Question 1

Operating systems – multiprogramming outcomes: Which of the following is an expected benefit of multiprogramming compared with single-program execution?

Accepted Answer

Introduction / Context:Multiprogramming lets the operating system keep several programs in memory at once, overlapping CPU and I/O so resources are not wasted. This increases system throughput and reduces idle time, a foundational concept for modern OS design and performance tuning.Given Data / Assumptions:There are multiple independent jobs with CPU and I/O bursts.The OS can context switch among jobs when one blocks for I/O.Memory management is sufficient to hold more than one process.Concept / Approach:When one job waits for I/O, the CPU can execute another. Overlapping work in this way reduces idle CPU cycles, improves throughput, and, with appropriate scheduling, leads to better utilization across the system. Therefore, several positive outcomes occur together.Step-by-Step Solution:Consider a workload with alternating CPU and I/O phases.In single-program mode, the CPU sits idle during I/O waits, increasing total elapsed time.In multiprogramming, the OS dispatches a ready process while another waits, keeping the CPU busy.More jobs complete per unit time (higher throughput) and idle time decreases.Schedulers (for example, round-robin, priority) improve responsiveness and fairness.Verification / Alternative check:Queueing models and OS metrics (CPU utilization, throughput, turnaround) consistently show higher values under multiprogramming compared to serial execution for mixed workloads.Why Other Options Are Wrong:a, b, c: Each is correct individually, but the most complete choice is all of the above.e: Incorrect because multiprogramming does deliver these benefits.Common Pitfalls:Confusing multiprogramming (multiple in memory) with multiprocessing (multiple CPUs); overlooking that high context-switch overhead or memory contention can limit gains if not tuned properly.Final Answer:All of the above

Question 2

ARP fundamentals: Which of the following statements about the Address Resolution Protocol (ARP) is incorrect?

Accepted Answer

Introduction / Context:The Address Resolution Protocol (ARP) maps IP addresses to link-layer addresses (for example, MAC) on a local network. While its operation is straightforward, exam questions often test nuanced behaviors about broadcasting, cache updates, and protocol layering. This item asks you to spot the incorrect statement.Given Data / Assumptions:ARP requests are broadcasts on the local link; ARP replies are typically unicasts.Each ARP packet carries the sender’s protocol and hardware addresses.Hosts maintain ARP caches with aging to avoid repeated broadcasts.Concept / Approach:Evaluate each statement against how ARP actually works. Pay attention to the words “every ARP broadcast” and “before processing,” which are red flags. ARP cache entries are usually updated upon reception/processing of a valid ARP packet, with implementation-specific safeguards (for example, gratuitous ARP handling).Step-by-Step Solution:Option a is true: ARP resolves an IP to a link-layer address on the same broadcast domain.Option b is problematic: although ARP requests include the sender’s binding, “every ARP broadcast” is not the whole story (replies are not broadcasts), and caches are updated upon validation/processing—not necessarily “before processing.”Option c is true conceptually: ARP abstracts hardware addressing, allowing independent IP assignment.Therefore, the incorrect statement is option b.Verification / Alternative check:Protocol traces (for example, tcpdump) show ARP requests as broadcasts and replies as unicasts. OS implementations document when ARP caches are refreshed (often after sanity checks).Why Other Options Are Wrong:a: Accurate description of ARP’s purpose.c: Correct high-level summary of ARP’s role in decoupling IP from MAC addressing.d: “All of the above” cannot be correct because at least one statement (a or c) is true.e: Not correct because one statement (b) is indeed incorrect.Common Pitfalls:Assuming all ARP packets are broadcasts, or that hosts blindly update caches without validation. Also, mixing ARP’s function with ICMP or DHCP responsibilities leads to confusion.Final Answer:The sender’s IP-to-physical binding is included in every ARP broadcast, and receivers update their ARP cache before processing the packet.

Question 3

Subnetting on a Class B network: With a subnet mask that uses 248 in the third octet (that is, 255.255.248.0), how many usable host addresses are available per subnet?

Accepted Answer

Introduction / Context:Knowing how to compute hosts per subnet is essential for IP addressing plans. This problem focuses on a Class B network using a subnet mask with 248 in the third octet, a common way to describe 255.255.248.0. You must convert the mask to prefix length and then compute usable hosts.Given Data / Assumptions:Subnet mask: 255.255.248.0.Class B addressing (traditional context); calculation is independent of “classful” rules once the mask is given.Usable hosts exclude the all-zeros (network) and all-ones (broadcast) addresses.Concept / Approach:Compute prefix length by counting mask bits. Then use the formula hosts per subnet = 2^(host_bits) - 2. Here, 248 corresponds to 11111000 in binary, providing 5 bits of subnetting in the third octet and leaving 3 bits in the third plus 8 bits in the fourth for hosts? Instead, treat it uniformly: a /21 mask leaves 11 host bits across the last two octets.Step-by-Step Solution:Convert mask: 255 (8 bits) + 255 (8 bits) + 248 (5 bits) + 0 (0 bits) = 8 + 8 + 5 + 0 = 21. So prefix is /21.Compute host bits: 32 - 21 = 11.Apply the formula: usable hosts = 2^11 - 2.Evaluate: 2^11 = 2048; 2048 - 2 = 2046.Thus, each /21 subnet supports 2,046 usable host addresses.Verification / Alternative check:Use a subnet calculator or quick table: /24 → 254 hosts; adding three more host bits (/21) multiplies by 2^3 = 8, giving 254 * 8 + small rounding to the precise 2046. Both methods agree.Why Other Options Are Wrong:a: 16,382 corresponds to /18 (14 host bits) not /21.b: 8,190 aligns with /19 (13 host bits) not this mask.c: 4,094 is /20 (12 host bits) not /21.e: 1,022 is /22 (10 host bits) not /21.Common Pitfalls:Mixing classful boundaries with CIDR math, miscounting bits in 248 (it is 11111000, five ones), or forgetting to subtract the network and broadcast addresses. Always recompute from the mask bits to avoid class-based confusion.Final Answer:2,046

Question 4

Signals and modulation: Demodulation is best described as which of the following processes?

Accepted Answer

Introduction / Context:Communication systems use modulation to impress information onto a carrier and demodulation to recover it. Recognizing the basic transmitter/receiver functions is essential whether dealing with analog voice, digital data, or hybrid systems such as modems and radio links.Given Data / Assumptions:At the transmitter, information is modulated onto a carrier suitable for the channel.At the receiver, the modulated signal must be demodulated to retrieve the baseband information.The baseband can be analog (audio) or digital (bitstream), depending on the system design.Concept / Approach:Modulation direction is “up” (baseband to passband); demodulation direction is “down” (passband to baseband). In data modems, demodulation recovers a digital bitstream from an analog carrier. In analog radio, it recovers an audio waveform. In all cases, demodulation extracts the original information from the received signal.Step-by-Step Solution:Identify the role of the transmitter: convert baseband into a modulated passband signal (modulation).Identify the role of the receiver: convert the received modulated signal back to baseband (demodulation).Map options: a is modulation, b is demodulation, c is multiplexing (specifically TDM/aggregation), d is frequency-division multiplexing.Select the definition matching demodulation: option b.Verification / Alternative check:Block diagrams of radios and modems consistently show the demodulator after the channel, producing the baseband output that feeds a decoder or audio stage.Why Other Options Are Wrong:a: Describes modulation (for example, DAC and modulator), not demodulation.c: Describes multiplexing/aggregation, unrelated to demodulation.d: Describes FDM (splitting bandwidth), not demodulation.e: Not applicable because option b is correct.Common Pitfalls:Assuming demodulation always produces analog output; in digital systems, demodulation yields a digital symbol or bitstream. Also, confusing multiplexing techniques with modulation/demodulation functions is common.Final Answer:Converting analog signals to the original digital (or baseband) information at the receiver

Question 5

IPv4 subnet design: For the private network 192.168.10.0, choose a subnet mask that supports up to 25 hosts per subnet while maximizing the number of subnets.

Accepted Answer

Introduction / Context:Choosing a subnet mask is a balancing act between hosts per subnet and number of subnets. The requirement here is to support at least 25 hosts while maximizing subnet count, so we should pick the smallest mask that still yields ≥25 usable addresses.Given Data / Assumptions:Network: 192.168.10.0 (a class C private block, but we use CIDR math).Need at least 25 usable hosts per subnet.Want the maximum number of subnets (choose the most restrictive mask that still meets hosts requirement).Concept / Approach:Usable hosts = 2^(host_bits) - 2. Evaluate candidate masks and find the smallest number of host bits that gives ≥25. The fewer host bits used, the more network (subnet) bits available, which increases the number of subnets.Step-by-Step Solution:Test /27 (255.255.255.224): host_bits = 5 → usable hosts = 2^5 - 2 = 30 (meets 25).Test /28 (255.255.255.240): host_bits = 4 → usable hosts = 2^4 - 2 = 14 (too small).Test /26 (255.255.255.192): host_bits = 6 → usable hosts = 62 (meets, but wastes addresses and yields fewer subnets than /27).Therefore, /27 (255.255.255.224) is the tightest mask that satisfies ≥25 hosts, maximizing subnet count.Choose 255.255.255.224.Verification / Alternative check:Subnet tables show /27 yields 8 subnets in a /24 space and 30 hosts each, which aligns with the requirement. A quick test block: 192.168.10.0/27 ranges 192.168.10.1–.30 with .31 broadcast.Why Other Options Are Wrong:a: /26 allows 62 hosts; it meets the host count but does not maximize subnets.c, d, e: /28, /29, /30 allow 14, 6, and 2 hosts respectively—insufficient for 25 hosts.Common Pitfalls:Forgetting to subtract network and broadcast addresses, or picking a larger host pool than needed, which reduces the number of subnets and wastes address space.Final Answer:255.255.255.224

Question 6

Parallel vs serial in microcomputers: In standard parallel data transmission used by microcomputers (for example, legacy printer ports), how many bits are typically transmitted simultaneously?

Accepted Answer

Introduction / Context:Parallel interfaces transfer multiple bits at the same time across several signal lines. Classic microcomputer peripherals—especially Centronics-style printer ports—used an 8-bit parallel data path, reflecting the byte-oriented architecture of most systems.Given Data / Assumptions:We refer to the typical microcomputer parallel port (for example, Centronics/IEEE 1284).Focus is on the data lines, not including status/control lines.Byte-oriented data transfer was standard for printers and some other peripherals.Concept / Approach:A “byte-wide” bus means 8 dedicated data lines (D0–D7) carry one full byte in a single strobe. Additional lines handle handshaking and status. Therefore, the number of simultaneous data bits equals eight.Step-by-Step Solution:Identify the physical interface: legacy printer port with D0–D7 data lines.Recognize that these eight lines convey a full byte per transfer cycle.Account for control lines (for example, STROBE, ACK) which do not carry data bits.Conclude that 8 bits are transmitted at once in typical microcomputer parallel transmission.Select option c (8).Verification / Alternative check:Examine pinouts of Centronics/IEEE 1284 connectors which clearly denote eight data pins plus several control/status pins. Interface documentation confirms byte-wide operation.Why Other Options Are Wrong:a: 16-bit parallel buses exist (for example, certain internal buses) but are not the typical microcomputer parallel printer interface.b: 9 is incorrect; confusion may arise from 9-pin serial connectors (DE-9 for RS-232).d: 4-bit nibble modes are fallbacks for compatibility but not the standard full-speed parallel transfer mode.e: Not applicable because 8 is standard.Common Pitfalls:Confusing serial connectors (DE-9) with parallel data width, or counting control lines as data lines. Also, IEEE 1284 has multiple modes, but the fundamental data path is byte-wide.Final Answer:8

Question 7

Network topologies – FDDI: Fiber Distributed Data Interface (FDDI) primarily uses which logical network topology?

Accepted Answer

Introduction / Context:FDDI (Fiber Distributed Data Interface) is a LAN/MAN technology standardized for high-speed optical networking. Recognizing its topology helps differentiate it from Ethernet bus/star designs and token-based rings like Token Ring, even though implementations may use star-wired physical layouts.Given Data / Assumptions:FDDI operates typically at 100 Mbps using fiber optic media.FDDI defines a dual-ring of trees architecture for resilience.Physical wiring can be star-like via concentrators, but the logical topology is ring.Concept / Approach:FDDI employs a token-passing mechanism over a primary and secondary ring for fault tolerance. Even when connected through concentrators, the logical data path forms a ring. Therefore, its primary logical topology is ring.Step-by-Step Solution:Recall that FDDI uses token passing, characteristic of ring topologies.Understand the “dual ring” concept: primary for traffic, secondary for backup or counter-rotating traffic.Note the distinction between physical and logical topologies; logical remains ring.Choose the option that states “Ring network.”Confirm with standard references on FDDI architecture.Verification / Alternative check:Topology diagrams in FDDI documentation show primary and secondary rings connecting stations or concentrators in a loop. Fault scenarios illustrate ring wrap for continuity.Why Other Options Are Wrong:b: Star describes physical wiring possible via concentrators, not the logical topology.c: Mesh is not the canonical FDDI topology.d: Bus aligns with early Ethernet (10BASE2/5), not FDDI.e: Not applicable because “Ring network” is correct.Common Pitfalls:Confusing physical cabling patterns with logical flow; overlooking the secondary ring’s role in resilience and misconstruing it as a different topology.Final Answer:Ring network

Question 8

Distance-vector control traffic: In a 50-router network, each router records delays as 10-bit numbers and exchanges its entire delay vector twice per second. Approximately how much bandwidth is consumed per full-duplex line by the routing exchanges …

Accepted Answer

Introduction / Context:Distance-vector routing protocols periodically exchange vectors of path metrics (for example, delays). Estimating the bandwidth consumed by these control messages is useful for capacity planning on low-speed links and for understanding protocol overheads relative to user traffic.Given Data / Assumptions:Number of routers (and destinations) = 50.Each delay is encoded as a 10-bit value.Each router sends its full vector twice per second.We ignore protocol headers and assume one neighbor per link for a simple per-link estimate.Concept / Approach:On a single link between two routers, each side sends a vector containing one entry per known destination. Bits per vector equals entries * bits per entry. Multiply by update frequency to get bits per second in one direction. A full-duplex line has independent capacities in each direction, but per-direction bandwidth is usually quoted.Step-by-Step Solution:Compute bits per vector: 50 destinations * 10 bits = 500 bits.Compute send rate per direction: 500 bits * 2 updates/second = 1000 bits/second.Therefore, on the link, A→B uses ~1000 bps for routing updates.B→A also uses ~1000 bps (full-duplex; symmetric load).The commonly reported per-direction bandwidth is 1000 bps; select 1000 bps.Verification / Alternative check:If you summed both directions, the pair would total ~2000 bps, but per-direction accounting (the usual convention) yields 1000 bps. Including headers would raise the figure slightly but not change the order of magnitude.Why Other Options Are Wrong:a: 500 bps accounts for one update per second, not two.b: 1500 bps does not follow directly from the provided numbers.c: 5 bps is far too low; it ignores vector length and frequency.e: Not applicable because 1000 bps is the consistent per-direction estimate.Common Pitfalls:Confusing per-direction (full-duplex) bandwidth with the sum of both directions; forgetting to multiply by update frequency; or miscounting destinations in the vector length.Final Answer:1000 bps

Question 9

Data communications – error detection fundamentals In data link communications, what is the term for the error-detecting code field that is inserted into each frame/block before transmission?

Accepted Answer

Introduction / Context:Error detection is a core concept in computer networks and serial links. At the data link layer, frames carry not only payload but also a special field designed to detect whether bits were corrupted in transit. Knowing the precise term for this field helps learners distinguish between general concepts (checksums) and the standardized field name used in data link protocols such as Ethernet and HDLC.Given Data / Assumptions:The question asks for the specific name of the error-detecting code field inserted into a transmitted block/frame.We are discussing data link layer framing, not higher-layer application checksums.Common implementations include CRC-based protection.Concept / Approach:Most link-layer protocols append a dedicated field called the Frame Check Sequence (FCS). An FCS typically contains a cyclic redundancy check (CRC) value computed over the header and payload. The receiver recomputes the CRC and compares it with the FCS to detect errors. While the word “checksum” is a broad class of error checks, the standardized field name in link-layer frames is FCS.Step-by-Step Solution:Identify the OSI layer: data link (framing and physical delivery).Recognize the standardized field name used in frames: Frame Check Sequence.Understand that FCS often stores a CRC value for robust error detection.Select “Frame check sequence” as the precise answer.Verification / Alternative check:Ethernet frames end with a 4-byte FCS. HDLC frames also include FCS. Textbooks and standards consistently use the term FCS for this field.Why Other Options Are Wrong:Checksum: generic term; not the specific standardized field name used in link frames.Flow control: regulates data rate; not an error-detection field.Error detecting code (generic): describes the concept, but the formal field name in frames is FCS.Common Pitfalls:Confusing checksums (generic) with the link-layer FCS (specific).Assuming FCS always uses the same polynomial; CRC parameters vary by protocol.Thinking FCS corrects errors; it only detects them.Final Answer:Frame check sequence

Question 10

Digital modulation – baud rate with M-ary PSK A 9600 bit/s digital stream is transmitted using 8-PSK (8-level phase shift keying). What is the resulting modulation rate in baud (symbols per second)?

Accepted Answer

Introduction / Context:Modulation schemes like M-ary PSK map multiple bits onto one symbol. The baud rate (symbols per second) is therefore lower than the bit rate when each symbol represents more than one bit. This question checks your ability to relate bit rate, constellation size, and baud rate for M-ary modulation.Given Data / Assumptions:Bit rate (Rb) = 9600 bit/s.Modulation = 8-PSK, which means M = 8 constellation points.Each symbol carries log2(M) bits.Concept / Approach:For M-ary modulation, bits per symbol = log2(M). Baud rate (Rs) = Rb / (bits per symbol). For 8-PSK, bits per symbol = log2(8) = 3. Therefore, Rs = 9600 / 3 = 3200 symbols/second (3200 baud).Step-by-Step Solution:Compute bits per symbol: log2(8) = 3.Relate baud and bit rate: Rs = Rb / (bits per symbol).Insert numbers: Rs = 9600 / 3.Calculate result: Rs = 3200 baud.Verification / Alternative check:If the scheme were BPSK (M=2), bits per symbol = 1, giving 9600 baud. For QPSK (M=4), bits per symbol = 2, giving 4800 baud. These comparisons validate that 8-PSK should yield 3200 baud.Why Other Options Are Wrong:1200 baud: would imply 8 bits per symbol, which is not 8-PSK.4800 baud: corresponds to QPSK (2 bits/symbol), not 8-PSK.9600 baud: corresponds to BPSK (1 bit/symbol), not 8-PSK.Common Pitfalls:Confusing bit rate with baud rate; they are equal only when 1 bit per symbol.Using log10 instead of log2 for bits per symbol.Ignoring coding/overhead; in basic calculations we assume ideal mapping without FEC overhead.Final Answer:3200 baud

Question 11

RS-232 / EIA-232 pinout (DB-25, DTE) For a standard EIA-232 (RS-232C) DB-25 DTE connector, which pin carries the Transmitted Data (TxD) signal from the terminal (DTE) to the modem (DCE)?

Accepted Answer

Introduction / Context:RS-232 (EIA-232) is a classic serial interface. Correctly identifying signal pins is essential for troubleshooting and cable fabrication. The DB-25 pinout distinguishes transmitted and received data as viewed from Data Terminal Equipment (DTE) such as a computer or terminal.Given Data / Assumptions:Connector type: DB-25 on a DTE device.We need the pin carrying Transmitted Data (TxD) from DTE to DCE.Conventions follow RS-232C standard assignments.Concept / Approach:On a DB-25 DTE connector: Pin 2 = TxD (transmitted data out of DTE), Pin 3 = RxD (received data into DTE), Pin 7 = Signal Ground, Pin 4 = RTS, Pin 5 = CTS. Note that DB-9 assignments are different (TxD is Pin 3 on DB-9), which often causes confusion.Step-by-Step Solution:Identify interface and viewpoint: DB-25 on DTE.Recall canonical mapping: Pin 2 is TxD on DB-25 (DTE side).Confirm that Pin 3 is RxD and Pin 4 is RTS; therefore Pin 2 is the answer.Select “Pin 2.”Verification / Alternative check:Consult any RS-232 DB-25 DTE pinout: 2-TxD, 3-RxD, 7-GND. Loopback plugs and breakout boxes also confirm signal directions in practice.Why Other Options Are Wrong:Pin 3: corresponds to RxD on DB-25 DTE; not transmitted data.Pin 4: corresponds to RTS control; not data.All of the above / None: incorrect because only one pin is used for TxD on DB-25 DTE.Common Pitfalls:Mixing DB-9 and DB-25 numbering (DB-9 TxD is Pin 3).Forgetting DTE vs. DCE perspective; the directions swap on DCE connectors.Assuming proprietary cables follow standards; always verify wiring diagrams.Final Answer:Pin 2

Question 12

Channel equalization – combating delay distortion In a digital receiver, which structure is used in a receive equalizer to compensate for frequency-dependent delay (delay distortion)?

Accepted Answer

Introduction / Context:Real channels do not transmit all frequencies equally; group delay can vary with frequency, causing intersymbol interference (ISI). Receive equalizers counteract these impairments so that symbol decisions are correct. The internal structure that makes many equalizers effective is a series of taps with adjustable weights.Given Data / Assumptions:We focus on receive-side equalization in digital communications.The impairment of interest is delay distortion leading to ISI.Standard linear equalizer structures are considered.Concept / Approach:Linear equalizers and adaptive equalizers (e.g., LMS, RLS) are commonly implemented as finite impulse response (FIR) filters. An FIR filter is realized using tapped delay lines: successive delayed versions of the input are weighted and summed to approximate the inverse channel response, thereby flattening amplitude and linearizing phase across the signal band.Step-by-Step Solution:Model the channel as a filter with amplitude/phase distortion.Design an FIR equalizer whose impulse response approximates the inverse channel.Implement FIR as tapped delay line: x[n], x[n-1], x[n-2], … multiplied by coefficients and summed.Adapt coefficients based on an error criterion to minimize ISI.Verification / Alternative check:Eye diagrams open as the equalizer converges; bit error rate decreases. Simulation or lab measurements confirm compensation of group delay ripple and amplitude tilt.Why Other Options Are Wrong:Gearshift mechanism: irrelevant to signal processing.Descrambler: removes data whitening; does not correct channel distortion.Difference engine: historical computational device; not used in modern equalizers.Common Pitfalls:Confusing scramblers/descramblers with equalizers; they address different issues.Overlooking phase (group delay) distortion; amplitude-only equalization is insufficient.Using too few taps, which cannot model the inverse channel effectively.Final Answer:Tapped delay lines

Question 13

Network topologies – active vs. passive media Which network topology is considered passive because it does not require electrically active central devices for signal forwarding?

Accepted Answer

Introduction / Context:Different topologies distribute signals in different ways. A passive topology does not rely on powered intermediary devices to repeat or switch signals. Understanding this distinction helps in analyzing cost, fault tolerance, and performance characteristics of legacy and modern LANs.Given Data / Assumptions:Passive means the medium is shared and does not actively regenerate signals.Classic examples include 10BASE-2/10BASE-5 bus Ethernet.Active topologies use hubs, switches, or token passing logic.Concept / Approach:In a bus topology, all stations tap into a common coaxial cable segment. The cable itself does not actively boost or direct the signal; it simply carries it, with proper termination at both ends. In contrast, star topologies use hubs/switches (active), and token ring uses active MAUs with signal regeneration and management.Step-by-Step Solution:Define passive: no active electronics required for distribution.Examine bus: a single backbone with terminators; passive medium.Contrast with star and ring: central devices or MAUs are active.Select “Bus.”Verification / Alternative check:Topology diagrams and IEEE 802.3 historical standards confirm that early Ethernet buses required only proper impedance matching and terminators, not active distribution hardware.Why Other Options Are Wrong:Ring: requires active token handling and MAUs.Star: relies on hubs/switches that regenerate and direct frames.Mesh: consists of many active links and devices.Cross: not a standard topology type in this context.Common Pitfalls:Confusing physical star with logical bus (e.g., 10BASE-T hubs act like multiport repeaters but are still active).Assuming “passive” means no power anywhere; end stations still require power.Forgetting about terminators; a passive bus must be properly terminated.Final Answer:Bus

Question 14

Combinatorics of link choices in a fully connected network Four routers are fully interconnected with point-to-point links. If each pair of routers can be connected by exactly one of three line types (high, medium, or low speed), how many distinct l…

Accepted Answer

Introduction / Context:Network design often involves selecting link capacities between every pair of nodes. When the physical connectivity is fixed (complete graph), the design problem reduces to counting the number of ways to assign link types. This is a straightforward application of combinatorics to a networking scenario.Given Data / Assumptions:Number of routers (nodes) = 4.Topology is fully connected (complete graph).Each link must be exactly one of three types: high, medium, or low speed.Concept / Approach:A fully connected undirected network of N nodes has E = N * (N - 1) / 2 distinct links. For each independent link, there are 3 choices. The multiplication rule for independent choices gives total configurations = 3^E. Here, N = 4, so E = 4 * 3 / 2 = 6; thus total = 3^6.Step-by-Step Solution:Compute number of links: E = C(4, 2) = 6.Assign choices per link: 3 types available per link.Apply product rule: total configurations = 3^6.Evaluate 3^6: 3^2 = 9, 3^3 = 27, 3^6 = 27 * 27 = 729.Verification / Alternative check:List smaller cases to verify the pattern: For N = 3, E = 3, total = 3^3 = 27. Extending to N = 4 yields 3^6 = 729, confirming consistency.Why Other Options Are Wrong:12: equals the number of directed arcs for a 4-node complete digraph without self-loops, not the count of assignments.81: equals 3^4, which would correspond to only 4 links.48: not a power of 3; does not match combinatorial structure.Common Pitfalls:Forgetting the network is undirected and using N * (N - 1) instead of N * (N - 1) / 2.Confusing “number of topologies” with “number of physical graphs”; connectivity is fixed here, only link types vary.Assuming order matters between line types on the same link; it does not.Final Answer:729

Question 15

Entry-level modem speeds Historically, the least expensive dial-up modems transmitted data at which of the following rates?

Accepted Answer

Introduction / Context:Dial-up modem technology progressed through a well-known sequence of standardized speeds: 300, 1200, 2400, 4800, 9600 bit/s, and beyond. Early, low-cost acoustic-coupler and direct-connect modems commonly operated at the slowest rates due to analog line quality and electronics limitations.Given Data / Assumptions:We are referencing historical consumer modem markets.“Cheapest” implies earliest, simplest models with minimal processing.Speeds are nominal bit rates over POTS lines.Concept / Approach:The earliest common standard was around 300 bit/s (Bell 103). Later standards such as V.22 (1200 bit/s) and V.22bis (2400 bit/s) improved speed but initially cost more. Therefore, entry-level and least expensive models in early eras were typically 300 bit/s units.Step-by-Step Solution:Identify historical baseline: 300 bit/s was the first widely adopted speed.Recognize later standards increased cost and complexity.Match “cheapest” with the earliest common standard.Select “300 bits per second.”Verification / Alternative check:Contemporary references and museum catalogs of computing hardware list 300 bit/s acoustic coupler modems as widely available and inexpensive compared to later, faster models when first released.Why Other Options Are Wrong:1,200 / 2,400 / 4,800 bit/s: later standards; initially more expensive on release.Common Pitfalls:Assuming present-day pricing applies to historical technology.Confusing nominal symbol rates with effective throughput (error correction/compression not considered here).Overlooking regional standard naming (Bell vs. CCITT/ITU-T).Final Answer:300 bits per second

Question 16

Encapsulation in layered networking Which option best describes encapsulation as used by layered protocols during data transmission?

Accepted Answer

Introduction / Context:Encapsulation is the fundamental mechanism that enables layered network models (such as OSI or TCP/IP) to work. Each layer provides services to the layer above by wrapping its data with appropriate headers (and sometimes trailers), producing a new protocol data unit for transmission across the next lower layer.Given Data / Assumptions:We are discussing general, layer-to-layer interactions.The lower layer provides a service interface to the higher layer.Headers/trailers are added at each step to convey control information.Concept / Approach:Encapsulation occurs when, for example, TCP hands a segment to IP, which places the TCP segment into the payload of an IP packet; IP then hands that to Ethernet, which places the IP packet into the payload of an Ethernet frame. This “wrapping” is how addressing, reliability, and physical delivery are layered and kept modular.Step-by-Step Solution:Identify the higher-layer data unit (e.g., application payload or TCP segment).Lower layer adds its header/trailer, creating a new PDU.Repeat down the stack until the physical layer transmits bits on the medium.At the receiver, decapsulation removes headers in reverse order.Verification / Alternative check:Protocol analyzers (e.g., Wireshark) display nested headers, visually confirming encapsulation: Ethernet > IP > TCP > application data.Why Other Options Are Wrong:Best-effort loop avoidance: relates to TTL/hop-limit, not encapsulation.IP fragmentation: creates smaller datagrams; still not the definition of encapsulation.“All of the above”: incorrect because only one statement defines encapsulation.Common Pitfalls:Confusing encapsulation with tunneling; tunneling encapsulates entire packets inside another protocol but is a specific use-case.Equating fragmentation with encapsulation; they solve different problems.Forgetting trailers (e.g., FCS) also participate at some layers.Final Answer:A process where a lower-layer protocol takes a higher-layer message and places it into the data field of its own frame

Question 17

Digital over analog telephone lines Computers use digital pulses, but the public switched telephone network carries analog audio frequencies. Which device converts between digital and analog for transmission over phone lines?

Accepted Answer

Introduction / Context:The classic telephone network was designed for analog voice, not digital data. To communicate over it, computers historically used devices that modulate digital signals into analog waveforms and demodulate them back into digital form at the other end. This is the origin of the ubiquitous term “modem.”Given Data / Assumptions:Source: digital computer data.Channel: analog voice-band telephone line.Requirement: convert digital bits to suitable analog signals and back again.Concept / Approach:MoDem stands for MOdulator-DEModulator. The transmitting modem converts a digital bit stream into analog tones (using schemes like FSK, PSK, QAM). The receiving modem demodulates the analog signal back into the original digital bits, handling filtering, equalization, and often error control/handshaking.Step-by-Step Solution:Identify the mismatch: digital source, analog channel.Select the device that modulates/demodulates: the modem.Recognize ancillary functions: training, adaptive equalization, compression, error control.Confirm: device name is “modem.”Verification / Alternative check:Industry standards (e.g., ITU-T V-series) define modem operation over PSTN lines. Legacy dial-up connections on PCs used internal or external modems to achieve negotiated link rates.Why Other Options Are Wrong:Interface: too generic; does not imply modulation/demodulation.Attenuation: a channel impairment, not a device.Teleprocessor: vague term; not the standard name for the converter.Common Pitfalls:Confusing modems with codecs; codecs convert between analog audio and digital PCM but are not necessarily for data modulation over PSTN.Assuming VoIP gateways are the same; they digitize voice but operate differently from dial-up data modems.Neglecting line quality issues that limit achievable data rates.Final Answer:Modem

Question 18

Logical network layouts – centralized vs. distributed A topology with one central computer surrounded by multiple satellite computers that connect back to it is most accurately called a:

Accepted Answer

Introduction / Context:Network topology describes how nodes are arranged and interconnected. A centralized pattern, in which many devices connect to a single focal point, is especially common in modern Ethernet LANs and wireless infrastructures, enabling straightforward management and fault isolation.Given Data / Assumptions:The problem mentions a single central computer and multiple satellites.All satellites communicate through the central node.We select the standard topology name that matches this description.Concept / Approach:In a star topology, each node has a point-to-point link to a central hub, switch, or controller. That central device mediates communication between nodes. By contrast, a bus uses a shared medium; a ring forms a closed loop where frames circulate; a mesh has multiple interconnections among nodes without a single point of concentration.Step-by-Step Solution:Match the description (one center with satellites) to known topologies.Recognize the defining feature of a star: centralized connectivity at one node.Exclude bus/ring/mesh because they do not have a single central node by definition.Select “Star network.”Verification / Alternative check:Modern Ethernet with switches is a physical star (often logical star as well). Topology diagrams consistently depict a central switch with spokes to each host.Why Other Options Are Wrong:Bus: uses a shared backbone; no single central node.Ring: nodes arranged in a loop; no central node.All of the above: only one topology matches the description.Common Pitfalls:Confusing logical and physical topology; some technologies use star wiring but logical bus behavior (old hubs).Equating “server-centric” application design with physical topology; the question is about cabling/layout.Assuming redundancy exists; a simple star has a single point of failure at the center.Final Answer:Star network

Question 19

In data communications, which summation operation over the bits is commonly used to verify an error-detecting code (i.e., to detect whether data was corrupted during transmission)?

Accepted Answer

Introduction / Context: In digital data communications, error detection ensures that the bits received are the same as the bits sent. Common schemes add a small check value to each frame so the receiver can test for corruption without re-sending the whole payload.Given Data / Assumptions: We are concerned with error detection, not error correction.The question asks which summation operation over bits is used to verify an error-detecting code.Typical link-layer mechanisms include parity, checksum, and cyclic redundancy check (CRC).Concept / Approach: A checksum is computed by performing arithmetic over a block of data (often ones-complement addition of words, or another defined sum) and appending the result as a short field. At the receiver, the same computation is performed and compared to the attached value. A mismatch signals an error.Step-by-Step Solution: 1) Partition the data stream into fixed-size words (implementation dependent).2) Sum all words using the scheme's defined arithmetic (e.g., ones-complement addition).3) Fold/carry as required by the checksum definition to keep the result in a fixed width.4) Transmit the data plus the checksum.5) Receiver repeats the same sum; if computed value matches the received checksum, the frame is assumed intact; otherwise it is flagged as corrupted.Verification / Alternative check: If any bit flips, re-summing at the receiver changes the total. While some error patterns can evade simple checksums, they still detect a large fraction of random errors and are lightweight to compute.Why Other Options Are Wrong: Codec: A general term for encoder/decoder components, not a specific summation-based error check. Coder–decoder: Same idea as codec; not an error check method. Attenuation: Physical-layer signal loss, not a logical integrity test. Parity totaling: Parity counts ones but is not the general summation operation referred to here; parity is weaker than checksums.Common Pitfalls: Confusing error detection (checksum/CRC) with error correction (Hamming, Reed–Solomon). Also assuming checksum guarantees correctness; it only reduces undetected error probability.Final Answer: Checksum

Question 20

In modem-based data transmission, demodulation refers to the process of converting the incoming analog carrier back into the original digital data signal at the receiver.

Accepted Answer

Introduction / Context: In data communication with modems, information is often sent over analog channels (such as telephone lines) by modulating a carrier. The receiver must reverse this process to recover the original digital bits. This reverse operation is called demodulation.Given Data / Assumptions: We are discussing modem-based communication over analog media.Modulation maps digital bits onto an analog waveform for transmission.Demodulation recovers the digital data from the received waveform.Concept / Approach: Modulation = digital-to-analog mapping at the transmitter. Demodulation = analog-to-digital recovery at the receiver. The demodulator estimates symbol values from the waveform (e.g., amplitude, frequency, phase) and decides which bits they represent.Step-by-Step Solution: 1) The transmitter modulates bits onto an analog carrier (e.g., PSK, FSK, QAM).2) The channel adds noise/impairments; the signal remains analog.3) The receiver's demodulator samples and filters the analog waveform.4) It estimates symbol parameters and maps them back to binary values.5) Output is a digital bitstream representing the original data.Verification / Alternative check: Loopback testing shows that applying modulation followed by demodulation reproduces the original digital data, within the error rate expected for the chosen scheme and channel quality.Why Other Options Are Wrong: Converting digital to analog signals: That is modulation, not demodulation. Multiplexing various signals: That is a separate function (e.g., FDM, TDM), not demodulation. Performing data description: Not a standard communication process term. Filtering high-frequency noise: Filtering may be part of receivers, but it is not the definition of demodulation.Common Pitfalls: Assuming demodulation always yields analog output; in data modems it yields digital bits. Also confusing demodulation with decoding of higher-layer codes.Final Answer: converting analog to digital signals

Networking Questions

Questions

Practice smarter