Question 1

Because a DHCP client receives its IP configuration dynamically, which Windows utility should you use to read and verify the current network configuration on that client?

Accepted Answer

Introduction / Context: DHCP environments assign IP parameters automatically, so technicians must inspect what the client actually received. Windows provides a built-in tool for this purpose. Given Data / Assumptions: OS: Windows workstation. DHCP assigns IP address, mask, gateway, and DNS. Goal: read and verify the live configuration. Concept / Approach: The ipconfig command displays current adapter settings. With the /all flag, it shows lease info, DHCP server, DNS servers, and whether DHCP is enabled. It is the definitive first check for configuration on Windows. Step-by-Step Solution: 1) Open Command Prompt.2) Run: ipconfig /all.3) Review IPv4 address, subnet mask, default gateway, DNS servers, DHCP server, and lease times.4) If incorrect or stale, use ipconfig /release and ipconfig /renew to refresh. Verification / Alternative check: After renew, run ipconfig /all again and compare. Optionally ping default gateway to verify reachability. Why Other Options Are Wrong: PING checks reachability only; it does not list configuration.TRACERT shows path to a destination, not local settings.ARP displays MAC-to-IP mappings cached locally, not full DHCP parameters.“None of the above” is wrong because IPCONFIG is correct. Common Pitfalls: Forgetting /all, misreading virtual adapters, and overlooking that VPNs or vSwitches may add multiple interfaces. Final Answer: IPCONFIG.

Question 2

A voice-grade line has SNR = 30.1 dB (power ratio ≈ 1023:1) and usable spectrum from 300 Hz to 4300 Hz (bandwidth = 4000 Hz). Using Shannon’s capacity formula, what is the maximum achievable data rate on this line?

Accepted Answer

Introduction / Context: Channel capacity places an upper bound on error-free data rate for a given bandwidth and signal-to-noise ratio. For telephony-grade channels, Shannon’s theorem provides the theoretical limit irrespective of modulation specifics. Given Data / Assumptions: SNR = 30.1 dB ≈ power ratio 1023:1. Frequency span: 300 Hz to 4300 Hz → bandwidth B = 4000 Hz. We apply Shannon capacity: C = B * log2(1 + SNR). Concept / Approach: Convert SNR from dB to linear: SNR_linear ≈ 10^(30.1/10) ≈ 1023. Then compute C with base-2 logarithm. Notation: use * for multiplication and / for division. Step-by-Step Solution: 1) SNR_linear ≈ 1023.2) 1 + SNR_linear ≈ 1024.3) log2(1024) = 10.4) C = B * log2(1 + SNR) = 4000 * 10 = 40000 bps. Verification / Alternative check: If one mistakenly uses 3000 Hz bandwidth, C would be 30000 bps, which is lower and does not reflect the given 4000 Hz span. The provided spectrum explicitly yields 40000 bps. Why Other Options Are Wrong: 34,000 bps and 31,000 bps result from using smaller bandwidth or rounding away from the exact log2(1024) = 10.9,600 bps and 28,800 bps are common modem rates subject to practical constraints, not the Shannon upper bound for the given parameters. Common Pitfalls: Mixing Nyquist and Shannon formulas, using dB directly without converting to linear ratio, or assuming 3000 Hz bandwidth by habit. Final Answer: 40,000 bps.

Question 3

In virtual circuit networks (e.g., X.25 or virtual circuits in frame relay/ATM), establishing a virtual connection before sending user data is most analogous to which everyday action?

Accepted Answer

Introduction / Context: Virtual circuit networks set up a logical path with identifiers and state before data transfer, ensuring ordered delivery and resource expectations similar to a phone call setup. Given Data / Assumptions: Virtual circuit requires a setup phase. After setup, data frames follow the established path. Teardown occurs when communication ends. Concept / Approach: The closest analogy is dialing a telephone: a signaling exchange allocates resources and assigns a path, then conversation (data) flows. This differs from datagram networks where each packet is routed independently without prior setup. Step-by-Step Solution: 1) Initiate call setup (signaling) → allocate a VC identifier.2) Transfer user data over the VC with guaranteed order.3) Release the VC when finished. Verification / Alternative check: Ordered delivery and fixed path semantics match the phone-call analogy far better than hardware cabling actions. Why Other Options Are Wrong: Virtual memory concerns OS memory, not networking.Physically connecting DTE/DCE is a hardware step, not a logical circuit setup.Placing a modem is equipment installation, not per-session signaling.“None of the above” is wrong because the telephone call analogy is correct. Common Pitfalls: Confusing logical setup with physical cabling; assuming setup implies dedicated leased lines (it does not). Final Answer: Placing a telephone call prior to a conversation.

Question 4

HDLC (High-level Data Link Control) is widely used at the data link layer. Which description best characterizes HDLC?

Accepted Answer

Introduction / Context: HDLC is a foundational, bit-oriented data link protocol used for point-to-point and multipoint links, influencing protocols such as PPP and LAPB. Given Data / Assumptions: Layer focus: Data Link (Layer 2). ISO standardization. Frame-based communication with control fields. Concept / Approach: HDLC defines framing, error detection, and control mechanisms (I/S/U frames). Being bit-oriented means control is defined at the bit level (flags, bit stuffing) rather than fixed byte boundaries for characters. Step-by-Step Solution: 1) Frames are delimited by flag 01111110.2) Bit-stuffing prevents accidental flag patterns in payload.3) CRC provides error detection.4) Control field supports reliable data transfer and link management. Verification / Alternative check: Compare with PPP, which borrows HDLC framing style for serial links. Why Other Options Are Wrong: Medium access control schemes (options A and B) describe contention/token systems, not HDLC's role.Circuit-switched access standards (option D) describe network access, not a data link framing protocol.“None of the above” is wrong since option C is correct. Common Pitfalls: Confusing HDLC with MAC methods like CSMA/CD or token passing; assuming byte-oriented protocols like PPP are unrelated. Final Answer: A very common bit-oriented data link protocol issued by ISO.

Question 5

In IP networking, the Time To Live (TTL) field prevents routing loops. Which description best explains how TTL is used?

Accepted Answer

Introduction / Context: TTL is a critical IPv4 header field (Hop Limit in IPv6) used to limit the lifetime of a packet in the network and thus prevent infinite circulation during routing anomalies. Given Data / Assumptions: Routers decrement TTL by 1 per hop. Packet is discarded when TTL reaches 0. ICMP Time Exceeded can be generated for diagnostics (e.g., traceroute). Concept / Approach: TTL bounds the maximum hop count a packet may traverse. This avoids persistent loops and aids troubleshooting by exposing path length characteristics. Step-by-Step Solution: 1) Sender sets initial TTL (e.g., 64 or 128).2) Each router decrements TTL by 1.3) If TTL becomes 0, router drops packet and may send ICMP Time Exceeded.4) Tools like traceroute vary TTL to map the path. Verification / Alternative check: Observe traceroute hop counts; increasing TTL reveals successive routers. Why Other Options Are Wrong: Encapsulation (option B) describes layering, not TTL.Fragmentation (option C) concerns MTU and datagram pieces, not loop prevention.“All of the above” is wrong because only option A describes TTL correctly. Common Pitfalls: Confusing TTL with time duration (it is hops), mixing fragmentation with loop control, and assuming TTL prevents all routing issues. Final Answer: A technique used to avoid endlessly looping packets.

Question 6

Distributed Queue Dual Bus (DQDB), defined by IEEE 802.6, is a standard primarily intended for which type of network?

Accepted Answer

Introduction / Context: DQDB (Distributed Queue Dual Bus) was developed to support metropolitan-scale networks with shared medium access and quality of service considerations across city-wide areas. Given Data / Assumptions: Standard: IEEE 802.6. Topology: dual unidirectional buses. Scope: metropolitan coverage. Concept / Approach: MANs bridge the gap between LAN (local building/campus) and WAN (long haul). DQDB’s dual-bus design and distributed queueing aim to coordinate access across multiple nodes over metropolitan distances. Step-by-Step Solution: 1) Recognize IEEE 802 family mappings: 802.3 (LAN Ethernet), 802.11 (WLAN), 802.6 (MAN DQDB).2) Match DQDB to MAN characteristics (scale, access method).3) Select the MAN option accordingly. Verification / Alternative check: Historical references consistently list DQDB under metropolitan technologies, not LAN/WAN categories. Why Other Options Are Wrong: LAN and Wireless LAN target smaller geographic areas.WAN spans much larger, carrier-grade distances and technologies (e.g., MPLS, SDH), distinct from DQDB's MAN scope.“LAN and MAN” conflates scopes; DQDB is specifically a MAN standard. Common Pitfalls: Assuming any IEEE 802 technology is LAN-only; overlooking the MAN category. Final Answer: MAN.

Question 7

User-interface dialogue techniques for terminal-based systems typically include structured interactions such as Q&A, forms, and menus. Which of the following is NOT considered a standard terminal dialogue technique?

Accepted Answer

Introduction / Context: Classic terminal UIs favor constrained interactions that are easy to parse and validate, such as forms and menus. Open-ended prompts are less deterministic and harder to process. Given Data / Assumptions: Terminal context with keyboard input and limited graphics. Need for predictable parsing and validation. Desire to minimize user errors. Concept / Approach: Structured dialogues—menus, forms, and controlled Q&A—reduce ambiguity. Open-ended questions (free text essays) are atypical in such systems because they require complex parsing and offer inconsistent data quality. Step-by-Step Solution: 1) Identify common terminal patterns: menus, form fields, yes/no prompts.2) Contrast with open-ended, unconstrained responses.3) Select the item that does not fit structured terminal dialogue. Verification / Alternative check: Examine classic CLI installers and admin tools; they rely on lists and forms, not essays. Why Other Options Are Wrong: Questions and answers in constrained form are standard.Form filling is a hallmark of data entry terminals.Menu display is the most common selection mechanism.“None of the above” is wrong because one item (open-ended questions) is indeed not standard. Common Pitfalls: Confusing open-ended with multi-line text fields in forms (which are still structured). Final Answer: Open-ended questions.

Question 8

Consider IPv4 addressing and routing behavior. Which statement below is incorrect?

Accepted Answer

Introduction / Context: IPv4 addressing encodes both network and host information. Routers forward based on network prefixes, while end hosts care about the full address for local delivery. Given Data / Assumptions: Classless routing (CIDR) is in use. Routers perform longest-prefix match on the destination network. Hosts may be multihomed with multiple interfaces and addresses. Concept / Approach: The key distinction is that forwarding decisions are made on network prefixes (the network portion), not on the host portion. A statement claiming the opposite is incorrect. Step-by-Step Solution: 1) Recognize that routers match destination network prefixes in FIB/RIB.2) Host-specific routes exist but are exceptional; general routing uses network portion.3) Evaluate each option and identify the one contradicting this behavior. Verification / Alternative check: Observe routing tables: entries are prefixes (e.g., 203.0.113.0/24), not individual hosts, confirming network-based decisions by default. Why Other Options Are Wrong: Option A is correct for typical IPv4 without mobility tunneling.Option C is correct: addresses bind to interfaces (network attachments).Option D is correct: different destination addresses can traverse different paths or hit different policies.“None of the above” is wrong because one statement (option B) is indeed incorrect. Common Pitfalls: Assuming routers examine host bits for path choice; conflating ARP/ND (local delivery) with inter-network routing. Final Answer: Routing uses the host portion of the IP address to choose the path (incorrect).

Question 9

In computer networking, which configured address on a host identifies the local router that forwards packets to other networks?

Accepted Answer

Introduction / Context: In a typical Internet Protocol (IP) local area network, hosts need a way to reach destinations that are not on their own subnet. The configured address that represents the local router for this purpose is known as the default gateway. Understanding what each common configuration field means prevents misconfiguration and connectivity loss. Given Data / Assumptions: Host participates in an IPv4 network. Standard host settings include IP address, subnet mask, and default gateway. No special static routes beyond default gateway are configured. Concept / Approach: The default gateway is the layer-3 (IP) next hop for traffic whose destination is outside the host's local subnet. When the host determines a packet is nonlocal (via subnet mask comparison), it forwards the packet to the default gateway's IP (the router's interface on the local subnet). Step-by-Step Solution: 1) Host compares destination IP with its own network using the subnet mask. 2) If the destination is outside the local subnet, the host selects the default route (0.0.0.0/0). 3) The next hop for the default route is the default gateway's IP address (the router interface). 4) The frame is sent to the router's MAC address, with the IP packet destined for the final remote address. Verification / Alternative check: On most systems, show the routing table (e.g., route print or ip route). The default route points to the gateway's IP on the local network. Why Other Options Are Wrong: The IP address: This is the host's own identity, not the router's next-hop setting. The TCP address: TCP is a transport-layer protocol; it has no addressing concept like an "address of the router". The subnet mask: This defines the boundary of the local network, not the router's address. None of the above: Incorrect because the correct term exists: default gateway. Common Pitfalls: Confusing the host's IP or DNS server with the gateway; using a gateway IP that is not on the same subnet as the host; leaving the gateway blank, which breaks off-subnet communication. Final Answer: The default gateway

Question 10

For the class C network 192.168.10.0, which subnet mask yields exactly four subnets while maximizing hosts per subnet?

Accepted Answer

Introduction / Context: Subnetting a class C network (default /24) to obtain a required number of subnets while maximizing hosts per subnet is a core IP addressing skill. Here we need four subnets with as many hosts as possible per subnet on 192.168.10.0. Given Data / Assumptions: Base network: 192.168.10.0/24. Required subnets: 4. Goal: maximize hosts per subnet. Concept / Approach: Number of subnets with borrowed bits: subnets = 2^n. Number of hosts per subnet = 2^(host_bits) - 2. For a class C (/24), there are 8 host bits before borrowing. Borrow as few bits as necessary to reach 4 subnets to keep more host bits (thus maximizing hosts per subnet). Step-by-Step Solution: 1) Need 4 subnets -> n such that 2^n = 4 -> n = 2 bits borrowed. 2) New prefix = /24 + 2 = /26. 3) Hosts per /26 = 2^(8 - 2) - 2 = 2^6 - 2 = 64 - 2 = 62 hosts. 4) /26 corresponds to mask 255.255.255.192. Verification / Alternative check: Subnets would start at 192.168.10.0, .64, .128, .192 with broadcast addresses .63, .127, .191, .255 respectively, each supporting 62 usable host IPs. This satisfies the requirement of exactly 4 subnets with the maximum host count per subnet. Why Other Options Are Wrong: 255.255.255.224 (/27): 8 subnets, only 30 hosts each. 255.255.255.240 (/28): 16 subnets, 14 hosts each. 255.255.255.248 (/29): 32 subnets, 6 hosts each. 255.255.255.252 (/30): 64 subnets, 2 hosts each (point-to-point). Common Pitfalls: Borrowing more bits than necessary (reduces hosts); forgetting to subtract 2 for network and broadcast; mixing up /26 vs /27 boundaries. Final Answer: 255.255.255.192

Question 11

In information systems, a teleprocessing system typically includes which categories of components working together end-to-end?

Accepted Answer

Introduction / Context: Teleprocessing refers to processing that involves terminals or user systems communicating with a central or distributed computer through communication facilities. Modern client-server and mainframe-terminal architectures are examples. Given Data / Assumptions: Teleprocessing spans end-user interaction, transmission, and centralized computing resources. We are looking for the complete set of categories involved. Concept / Approach: Any teleprocessing arrangement must include endpoints (user systems), a transmission path or network (communications systems), and computing resources (computer center systems). Removing any one of these breaks the end-to-end design. Step-by-Step Solution: 1) Identify the endpoint: user systems such as terminals, PCs, or thin clients. 2) Identify the transport: communication lines, modems, switches, and protocols. 3) Identify the processing hub: computer center systems such as mainframes, servers, databases, and applications. 4) Conclude that all three categories are integral. Verification / Alternative check: Classic mainframe CICS or modern web apps both fit this triad: users, network, servers. If any is missing, teleprocessing cannot occur. Why Other Options Are Wrong: User systems alone do not enable remote processing. Communications systems alone cannot process or originate business logic. Computer center systems alone cannot be reached by users without a communication path. None of the above fails because all are indeed required. Common Pitfalls: Confusing teleprocessing with batch-only systems; assuming peer-to-peer negates the need for a central system (it still requires user endpoints and communications); overlooking that modern cloud still embodies the same triad. Final Answer: All of the above

Question 12

Home information services once offered an interactive, graphics-rich selection experience. What was this user-selectable on-demand system called?

Accepted Answer

Introduction / Context: Before the modern web, interactive information services for homes and kiosks emerged that allowed users to browse pages of text and graphics over phone lines or cable. These systems laid groundwork for later online experiences. Given Data / Assumptions: Interactive, user-selectable content at home. Graphics-rich presentation compared to plain text services. Menu-driven navigation over telecommunications links. Concept / Approach: The term for that family of services is Videotex (including variants like Viewdata and Minitel). Videotex offered two-way interactivity, menu navigation, and graphical pages transmitted digitally to a terminal or TV adapter. Step-by-Step Solution: 1) Identify interactivity and user choice as key features. 2) Note the graphics-rich nature versus simple teletype text. 3) Match the description to Videotex, not Teletex or fax. Verification / Alternative check: Historic examples like France's Minitel are classic Videotex platforms: searchable directories, messaging, and information retrieval with menu graphics. Why Other Options Are Wrong: Teletex system: Enhanced Telex for office messaging, primarily text-oriented, not the consumer interactive graphics system described. Fax system: Transmits images of documents point-to-point, not menu-based interactive services. Microwave system: A radio transmission method, not an end-user information service. None of the above: Incorrect because Videotex precisely fits. Common Pitfalls: Confusing Videotex with Teletext (broadcast one-way) or Teletex (office messaging). Only Videotex provided interactive, user-selectable pages over two-way links for home users. Final Answer: Videotex system

Question 13

For network 172.16.0.0, choose a subnet mask that supports eight subnets while maximizing hosts per subnet.

Accepted Answer

Introduction / Context: Subnet design for a class B private network (172.16.0.0/16) often optimizes for a required number of subnets while preserving large host counts per subnet. Here, exactly eight subnets are required, and host capacity should be maximized. Given Data / Assumptions: Base network: 172.16.0.0/16 (class B). Required subnets: 8. Goal: maximize hosts per subnet. Concept / Approach: For a class B /16, there are initially 16 host bits. Borrow n bits for subnetting where 2^n = number of subnets. Keep n minimal to maximize remaining host bits. Hosts per subnet = 2^(remaining host bits) - 2. Step-by-Step Solution: 1) Need 8 subnets -> 2^n = 8 -> n = 3 bits. 2) New prefix = /16 + 3 = /19. 3) Hosts per /19 = 2^(32 - 19) - 2 = 2^13 - 2 = 8192 - 2 = 8190. 4) /19 corresponds to mask 255.255.224.0. Verification / Alternative check: Subnet increments in the third octet are 32 (224 mask), yielding subnets starting at 172.16.0.0, 172.16.32.0, 172.16.64.0, etc., providing eight or more distinct subnets with large host ranges, which satisfies the requirement. Why Other Options Are Wrong: 255.255.192.0 (/18): 4 subnets if counting by borrowed bits from /16, not enough for 8. 255.255.240.0 (/20): 16 subnets, but hosts per subnet are smaller than with /19; not maximizing hosts. 255.255.248.0 (/21): 32 subnets with even fewer hosts per subnet. 255.255.255.0 (/24): 256 subnets but very small host capacity per subnet compared to /19. Common Pitfalls: Forgetting classful starting prefix; miscounting borrowed bits; mixing up /18 vs /19; ignoring the maximize-hosts requirement and choosing a longer prefix than needed. Final Answer: 255.255.224.0

Question 14

Within the OSI model, what does the Physical layer define and provide?

Accepted Answer

Introduction / Context: The OSI model separates networking into seven layers to clarify roles and interoperability. The Physical layer is the lowest layer and deals with transmission of raw bits over a medium, not higher-level framing or virtual circuit services. Given Data / Assumptions: Physical layer handles electrical/optical/Radio Frequency characteristics, connectors, pinouts, signaling, and bit timing. Data link framing and packet services belong to higher layers. Concept / Approach: Match each option to its OSI layer. Anything involving frames belongs to Data Link (Layer 2). Interfaces to packet services (like X.25 virtual circuits) are Layer 2/3 functions, not Layer 1. Therefore, none of the listed options accurately describes the Physical layer's responsibilities. Step-by-Step Solution: 1) Identify that frames imply Layer 2, not Layer 1. 2) Recognize X.25 interfaces and virtual circuit concepts as Layer 2/3. 3) Conclude the Physical layer is instead about media, voltages, wavelengths, and bit-level transmission. Verification / Alternative check: Standards like IEEE 802.3 (physical signaling portions), RS-232, and fiber specifications cover Physical layer parameters: line coding, signal levels, connector types, and timing, not framing procedures or VC interfaces. Why Other Options Are Wrong: Data link procedures for frames: Layer 2 (e.g., MAC, LLC). Interface between X.25 and packet-mode device: X.25 operates at Layers 2/3. Virtual circuit interface to packet-switched service: Not Layer 1. All of the above: Fails because none match Layer 1. Common Pitfalls: Assuming anything about "packets" or "frames" is Physical; confusing cable specifications (Layer 1) with link protocols (Layer 2). Final Answer: None of the above

Question 15

Identify the odd one out: which item is not a guided transmission medium used for data communications?

Accepted Answer

Introduction / Context: Transmission media for data communications can be guided (signals confined within a physical path) or unguided (signals propagate through free space). Distinguishing the two categories is foundational to network design. Given Data / Assumptions: Guided media examples: twisted pair, coaxial cable, optical fibre. Unguided media examples: radio, microwave, infrared, satellite. Concept / Approach: Identify which option represents an unguided medium. Cables and fibre strands physically guide signals; microwaves travel through space without a physical conduit. Step-by-Step Solution: 1) Coaxial cable: guided; signal confined within dielectric and shield. 2) Optical fibre: guided; light travels within core via total internal reflection. 3) Twisted pair: guided; copper conductors carry electrical signals. 4) Microwaves: unguided; radio waves propagate in air or vacuum. Verification / Alternative check: Standards and textbooks consistently classify microwave links as wireless (unguided) media used for point-to-point backhaul, satellite, or WLAN, not as a cable medium. Why Other Options Are Wrong: Coaxial cable is clearly a guided medium. Optical fibre is a guided optical waveguide. Twisted pair is a guided copper medium. None of the above is incorrect since one option (microwaves) is unguided. Common Pitfalls: Confusing waveguides with "guided" in the generic sense; while the term waveguide exists, common microwave links in networking are not physical cables and thus are unguided in this context. Final Answer: Microwaves

Question 16

Which device or technique splits a high-speed signal into multiple frequency bands so several channels can share the same medium simultaneously?

Accepted Answer

Introduction / Context: Multiplexing enables multiple independent signals to share one medium efficiently. Two classic families are Frequency-Division Multiplexing (FDM) and Time-Division Multiplexing (TDM), each using a different separation axis (frequency vs time). Given Data / Assumptions: High-speed composite signal needs to be split into parallel channels. Separation is done by frequency bands, not time slots. Concept / Approach: An FDM device (frequency-division multiplexer) divides available bandwidth into non-overlapping frequency bands, assigning each channel a separate band. Guard bands reduce interference between adjacent channels. Step-by-Step Solution: 1) Identify requirement: concurrent channels using different frequencies. 2) Map requirement to FDM rather than TDM. 3) Select the device name: frequency-division multiplexer. Verification / Alternative check: Analog cable systems and some radio backhauls use FDM by modulating each channel onto a distinct carrier frequency. Spectrum analyzers show multiple peaks corresponding to channels. Why Other Options Are Wrong: t-switch: Not a standard multiplexing device term. modem: Modulates/demodulates a single data stream, not a multi-channel frequency splitter by itself. time-division multiplexer: Separates channels into time slots, not frequency bands. None of the above: Incorrect because FDM exactly fits. Common Pitfalls: Confusing TDM with FDM; assuming a modem inherently multiplexes many user channels in frequency; overlooking guard bands and channel spacing in FDM design. Final Answer: frequency-division multiplexer

Question 17

When one terminal sends a typed message to another over data networks, the content is converted to digital signals, transmitted, and reconstructed at the destination. What is this system commonly called?

Accepted Answer

Introduction / Context: Exchanging messages between computer users is a fundamental network application. The process converts text into digital signals, transports them over various media (cable, telephone, satellite), and reconstructs the message at the destination inbox. Given Data / Assumptions: Point-to-point user messaging over networks. Store-and-forward delivery with mail servers is typical. Concept / Approach: The well-known term for this is electronic mail (email). It comprises user agents (mail clients), message transfer agents (servers), and protocols for submission, relay, and retrieval. Step-by-Step Solution: 1) User composes a message in a mail client. 2) Client submits the message to a server using SMTP or related submission protocol. 3) Servers relay and deliver to the recipient's server. 4) Recipient retrieves via POP/IMAP or webmail; the original text is reconstructed for display. Verification / Alternative check: Inspect message headers to see SMTP hops; observe that the content was transported and restored exactly or with defined encodings (e.g., quoted-printable, base64 for attachments). Why Other Options Are Wrong: Paperless office: A broad concept reducing paper, not specifically messaging. Global network: A generic term, not a messaging system. Electronic newspaper: One-to-many publishing, not interpersonal messaging. None of the above: Incorrect because "Electronic mail" precisely describes it. Common Pitfalls: Confusing chat/instant messaging with email; both send text, but email is store-and-forward and uses distinct protocols and addressing. Final Answer: Electronic mail

Question 18

On a Windows or UNIX-like system, which utility directly displays the local computer’s host name?

Accepted Answer

Introduction / Context: Knowing a machine's local host name is useful for identification, scripting, and network diagnostics. Different tools expose different aspects of the networking stack; selecting the direct one prevents confusion. Given Data / Assumptions: Command-line environment available (Windows, Linux, macOS, BSD). User wants the local system's configured host name, not routing or session stats. Concept / Approach: The utility that directly prints the system's host name is commonly named "hostname". On Windows PowerShell and Command Prompt, "hostname" prints the NetBIOS/host name; on UNIX-like systems, the same command reports the configured host name per system settings. Step-by-Step Solution: 1) Open a terminal or command prompt. 2) Run: hostname 3) Read the output line; it is the local host's name. Verification / Alternative check: On UNIX-like systems, you can also check files or commands such as /etc/hostname, scutil --get ComputerName (macOS), or uname -n. On Windows, echo %COMPUTERNAME% also shows the system name, but "hostname" is the direct utility across platforms. Why Other Options Are Wrong: NBTSTAT: Displays NetBIOS over TCP/IP statistics and name tables, not a simple universal host name printout. Netstat: Shows network connections, routing tables, and interface statistics. PING: Sends ICMP echo requests; while you can ping a host name, it does not reveal the local host name directly. None of the above: Incorrect because the dedicated command exists: hostname. Common Pitfalls: Confusing environment variables (e.g., COMPUTERNAME) with a cross-platform utility; relying on tools that show remote names rather than the local system's configured name. Final Answer: Hostname

Question 19

In signals and systems, which statement best defines a bandlimited signal in practical communication engineering terms?

Accepted Answer

Introduction: A core idea in signals and systems and communication engineering is the notion of a bandlimited signal. This concept explains when and why sampling, filtering, and spectrum allocation work as expected. The question asks for the most accurate engineering definition of a bandlimited signal. Given Data / Assumptions: We consider deterministic signals with well-defined Fourier transforms. “Finite frequency interval” means there exists B > 0 such that the spectrum is zero for |f| > B. Real-world hardware approximates bandlimiting via filters with finite roll-off. Concept / Approach: A signal x(t) is bandlimited if its Fourier transform X(f) = 0 for |f| > B. This property underlies ideal sampling: if the highest frequency is B, perfect reconstruction is possible when sampling at rate > 2B (Nyquist rate), assuming ideal conditions. Step-by-Step Solution: 1) Identify the concept: “bandlimited” is a spectral (frequency-domain) description.2) State the formal condition: there exists B such that X(f) = 0 for |f| > B.3) Map to options: the only option that states this spectral confinement is the one describing finite frequency interval.4) Eliminate distractors not describing spectral confinement. Verification / Alternative check: The definition does not mention modulation, number of receivers, or network topology. It focuses purely on spectral support of the signal. Therefore, the correct choice must reference finite bandwidth. Why Other Options Are Wrong: Transmission without modulation: possible for short links but unrelated to the spectral definition. Simultaneous transmission to many stations: describes a distribution method, not spectrum confinement. All of the above: false because the prior two are not definitions of bandlimitedness. None of the above: incorrect because one option is correct. Common Pitfalls: Confusing implementation methods (baseband, broadcast) with spectral properties; assuming “finite energy” implies “finite bandwidth” (it does not). Final Answer: A bandlimited signal has zero spectral content outside a finite frequency band.

Question 20

In IPv4 addressing, what is the valid first-octet range that identifies a Class C network address block?

Accepted Answer

Introduction: Classical IPv4 “classful” addressing divides the address space into Classes A, B, C, D, and E based on the value of the first octet. Although classless routing (CIDR) is dominant today, understanding classful ranges remains a common exam topic. Given Data / Assumptions: IPv4 uses four octets, each 0–255. Class C networks historically used smaller subnets with many networks and few hosts. We are asked only for the first-octet range that denotes Class C. Concept / Approach: In classful IPv4: Class A = 1–126, Class B = 128–191, Class C = 192–223, Class D (multicast) = 224–239, Class E (experimental) = 240–255. Loopback uses 127.x.x.x and is not part of these usable network ranges. Step-by-Step Solution: 1) Recall class boundaries by first octet.2) Identify the Class C window: 192 through 223 inclusive.3) Match the option that exactly reflects this window. Verification / Alternative check: Binary prefix: Class C addresses begin with 110 in binary for the first octet, i.e., decimal 192–223. Why Other Options Are Wrong: 192–255: includes multicast (224–239) and experimental (240–255) ranges. 192–226: arbitrarily truncates the valid upper bound (should be 223). 128–191: this is Class B. 1–126: this is Class A (excluding 127 loopback). Common Pitfalls: Mixing classful ranges with CIDR masks; assuming 127 is Class A usable space (it is reserved for loopback). Final Answer: 192–223.

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