Domain Name System).

Question 14

Conceptually, how does the four-layer DoD (TCP/IP) model differ from the seven-layer OSI model in terms of layer grouping and typical protocol placement?

Accepted Answer

Introduction / Context: The DoD (TCP/IP) model and the OSI model are both conceptual frameworks used to describe how network protocols interact. Cisco exams expect you to understand both models and how they relate. This question asks you to explain the main conceptual difference between the two. Given Data / Assumptions: The OSI model defines seven layers: Physical, Data Link, Network, Transport, Session, Presentation, and Application. The DoD model defines four layers: Network Access, Internet, Host-to-Host, and Application. Both models can be used to map real protocols such as Ethernet, IP, TCP, HTTP, and others. Concept / Approach: The OSI model is more granular, splitting functionality into seven layers, especially distinguishing between Session, Presentation, and Application at the top and between Physical and Data Link at the bottom. The DoD model reflects practical TCP/IP implementations by grouping several OSI layers together. In the DoD model, the Application layer covers what OSI calls Application, Presentation, and Session. The Host-to-Host layer corresponds roughly to OSI Transport. The Internet layer corresponds to OSI Network. The Network Access layer groups the OSI Data Link and Physical layers into one. Step-by-Step Solution: Step 1: Identify that the DoD Application layer includes application protocols (HTTP, FTP, SMTP), as well as data representation and session management functions that OSI separates into three layers.Step 2: Recognize that the DoD Host-to-Host layer is where TCP and UDP operate, providing transport services similar to OSI Layer 4.Step 3: Note that the DoD Internet layer contains IP and routing protocols and corresponds to the OSI Network layer.Step 4: See that the DoD Network Access layer handles both physical media and data link framing, which OSI divides into the Physical and Data Link layers.Step 5: Conclude that the main difference is the grouping of functions into four broader layers versus seven more detailed layers. Verification / Alternative check: Standard mapping diagrams in textbooks show OSI layers 5 to 7 collapsing into the DoD Application layer and OSI layers 1 and 2 collapsing into the Network Access layer. This visual comparison supports the explanation that the DoD model is a simplified, implementation oriented view of the same stack. Why Other Options Are Wrong: Option B is incorrect because the DoD model has fewer layers than OSI and is widely used in real networks; OSI is largely a teaching model.Option C incorrectly claims that OSI is obsolete; while TCP/IP is dominant, OSI remains valuable as a conceptual reference.Option D is wrong because the models clearly use different names and layer boundaries. Common Pitfalls: Learners sometimes attempt to memorize both models without understanding the mapping, leading to confusion. Focusing on how OSI layers collapse into the more compact DoD model makes it easier to recall and apply both frameworks in exam scenarios. Final Answer: The key difference is that The DoD model has four layers that combine OSI upper layers into a single Application layer and merge the lower layers into a Network Access layer, while the OSI model separates functions into seven distinct layers with more granular roles.

Question 15

An IPv4 host has the address 172.16.45.14 with a /30 subnet mask. To which subnetwork address does this host belong on the 172.16.45.0 network?

Accepted Answer

Introduction / Context: In IPv4 subnetting, a very common task is to determine the network or subnetwork address to which a particular host belongs. This question uses a private Class B range with a Classless Inter Domain Routing (CIDR) mask of /30, which is frequently used for point to point links in Cisco certification exams and real networks. Understanding how to calculate block sizes and network boundaries for a /30 subnet is essential for routing, access control lists, and troubleshooting connectivity problems. Given Data / Assumptions: Host IPv4 address: 172.16.45.14. Prefix length: /30, which corresponds to a subnet mask of 255.255.255.252. We assume standard classless addressing with no special summarisation tricks. We need the network (subnetwork) address that this host falls into on the 172.16.45.0 network. Concept / Approach: To find the subnetwork, we focus on the octet where subnetting occurs. A /30 mask means 30 bits are network bits and 2 bits are host bits. In dotted decimal, /30 is 255.255.255.252. The block size in the affected octet is calculated as 256 minus the mask value in that octet. For 252 this block size is 4. Each subnet therefore spans 4 addresses, with the first being the network address and the last being the broadcast address. The usable host addresses are the two addresses between them. We find the largest multiple of the block size that is less than or equal to the host value in that octet, which gives the network address range. Step-by-Step Solution: 1. Write the subnet mask: /30 = 255.255.255.252. 2. Identify the varying octet. The first three octets 172.16.45 are fixed by the mask. The subnetting happens in the fourth octet. 3. Compute the block size in the fourth octet: 256 - 252 = 4. 4. List the possible network starts in the fourth octet: 0, 4, 8, 12, 16, 20, and so on. 5. Find where 14 fits. It lies between 12 and 16, so the subnet starts at 12. 6. Therefore the subnetwork address is 172.16.45.12. The usable host addresses in this subnet are 172.16.45.13 and 172.16.45.14, and the broadcast address is 172.16.45.15. Verification / Alternative check: We can verify by converting the last octet to binary and applying the mask. The last octet 14 in binary is 00001110. The mask 252 in binary is 11111100. A bitwise AND of these two values is 00001100 which is decimal 12. Therefore, the network portion for the host 172.16.45.14 with mask 255.255.255.252 is indeed 172.16.45.12, which confirms our earlier block size method. Both decimal block method and binary method agree, so the answer is consistent and reliable. Why Other Options Are Wrong: 172.16.45.0: This is the first network of the /30 ranges on that third octet, but it covers hosts .1 and .2, not .14. 172.16.45.4: This subnet covers addresses 172.16.45.4 to 172.16.45.7. The host 14 is not inside this range. 172.16.45.8: This subnet covers addresses 172.16.45.8 to 172.16.45.11 only. Again, 14 is outside this range. 172.16.45.16: This would be the start of the next subnet after the one beginning at 12, and would contain host addresses 17 and 18, not 14. Common Pitfalls: Learners often forget to compute the block size correctly and may treat the subnet as if it uses a larger host range. Another frequent mistake is to misinterpret the broadcast address as the network address. Some candidates also mix up hexadecimal or binary conversions, or assume that a /30 subnet offers more than two usable hosts. Remember that /30 gives a block size of 4 addresses, so there are always only two usable host addresses. Keeping the block size rule clear in your mind prevents many such miscalculations during exams and troubleshooting. Final Answer: 172.16.45.12

Question 16

You must subnet a network into at least 5 subnets, and each subnet must support at least 16 usable hosts. Which classful subnet mask should you choose to meet these requirements?

Accepted Answer

Introduction / Context: Subnet design is a core topic in Cisco certification and practical networking. Network designers often have to balance the number of subnets against the number of hosts per subnet. This question asks for a single classful subnet mask that produces at least five subnets, each with at least sixteen usable host addresses. Understanding the host and network bit trade off, as well as the formula for usable hosts per subnet, is essential for correct design and troubleshooting of IPv4 networks. Given Data / Assumptions: We need at least 5 subnets. Each subnet must have at least 16 usable host addresses. We are working with classful style subnet masks such as 255.255.255.192, 255.255.255.224, 255.255.255.240, and similar. We assume a Class C style starting point of 255.255.255.0 for conceptual simplicity. Concept / Approach: The number of usable hosts in a subnet is calculated as 2^h - 2, where h is the number of host bits and the minus two accounts for the network and broadcast addresses. To obtain at least 16 hosts, we set 2^h - 2 greater than or equal to 16. We also know that subnetting a Class C network uses host bits from the last octet. Each additional network bit doubles the number of subnets while reducing host bits. We must therefore find a mask that has enough host bits left, but also creates at least five subnets, which requires at least three subnet bits (2^3 = 8 subnets). Step-by-Step Solution: 1. For at least 16 hosts per subnet, find h such that 2^h - 2 ≥ 16. 2. Try h = 4: 2^4 - 2 = 16 - 2 = 14, which is too small. 3. Try h = 5: 2^5 - 2 = 32 - 2 = 30, which is enough to support 16 hosts. 4. If 5 bits are used for hosts in the last octet, then 3 bits are used for subnetting that octet, because an octet has 8 bits. 5. Three subnet bits yield 2^3 = 8 subnets, which is more than the required 5 subnets. 6. A mask that uses 3 subnet bits in the last octet of a Class C network is /27, which corresponds to 255.255.255.224. 7. Therefore, 255.255.255.224 meets both constraints: at least 5 subnets and at least 16 hosts per subnet. Verification / Alternative check: We can verify by listing the host count for each candidate mask. For 255.255.255.192 (prefix /26) there are 6 host bits, so 2^6 - 2 = 62 usable hosts, which is more than enough, and 2 subnet bits provide 4 subnets which is fewer than the required 5. For 255.255.255.224 (prefix /27) there are 5 host bits, giving 30 usable hosts, and 3 subnet bits provide 8 subnets which satisfies the requirement. For 255.255.255.240 (prefix /28), there are 4 host bits and 2^4 - 2 = 14 usable hosts, which is insufficient. For 255.255.255.248 (prefix /29) there are 3 host bits, only 6 usable hosts. Thus only the /27 mask satisfies both constraints. Why Other Options Are Wrong: 255.255.255.192: Provides enough hosts but only 4 subnets, which is less than the required 5 subnets. 255.255.255.240: Provides only 14 usable hosts, so it fails the requirement of at least 16 hosts per subnet. 255.255.255.248: Provides only 6 usable hosts, which is much too small for the requirement. 255.255.255.128: This mask yields 126 hosts but only 2 subnets, so it also fails the subnet requirement. Common Pitfalls: A common mistake is to focus only on the host count and forget the minimum number of subnets, or to focus only on subnets and ignore the host requirement. Some learners forget to subtract 2 for the network and broadcast addresses when calculating usable hosts. Others confuse prefix length and dotted decimal masks. To avoid these issues, always apply the formula hosts = 2^h - 2 and subnets = 2^s, and carefully check both conditions before finalising your choice. Final Answer: 255.255.255.224

Question 17

You Telnet into a remote Cisco router and enter the command debug ip rip, but no debug output appears in your session. What is the most likely reason for this, and what should you do?

Accepted Answer

Introduction / Context: Cisco IOS debug commands are powerful tools for troubleshooting routing protocols such as Routing Information Protocol (RIP). However, many network engineers are surprised when they enable a debug command during a Telnet or Secure Shell session and see no output. This question focuses on how Cisco IOS handles debug output and which additional command is required to display that output on a virtual terminal line. Understanding this behaviour is vital for real time troubleshooting in production networks. Given Data / Assumptions: Access to a Cisco router via Telnet, not through the console port. The command debug ip rip has been entered successfully in privileged EXEC mode. No debug output is visible in the Telnet session even though RIP is assumed to be operating. The goal is to understand why and to identify the correct corrective command. Concept / Approach: By default, Cisco IOS sends debug output to the console line only. Virtual terminal lines such as Telnet and Secure Shell do not receive debug output unless they are explicitly enabled to do so. The command that instructs the router to display logging and debug messages on the current virtual terminal session is terminal monitor. Without this command, the debug messages are still generated and can be seen on the console, but they do not appear in the Telnet session, which leads many administrators to think the debug is not working. Step-by-Step Solution: 1. Confirm that the user is connected through a Telnet session, not via the console port. 2. Enter privileged EXEC mode and issue the command debug ip rip. 3. Observe that no output appears in the Telnet session, even though the router is exchanging RIP updates. 4. Recall that default debug output is directed to the console line only. 5. On the same Telnet session, enter the command terminal monitor. 6. After enabling terminal monitor, debug ip rip output will start to appear in the Telnet session as RIP packets are processed. Verification / Alternative check: As an alternative check, you can connect to the router through the console port and enable debug ip rip there. You will see immediate debug messages on the console without needing terminal monitor. This confirms that the debug feature itself is working. Another verification method is to use the show logging command to review logging configuration and ensure that logging to the monitor is enabled by default. In almost all standard configurations, using terminal monitor in the Telnet session is sufficient to view debug messages. Why Other Options Are Wrong: You must enable RIP routing with the router rip command before using debug ip rip: RIP is already assumed to be running, and even if it were not, the debug command would still show some messages such as error or status information. Lack of output in a Telnet session is due to output destination, not protocol activation. IP addressing on the network is incorrect so no debug messages are generated: Incorrect addressing could reduce the number of updates, but usually some debug messages still appear. Moreover, the question specifically focuses on Telnet behaviour. You must first run show ip rip for debug output to appear: Show commands and debug commands are independent. Show commands do not enable or unlock debug output. Debug ip rip can only be executed from privileged EXEC mode on the console: Debug commands can be executed from privileged EXEC mode on any line, including virtual terminal lines, as long as the user has the correct privilege level. Common Pitfalls: Many learners forget that terminal monitor is required for Telnet and Secure Shell sessions, which leads to unnecessary troubleshooting of routing protocols instead of the session. Some administrators also leave debug commands running for too long, which can overload the router CPU. Best practice is to use debug selectively, view the needed output, and then disable debugging with undebug all or no debug all to maintain performance. Always remember where debug output is being directed: console, terminal monitor, or logging buffer. Final Answer: You must use the terminal monitor command so that debug output is sent to the Telnet session

Question 18

On a Cisco router, which command copies the startup configuration file from nonvolatile memory into DRAM so that it becomes the running configuration?

Accepted Answer

Introduction / Context: Cisco routers and switches maintain two main configuration files. The startup configuration is stored in nonvolatile Random Access Memory and is used when the device boots. The running configuration resides in Dynamic Random Access Memory and controls the current behaviour of the device. Many administrative tasks require copying configuration data from one location to another. This question focuses on the exact command that copies the startup configuration file into DRAM, effectively replacing or merging it with the running configuration. Given Data / Assumptions: We are working on a Cisco IOS based router or switch. There is a valid startup configuration saved in nonvolatile memory. The administrator wants to load that saved configuration into DRAM as the running configuration. We assume access to privileged EXEC mode where copy commands are available. Concept / Approach: Cisco IOS uses the copy source destination syntax for moving configuration data. The most common operations are copying running-config to startup-config to save current settings permanently, and copying startup-config to running-config to restore a saved configuration into active memory. When a device boots, it automatically reads the startup-config into running-config. However, an administrator can manually repeat this action at any time without rebooting by issuing the correct copy command. This is frequently done after testing configuration changes and wanting to revert to the last saved version. Step-by-Step Solution: 1. Identify the source file: the stored configuration in nonvolatile memory is called startup-config. 2. Identify the destination file: the active configuration file in DRAM is called running-config. 3. Apply the command syntax copy source destination, replacing source with startup-config and destination with running-config. 4. The resulting full command is copy startup-config running-config. 5. After executing this command in privileged EXEC mode, the router prompts for confirmation and then reads the saved configuration into the running configuration. Verification / Alternative check: After issuing copy startup-config running-config, you can use the show running-config command to confirm that the active configuration matches the expected saved configuration. Another check is to compare show startup-config and show running-config before and after the copy. In many cases, administrators perform this command when they have accidentally made unwanted changes in the running configuration and want to revert to the last saved state without restarting the device. Why Other Options Are Wrong: copy running-config startup-config: This command does the reverse operation. It saves the current active configuration into nonvolatile memory so that it is used at the next reload. It does not copy from startup to running. copy nvram flash: This command would copy data from nonvolatile memory into flash storage, not into DRAM as the running configuration. write erase: This command erases the startup configuration and is used when resetting a router to factory defaults. It does not copy anything into DRAM. reload startup-config: There is no reload startup-config command in Cisco IOS. The reload command alone reboots the device, and the router then automatically loads startup-config, but this is not an immediate copy into DRAM without a reboot. Common Pitfalls: A frequent error is to mix up the direction of the copy command. Many new administrators intend to save changes permanently but accidentally use copy startup-config running-config, which instead restores old settings on top of new ones. Another mistake is to expect write memory to load configuration from nonvolatile memory, when in fact it saves the current running configuration. Always read the command from left to right as copy source to destination, and be very careful about which file is the source and which is the destination to avoid losing important configuration changes. Final Answer: copy startup-config running-config

Question 19

A network administrator runs show ip route and notices that the same destination network is being advertised by both RIP and EIGRP. In the routing table, the route is installed as an EIGRP route. Why is the RIP route not used?

Accepted Answer

Introduction / Context: Cisco routers can run multiple routing protocols at the same time, such as Routing Information Protocol and Enhanced Interior Gateway Routing Protocol. When more than one protocol learns a route to the same destination network, the router must choose which route to install in the routing table. Understanding the concept of administrative distance is crucial, because it determines which protocol is trusted more when there is competing route information. This question examines why an EIGRP route is chosen over a RIP route for the same destination. Given Data / Assumptions: The router is running both RIP and EIGRP. The same destination network is learned via both protocols. The show ip route output shows the route marked as an EIGRP route. There are no hints of misconfiguration such as obvious routing loops in the description. Concept / Approach: Administrative distance is a Cisco specific value that expresses how trustworthy a route source is. A lower administrative distance indicates a more trusted source. When a router learns the same prefix from multiple sources, it compares their administrative distances before considering metrics inside each protocol. Default administrative distances are well known: connected routes are 0, static routes are 1, EIGRP is 90 for internal routes, OSPF is 110, and RIP is 120. Therefore, when both EIGRP and RIP provide a route to the same destination, the router prefers EIGRP because 90 is lower than 120. Step-by-Step Solution: 1. Identify that the same network prefix appears in advertisements from both RIP and EIGRP. 2. Recall default administrative distances: EIGRP internal routes use distance 90 and RIP uses distance 120. 3. When multiple protocols advertise the same route, Cisco routers choose the one with the lowest administrative distance. 4. Compare the distances: 90 for EIGRP is lower than 120 for RIP. 5. Therefore, the router installs the EIGRP route in the routing table and the RIP route is ignored, although it can still be seen in debugging or protocol specific outputs. Verification / Alternative check: You can verify this behaviour by using show ip route and show ip protocols. Show ip route will mark the active route with the code D for EIGRP and R for RIP, but only the D route will appear as active for that prefix. Show ip protocols displays the default administrative distances for each protocol. Additionally, you can manually change administrative distances with distance commands under routing process configuration to observe how the chosen route changes if EIGRP is made less trusted than RIP. This confirms that administrative distance, rather than metrics, is the primary factor when routes from different protocols compete. Why Other Options Are Wrong: EIGRP has a faster routing update timer than RIP: Update frequency does not control which route is placed in the table when the same prefix is present from both protocols. RIP has a higher metric for that network and metrics always override administrative distance: Metrics are compared only within a single protocol. When choosing between different protocols, administrative distance takes priority. The RIP path is rejected because it contains a routing loop: The question does not mention any loop conditions, and routing loops are not automatically assumed when multiple protocols advertise the same route. RIP cannot advertise routes that overlap with EIGRP routes in the same router: A router can receive and store routes from multiple protocols for the same destination. Only one becomes active in the table based on administrative distance and other selection rules. Common Pitfalls: Learners often confuse administrative distance with metric values such as hop count or composite metrics. Another common issue is forgetting the default distance values for major protocols, which leads to incorrect assumptions about which protocol will win. To avoid mistakes, memorise the key defaults and remember that administrative distance decides between protocols first, and only after that does the router examine metrics within a single protocol to choose the best route. Final Answer: EIGRP has a lower administrative distance than RIP, so it is preferred

Question 20

If a routing table contains a static route, a RIP route, and an EIGRP route to the same destination network, which route will the Cisco router use by default to forward packets?

Accepted Answer

Introduction / Context: When multiple routes to the same destination exist in a Cisco router, the device must decide which one to install in the routing table. This decision is made first using administrative distance, which represents how trustworthy a source of routing information is. Static routes, dynamic distance vector protocols such as RIP, and advanced protocols such as EIGRP all have different default administrative distances. Understanding this hierarchy is crucial for designing deterministic routing behaviour and for troubleshooting unexpected paths in a network. Given Data / Assumptions: The router has three routes for the same destination prefix. One route is a static route. One route is learned from RIP. One route is learned from EIGRP. No special administrative distance changes have been configured manually. Concept / Approach: Cisco uses administrative distance to choose between routes learned from different sources. The lower the administrative distance, the more preferred the route. Default values are well known: connected routes are 0, static routes are 1, Enhanced Interior Gateway Routing Protocol internal routes are 90, Open Shortest Path First routes are 110, and Routing Information Protocol routes are 120. Therefore, if the router has a static route and dynamic routes to the same destination and no distances have been changed, the static route is chosen because it has the lowest administrative distance value of 1. Step-by-Step Solution: 1. List the administrative distances for each route source: - Static route: distance 1. - EIGRP internal route: distance 90. - RIP route: distance 120. 2. The router compares the administrative distance values for the three competing routes. 3. The smallest value represents the most preferred route. In this case, 1 is smaller than 90 and 120. 4. Therefore, the static route is installed in the routing table and used to forward traffic to the destination network by default. Verification / Alternative check: You can confirm this behaviour by configuring a static route alongside RIP and EIGRP on a lab router and then using show ip route. The route for the destination will be marked with the code S for static. If you change the administrative distance of the static route using the ip route command with an added distance parameter that is greater than 90, the router will then prefer the EIGRP route. This experiment demonstrates that administrative distance is the primary factor when choosing between different routing sources. Why Other Options Are Wrong: Any available route because all are equal: They are not equal, because each routing source has a different default administrative distance value. RIP route because distance vector protocols are preferred: There is no general preference for distance vector over other protocols. RIP also has a higher distance value than EIGRP and static routes. EIGRP route because it has the most advanced metric: While EIGRP uses an advanced composite metric, the router only considers the metric after it has chosen the protocol based on administrative distance. Static routes with distance 1 override both dynamic protocols by default. None, the router will drop the traffic: A route is present, so the router will forward packets according to the best available route rather than dropping them. Common Pitfalls: Many learners initially assume that the most sophisticated protocol, such as EIGRP or OSPF, will always be chosen. Others mistakenly believe that path metric alone decides between routing protocols. In reality, administrative distance is applied first across protocols, and then metrics and other tie breakers are applied within the selected protocol. To avoid confusion, remember the key default values and understand that static routes are generally preferred over dynamic routes unless you intentionally alter their administrative distance. Final Answer: Static route because it has the lowest administrative distance

CISCO Certification Questions