Question 1

Within the OSI reference model, which layer is responsible for providing communication modes such as simplex, half duplex and full duplex between applications on different hosts?

Accepted Answer

Introduction / Context: The OSI reference model divides network communication into seven conceptual layers, each with distinct responsibilities. While lower layers handle physical transmission and framing, the upper layers deal with how applications communicate and manage their dialogues. Simplex, half duplex and full duplex describe how data can flow between two endpoints: in one direction only, in both directions but not simultaneously, or in both directions at the same time. The layer responsible for managing these communication modes is often examined in Cisco certification questions. Given Data / Assumptions: We are using the standard seven layer OSI model: Physical, Data Link, Network, Transport, Session, Presentation and Application. Simplex, half duplex and full duplex here refer to dialog control between endpoints, not just electrical properties of a physical link. The question associates these modes with a particular OSI layer. Concept / Approach: The Session layer (Layer 5) is responsible for establishing, managing and terminating sessions between applications. It provides dialog control and synchronization, which includes determining who can send data when and in which direction. It can support three dialog modes: simplex (one way), half duplex (two way but one direction at a time) and full duplex (two way simultaneously). Lower layers like Data Link and Physical do implement duplex modes at the hardware level, but in the context of OSI exam questions about dialogue between processes, these modes are typically attributed to the Session layer. Step-by-Step Solution: Recall the main responsibilities of the Session layer: establishing sessions, managing dialog control and handling synchronization points. Dialog control includes deciding whether communication will be one way at a time or both ways at once, which corresponds to simplex, half duplex and full duplex modes. The Presentation layer focuses on data representation, encryption and compression, not dialog modes. The Transport layer provides end to end transport services, segmentation, reassembly and reliability, but it does not explicitly manage dialog control modes in the OSI description. Therefore, among the given OSI layers, the Session layer is the correct answer for supporting these communication modes. Verification / Alternative check: OSI model explanations in many networking textbooks specifically mention that the Session layer provides dialog control, allowing for half duplex and full duplex sessions. Example descriptions explain that the Session layer "controls which side transmits, and when" and can support different dialog modes. While in real world implementations these responsibilities may be distributed differently, in exam oriented theory, the Session layer is the expected answer. Why Other Options Are Wrong: Option a (Presentation layer) deals with syntax and semantics of the information transmitted, such as character set conversion and encryption, not with dialog control. Option b (Transport layer) ensures reliable or best effort delivery and may provide flow control, but duplex mode selection is not its primary OSI function. Option c (Application layer) provides services directly to end user applications, not low level dialog management. Option e (Data Link layer) does handle physical duplex modes at the hardware level, but in classic OSI exam terminology, simplex/half/full duplex as communication modes between processes are associated with the Session layer. Common Pitfalls: Students often confuse physical duplex settings such as configuring a network interface card for full duplex with the OSI Session layer's conceptual dialog control. Remember that OSI is an abstract model and that exam questions may emphasize the theoretical responsibilities rather than exact implementation details seen on modern networks. Final Answer: The OSI layer that offers simplex, half duplex and full duplex communication modes is the Session layer.

Question 2

A receiving host in a TCP communication session has failed to receive some of the segments that it should acknowledge. To improve the reliability of this session and reduce further loss, what can the host do?

Accepted Answer

Introduction / Context: Transmission Control Protocol (TCP) is a reliable, connection oriented transport protocol that uses sequence numbers, acknowledgements and flow control to ensure ordered delivery. When a receiving host detects that it has not received all of the segments it should acknowledge, this indicates possible congestion or loss in the network. Adjusting the TCP window size is one of the key mechanisms used to improve reliability and adapt to network conditions. Given Data / Assumptions: We have an established TCP connection between a sender and a receiver. The receiver has noticed gaps in the sequence numbers, meaning some segments are missing. The question asks which action the receiving host can take to improve reliability during the existing session. Concept / Approach: TCP uses a sliding window mechanism for flow control. The window size tells the sender how many bytes it can send before waiting for an acknowledgment. If the receiver is missing segments or the network is experiencing congestion, reducing the window size results in fewer unacknowledged segments in transit, reducing the chance of further loss and retransmission storms. Changing the source port or decreasing sequence numbers is not part of normal TCP behavior during a session; sequence numbers always increase and ports remain fixed for that connection. Step-by-Step Solution: Recognize that missing segments often indicate congestion or buffer issues in the network or at the receiver. Recall that TCP's advertised receive window is used by the sender to limit the volume of outstanding data. By decreasing the window size, the receiver effectively asks the sender to slow down, reducing the number of in flight segments and giving the network a chance to clear congestion. This helps reduce additional loss and leads to more reliable delivery without tearing down the connection. Therefore, the correct action in the context of TCP's built in mechanisms is to decrease the window size. Verification / Alternative check: TCP congestion control algorithms such as slow start and congestion avoidance also adjust the effective sending window on the sender side, but the receiver can influence flow control via the advertised window. In both cases, reducing the window improves resilience when loss is detected. Standard TCP implementations do not change source ports mid connection, nor do they decrease sequence numbers once the session is established, which confirms that those options are invalid. Why Other Options Are Wrong: Option a (send a different source port number) would actually define a new TCP connection, not fix the current one; source ports remain constant for a given session. Option b (restart the virtual circuit) is heavy handed and not necessary; TCP is designed to recover from loss without tearing down the connection. Option c (decrease the sequence number) violates TCP rules, as sequence numbers are monotonically increasing and used to detect duplicates and maintain order. Option e (increase the MTU) can actually make things worse if fragmentation or MTU issues are part of the problem, and it does not address the missing segments directly. Common Pitfalls: A common misunderstanding is to think that any loss requires resetting the connection. In reality, TCP is robust against occasional losses and uses retransmission and window adjustment to recover. Another pitfall is to confuse flow control (receiver driven, via window size) with congestion control (sender driven). Both mechanisms can reduce the amount of in flight data, but the question focuses on what the receiving host can do directly. Final Answer: To improve reliability, the host should decrease the TCP window size so that fewer unacknowledged segments are in flight at any time.

Question 3

When a station on an Ethernet network sends a frame to the destination MAC address ff:ff:ff:ff:ff:ff, what kind of Layer 2 transmission is being performed?

Accepted Answer

Introduction / Context: Ethernet uses MAC (Media Access Control) addresses to deliver frames at the data link layer. There are different types of MAC addresses: unicast, multicast and broadcast. Each type determines how the frame is treated by devices on the local segment. The special MAC address ff:ff:ff:ff:ff:ff is reserved for broadcast and is important for protocols such as ARP, DHCP and others that need to reach all hosts on a local network. Given Data / Assumptions: We are working on an Ethernet or IEEE 802.3 network segment. The destination MAC address is ff:ff:ff:ff:ff:ff, which in binary is all ones (48 bits set to 1). We are asked to identify the type of transmission represented by this destination address. Concept / Approach: In Ethernet, a unicast address identifies a single NIC, typically using a unique MAC. A multicast address identifies a group of receivers that have joined a certain multicast group. The broadcast address, with all bits set to 1, is treated as "send to everyone" on the local broadcast domain. All hosts on the LAN receive the frame, pass it up to their network stacks and then decide whether to process it based on the upper layer protocol. Step-by-Step Solution: Recognize that ff:ff:ff:ff:ff:ff is the all ones MAC address, defined by the Ethernet standard as the broadcast address. When a switch or hub sees a frame with this destination MAC, it forwards the frame out all ports in that VLAN or collision domain, except the one on which it was received. All hosts on that LAN segment receive the frame and pass it up the stack. This behavior is characteristic of a broadcast, not a unicast, multicast or anycast transmission. Therefore, the correct classification is a broadcast transmission. Verification / Alternative check: You can confirm this by observing ARP requests in a protocol analyzer such as Wireshark. The ARP request "Who has 192.168.1.10?" is typically sent to the destination MAC ff:ff:ff:ff:ff:ff. All devices on the LAN see the request, but only the device with the matching IP responds. This is a classic example of broadcast behavior at Layer 2. Why Other Options Are Wrong: Option a (unicast) would use a specific MAC address assigned to a single NIC, not the all ones address. Option b (multicast) uses special group addresses that start with a specific bit pattern, not all ones; multicast is directed to a subset of hosts, not all of them. Option c (anycast) is a Layer 3 concept in IP routing where multiple devices share the same IP address, and routing delivers to the nearest instance; it is not a Layer 2 MAC address concept in classic Ethernet. Option e (hubcast) is not an official networking term and is not used in IEEE standards. Common Pitfalls: Some learners confuse broadcast and multicast, assuming both always reach all hosts. In reality, multicast is intended for a selected group, while broadcast is truly "to everyone" on the broadcast domain. Another pitfall is to think that switches block broadcasts by default; they forward them across all ports in the same VLAN, which is why broadcast storms can be a problem in poorly designed networks. Final Answer: A frame sent to ff:ff:ff:ff:ff:ff is a broadcast transmission to all hosts on the local network segment.

Question 4

If a hub has three computers connected to it, how many broadcast domains and how many collision domains exist on this network segment?

Accepted Answer

Introduction / Context: Understanding broadcast domains and collision domains is fundamental to Ethernet network design and is tested heavily in Cisco certifications. Hubs are legacy devices that operate at Layer 1 of the OSI model and simply repeat electrical signals out all ports. This behavior affects both how broadcasts propagate and how collisions occur on the network segment. Given Data / Assumptions: We have a single hub with three computers connected to it by Ethernet cables. The hub is an unmanaged, Layer 1 device that does not perform any MAC based switching or segmentation. There are no routers or switches dividing this segment into multiple broadcast or collision domains. Concept / Approach: A broadcast domain is the set of devices that receive each other's Layer 2 broadcasts. A collision domain is the set of devices where Ethernet frames can collide with each other because they share the same physical medium. Hubs do not segment collision domains; all ports on a hub belong to the same collision domain. Similarly, because there is no Layer 3 boundary or VLAN separation, the entire hub based segment is a single broadcast domain. Step-by-Step Solution: Consider broadcasts first: if any one of the three computers sends a broadcast frame, the hub repeats it to all other ports. All three computers are therefore in the same broadcast domain; there is one broadcast domain on this segment. Now consider collisions: when two computers transmit at the same time, their signals collide on the shared medium, and the hub repeats the collision. Because the hub provides a shared medium, all three computers share the same collision domain. Thus, for three computers connected to a single hub, there is one broadcast domain and one collision domain. Verification / Alternative check: Compare this with a switch: a switch creates one collision domain per port, but still a single broadcast domain per VLAN. If these three computers were connected to a switch in the same VLAN, there would be one broadcast domain and three collision domains. The fact that a hub behaves differently (one collision domain across all ports) confirms that the answer is one broadcast domain and one collision domain. Why Other Options Are Wrong: Option a and b claim multiple broadcast domains, which would require routers or VLAN boundaries, not present in a simple hub configuration. Option c suggests one broadcast domain and three collision domains, which describes a switch, not a hub. Option e introduces two broadcast domains, which again would require routing or VLANs, not a single hub. Common Pitfalls: A common mistake is to assume that each port on any device is a separate collision domain. This is true for switches but not for hubs. Another pitfall is to confuse broadcast domains with collision domains: broadcasts are limited by routers and VLAN boundaries, while collisions are limited by physical or shared media connections. Remember that hubs repeat signals at Layer 1 and do not provide any segmentation. Final Answer: A hub with three connected computers creates one broadcast domain and one collision domain.

Question 5

In the OSI reference model, which layer is responsible for reliable communication between end nodes over the network, including establishing, maintaining and terminating virtual circuits, providing transport fault detection and recovery, and control…

Accepted Answer

Introduction / Context: The OSI reference model describes seven conceptual layers that divide network functions. Reliable end to end delivery, including connection establishment, flow control and error recovery between hosts, is a key responsibility of the Transport layer. Knowing which layer provides these services is an important part of Cisco and general networking theory. Given Data / Assumptions: The question mentions establishing, maintaining and terminating virtual circuits between end nodes. It also refers to fault detection, recovery and flow control for transport of information. We are asked to map these responsibilities to one OSI layer. Concept / Approach: The Transport layer (Layer 4) sits between the Network layer (Layer 3) and the Session layer (Layer 5). It is responsible for providing transparent data transfer between end systems, isolating upper layers from network details. Protocols such as TCP and UDP operate at this layer. TCP provides connection oriented, reliable service with sequencing, acknowledgements, retransmissions and flow control, which matches the description in the question. Step-by-Step Solution: Recognize that "virtual circuits" in this context refers to logical connections between endpoints, as implemented by connection oriented transport protocols like TCP. Fault detection and recovery at this level include sequence numbering, acknowledgements and retransmission of lost segments. Flow control mechanisms, such as sliding windows, are classic features of the Transport layer. The Network layer focuses on routing packets between devices, not end to end transport reliability. Therefore, the OSI layer that matches the description is the Transport layer. Verification / Alternative check: Reviewing OSI model descriptions confirms that Layer 4 responsibilities include segmentation, reassembly, connection control and reliability. Meanwhile, the Network layer provides logical addressing and routing, the Session layer manages dialog and synchronization, the Data Link layer handles frames on a local segment and the Physical layer deals with raw bits. This cross check aligns with selecting the Transport layer. Why Other Options Are Wrong: Option a (Network layer) handles addressing and routing but does not guarantee reliable transport between end nodes; IP is a best effort protocol. Option c (Session layer) manages dialogs and sessions but relies on Transport layer services for reliable delivery. Option d (Data link layer) provides reliability only over a single physical link, not end to end across a network. Option e (Physical layer) simply transmits bits and is unaware of higher level concepts like virtual circuits or flow control. Common Pitfalls: Learners sometimes confuse reliability at the Data Link layer (for example, link level retransmissions in wireless) with end to end reliability. Remember that the Transport layer focuses on host to host reliability, whereas Data Link reliability is limited to a single hop. Another pitfall is to assume that all applications require TCP; some use UDP and implement reliability at the application layer if needed. Final Answer: The layer described is the Transport layer of the OSI model.

Question 6

In the OSI reference model, which layer ensures trustworthy transmission of data across a physical link and is primarily concerned with physical addressing, line discipline, network topology, error notification, ordered delivery of frames and flow c…

Accepted Answer

Introduction / Context: The OSI model separates network functions into multiple layers to clarify responsibilities. While the Physical layer deals with the raw transmission of bits, the Data Link layer builds those bits into frames, handles addressing on the local link and provides basic reliability mechanisms. The description in the question emphasizes physical addressing, line discipline and frame delivery, which are trademarks of the Data Link layer. Given Data / Assumptions: We are focusing on communication across a single physical link or local network segment. The layer mentioned uses physical addressing (such as MAC addresses), manages line discipline and network topology, and handles ordered delivery of frames and error notification. We must choose the OSI layer that best matches these responsibilities. Concept / Approach: The Data Link layer (Layer 2) sits directly above the Physical layer and is responsible for framing, physical addressing, error detection, and sometimes correction, and flow control on the local link. It includes sublayers such as Logical Link Control (LLC) and Media Access Control (MAC). This layer understands how devices are arranged on the local network and ensures that frames are delivered reliably to the next hop. Step-by-Step Solution: Recognize that "physical addressing" refers to MAC addresses and similar link layer identifiers, not IP addresses. Line discipline and network topology awareness (for example, bus, ring, star) are provided by link layer protocols like Ethernet and token ring. Error notification and ordered delivery of frames at this level involve mechanisms such as frame check sequences and, in some technologies, link level acknowledgements. Flow control on the local link, such as pause frames in Ethernet, also belongs to the Data Link layer. These functions clearly map to the Data Link layer rather than Physical, Network or higher layers. Verification / Alternative check: According to OSI model descriptions, Layer 1 handles electrical, optical or radio signals and bit timing, but says nothing about frames or addresses. Layer 2, by contrast, is described as responsible for node to node delivery, framing, addressing and error detection. This matches the detailed list of responsibilities in the question, confirming that the Data Link layer is correct. Why Other Options Are Wrong: Option a (Physical layer) deals purely with bit level transmission and does not know about addresses, frames or error notification beyond basic signal integrity. Option c (Network layer) uses logical addressing (for example IP addresses) and routing between networks, rather than local link addressing and line discipline. Option d (Transport layer) provides end to end services such as segmentation and flow control between hosts, not across a single physical link. Option e (Session layer) manages dialogs between applications and is unrelated to frames and physical addressing. Common Pitfalls: Students sometimes confuse "physical addressing" with the Physical layer because of the word "physical" in both. Remember that physical addressing (MAC addresses) belongs to the Data Link layer, while the Physical layer handles the medium and signal encoding itself. Another pitfall is to assume that ordered delivery is always provided by higher layers, but many link layer technologies provide some level of ordering and error handling on each hop. Final Answer: The layer described is the Data link layer of the OSI model.

Question 7

On an Ethernet network, which medium access control method is used to control how devices share the transmission medium?

Accepted Answer

Introduction / Context: Ethernet is the dominant LAN technology and historically used a specific medium access control (MAC) method to manage how multiple devices share a common medium. On classic shared Ethernet segments, collisions could happen when two devices transmitted simultaneously. Carrier Sense Multiple Access with Collision Detection (CSMA/CD) was designed to reduce and manage these collisions. Given Data / Assumptions: We are referring to traditional Ethernet as defined by IEEE 802.3. Devices share a common collision domain, such as on a coaxial bus or a hub based star topology. The question asks specifically which connection or access mechanism Ethernet uses. Concept / Approach: CSMA/CD stands for Carrier Sense Multiple Access with Collision Detection. "Carrier sense" means a device listens before transmitting to check whether the medium is idle. "Multiple access" means many devices share the same medium. "Collision detection" means that if a device detects a collision while transmitting, it stops, sends a jam signal and backs off for a random time before trying again. This mechanism is central to classic half duplex Ethernet operation. Although modern switched full duplex Ethernet largely eliminates collisions, CSMA/CD remains part of the original Ethernet specification. Step-by-Step Solution: Recall that token passing is associated with technologies like Token Ring and FDDI, not Ethernet. CSMA/CA is used by wireless LAN technologies such as IEEE 802.11, which cannot reliably detect collisions on the air and therefore try to avoid them. Host polling and TDMA are used in other networking systems but are not part of the standard Ethernet MAC method. CSMA/CD is explicitly named in Ethernet and IEEE 802.3 standards as the access method, making option b the correct answer. Therefore, Ethernet uses CSMA/CD as its connection and access mechanism on shared media networks. Verification / Alternative check: Textbooks and Cisco training material describe the historical "Ethernet bus" or "hub based Ethernet" networks as CSMA/CD systems. You can also see CSMA/CD referenced in configuration outputs and documentation describing half duplex Ethernet behavior. This cross checks the exam style statement that Ethernet uses CSMA/CD. Why Other Options Are Wrong: Option a (token passing) relates to Token Ring LANs where a token circulates and only the holder may transmit, which is a different MAC approach. Option c (CSMA/CA) is specific to wireless LAN technologies and is not used by wired Ethernet. Option d (host polling) is a centralized method where a controller polls each device, which is not how Ethernet works. Option e (TDMA) divides the channel into fixed time slots, used in some WAN and cellular systems, not in classic Ethernet. Common Pitfalls: Learners sometimes confuse CSMA/CD and CSMA/CA, especially when studying both wired and wireless networking. Remember the simple association: "D" for "detection" in wired Ethernet, "A" for "avoidance" in Wi-Fi. Also note that CSMA/CD is practically obsolete on modern full duplex switched Ethernet, but it remains an important concept for understanding legacy networks and exam questions. Final Answer: Ethernet uses CSMA/CD (Carrier Sense Multiple Access with Collision Detection) as its medium access control method.

Question 8

In an IEEE 48-bit MAC address, which portion is known as the organizationally unique identifier (OUI)?

Accepted Answer

Introduction / Context: MAC addresses are 48-bit hardware addresses used at the Data Link layer to uniquely identify network interface cards on Ethernet and other LAN technologies. To ensure global uniqueness, IEEE allocates blocks of MAC addresses to manufacturers. The organizationally unique identifier (OUI) is the portion of the MAC address that identifies the vendor or organization that manufactured the device. Given Data / Assumptions: A standard MAC address is 48 bits long, usually written as six groups of two hexadecimal digits (for example, 00:1A:2B:3C:4D:5E). The OUI identifies the vendor and is assigned by the IEEE Registration Authority. We need to determine which part of the 48-bit address constitutes the OUI. Concept / Approach: The MAC address can be split into two 24-bit halves. The first 24 bits (the first three bytes) are the OUI, which is unique to a manufacturer or organization. The remaining 24 bits (the last three bytes) are assigned by the vendor to identify individual interfaces or devices. This scheme allows a large number of unique MAC addresses to be generated under each OUI. Step-by-Step Solution: Write a MAC address in the common format: aa:bb:cc:dd:ee:ff. The first three octets (aa, bb, cc) together form 24 bits, known as the organizationally unique identifier (OUI). The last three octets (dd, ee, ff) form the device specific portion, assigned by the vendor to individual NICs. Therefore, the OUI is the first 24 bits (3 bytes) of the address. Among the options, only option a correctly describes this structure. Verification / Alternative check: If you search for OUI databases or vendor lookup tools, they typically accept the first three bytes of a MAC address (for example 00-1A-2B) to identify the manufacturer. This practical usage confirms that the OUI is exactly the first 24 bits of the 48-bit MAC address and not any other portion. Why Other Options Are Wrong: Option b mentions the first 12 bits but then incorrectly states "3 bytes"; 12 bits is only 1.5 bytes and does not match the actual OUI size. Option c says the first 24 bits or 6 bytes, which is inconsistent; 6 bytes would be the entire MAC address, not just the OUI. Option d refers to the first 32 bits, which would be 4 bytes, not 3, and is not how OUIs are defined. Option e incorrectly identifies the last 24 bits as the OUI; these bits are vendor assigned device identifiers, not the organizational identifier. Common Pitfalls: Students sometimes misinterpret the term "organizationally unique" and assume it refers to the entire MAC address. In reality, uniqueness is ensured by combining a unique OUI with a vendor assigned device portion. Another pitfall is confusing MAC addresses with IPv6 interface identifiers, which have different formats and structures. Final Answer: The OUI is the first 24 bits, or 3 bytes, of the MAC address.

Question 9

You need to implement a network cabling medium that is not susceptible to electromagnetic interference (EMI). Which type of cable should you choose?

Accepted Answer

Introduction / Context: Electromagnetic interference (EMI) and radio frequency interference (RFI) can degrade the performance of copper based network cabling, especially in environments with strong electrical noise such as factories or near heavy machinery. Choosing the right cabling medium can significantly improve signal integrity. Fiber optic cabling is well known for its immunity to EMI because it uses light rather than electrical signals. Given Data / Assumptions: The environment where the network will be deployed has significant EMI or the risk of such interference. The goal is to choose a medium that is inherently immune to EMI, not just resistant. Options include common copper based cables and fiber optic cable. Concept / Approach: Copper based cabling (coaxial and twisted pair) carries electrical signals and therefore can be affected by external electromagnetic fields. While shielding and twisted pairs can reduce susceptibility, they cannot completely eliminate it. Fiber optic cable uses strands of glass or plastic to transmit light pulses; since there is no electrical signal traveling through metal conductors, fiber is immune to EMI and RFI. This makes it the best choice for noisy environments or long distance high speed links. Step-by-Step Solution: Evaluate coaxial cables (thicknet and thinnet): both are copper based and carry electrical signals, so they can be affected by EMI, despite some shielding. Consider unshielded twisted pair (UTP) cables such as Category 5 and Category 3: twisting reduces interference, but the cables are still copper and not immune to strong EMI. Recognize that fiber optic cables transmit data as light, not electricity, and therefore are immune to electromagnetic interference. Among the options given, only fiber optic cable fully meets the requirement of not being susceptible to EMI. Thus, the correct choice is fiber optic cabling. Verification / Alternative check: Network design guides often recommend fiber optic cabling for industrial environments, areas near high voltage equipment and for long distance runs, precisely because it is immune to EMI and offers high bandwidth. This consistent recommendation across industry sources confirms that fiber is the right answer when EMI immunity is the primary concern. Why Other Options Are Wrong: Option a (thicknet coax) and option b (thinnet coax) both use copper conductors and are susceptible to EMI, even though coaxial design offers some shielding. Option c (Category 5 UTP) and option e (Category 3 UTP) use unshielded twisted pair, which reduces but does not eliminate interference; they are not immune to EMI. Only fiber optic cable avoids EMI entirely because it carries light signals through non metallic conductors. Common Pitfalls: A pitfall is to think that any shielded or twisted copper cable is completely immune to interference. In reality, strong electromagnetic fields can still induce noise in copper conductors. Another confusion arises between EMI immunity and lightning or surge immunity; while fiber is excellent for EMI, physical installation practices must still consider other hazards. Final Answer: To avoid EMI, you should use fiber-optic cable as the network medium.

Question 10

Which of the following physical layer standards is part of the IEEE 802.3u Fast Ethernet specification?

Accepted Answer

Introduction / Context: IEEE 802.3u defines Fast Ethernet, which operates at 100 Mbps. It includes several physical layer standards that describe how 100 Mbps Ethernet can run over different media types, such as twisted pair and fiber. Knowing which specific variants belong to 802.3u helps in selecting the correct cabling and hardware for a network. Given Data / Assumptions: IEEE 802.3u is the Fast Ethernet standard, operating at 100 Mbps. Common Fast Ethernet physical layer types include 100Base-TX, 100Base-T4 and 100Base-FX. The options list several Base-X styles, some of which correspond to other Ethernet families. Concept / Approach: Historically, 10Base5 and 10Base2 are 10 Mbps Ethernet standards (thicknet and thinnet coax respectively) under older 802.3 specifications, not 802.3u. 1000Base-T is a 1 Gbps standard under 802.3ab, referred to as Gigabit Ethernet. 100Base-TX is a 100 Mbps Fast Ethernet standard that uses two pairs of Category 5 or better twisted pair cabling. Step-by-Step Solution: Identify which options are 10 Mbps: 10Base5 and 10Base-T clearly operate at 10 Mbps, not 100 Mbps, so they cannot be 802.3u. Recognize 1000Base-T as a 1 Gbps (1000 Mbps) standard, part of Gigabit Ethernet, defined in 802.3ab, not 802.3u. 100Base2 is not a standard Fast Ethernet designation; 10Base2 is, for 10 Mbps coax. 100Base-TX is known as a common Fast Ethernet physical layer that belongs to IEEE 802.3u. Therefore, 100Base-TX is the correct answer. Verification / Alternative check: Reference materials for Ethernet standards show that 802.3u includes 100Base-TX, 100Base-T4 and 100Base-FX. 100Base-TX is the most common and widely deployed for Fast Ethernet over copper. In contrast, 802.3ab specifies 1000Base-T for Gigabit Ethernet, and 10Base5 and 10Base-T belong to earlier 10 Mbps standards. This confirms that 100Base-TX is the only option that fits 802.3u. Why Other Options Are Wrong: Option a (100Base2) does not correspond to a standard Ethernet physical layer; the correct 10 Mbps coax standard is 10Base2. Option b (10Base5) is a 10 Mbps thick coax standard, part of older Ethernet, not Fast Ethernet. Option d (1000Base-T) is a Gigabit Ethernet standard under 802.3ab, operating at 1000 Mbps, not 802.3u. Option e (10Base-T) is 10 Mbps Ethernet over twisted pair and predates Fast Ethernet. Common Pitfalls: Students sometimes treat any "100Base" standard as Fast Ethernet without checking the associated IEEE clause. It is important to tie the correct speed and media type to the right 802.3 extension. Another pitfall is mixing up 100Base-TX and 100Base-T4; the former uses two pairs of Cat5, while the latter uses four pairs and is far less common. Final Answer: The Fast Ethernet standard in IEEE 802.3u listed here is 100Base-TX.

Question 11

When making a console connection to a Cisco router or switch using HyperTerminal (or a similar terminal program), what is the standard default bit rate (baud rate) that the terminal should be configured to?

Accepted Answer

Introduction / Context: Cisco routers and switches are often configured and managed via a console connection, especially when no network connectivity is yet available. Terminal emulation programs such as HyperTerminal, PuTTY or Tera Term are used to connect to the console port. For the connection to work properly, the terminal settings must match the device's console line configuration, including the bit rate (baud rate). Given Data / Assumptions: We are using the default console settings on a Cisco device that has not been modified. We are connecting via a serial console cable using a terminal program. We need to know the default bit rate for this console connection. Concept / Approach: By default, Cisco devices use a console configuration of 9600 bps, 8 data bits, no parity, 1 stop bit and no flow control (often written as 9600 8N1). Although it is possible to change these settings, the exam and many real world scenarios assume the default. Therefore, configuring HyperTerminal or any other terminal emulator to 9600 bps is standard practice. Step-by-Step Solution: Open the terminal emulator and choose the correct COM port for the console cable. Set the bit rate (baud rate) to 9600 bps, which is the Cisco default. Configure 8 data bits, no parity, 1 stop bit and no hardware or software flow control. Initiate the connection and press Enter; you should see the device prompt or boot messages if the settings match. Thus, among the options, 9600 bps is the correct default bit rate. Verification / Alternative check: Cisco documentation and configuration guides consistently show examples where "line console 0" is configured with speed 9600 by default. Many terminal emulator presets for "Cisco console" also default to 9600 bps. While some environments use higher speeds for specialized needs, the exam expects the commonly used default value. Why Other Options Are Wrong: Option a (2400 bps) and option b (1200 bps) are much slower legacy speeds rarely used for modern Cisco console connections. Option d (6400 bps) is not a common standard bit rate for console lines. Option e (19200 bps) is a valid serial speed but not the default used by Cisco consoles. Common Pitfalls: An incorrect bit rate setting is a common reason that users see garbled characters or no output at all in their terminal emulator. Always verify that your settings are 9600 8N1 with no flow control when first connecting. If the device has been reconfigured to a different speed, you may need to try other common rates, but 9600 is the correct starting point. Final Answer: The standard default bit rate for a Cisco console connection is 9600 bps.

Question 12

In copper cabling such as twisted pair Ethernet, what is the specific term for unwanted signal interference that is induced from adjacent wire pairs within the same cable?

Accepted Answer

Introduction / Context: When transmitting data over copper cables, several types of interference and signal degradation can occur. Understanding the terminology helps in diagnosing cable problems and choosing appropriate cabling for a given environment. The question focuses on interference between adjacent pairs within the same cable, which is a common concern for twisted pair Ethernet cabling. Given Data / Assumptions: We are dealing with twisted pair copper cabling, such as Category 5 or Category 6 Ethernet cables. Interference originates from signals on neighboring wire pairs inside the same physical cable, not from outside sources. We need to identify the technical term for this kind of interference. Concept / Approach: Crosstalk is the term used to describe unwanted coupling of signals from one pair of wires to another pair within the same cable. It occurs because electromagnetic fields generated by a signal on one pair induce currents on adjacent pairs. EMI and RFI generally refer to interference from external sources, while attenuation refers to signal loss over distance and reflection relates to impedance mismatches. Step-by-Step Solution: Identify that the interference originates from adjacent wire pairs within the same cable jacket. Recognize that this pattern of interference is called crosstalk, specifically near end crosstalk (NEXT) or far end crosstalk (FEXT) depending on where it is measured. EMI and RFI refer to external sources like motors, fluorescent lights or radio transmitters, which is not the scenario described here. Attenuation is a gradual loss of signal strength as it travels along the cable, not interference from another pair. Thus, the correct term for interference from adjacent pairs is crosstalk. Verification / Alternative check: Cable certification tools measure and report crosstalk values, often labeling them as NEXT or FEXT in their reports. Standards for twisted pair cabling specify maximum allowable crosstalk levels at various frequencies. This confirms that the industry uses the term crosstalk for internal pair to pair interference. Why Other Options Are Wrong: Option a (EMI) refers broadly to interference from external electromagnetic sources, not specifically from neighboring pairs inside the same cable. Option b (RFI) is a subset of EMI related to radio frequencies from external sources, not internal cable pair coupling. Option d (attenuation) is signal loss over distance and does not describe interference from other signals. Option e (reflection) occurs when a signal encounters an impedance mismatch and part of it is reflected back; it is not the same as pair to pair interference. Common Pitfalls: A common mistake is to use EMI as a catch all term for any interference, but exam questions frequently require the more precise term crosstalk for within cable interference. Proper cable installation practices, such as maintaining twists up to the termination and avoiding kinks, help reduce crosstalk and maintain signal quality. Final Answer: The unwanted interference from adjacent pairs in the cable is called crosstalk.

Question 13

In the operation of CSMA/CD on a shared Ethernet segment, after the backoff algorithm has completed following a collision, which host or hosts have priority to attempt transmission?

Accepted Answer

Introduction / Context: CSMA/CD (Carrier Sense Multiple Access with Collision Detection) was designed to manage access to a shared Ethernet medium and handle collisions. When a collision occurs, all transmitting hosts stop, send a jam signal and then wait for a random backoff time before attempting to retransmit. Understanding who gets priority after the backoff process is essential for grasping how CSMA/CD ensures fairness. Given Data / Assumptions: We are dealing with a traditional half duplex Ethernet network that uses CSMA/CD, such as a hub based or coaxial bus network. A collision has occurred, and all transmitting hosts have detected it and invoked the backoff algorithm. The question asks which host or hosts have priority to transmit after the backoff period expires. Concept / Approach: CSMA/CD is a decentralized access method. After a collision, each host involved chooses a random backoff interval based on the binary exponential backoff algorithm. No host is granted special priority; instead, each host independently waits its random backoff time, then senses the carrier again and transmits if the medium is idle. Because of the randomization, statistically, one host will likely win the next opportunity to transmit, but all hosts are treated equally by the algorithm. Step-by-Step Solution: When a collision is detected, all transmitting hosts send a jam signal to ensure that every host on the segment is aware of the collision. Each host then computes a random backoff time, often measured in slot times, according to the binary exponential backoff rules. During backoff, each host waits and continues to sense the medium. After its individual backoff timer expires, any host that finds the medium idle may attempt to transmit again. Because backoff times are random and independently chosen, no inherent priority is given; all hosts have equal opportunity. Verification / Alternative check: Descriptions of CSMA/CD in networking textbooks and IEEE 802.3 specifications emphasize its probabilistic nature and fairness. There is no mention of special treatment for the host that sent the jam signal or for hosts with particular MAC addresses. Over many transmissions, each host sees a fair share of the medium, assuming similar traffic patterns, which aligns with the idea of equal priority. Why Other Options Are Wrong: Option b wrongly suggests that only the two hosts involved in the collision have priority; in reality, any host on the segment can attempt transmission after its backoff expires. Option c gives special priority to the host that sent the jam signal, but all transmitting hosts send jam signals and no priority is defined for them. Option d and option e propose priority based on MAC address ordering, which is not a mechanism defined in CSMA/CD or the Ethernet standard. Common Pitfalls: A common misconception is to think of CSMA/CD as assigning turns or priorities, but in fact it relies on random backoff and carrier sensing to reduce the probability of repeated collisions. Another pitfall is to assume that CSMA/CD still applies on modern switched full duplex networks; in those environments, there is no shared medium and collisions are essentially eliminated, so CSMA/CD is not used. Final Answer: After the backoff algorithm, all hosts have equal priority and may attempt to transmit once their individual backoff timers expire and the medium is idle.

Question 14

In Ethernet networking, which of the following statements about full-duplex and half-duplex operation is correct?

Accepted Answer

Introduction / Context: Ethernet local area networks can operate in either half-duplex or full-duplex mode. Understanding how many wire pairs are used and how traffic flows in each direction is important for troubleshooting collisions, performance issues, and cabling design in modern switched networks. Given Data / Assumptions: We are considering standard twisted-pair Ethernet links, such as Fast Ethernet or Gigabit Ethernet over copper. Full-duplex allows simultaneous sending and receiving of frames. Half-duplex allows sending or receiving at one time, but not both simultaneously. We assume a typical point-to-point link between a switch port and a host network interface card. Concept / Approach: On twisted-pair Ethernet, the physical medium uses separate wire pairs for transmit and receive. In full-duplex mode, one pair is dedicated to sending data and another pair is dedicated to receiving data. This separation removes the need for collision detection and improves throughput. In half-duplex mode, the medium is logically shared for send and receive, and the device must use carrier sense multiple access with collision detection to avoid and recover from collisions. The key idea is that full-duplex relies on two independent paths, one in each direction. Step-by-Step Solution: 1. Recall that a twisted-pair Ethernet cable contains multiple wire pairs.2. In standard full-duplex Ethernet over copper, one pair is used to transmit data from device A to device B.3. Another separate pair is used to transmit data from device B back to device A.4. Because the transmit and receive paths are independent, both devices can send and receive at the same time.5. Therefore, the correct statement is that full-duplex Ethernet uses two separate pairs of wires, one for each direction of communication. Verification / Alternative check: You can verify this by looking at standard Ethernet pinouts. For example, on a typical RJ-45 connector for 100 Mbps Ethernet, one pair of pins is used for transmit and another pair for receive. Network documentation and cabling diagrams also show separate transmit and receive pairs for full-duplex links. Why Other Options Are Wrong: Option A is incorrect because using only one pair for both sending and receiving would force half-duplex operation. Option C is incorrect because half-duplex cannot send and receive at the same time and does not typically use two independent pairs for simultaneous traffic. Option D is incorrect because full-duplex Ethernet does not use three pairs; the common implementation uses two pairs for data, and any additional pairs are unused or reserved for other purposes depending on the standard. Common Pitfalls: A common mistake is to confuse the number of pairs in the cable with how many are actually used for data transmission. Many Ethernet cables have four pairs, but not all pairs are used by every Ethernet standard. Another pitfall is assuming that full-duplex only changes software behavior; in reality, it also depends on how the physical medium is wired and used. Final Answer: Full-duplex Ethernet uses two separate pairs of wires, one for transmitting and one for receiving.

Question 15

Convert the IPv4 address 192.168.10.15 from dotted-decimal notation to its 32-bit binary representation.

Accepted Answer

Introduction / Context: In computer networking, IPv4 addresses are often written in dotted-decimal form for human readability, but routers and hosts work with binary values. Being able to convert between dotted-decimal and binary notation is a fundamental skill for understanding subnetting, masks, and routing behavior in TCP/IP networks. Given Data / Assumptions: Given IPv4 address: 192.168.10.15. An IPv4 address consists of four octets, each 8 bits. Each octet in decimal must be converted to an 8-bit binary number. We maintain leading zeros in each octet to show all 8 bits. Concept / Approach: The approach is to convert each octet from decimal to binary using powers of two. Each octet ranges from 0 to 255 and is represented by bits with place values 128, 64, 32, 16, 8, 4, 2, and 1. For each octet, we repeatedly subtract the largest power of two that fits and mark a 1 in that position, leaving 0 where the power of two does not fit. We then join the octets with dots to represent the full 32-bit binary address. Step-by-Step Solution: 1. Convert 192 to binary: 192 = 128 + 64, so bits are 11000000.2. Convert 168 to binary: 168 = 128 + 32 + 8, so bits are 10101000.3. Convert 10 to binary: 10 = 8 + 2, so bits are 00001010.4. Convert 15 to binary: 15 = 8 + 4 + 2 + 1, so bits are 00001111.5. Combine the four 8-bit groups with dots: 11000000.10101000.00001010.00001111. Verification / Alternative check: You can verify correctness by converting each binary octet back to decimal. For example, 11000000 is 128 + 64 = 192, 10101000 is 128 + 32 + 8 = 168, 00001010 is 8 + 2 = 10, and 00001111 is 8 + 4 + 2 + 1 = 15. This round-trip check confirms that the binary representation matches the original dotted-decimal address. Why Other Options Are Wrong: Option B swaps the last two octets, giving the wrong host portion. Option C changes one bit in the second octet, altering 168 to a different decimal value. Option D changes bits in the last octet, turning 15 into another number. Any such change results in a different IP address and therefore an incorrect conversion. Common Pitfalls: Common mistakes include forgetting to pad each octet to 8 bits with leading zeros, mixing up octet order, or miscalculating the contribution of each power of two. Another pitfall is writing binary without dots, which makes it harder to see the four-octet structure of the IPv4 address. Final Answer: 11000000.10101000.00001010.00001111

Question 16

Convert the 32-bit binary IPv4 address 11001010.11110101.10000011.11101011 to dotted-hexadecimal notation.

Accepted Answer

Introduction / Context: Network engineers sometimes represent IPv4 addresses in hexadecimal form, especially in low-level debugging or when working with certain protocols and configuration tools. Converting between binary and hexadecimal is faster and more compact than working only with binary, but it still accurately represents the same underlying bits. Given Data / Assumptions: Binary IPv4 address: 11001010.11110101.10000011.11101011. Each dotted group is an 8-bit octet. Each octet will be converted to hexadecimal. Two hexadecimal digits represent one 8-bit binary value. Concept / Approach: To convert from binary to hexadecimal, group the bits of each octet into two groups of four bits (nibbles). Each nibble corresponds directly to one hexadecimal digit using the mapping 0000 to 0 through 1111 to F. Once each octet is converted into two hexadecimal digits, we join the four hexadecimal octets with dots to form dotted-hexadecimal notation. This preserves the structure of the original IPv4 address while shortening the representation. Step-by-Step Solution: 1. First octet: 11001010 splits into 1100 and 1010. 1100 is C, 1010 is A, so the first octet is CA.2. Second octet: 11110101 splits into 1111 and 0101. 1111 is F, 0101 is 5, so the second octet is F5.3. Third octet: 10000011 splits into 1000 and 0011. 1000 is 8, 0011 is 3, so the third octet is 83.4. Fourth octet: 11101011 splits into 1110 and 1011. 1110 is E, 1011 is B, so the fourth octet is EB.5. Combine the hexadecimal octets: CA.F5.83.EB. Verification / Alternative check: As an additional check, you can convert each hexadecimal octet back to decimal via binary. For example, CA is 11001010 in binary, which is 202 in decimal. Performing this reverse conversion for all octets will confirm that the original binary sequence and the hexadecimal representation are equivalent. Why Other Options Are Wrong: Option B changes the first octet from CA to C9, which corresponds to different high-order bits. Option C uses F4 instead of F5 in the second octet, altering four bits. Option D changes the final octet from EB to EE, which again modifies the underlying binary bits. Any such change means the hexadecimal does not match the original binary address. Common Pitfalls: Typical errors include grouping bits incorrectly, forgetting that four bits equal one hexadecimal digit, or misreading binary values when referencing the hexadecimal conversion table. Another pitfall is accidentally converting the dotted-decimal representation instead of the provided binary, leading to a completely different result. Final Answer: CA.F5.83.EB

Question 17

In a Frame Relay WAN, you must configure a static map to the remote S-AMER site at IP address 172.16.1.3 over DLCI 704. Which frame-relay map command correctly creates this mapping and forwards broadcasts?

Accepted Answer

Introduction / Context: Frame Relay is a legacy wide area network technology that uses virtual circuits identified by Data Link Connection Identifiers, commonly called DLCIs. When dynamic Inverse ARP is not available or not desired, administrators can configure static frame-relay map statements to associate remote IP addresses with local DLCIs. Understanding the correct syntax of the frame-relay map command is essential for ensuring that traffic reaches remote sites and that broadcasts and multicast packets are forwarded when needed. Given Data / Assumptions: Remote site S-AMER has IP address 172.16.1.3. The local Frame Relay interface uses DLCI 704 to reach S-AMER. Broadcast and multicast traffic should be forwarded over this virtual circuit. We are configuring a static mapping on a Cisco router interface. Concept / Approach: The frame-relay map command associates a Layer 3 protocol address with a Layer 2 DLCI. The basic syntax for IPv4 is frame-relay map ip remote-ip dlci broadcast. The broadcast keyword tells the router to forward broadcast and multicast frames, which is important for protocols like routing updates or DHCP that rely on broadcast traffic. Therefore, the correct command must use the given remote IP address, the correct DLCI, and the broadcast keyword in the proper order. Step-by-Step Solution: 1. Identify the remote IP address of the S-AMER site: 172.16.1.3.2. Identify the correct DLCI for reaching S-AMER: 704.3. Apply the Cisco syntax: frame-relay map ip remote-ip dlci broadcast.4. Substituting values gives: frame-relay map ip 172.16.1.3 704 broadcast.5. Compare this to the options and select the one that exactly matches. Verification / Alternative check: You can verify this choice by checking the interface configuration in Cisco documentation. The remote IP address must match the far end router interface, and the DLCI must match the value assigned by the Frame Relay provider. Verifying these details against a network diagram or service provider information confirms that DLCI 704 is correct for S-AMER and that the broadcast keyword is required for dynamic routing protocols. Why Other Options Are Wrong: Option B uses DLCI 196, option C uses 702, and option D uses 344. These DLCIs represent different virtual circuits and would send traffic to the wrong remote endpoint. Even though the syntax and broadcast keyword look correct, using an incorrect DLCI prevents packets from reaching the S-AMER site. Therefore, only the mapping with DLCI 704 is valid for this scenario. Common Pitfalls: A common mistake is to forget the broadcast keyword, which can cause routing protocols to fail across the Frame Relay cloud. Another pitfall is assuming that the DLCI is globally significant; in reality, it is often locally significant to each router, which means you must always check the provider documentation or network diagram. Confusing DLCI values between different remote sites can break connectivity in complex hub-and-spoke topologies. Final Answer: frame-relay map ip 172.16.1.3 704 broadcast

Question 18

In Cisco's Enhanced Interior Gateway Routing Protocol (EIGRP), what is a successor route?

Accepted Answer

Introduction / Context: EIGRP is a Cisco proprietary advanced distance-vector routing protocol that uses concepts like successor and feasible successor routes to ensure loop-free and efficient path selection. Understanding what a successor route is helps you interpret EIGRP topology and routing tables and troubleshoot how the protocol chooses paths for forwarding user traffic. Given Data / Assumptions: EIGRP maintains a neighbor table, a topology table, and a routing table. For each destination network, EIGRP selects one or more routes based on metrics like bandwidth and delay. The term successor is specific to EIGRP and relates to the best route to a destination. EIGRP may also keep backup routes known as feasible successors. Concept / Approach: A successor route in EIGRP is the route with the lowest metric, called the feasible distance, to a particular destination network. This route is considered loop-free and is placed directly into the EIGRP routing table, where it is used to forward packets. A feasible successor, if present, is a backup route that meets the feasibility condition. It is stored in the topology table and can be promoted to successor if the primary path fails. Therefore, the correct definition emphasizes that a successor route is the active, primary forwarding route for a destination. Step-by-Step Solution: 1. Recall that EIGRP calculates a composite metric for each path to a destination.2. Among all the candidate paths, EIGRP chooses the path with the lowest feasible distance as the primary route.3. This chosen primary route is called the successor route and is installed in the IP routing table.4. EIGRP may also identify other loop-free backup paths that meet the feasibility condition, called feasible successors.5. Thus, a correct statement must describe the successor as the primary forwarding route, possibly backed up by a feasible successor. Verification / Alternative check: You can confirm this by examining EIGRP show commands on a Cisco router. The show ip eigrp topology command lists successor and feasible successor routes. The successor appears as the primary entry, and the same network appears in the routing table via show ip route eigrp, confirming that the successor is used for actual data forwarding. Why Other Options Are Wrong: Option A describes a feasible successor, not the successor route. Option C refers to an active route, which indicates that EIGRP is currently performing a diffusing computation and is not the stable primary route. Option D confuses the concept with external routes; external routes can still have successors, but being external does not define the term successor. Common Pitfalls: Learners often confuse successor and feasible successor, assuming both are used for load sharing by default. Another pitfall is assuming that any route in the topology table is automatically in the routing table, which is incorrect. Only successor routes are installed in the routing table for forwarding traffic, while feasible successors stay as backups until needed. Final Answer: The primary route selected by EIGRP to forward traffic to a destination network, possibly backed up by a feasible successor.

Question 19

On router R1, users in the 192.168.10.0/24 LAN must reach a file server on a remote network via next-hop WAN address 209.165.200.226. Which static route on R1 will allow these file requests to reach the server?

Accepted Answer

Introduction / Context: Static routes are manually configured paths that tell a router how to reach remote networks when dynamic routing is not used or when specific paths are preferred. In many small office scenarios, a default route is configured on the edge router to send all non-local traffic toward an upstream service provider or another router that knows the rest of the network topology. Given Data / Assumptions: Local LAN: 192.168.10.0 with mask 255.255.255.0. Router R1 connects this LAN to a WAN link with next-hop IP 209.165.200.226. The file server is located on a remote network reachable beyond that next hop. We must choose one static route command on R1 that permits users to reach the remote server. Concept / Approach: When the exact remote network address is not specified or multiple external networks must be reachable, a default route is often configured. In Cisco IOS, a default route is written as ip route 0.0.0.0 0.0.0.0 next-hop, which means that any destination not found in the more specific routes will be forwarded to the specified next-hop address. Here, the upstream next-hop toward the file server is 209.165.200.226, so the default route must point to that address rather than an interface or an internal IP. Step-by-Step Solution: 1. The LAN 192.168.10.0/24 is directly connected, so R1 already knows this network.2. Remote networks, including the one containing Server 1, are reached through the WAN next hop 209.165.200.226.3. Configure a default route so that any unknown destination is forwarded to 209.165.200.226.4. The correct Cisco IOS syntax is ip route 0.0.0.0 0.0.0.0 209.165.200.226.5. This route will send all non-local traffic from the 192.168.10.0/24 LAN to the remote file server through the next hop. Verification / Alternative check: After configuration, you can verify the route with show ip route on R1. The default route should appear with a 0.0.0.0/0 entry pointing to 209.165.200.226. A ping or traceroute from a LAN host to the server address can further confirm end-to-end connectivity, provided that return routes exist on the remote side. Why Other Options Are Wrong: Option A points the default route to an outgoing interface only; while this can work in some setups, the question specifically highlights the next-hop IP, and relying solely on an interface can cause issues with multipoint networks. Option C defines a static route for network 209.165.200.0 with the next hop inside the LAN, which is not valid and does not point to the WAN. Option D incorrectly states that 192.168.10.0 is reached via 209.165.100.250, which conflicts with the fact that 192.168.10.0 is directly connected and should not point to a remote next hop. Common Pitfalls: Students often forget that default routes must use 0.0.0.0 0.0.0.0 as the destination and mask. Another pitfall is pointing a route to the wrong next-hop IP or to an address that is not directly reachable, which results in routing loops or dropped packets. Always verify that the next hop is on a directly connected network of the router where the static route is configured. Final Answer: ip route 0.0.0.0 0.0.0.0 209.165.200.226

Question 20

All hosts in a network have already received IP configuration from a DHCP server. Several hours later the DHCP server fails. What happens to these hosts?

Accepted Answer

Introduction / Context: Dynamic Host Configuration Protocol servers assign IP addresses and related configuration to client devices on a network. Understanding what happens if the DHCP server becomes unavailable is important for diagnosing connectivity issues and planning network redundancy. Many administrators worry that a DHCP outage will instantly disconnect users, but the behavior is controlled by the lease mechanism. Given Data / Assumptions: All hosts obtained their IP settings earlier from a functioning DHCP server. After several hours, the DHCP server goes down and cannot respond to requests. No configuration changes are made on the clients when the server fails. DHCP clients are using standard lease renewal behavior defined by the protocol. Concept / Approach: When a DHCP client obtains an IP address, it also receives a lease time indicating how long it may use that address. Until the lease expires, the client treats the configuration as valid and continues normal operation. The client begins to renew the lease before expiration, usually at half the lease duration. If the server is unreachable, renewal attempts fail, but the client still retains the address until the lease finally expires. Therefore, the client does not lose connectivity immediately when the server goes down. Step-by-Step Solution: 1. A DHCP client obtains an IP address, subnet mask, gateway, and DNS information, along with a lease time.2. The client uses this configuration for normal communication on the network.3. When the DHCP server fails, the client is not automatically notified.4. The client keeps using the leased address; only when the renewal time is reached does it attempt to contact the server.5. If the server remains down until the lease expires, the client may then lose its address; until then, communication continues as usual. Verification / Alternative check: You can verify this behavior by inspecting a client's IP configuration and lease times. On many operating systems, commands such as ipconfig or similar tools show the lease obtained and expiry time. Observing a running network during a planned DHCP outage confirms that existing clients continue to work until their leases age out. Why Other Options Are Wrong: Option A is incorrect because clients do not lose connectivity immediately when the server fails. Option C is incorrect because the ability to communicate inside or outside the network is unrelated to the server's current status; it depends on the IP configuration already in place. Option D is incorrect because name resolution continues to work as long as DNS settings are valid and DNS servers remain reachable, independent of the DHCP server itself. Common Pitfalls: Many learners assume that DHCP acts like a continuous control plane, where server failure instantly breaks client connectivity. In reality, DHCP is mostly used during initial configuration and renewal. Another pitfall is forgetting about the lease time, leading to confusion when some devices lose connectivity later while others continue to function, depending on when each lease expires. Final Answer: They continue to communicate normally until their DHCP leases expire.

CISCO Certification Questions