Question 1

In the standard data communication model, the channel (transmission medium) between sender and receiver can take which of the following forms?

Accepted Answer

Introduction / Context: A communication system comprises a source, channel, and destination, with possible noise and encoding/decoding steps. The “channel” is any medium capable of carrying messages from sender to receiver, whether physical or electromagnetic, synchronous or asynchronous, digital or analog.Given Data / Assumptions: The question frames “channel” broadly, not limited to digital links.Examples range from physical delivery to electrical and RF media.We must select the most comprehensive answer.Concept / Approach: Channels may be tangible (postal mail, couriers), guided electrical/optical (twisted pair, fiber, telephone lines), or unguided electromagnetic (radio, microwave, satellite). All listed options are valid channel types in an abstract communication model.Step-by-Step Solution: Assess each candidate: mail (physical channel), telephony (guided electrical/optical), radio (unguided EM).All qualify as channels moving messages from sender to receiver.Choose the inclusive option aggregating all valid examples.Verification / Alternative check: Shannon’s model and practical system diagrams treat the channel generically; modulation/coding are chosen based on the channel’s characteristics (delay, noise, bandwidth).Why Other Options Are Wrong: Single-medium options are incomplete views. None of the above: Incorrect because each listed example is a legitimate channel type.Common Pitfalls: Restricting “channel” to only electronic media; the communications abstraction encompasses any reliable conveyance of symbols.Final Answer: All of the above

Question 2

You administer the Class B network 134.57.0.0 and must create exactly 8 subnets while maximizing the number of usable host IDs per subnet. Which subnet mask should you assign?

Accepted Answer

Introduction / Context: Subnetting a Class B (/16) network involves borrowing bits from the host portion to create subnets. The design goal here is to produce exactly 8 subnets and keep each subnet as large as possible (maximum hosts per subnet).Given Data / Assumptions: Base network: Class B 134.57.0.0 with default /16.Required subnets: 8.Objective: largest possible hosts per subnet.Use contemporary practice that allows all-zeros and all-ones subnets.Concept / Approach: Number of subnets = 2^(borrowed bits). To get exactly 8 subnets, borrow 3 bits (since 2^3 = 8). New prefix = /16 + 3 = /19. The /19 mask in dotted decimal is 255.255.224.0 (since 224 = 11100000 in the third octet). Hosts per /19 subnet = 2^(32 − 19) − 2 = 8192 − 2 = 8190 usable addresses.Step-by-Step Solution: Borrow 3 bits from the host field: /16 → /19.Convert /19 to dotted decimal: 255.255.224.0.Confirm subnets: 2^3 = 8 distinct subnets.Confirm host capacity: 2^(13) − 2 = 8190 usable per subnet.Verification / Alternative check: Subnet increments occur every 32 in the third octet (.0, .32, .64, .96, .128, .160, .192, .224), yielding exactly eight /19 blocks—matching the requirement.Why Other Options Are Wrong: 255.255.240.0 (/20), 255.255.248.0 (/21), 255.255.252.0 (/22): Borrow 4, 5, or 6 bits → 16, 32, or 64 subnets (more than needed) and smaller host ranges. 255.255.255.255: Host mask, not usable for subnets.Common Pitfalls: Forgetting to maximize hosts (thus borrowing too many bits) or using outdated “no all-zeros/all-ones subnet” rules which would alter counts.Final Answer: 255.255.224.0

Question 3

A 2000-character text file must be sent through a 1,200-baud modem over an asynchronous serial link. Assuming standard 8-N-1 framing (1 start, 8 data, 1 stop = 10 bits/char), how long will the transfer take (ignoring protocol overhead)?

Accepted Answer

Introduction / Context: Serial modem throughput depends on baud rate (symbols/s) and framing. In typical asynchronous links, each character is wrapped with start/stop bits. Calculating transfer time requires converting characters to bits, dividing by the effective bit rate, and comparing against the choices.Given Data / Assumptions: File length: 2000 characters.Line rate: 1,200 baud (assume 1 bit/symbol as with simple FSK for baseline).Framing: 8-N-1 → 10 bits per character (1 start + 8 data + 1 stop).Ignore higher-layer overhead (no parity, no flow-control gaps, no retransmissions).Concept / Approach: Total bits = characters * bits/character. Time = total bits / bit rate. With 8-N-1, each character consumes 10 line bits. Therefore, 2000 chars require 20,000 bits. At 1,200 bit/s, time ≈ 16.67 s. None of the listed numeric options equals 16.67 exactly, so the correct choice among those presented is “None of the above.”Step-by-Step Solution: bits_per_char = 10 (8-N-1).total_bits = 2000 * 10 = 20,000 bits.time_seconds = 20,000 / 1,200 ≈ 16.67 s.Compare to options: 2, 12, 20, 120 → none matches ≈16.67 s.Verification / Alternative check: If parity were used (8-E-1 → 11 bits/char), time would be ≈ 18.33 s; if one incorrectly ignores start/stop (8 bits only), time ≈ 13.33 s. These also do not match the provided options, reinforcing “None of the above.”Why Other Options Are Wrong: 12 s: Implies only 8 bits/char at 1,333 bit/s or similar incorrect assumptions. 20 s: Would require 24,000 bits transmitted or a lower effective rate. 2 s and 120 s: Orders of magnitude off for the stated parameters.Common Pitfalls: Forgetting framing overhead in asynchronous links and equating baud to characters per second. Always multiply by bits per framed character.Final Answer: None of the above

Question 4

Within the OSI reference model, which layer is Layer 1 and deals with the physical means of transmitting raw bits over a medium?

Accepted Answer

Introduction / Context: The OSI model splits networking into seven layers. The lowest layer defines the mechanical, electrical, optical, and procedural characteristics for moving bits across a link. Identifying this layer is foundational for troubleshooting cabling, signaling, and connectors.Given Data / Assumptions: We are mapping the layer number (1) to its name and responsibility.Focus is on raw bit transmission over the physical medium.Concept / Approach: Layer 1 is the Physical layer. It specifies voltage levels, timing, pinouts, modulation, line coding, wavelengths, and connectors. It does not include framing or addressing (Layer 2) nor routing (Layer 3).Step-by-Step Solution: Recall OSI stack from bottom up: Physical, Data Link, Network, Transport, Session, Presentation, Application.Match Layer 1 to “Physical.”Confirm scope: signaling and media only.Verification / Alternative check: Ethernet physicals (e.g., 1000BASE-T, 10GBASE-SR) and fiber standards are classic Layer-1 examples; MAC addressing belongs to Layer 2.Why Other Options Are Wrong: link layer: This is Layer 2 (Data Link). transport / network: Layer 4 and 3 respectively; not physical signaling. None of the above: Physical layer fits exactly.Common Pitfalls: Attributing error detection or media access to Layer 1; those are Layer-2 responsibilities.Final Answer: physical layer

Question 5

In classical TCP/IP, when a host knows its hardware (MAC) address but not its own IP address, which protocol can it use to discover the IP address?

Accepted Answer

Introduction / Context: Early diskless workstations and embedded devices sometimes booted knowing only their MAC address. They needed a way to obtain an IP address from a server on the local network before higher-level configuration could proceed.Given Data / Assumptions: Host starts with a known MAC address.IP address is unknown and must be discovered.Legacy TCP/IP environment (pre-DHCP or alongside BOOTP/DHCP).Concept / Approach: RARP (Reverse Address Resolution Protocol) maps a known hardware address to an IP address, essentially the inverse of ARP. A RARP server listens on the LAN, receives a RARP request containing the client’s MAC, and replies with the corresponding IP. Modern networks typically use BOOTP/DHCP instead, but RARP is the direct answer to the legacy question.Step-by-Step Solution: Identify the need: MAC → IP mapping.Match protocol to function: that is RARP.Note that ARP does the opposite (IP → MAC) and is not applicable here.Verification / Alternative check: DHCP/BOOTP workflows can also assign IPs using broadcast discovery, but the specific “reverse” of ARP was historically RARP.Why Other Options Are Wrong: ARP: Resolves IP to MAC, not MAC to IP. IGMP: Manages multicast group membership. ICMP: Control/diagnostic messaging (e.g., echo). None of the above: Incorrect because RARP fits exactly.Common Pitfalls: Mixing up ARP and RARP directions; remembering that “Reverse” in RARP indicates MAC → IP assistance.Final Answer: RARP

Question 6

Most data communications over traditional telegraph or teletype lines operate over which type of channel?

Accepted Answer

Introduction / Context: Telegraphy and teletype (TTY) systems predate modern broadband data services and operated with very limited bandwidth and symbol rates. Understanding their channel characteristics helps contextualize early data networks and low-speed serial communication.Given Data / Assumptions: We are discussing legacy telegraph/TTY lines.Throughput and frequency ranges were modest (tens to hundreds of baud historically).We must choose the best descriptor of channel bandwidth.Concept / Approach: Telegraph/TTY services are classic examples of narrowband communications: low symbol rates and constrained frequency response suitable for simple modulation schemes. Wideband describes channels supporting much higher rates or broader spectra and does not fit legacy telegraph services. “Dialed service” refers to connection establishment rather than bandwidth classification.Step-by-Step Solution: Identify historical data rates: low, often measured in baud for teleprinters.Map low data rate to narrowband classification.Select “narrowband channels.”Verification / Alternative check: Early modems and teleprinters (e.g., 45.45 baud RTTY) operated over limited audio bandwidth lines; these are quintessential narrowband systems.Why Other Options Are Wrong: simplex lines: Directionality term, not bandwidth; many TTY links were half-duplex or full-duplex variants. wideband channels: Too much capacity relative to telegraph systems. dialed service: Describes access method, not bandwidth class. None of the above: Incorrect because narrowband is apt.Common Pitfalls: Confusing duplex terms with bandwidth classes; they describe different aspects of a link.Final Answer: narrowband channels

Question 7

On Microsoft networks, which local file is specifically used to resolve NetBIOS names to IP addresses for NetBIOS name resolution on a host?

Accepted Answer

Introduction / Context: Name resolution on TCP/IP hosts can involve multiple databases and protocols. While the HOSTS file maps DNS hostnames to IP addresses, classic Windows networking also used NetBIOS names. A different local file supported static mappings for those NetBIOS names.Given Data / Assumptions: We are resolving NetBIOS names (not DNS hostnames).Local, file-based resolution is in scope.Operating context is classic Windows/NetBIOS over TCP/IP.Concept / Approach: The LMHOSTS file provides static mappings of NetBIOS names to IP addresses for NetBIOS name resolution. By contrast, the HOSTS file maps DNS hostnames. ARP is a protocol used to resolve IP addresses to MAC addresses on a LAN, and FQDN is a DNS concept (fully qualified domain name), not a file.Step-by-Step Solution: Identify the namespace: NetBIOS names.Select the file that maps NetBIOS name → IP: LMHOSTS.Exclude HOSTS (DNS), ARP (IP→MAC protocol), and FQDN (DNS term).Verification / Alternative check: Administrators historically edited LMHOSTS to work around WINS/DNS issues for NetBIOS name resolution, especially in small networks without WINS.Why Other Options Are Wrong: HOSTS: Used for DNS hostname mappings, not NetBIOS. ARP: Dynamic protocol, not a file or NetBIOS resolver. FQDN: A DNS naming construct; not a file. None of the above: Incorrect because LMHOSTS is correct.Common Pitfalls: Assuming HOSTS also covers NetBIOS; in classic stacks, LMHOSTS was separate for NetBIOS while HOSTS supported DNS.Final Answer: LMHOSTS

Question 8

In the OSI reference model, which layer transforms data from machine-dependent representations into forms suitable for users and applications (e.g., translation, encryption, and compression)?

Accepted Answer

Introduction:Understanding the Open Systems Interconnection (OSI) model helps diagnose networking problems and design interoperable systems. A frequent point of confusion is which layer “makes data understandable”—that is, which layer handles representation issues like character set translation, data compression, and encryption/decryption.Given Data / Assumptions:We are mapping a functional description to a specific OSI layer.“Transforms information from machine format into that understandable by user” implies data representation services.We assume the canonical seven-layer OSI stack: Physical, Data Link, Network, Transport, Session, Presentation, Application.Concept / Approach:The Presentation layer (Layer 6) is responsible for data representation. It provides translation between different character sets (e.g., ASCII vs. EBCDIC), manages data structure serialization, performs encryption/decryption, and applies compression. The Application layer (Layer 7) provides user-facing services (e.g., HTTP, SMTP), while the Session layer (Layer 5) coordinates dialogues and synchronization. Lower layers handle transport and forwarding, not representation.Step-by-Step Solution:1) Identify keywords: translation, formatting, encryption, compression.2) Associate these functions with OSI Layer 6 (Presentation).3) Exclude Application (Layer 7) because it offers service interfaces, not the generic representation transformations.4) Confirm that Physical (Layer 1) only transmits bits and does not alter data semantics.Verification / Alternative check:Many textbooks summarize Layer 6 as “the syntax layer,” bridging differences in representation so Applications can interoperate regardless of underlying machine formats.Why Other Options Are Wrong:application: offers end-user services, not general representation functions.session: manages dialogue control and checkpoints, not format translation.physical: defines signaling and connectors; no data formatting semantics.None of the above: invalid because Presentation is correct.Common Pitfalls:Attributing encryption to the Application layer exclusively; while applications can encrypt, the OSI model attributes generic encryption/decryption capability to the Presentation layer.Final Answer:presentation

Question 9

In cyclic redundancy check (CRC) encoding at the sender, what happens to the quotient obtained during the modulo-2 division by the generator polynomial?

Accepted Answer

Introduction:CRC (cyclic redundancy check) is a robust error-detection method used in networking and storage. Understanding each artifact of the division process—dividend, divisor (generator), quotient, and remainder—clarifies how transmitters and receivers implement CRC efficiently.Given Data / Assumptions:The sender divides an augmented data polynomial by a generator polynomial using modulo-2 arithmetic.The process yields a quotient and a remainder (the CRC).The receiver re-checks integrity using the same generator.Concept / Approach:In CRC, only the remainder is significant for transmission integrity; it is appended to the data as the Frame Check Sequence (FCS). The quotient is an intermediate computational result with no role in verification and is therefore discarded. At the receiver, the concatenated data+CRC is divided by the same generator; a zero remainder indicates no detectable error.Step-by-Step Solution:1) Sender forms the message polynomial (data shifted by k bits where k = degree of generator).2) Compute modulo-2 division: message / generator → quotient and remainder.3) Append the remainder (CRC) to the original data; ignore the quotient.4) Receiver divides (data + CRC) by the generator and expects remainder = 0.Verification / Alternative check:Standards such as Ethernet specify a 32-bit CRC appended to frames; no field carries the quotient—confirming it is not transmitted.Why Other Options Are Wrong:becomes the dividend/divisor: incorrect; the receiver’s dividend is the received frame; the divisor remains the agreed generator polynomial.is the remainder: wrong; the remainder is the FCS, not the quotient.None of the above: invalid because the quotient is discarded.Common Pitfalls:Confusing intermediate math artifacts with transmitted fields; only the remainder (CRC) is sent, not the quotient.Final Answer:is discarded

Question 10

Identify the communications medium that transports data as pulses of laser light through strands of glass or plastic fibers.

Accepted Answer

Introduction:Choosing the correct physical medium is fundamental to network performance. One medium stands out for its extremely high bandwidth, low attenuation, and immunity to electromagnetic interference: the optical fiber link that carries digitally modulated light pulses over long distances.Given Data / Assumptions:The description references laser light and glass fibers.We are selecting the medium, not a service, device class, or software.Modern networks use both single-mode and multi-mode fibers.Concept / Approach:Fiber optic cables guide light through a core using total internal reflection. Data is encoded by modulating light sources (lasers or LEDs), enabling very high symbol rates and long-reach transmission. Compared to copper, fiber offers greater bandwidth, lower signal loss, and resilience to electromagnetic noise, which is why it dominates backbones, data centers, and submarine cables.Step-by-Step Solution:1) Match “pulses of laser light” and “glass fibers” to optical fiber technology.2) Exclude cellular (a radio access network, not a cabled medium).3) Exclude processors and software (they are components and programs, not transmission media).4) Select Fiber optic cables.Verification / Alternative check:Typical standards include ITU-T G.652 (single-mode) and OM3/OM4 multi-mode fibers used with high-speed Ethernet optics.Why Other Options Are Wrong:Cellular phone systems: wireless RF service, not a fiber medium.Telecommunications processors/software: functional components, not physical links.None of the above: invalid because fiber is correct.Common Pitfalls:Confusing “optical network” services with the underlying medium; here the medium itself is the subject.Final Answer:Fiber optic cables

Question 11

Transport protocols: Which of the following is connection-oriented and establishes a reliable end-to-end session before data transfer?

Accepted Answer

Introduction:Networking stacks separate connectionless delivery from connection-oriented, reliable transport. Identifying which protocol provides ordered delivery, retransmissions, and flow control is vital for application design and troubleshooting latency or loss issues.Given Data / Assumptions:The options mix transport and legacy protocol families.“Connection-oriented” implies a handshake, sequencing, acknowledgments, and flow/congestion control.We assume common textbook definitions.Concept / Approach:Transmission Control Protocol (TCP) establishes a connection using a three-way handshake, numbers bytes, acknowledges receipt, retransmits lost data, and manages flow and congestion to provide a reliable stream. UDP is connectionless and unreliable by design. IPX and NetBEUI are legacy network/transport family protocols that do not provide TCP-like reliable, connection-oriented semantics in the standardized Internet stack.Step-by-Step Solution:1) Map “connection-oriented and reliable” to TCP features.2) Exclude UDP (no connections or reliability guarantees).3) Exclude IPX/NetBEUI since they are not the standard connection-oriented transport layer like TCP in today’s networking model.4) Select TCP.Verification / Alternative check:Packet traces show SYN, SYN-ACK, ACK handshake followed by sequenced payloads and ACKs—canonical TCP behavior.Why Other Options Are Wrong:IPX: legacy Novell protocol suite; not the Internet’s reliable transport.UDP: connectionless datagrams without built-in reliability.NetBEUI: non-routable protocol; not a TCP-like transport.None of the above: invalid because TCP is correct.Common Pitfalls:Assuming “connection-oriented” means stateful firewall sessions; here it specifically refers to transport protocol semantics.Final Answer:TCP

Question 12

Dial-up data communications: What is the name given to the exchange of control signals required to establish a connection between the modem+computer at one end of a line and the modem+computer at the other end?

Accepted Answer

Introduction:Before data transfer can occur over dial-up or other circuit-oriented links, endpoints must synchronize their parameters. This preliminary exchange ensures both sides agree on speeds, modulation, compression, and error control. Recognizing the term for this process is essential in legacy networking and still relevant conceptually in modern link negotiation.Given Data / Assumptions:Two modems and their attached computers attempt to form a reliable link.Control signals and messages precede payload data.We want the standard term for this negotiation.Concept / Approach:Handshaking refers to the initial signal exchange that sets up communication parameters. In classic modems, this includes carrier detection, training sequences, and feature negotiation (e.g., error correction, compression). While “protocol” is the broader set of rules governing communication, handshaking is the specific connection-establishment phase. “Duplexing” describes simultaneous send/receive capability, not the negotiation itself.Step-by-Step Solution:1) Identify the need: agree on link parameters prior to data.2) Name the process: handshaking.3) Distinguish from generic terms like protocol or duplexing.4) Choose “Handshaking.”Verification / Alternative check:Audible modem “tones” during dial-up were the handshaking and training sequences setting modulation profiles and speeds.Why Other Options Are Wrong:Modem options: configuration choices, not the negotiation process.Protocol: too broad; encompasses the entire rule set, not just connection establishment.Duplexing: describes directionality, not the setup handshake.None of the above: invalid because handshaking is correct.Common Pitfalls:Using “protocol” as a catch-all; precise terminology distinguishes phases like handshaking from ongoing data transfer.Final Answer:Handshaking

Question 13

Core responsibilities of an operating system (OS): which selection best summarizes standard OS functions used in general-purpose computing?

Accepted Answer

Introduction:An operating system provides the essential software layer that manages hardware and offers services to applications. Recognizing the breadth of OS responsibilities helps students appreciate resource management and performance trade-offs across different platforms.Given Data / Assumptions:The question lists three canonical OS capabilities.We assume a multi-user or general-purpose OS context.We must determine whether each listed function belongs to typical OS duties.Concept / Approach:Typical OS functions include: I/O control (device drivers, buffering, scheduling), memory management (paging/segmentation; “virtual storage” exposes large logical address spaces), and multiprogramming (scheduling multiple processes to share CPU). These, along with file systems, security, inter-process communication, and networking stacks, represent core OS services.Step-by-Step Solution:1) Evaluate I/O control: the OS abstracts devices via drivers and I/O subsystems.2) Evaluate virtual storage: OS implements paging/virtual memory mapping between processes and physical RAM.3) Evaluate multiprogramming: OS schedulers interleave processes to maximize CPU utilization.4) Since all three are bona fide OS functions, select “All of the above.”Verification / Alternative check:Standard OS texts (e.g., modern UNIX, Windows, Linux documentation) enumerate these features as foundational components.Why Other Options Are Wrong:input/output control, virtual storage, or multiprogramming alone understate the OS role.None of the above: incorrect because all listed items are valid OS functions.Common Pitfalls:Assuming virtual memory is optional; while embedded systems may omit it, general-purpose OS designs include it as a core capability.Final Answer:All of the above

Question 14

Teleprinters/Teletypes (TTY): which single statement best describes their functionality and common terminology in classic data communications?

Accepted Answer

Introduction:Before graphical terminals and PCs, electromechanical terminals played a central role in human–computer interaction over serial lines. Knowing what a teleprinter is—and how it relates to the term “teletype” (TTY)—clarifies many historical references in operating systems and networking literature.Given Data / Assumptions:Teleprinters were used as terminals on serial links and telex networks.They supported both input and hard-copy output.We are choosing the single best descriptive statement.Concept / Approach:A teleprinter (often branded Teletype, hence TTY) integrates a typewriter-like keyboard with a printing mechanism. Users type commands; responses print on paper. TTY conventions persist today as a term for terminals in UNIX-like systems. Teleprinters are not electronic mail systems, displays only, or output-only devices—they are interactive text terminals with integrated printing.Step-by-Step Solution:1) Identify core functions: keyboard input + printed output on paper.2) Recognize synonym: “teletype” → TTY.3) Exclude claims that they are output-only or that they equal X.400 email.4) Select the composite description capturing name and functionality.Verification / Alternative check:Historical Teletype models (e.g., ASR-33) provided both a keyboard and a printer; “/dev/tty” device files in UNIX trace back to these terminals.Why Other Options Are Wrong:Only printing, no input: incorrect—teleprinters include keyboards.Same as X.400: X.400 is an email standard, unrelated to electromechanical terminals.Display only: the opposite; teleprinters print hard copy.None of the above: invalid because option B is correct.Common Pitfalls:Confusing teleprinters with later CRT/VDU terminals or with messaging protocols; teleprinters are physical, hard-copy terminals.Final Answer:Teleprinters are commonly called teletypes (TTY) and combine a keyboard for input with a printer for output.

Question 15

In data networks, the simultaneous one-to-many transmission of the same data to all stations (or to all in a defined domain) is termed what?

Accepted Answer

Introduction:Network transmission scopes include unicast (one-to-one), multicast (one-to-many, group-scoped), and broadcast (one-to-all on a local domain). Recognizing these terms is essential for configuring switches, routers, and wireless networks, and for understanding traffic behavior and security implications.Given Data / Assumptions:The phrase “simultaneous transmission to a number of stations” indicates a one-to-many pattern.We focus on the conventional local network notion where all stations on a segment may receive the frame.Terminology should match standard data communications usage.Concept / Approach:Broadcast means sending a frame or signal such that all nodes in the broadcast domain receive it (e.g., Ethernet FF:FF:FF:FF:FF:FF, IPv4 255.255.255.255 or subnet broadcast). This differs from multicast (group-scoped) and from protocol names like ALOHA (a random access method). “Bandwidth” is a capacity metric, not a transmission scope. “Analog transmission” indicates signal form, not audience scope.Step-by-Step Solution:1) Map “one sender → many/all receivers” to broadcast.2) Exclude unrelated terms: bandwidth (capacity), ALOHA (MAC protocol), analog (signal type).3) Select “broadcast.”4) Note that on routed networks, broadcasts are confined to a local domain for scalability.Verification / Alternative check:Switch behavior (flooding unknown unicast and propagating broadcasts) illustrates how broadcast reaches all ports within a VLAN unless filtered.Why Other Options Are Wrong:Band width: not a delivery scope.Aloha: contention protocol, not a transmission scope definition.Analog transmission: describes signaling, not audience.None of the above: invalid because broadcast is correct.Common Pitfalls:Using “broadcast” when “multicast” is intended; multicast targets subscribed members, not every station.Final Answer:broadcast

Question 16

Communication interconnects: which transmission system typically provides the highest data rate to a single attached device?

Accepted Answer

Introduction:Not all links are created equal. Some are engineered for long distance and sharing, others for ultra-high local bandwidth. Understanding which interconnect class delivers the highest per-device data rate clarifies why internal computer fabrics look very different from wide-area circuits.Given Data / Assumptions:We compare a local computer bus with various telecom links.“Highest data rate to an individual device” means peak local attachment bandwidth, not aggregate network capacity.Assume modern examples: PCIe lanes vs. dial-up modems vs. copper leased lines.Concept / Approach:A computer bus (e.g., PCI Express) connects components at chip or board level, offering gigabytes-per-second throughput per device with very low latency. In contrast, telephone lines and voice band modems are limited by voice-bandwidth channels (kilobits per second historically). Leased lines deliver higher WAN speeds than dial-up but remain far below on-board interconnect bandwidth and typically serve shared network segments rather than a single internal device link.Step-by-Step Solution:1) Define the metric: peak per-device attachment rate.2) Evaluate options: computer bus (chip/board interconnect) vastly outstrips telecom circuits in raw throughput.3) Exclude telephone/modem as low-rate legacy media.4) Exclude leased lines due to WAN nature and lower per-device rates.Verification / Alternative check:PCIe Gen4/Gen5 per-lane rates (multi-GB/s) eclipse typical WAN circuits by orders of magnitude, confirming the bus’s advantage.Why Other Options Are Wrong:Telephone lines/Voice band modem: severely bandwidth-limited.Leased lines: higher than dial-up but still far lower than local buses; also designed for site-to-site networking.None of the above: invalid because the computer bus clearly wins.Common Pitfalls:Comparing aggregate WAN capacity with dedicated device links; the question asks about rate to an individual device.Final Answer:Computer bus

Question 17

Early data communications history: which organization is most closely associated with inventing and first deploying the modem for digital data over telephone networks?

Accepted Answer

Introduction:The modem—short for modulator-demodulator—enabled digital data to traverse analog telephone circuits and catalyzed remote computing and early Internet access. Identifying the organization behind the first practical modems anchors the history of data communications technology.Given Data / Assumptions:We focus on the earliest practical development and deployment for telephone networks.Choices include several tech firms; only one aligns with the origin of modems for public network use.Terminology maps AT&T corporate structures to the Bell System and Bell Labs heritage.Concept / Approach:Modem technology traces to the Bell System in the 1950s–1960s. Bell Labs and AT&T developed early data sets (e.g., Bell 101 modem), originally used in defense and airline reservation systems before widespread commercial adoption. While other companies later produced influential modems or computer systems, the invention and initial standardization are historically attributed to the Bell/AT&T ecosystem.Step-by-Step Solution:1) Associate “first modem deployments” with Bell 101/103 series from the Bell System.2) Map organizational naming: AT&T Information Systems is the commercial arm linked to these developments.3) Exclude later or unrelated vendors.4) Select AT & T Information Systems, USA.Verification / Alternative check:Historical references to Bell 101 (circa 1958) and Bell 103 (300-bps) cite Bell Labs/AT&T as originators for public network modems.Why Other Options Are Wrong:Wang Laboratories/Digital Equipment: important computing firms but not credited with inventing the modem.Apple Computers Inc.: a later PC company, not involved in the origin of telephone modems.None of the above: invalid because the Bell/AT&T lineage is correct.Common Pitfalls:Confusing popular consumer modem brands of the 1990s with the earlier invention period dominated by the Bell System.Final Answer:AT & T Information Systems, USA

Question 18

IPv4 classes and defaults: given the IP address 201.142.23.12 (a Class C address), what is the default subnet mask that applies before any custom subnetting?

Accepted Answer

Introduction / Context: In classical IPv4 addressing (pre–Classless Inter-Domain Routing), each address fell into a class with a default subnet mask. Recognizing the default mask for a given dotted-decimal address helps with quick network/broadcast calculations and legacy exam questions. Here, we analyze address 201.142.23.12 and identify its default mask prior to any subnetting changes.Given Data / Assumptions:Address: 201.142.23.12We assume traditional classful defaults (no VLSM/CIDR overrides).We seek the default mask only, not a designed subnet mask.Concept / Approach: Classful ranges by first octet are: Class A = 1–126 → default mask 255.0.0.0; Class B = 128–191 → default mask 255.255.0.0; Class C = 192–223 → default mask 255.255.255.0. Since 201 falls within 192–223, 201.x.y.z is Class C, so the default mask is 255.255.255.0.Step-by-Step Solution:Identify first octet: 201.Map 201 to class ranges: 192–223 → Class C.Recall default mask for Class C: 255.255.255.0.Select 255.255.255.0 as the correct answer.Verification / Alternative check: Convert 201 to binary (11001001). The leading bits for Class C are 110, which matches the 192–223 range and therefore the /24 (255.255.255.0) default network mask in classful notation.Why Other Options Are Wrong:0.0.0.0: not a default mask; often represents a default route or unspecified address context.255.0.0.0: Class A default; does not apply to 201.x.x.x.255.255.0.0: Class B default; does not apply to 201.x.x.x.None of the above: incorrect because 255.255.255.0 is correct.Common Pitfalls: Confusing classful defaults with designed subnet masks; mixing up Class B and C boundaries (191 vs 192). Remember, 201 clearly lies in Class C.Final Answer: 255.255.255.0.

Question 19

Media access control basics: which item below names a multiple-access channel access technique (MAC) rather than a signal property or impairment?

Accepted Answer

Introduction / Context: On shared media, multiple stations contend to transmit. Media Access Control (MAC) techniques define rules for who can send and when. Distinguishing a MAC method from unrelated physical-layer terms or impairments is a foundational networking skill. This question asks you to pick the item that is actually a MAC technique for multiple-access media.Given Data / Assumptions:We compare one MAC method with physical-layer concepts.Only one option corresponds to a contention-access protocol.We are not evaluating modulation quality or link loss.Concept / Approach: Aloha is a random-access MAC protocol originating from early packet radio networks. Stations transmit when ready; collisions may occur and are recovered by retransmission. Slotted Aloha improves efficiency by forcing transmissions to time slots. By contrast, amplitude is a waveform magnitude; angle modulation is a family of analog modulation methods (frequency/phase); attenuation describes signal loss over distance or media—not MAC behavior.Step-by-Step Solution:Identify which option governs medium access among multiple users.Recognize Aloha as a contention-based access protocol.Eliminate physical-layer terms (amplitude, angle modulation) and impairments (attenuation).Select Aloha as the correct MAC technique.Verification / Alternative check: Many textbooks trace Ethernet’s early design lineage through Aloha concepts, demonstrating Aloha’s role as a seminal MAC approach.Why Other Options Are Wrong:Amplitude: a signal magnitude, not an access rule.Angle modulation: FM/PM analog modulation, not MAC.Attenuation: loss phenomenon, not an access method.None of the above: incorrect because Aloha is valid.Common Pitfalls: Confusing modulation or channel impairments with MAC algorithms; MAC operates at the link layer, while modulation/attenuation are physical-layer concerns.Final Answer: Aloha.

Question 20

Telecommunications capabilities: networks come in a wide range of speeds and capacities. What is the best term for these differing speed/capacity choices available to designers and users?

Accepted Answer

Introduction / Context: When planning a network, you choose among many service options that offer different maximum data rates and capacities (for example, DSL tiers, fiber uplinks, leased lines, wireless links). The question asks for the phrase that best captures this variety of throughput choices available to adopters.Given Data / Assumptions:We focus on speed/capacity options, not topology or organizational scope.We do not evaluate switching types or inter-company connectivity models.We want a generic term describing selectable data rate options.Concept / Approach: The phrase “bandwidth alternatives” concisely describes selectable offerings with different bit-rate ceilings and service characteristics. “Switching alternating” is not a standard term; “inter organizational networks” and “extranets” describe who connects with whom, not the range of speed/capacity options themselves.Step-by-Step Solution:Identify that the stem emphasizes speed/capacity ranges.Map this to the general term “bandwidth alternatives.”Exclude organizational or nonstandard phrases.Select “Bandwidth alternatives.”Verification / Alternative check: Service catalogs commonly list bandwidth tiers (e.g., 50 Mbps, 100 Mbps, 1 Gbps), illustrating the notion of “alternatives” in capacity.Why Other Options Are Wrong:Switching alternating: not recognized terminology.Inter organizational networks: describes scope/ownership, not speed options.Extranets: partner-facing private networks; unrelated to capacity ranges.None of the above: incorrect because “Bandwidth alternatives” fits best.Common Pitfalls: Confusing who connects (extranet/intranet) with how fast you can connect (bandwidth tiers).Final Answer: Bandwidth alternatives.

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