Question 1

Which system program is responsible for placing an already linked (or relocatable) program into main memory and preparing it to begin execution?

Accepted Answer

Introduction / Context: After code is compiled and linked, the resulting executable must be mapped into memory so the CPU can run it. This final step assigns memory, fixes addresses if needed, and transfers control to the program's entry point. The component that performs these tasks is the loader. Given Data / Assumptions: Executable or relocatable image is available. Operating system provides loading facilities. Relocation and initialization may be required. Concept / Approach: The loader reads the program image, allocates memory, resolves relocations (if not already fixed), links with shared libraries at load time as necessary, sets up the stack and environment, and jumps to the entry point. Linkers combine modules; assemblers and compilers translate source code but do not run the program. Step-by-Step Solution: 1) Open the executable image and parse headers to determine segments.2) Reserve and map code/data segments into main memory.3) Apply relocations and resolve dynamic libraries if applicable.4) Initialize runtime state and transfer control to the entry point. Verification / Alternative check: Tracing a program start shows the OS loader reading the binary, mapping sections, and invoking the program's main entry after runtime initialization. Why Other Options Are Wrong: Linker: Produces executables; does not execute them. Assembler / Compiler: Translate code; not responsible for loading. None of the above: Incorrect because the loader performs the setup. Common Pitfalls: Confusing static linking with loading; even statically linked programs require a loader to map and start execution. Final Answer: Loader

Question 2

In virtual memory management, which condition most effectively prevents thrashing by respecting process working sets?

Accepted Answer

Introduction / Context: Thrashing occurs when the system spends more time paging than executing, typically because active pages of processes do not fit in physical memory. The working-set model provides a principled way to allocate frames to keep each process's most frequently used pages resident, reducing page faults. Given Data / Assumptions: Demand paging is used with finite physical memory. Processes have identifiable working sets that vary over time. Replacement and admission policies influence residency. Concept / Approach: Ensuring that each process's working set is resident stabilizes locality and minimizes fault frequency. Increasing CPU or I/O speed does not address the root cause; the bottleneck is excessive page faults due to insufficient resident locality, not raw compute or I/O throughput. Step-by-Step Solution: 1) Measure or estimate each process's working set size over a window.2) Allocate frames to cover these sets (or reduce degree of multiprogramming).3) Monitor page-fault rate; if high, adjust allocations or suspend processes.4) Maintain stability by avoiding oscillation in replacement decisions. Verification / Alternative check: After ensuring working sets are resident, CPU utilization rises and page-fault rate drops sharply, confirming thrashing mitigation independent of CPU/I-O speeds. Why Other Options Are Wrong: Increase CPU speed: A faster CPU still stalls on page faults. Increase I/O processor speed: Faster I/O helps marginally but does not fix locality mismatch. All of the above: Incorrect; only working-set residency tackles the cause. None of the above: Incorrect because working-set residency is effective. Common Pitfalls: Tuning disk hardware without adjusting memory allocation or degree of multiprogramming, which leaves the fundamental locality problem unresolved. Final Answer: The pages belonging to the working set of the programs are in main memory

Question 3

In instruction design, what does the term ‘‘addressing structure’’ primarily define with respect to computing effective operand addresses?

Accepted Answer

Introduction / Context: Addressing determines how instructions locate their operands. Clear understanding of addressing structures and their variations is essential for assembly programming, compiler backends, and microarchitecture optimization. Given Data / Assumptions: ISAs provide base addressing forms (for example, immediate, direct, indirect, indexed, base+offset). Some syntactic or semantic variations exist (auto-increment, auto-decrement). We must distinguish fundamental structures from unrelated execution semantics. Concept / Approach: Addressing structure refers to the fundamental scheme by which an instruction computes an effective address from fields such as base register, index, displacement, or immediate. Variations (addressing modes) extend these basics (for example, auto-increment). Arithmetic on registers is not an addressing structure—it is an operation type. Step-by-Step Solution: 1) Identify core address computation patterns (immediate, direct, indirect, indexed, base+offset).2) Recognize enhancements like auto-increment as variations derived from core structures.3) Separate addressing from operation semantics (ALU on registers is not addressing).4) Conclude that addressing structure primarily defines how to compute the effective address. Verification / Alternative check: ISA documentation sections titled “Addressing Modes” detail computations of effective addresses, confirming the definition aligns with option (a). Why Other Options Are Wrong: Variations description: This is secondary (modes), not the primary definition. Two-register operation: Describes instruction format/ALU, not addressing. All of the above: Incorrect because (c) is unrelated to addressing. None of the above: Incorrect as (a) is correct. Common Pitfalls: Confusing instruction format (register counts) with addressing, and conflating addressing with data-path operations. Final Answer: Defines the fundamental method of determining effective operand addresses

Question 4

Which statement best characterizes associative memory (content-addressable memory) in terms of how data is retrieved?

Accepted Answer

Introduction / Context: Associative memory, more formally called content-addressable memory (CAM), enables lookups based on content rather than explicit addresses. This capability is valuable in networking (for example, MAC tables), caches (tags), and translation lookaside buffers (TLBs). Given Data / Assumptions: Goal is to retrieve or match entries by value (key) rather than numeric address. Hardware can compare a search key against many entries in parallel. Result may return the location or a hit signal indicating a match. Concept / Approach: Unlike RAM, where the CPU supplies an address, CAM takes a search word and simultaneously compares it against stored words. A match returns the index or associated data, eliminating the need to specify an address beforehand. Hence, “the data is used as an address” captures the essence of associative access. Step-by-Step Solution: 1) Provide a search key to the associative array.2) Hardware compares the key with all stored entries in parallel.3) On match, assert a hit and optionally output the matched index or data.4) Use the match to route packets, translate addresses, or verify cache tags. Verification / Alternative check: Observing TLB behavior shows that virtual-page numbers are matched against stored tags in parallel; on a hit, the frame number is returned without a CPU-provided RAM address. Why Other Options Are Wrong: Sequential access: Describes tapes or list scans, not CAM. CPU-generated address / user-supplied address: RAM paradigms, not associative lookup. None of the above: Incorrect because associative lookup is accurately described by option (a). Common Pitfalls: Assuming associative memory stores full data payloads; often it stores tags/keys with small associated fields to keep hardware practical and fast. Final Answer: There is no need for an address i.e. the data is used as an address

Question 5

Which addressing mode uses the contents of a specified register as the effective address and then automatically increments that register to point to the next element in a list?

Accepted Answer

Introduction / Context: Addressing modes extend instruction flexibility and can accelerate common patterns like iterating through arrays or stacks. Auto-increment addressing is particularly useful for linear scans of buffers because it eliminates explicit instructions to update the pointer after each access. Given Data / Assumptions: A register holds the address of the current element. Accessing the operand should also advance the pointer. Increment amount may depend on operand size (for example, +1 for byte, +4 for word). Concept / Approach: In auto-increment mode, the effective address is taken from a register (like Rn). After the read or write, the CPU increments Rn automatically to the next element, reducing instruction count and improving pipeline efficiency for sequential access patterns. Step-by-Step Solution: 1) Let Rn contain the address of A[i].2) Execute an instruction using [Rn] as the effective address (for example, load).3) The hardware completes the memory access.4) The CPU automatically increments Rn to point to A[i+1] (size-based increment). Verification / Alternative check: Assembly listings on architectures supporting this mode (for example, some ARM or DSP variants) show that pointer updates are implicit, saving an extra add instruction per iteration. Why Other Options Are Wrong: Indirect addressing: Uses a register for address but does not auto-update. Index addressing: Adds index/offset; no implicit increment of the base register. Auto decrement: Similar pattern but decrements after access (useful for stacks). None of the above: Incorrect because auto increment matches the description. Common Pitfalls: Ignoring element size when reasoning about increment amounts, which can yield misaligned accesses or skipped elements. Final Answer: Auto increment

Question 6

In system software (linkers and loaders), which operations are performed by the loader during program load and start-up?

Accepted Answer

Introduction / Context: In operating systems and system programming, a loader is responsible for preparing a compiled program for execution. It takes the output of the linker (often relocatable or executable object files) and ensures the program occupies appropriate memory with all addresses adjusted correctly so that the CPU can begin execution without faults. Given Data / Assumptions: The prompt lists distinct responsibilities a loader may perform. We assume a conventional environment where relocation and symbol resolution may still be required at load time (static or dynamic scenarios). The end goal is a runnable image in memory with a proper entry point. Concept / Approach: A classical loader will: (1) allocate memory for the program's segments, (2) resolve external symbols not fixed at link time (or finalize relocations), and (3) copy or map program text/data into RAM, then transfer control to the entry point. These tasks correspond to space allocation, relocation/adjustment of address-dependent locations, and physical placement (loading) of code and data. Step-by-Step Solution: Identify loader responsibilities: memory allocation, relocation, and placement.Match each responsibility to the listed options.Recognize that all listed functions are canonical loader tasks.Therefore, select 'All of the above'. Verification / Alternative check: Standard OS texts define loader duties exactly as memory allocation, relocation, binding, and loading, often followed by setting up the initial process state and jumping to the program's entry point. Why Other Options Are Wrong: Options a–c are correct individually but incomplete alone. None of the above is false because each listed function is indeed a loader activity. Common Pitfalls: Confusing linker vs. loader roles; linkers produce combined objects, loaders finalize and place in memory. Assuming demand-paged systems eliminate loaders; even then, a loader (or the kernel's exec) performs these steps via mapping. Final Answer: All of the above.

Question 7

In process management for operating systems, what does the term 'fork' specifically refer to?

Accepted Answer

Introduction / Context: Fork is a classic UNIX/POSIX system call central to process management. Understanding it is crucial for systems programming, shell behavior, and concurrent execution models on UNIX-like operating systems. Given Data / Assumptions: The context is operating system terminology. 'Task' and 'job' may have broader or scheduler-specific meanings, but we focus on the POSIX definition. Fork duplicates the calling process to create a new, separate process context. Concept / Approach: In POSIX, fork() creates a child process by duplicating the parent's address space and execution context (with copy-on-write optimizations typically). The new process receives its own PID and resources as defined by the OS. It is not merely dispatching or priority management; it is process creation. Often, exec follows fork to overlay the child process with a new program image. Step-by-Step Solution: Map 'fork' to POSIX semantics: duplication of the calling process.Differentiate from job creation (batch systems) and dispatching (scheduling ready-to-run tasks).Identify the correct definition: creation of a new process.Select the option that states this explicitly. Verification / Alternative check: Man pages for fork(2) and OS textbooks consistently define fork as process creation with a duplicated execution context. Why Other Options Are Wrong: Dispatching of a task: that is a scheduler action, not creation. Creation of a new job: batch/job control concept; not specific to POSIX fork. Increasing priority: unrelated to fork. None of the above: incorrect because one option matches precisely. Common Pitfalls: Confusing threads with processes; fork creates a process, not a thread (although threads may be duplicated subject to semantics). Final Answer: the creation of a new process.

Question 8

Which memory allocation approach is particularly susceptible to external fragmentation in operating systems?

Accepted Answer

Introduction / Context: Fragmentation in memory management appears as either internal or external. External fragmentation occurs when free space exists but is divided into small noncontiguous blocks that are insufficient for a request, even though the total free space is adequate. Recognizing which schemes suffer most helps explain design choices like paging or hybrid strategies. Given Data / Assumptions: Segmentation allocates variable-size segments. Pure paging uses fixed-size frames/pages. Multiple fixed partitions allocate fixed-size blocks (leading to internal, not external, fragmentation). Swapping is a mechanism, not an allocation scheme by itself. Concept / Approach: Variable-size allocations (e.g., segmentation or dynamic partitions) create gaps as segments are loaded and removed; over time, free memory becomes fragmented externally. Paging, by contrast, eliminates external fragmentation by using uniform page/frame sizes (though it can suffer internal fragmentation in the last page). Fixed partitions also avoid external fragmentation but suffer internal waste when processes do not perfectly fill a partition. Step-by-Step Solution: Identify schemes with variable-sized allocations: segmentation.Assess others: pure demand paging uses fixed frames; fixed partitions are fixed-size; swapping is orthogonal.Conclude that segmentation is most prone to external fragmentation.Select 'segmentation'. Verification / Alternative check: OS textbooks emphasize that paging eliminates external fragmentation while segmentation, dynamic partitioning, and heap-like allocators suffer from it. Why Other Options Are Wrong: pure demand paging: fixed-size pages avoid external fragmentation. multiple contiguous fixed partitions: mostly internal fragmentation. swapping: a process placement mechanism, not the allocation granularity determining fragmentation type. Common Pitfalls: Confusing internal with external fragmentation or conflating swapping with allocation policy. Final Answer: segmentation.

Question 9

In CPU scheduling, what does the FIFO (First-In, First-Out) / FCFS policy run first?

Accepted Answer

Introduction / Context: First-Come, First-Served (FCFS), also called FIFO, is one of the simplest CPU scheduling algorithms. It orders processes strictly by arrival time, providing a baseline for understanding more advanced schedulers like SJF, priority, and round-robin. Given Data / Assumptions: Processes are queued as they arrive. Preemption is not used in basic FCFS. Service begins with the head of the queue. Concept / Approach: FCFS services requests in arrival order. The first job to enter the ready queue is the first to be executed. While simple, FCFS can lead to the convoy effect, where short jobs wait behind long ones, increasing average waiting time compared to SJF. Step-by-Step Solution: Define FCFS policy: arrival-order execution.Identify the first-arriving job as the first to run.Select the option that states this explicitly.Therefore, choose 'executes first the job that first entered the queue'. Verification / Alternative check: Scheduling theory and OS curricula consistently describe FCFS as non-preemptive, arrival-order servicing. Why Other Options Are Wrong: last entered: LIFO, not FCFS. longest in queue: equivalent to first entered, but the phrasing in option c is ambiguous and not canonical; option b is precise. least processor needs: describes SJF or Shortest Remaining Time First, not FCFS. Common Pitfalls: Confusing FCFS with priority-based or length-based scheduling. Final Answer: executes first the job that first entered the queue.

Question 10

Language translators in software engineering A program that reads an entire source file written in a high-level language and produces machine code is called a:

Accepted Answer

Introduction / Context: Understanding how source code becomes executable is foundational for both software and IT professionals. Translators include assemblers, compilers, and interpreters, each with distinct roles and workflows. Given Data / Assumptions: Source language is a high-level language (e.g., C, C++, Pascal, Rust). Output is machine code (object files or an executable binary). The process consumes the entire program rather than executing line-by-line. Concept / Approach: A compiler translates high-level source code into machine code, often generating intermediate representation and object files which are then linked. Assemblers convert assembly language (low-level mnemonic form) into machine code. Interpreters execute high-level code statements directly without producing a standalone machine binary. Step-by-Step Solution: Identify the input: high-level language source.Identify the output: machine language program.Match to the translator type: compiler. Verification / Alternative check: Toolchains like GCC/Clang (compilers) read the entire translation unit, emit object code, and link to produce executables, matching the definition. Why Other Options Are Wrong: Assembler works from assembly, not high-level languages. “Translator” is too generic and not a specific tool class. “System software” is a category (OS, drivers), not a translator. Common Pitfalls: Mixing up compiler vs interpreter; overlooking JIT compilers that compile at runtime but still compile rather than interpret line-by-line semantics. Final Answer: compiler.

Question 11

Early IBM PC operating systems IBM released its first PC in 1981. Which operating system was most popularly associated with that machine at launch?

Accepted Answer

Introduction / Context: The IBM PC (Model 5150) established a de facto standard platform. Understanding its original operating system clarifies many historical compatibility decisions still echoing in today’s PCs. Given Data / Assumptions: Year: 1981 release of the IBM PC. IBM distributed its own branded operating system for this machine. Similar codebase was also licensed and sold under a different name by Microsoft to other OEMs. Concept / Approach: IBM marketed the OS as PC-DOS, licensed from Microsoft’s MS-DOS. While CP/M was a popular OS on earlier 8-bit systems, IBM’s PC-DOS rapidly became the dominant OS on the IBM PC platform. OS/360 was an IBM mainframe operating system, unrelated to the PC. Step-by-Step Solution: Link the IBM brand to its OEM-licensed DOS → PC-DOS.Recognize MS-DOS is the Microsoft-branded variant for non-IBM OEMs.Select “PC-DOS” as the historically correct association. Verification / Alternative check: IBM documentation for the 5150 references PC-DOS versions 1.x. Press materials and technical manuals from the era confirm. Why Other Options Are Wrong: MS-DOS is closely related but not the IBM-branded OS for the IBM PC at launch. CP/M was considered, but IBM shipped PC-DOS. OS/360 is a mainframe OS. Common Pitfalls: Conflating PC-DOS and MS-DOS naming; assuming CP/M carried over from earlier microcomputers to the IBM PC default. Final Answer: PC-DOS.

Question 12

Virtual Memory and Page Faults  In operating systems that implement virtual memory, what best describes a page fault event when a running program accesses memory?

Accepted Answer

Introduction / Context: A page fault is a fundamental concept in virtual memory systems used by modern operating systems. Understanding what triggers a page fault clarifies how the OS transparently extends memory using disk-backed pages and on-demand loading. Given Data / Assumptions: Virtual memory is enabled and pages are the unit of memory management. Processes reference virtual addresses that must be translated to physical frames. The OS maintains page tables with presence/valid bits. Concept / Approach: When a process references a virtual address, the memory management unit (MMU) checks the corresponding page table entry. If the valid/present bit indicates the page is not in RAM, the MMU raises a page-fault exception. The OS then services the fault by bringing the needed page into a free frame (possibly evicting another page) and resumes the faulting instruction. Step-by-Step Solution: Step 1: Process issues a load/store to a virtual address.Step 2: MMU consults the page table entry (PTE) for that page.Step 3: If PTE.present = 0, a page-fault trap is generated.Step 4: OS page-fault handler locates the page on disk (swap or file), selects a free/eviction frame, and schedules I/O.Step 5: After I/O completes, OS updates the PTE (present = 1, frame number set), then restarts the faulting instruction. Verification / Alternative check: A minor (soft) page fault occurs when the page is already in memory (e.g., in page cache) but not mapped; a major (hard) page fault requires disk I/O. In both cases, the trigger is that the page was not currently resident or mapped for the requesting process. Why Other Options Are Wrong: Option A: A page fault is not inherently an error or data corruption; it is a normal control-flow event.Option B: Not every access causes a fault—only accesses to non-resident or unmapped pages.Option D: Accessing another process’s memory would normally cause a protection fault, not a page fault.Option E: Not applicable because Option C is correct. Common Pitfalls: Confusing page faults with segmentation/protection faults. Assuming all page faults indicate performance problems; many are expected during normal execution. Believing a page fault always implies disk access; soft faults do not. Final Answer: It is an access to a page that is not currently resident in physical memory (RAM).

Question 13

Precise Interrupt Handling  After a CPU receives an interrupt request from an I/O device, when does control transfer to the interrupt service routine (ISR)?

Accepted Answer

Introduction / Context: Interrupts enable responsive I/O by letting devices signal the CPU. Precise interrupt semantics define when and how control switches from the running program to the interrupt service routine (ISR). Given Data / Assumptions: Conventional, precise interrupts as implemented by mainstream CPUs. The current instruction is atomic from the programmer’s perspective. Interrupts are maskable unless disabled. Concept / Approach: With precise interrupts, the CPU finishes the current instruction, saves minimal context (program counter, flags, sometimes registers), and vectors to the ISR using an interrupt vector table. This guarantees a consistent architectural state at the ISR entry point. Step-by-Step Solution: Step 1: Device asserts an interrupt request (IRQ).Step 2: CPU acknowledges the interrupt at an instruction boundary.Step 3: CPU pushes return state (PC/flags) to a stack or dedicated area.Step 4: CPU loads the ISR address from the vector and jumps to it.Step 5: ISR services the device, then executes a return-from-interrupt instruction to resume the preempted task. Verification / Alternative check: Architectures may support non-maskable interrupts (NMI) and precise exceptions; the common property is transfer at a clean boundary after completing the current instruction, not in the middle of one. Why Other Options Are Wrong: Option A: CPUs do not arbitrarily halt for fixed durations on interrupts.Option B: Bus control is not simply handed to the device; the CPU still orchestrates transfers.Option C: Mid-instruction preemption would break precise state guarantees.Option E: Not applicable because Option D is correct. Common Pitfalls: Confusing DMA bus mastering with interrupt handling. Assuming interrupts always preempt instantly mid-instruction. Forgetting that interrupts can be masked or prioritized. Final Answer: The CPU completes the current instruction, then branches to the ISR.

Question 14

Disk Storage vs. Main Memory  Which statement is false when comparing magnetic/SSD disk storage to main memory (RAM)?

Accepted Answer

Introduction / Context: Choosing appropriate storage requires understanding trade-offs among speed, capacity, volatility, and cost. Comparing disk storage and RAM highlights why systems use hierarchical memory. Given Data / Assumptions: Disk may refer to HDD or SSD; RAM refers to volatile DRAM. We compare typical characteristics, not exotic hardware. Concept / Approach: RAM provides much lower latency and higher bandwidth than disk but is volatile and more expensive per bit. Disk (especially HDD) is far slower but cheaper and non-volatile. SSDs improve speed but still exhibit higher latency than DRAM. Step-by-Step Solution: Evaluate non-volatility: Disk is non-volatile; RAM loses data on power-off.Evaluate capacity: Disks commonly provide far larger capacities than RAM.Evaluate cost per bit: Disk cost/bit is lower than RAM.Evaluate performance: RAM access latency/bandwidth is superior; therefore “disk is faster” is false. Verification / Alternative check: Typical DRAM latency is on the order of tens of nanoseconds; SSD latency is microseconds; HDD access involves milliseconds—orders of magnitude slower than RAM. Why Other Options Are Wrong: Option A: Correct—disks are non-volatile.Option B: Correct—disks usually provide much larger capacity.Option C: Correct—cost per bit is lower for disks than for DRAM.Option E: Not applicable because Option D is the false statement. Common Pitfalls: Assuming SSD speed matches RAM speed; SSDs are faster than HDDs but still slower than DRAM. Confusing throughput (MB/s) with latency; both favor RAM. Equating cache memory with main memory; CPU caches are even faster than DRAM. Final Answer: Disk is faster than main memory.

Question 15

Register-to-Register (RR) Instruction Class  Which description best characterizes RR instructions in a typical instruction set architecture?

Accepted Answer

Introduction / Context: Instruction formats are often categorized by where operands reside: register-to-register (RR), register-to-memory (RM), and memory-to-memory (MM). Recognizing RR instructions is core to understanding CPU datapaths and pipeline throughput. Given Data / Assumptions: General-purpose registers are present. Opcodes specify operand sources and destination. Concept / Approach: RR instructions fetch operands exclusively from registers and write the result back to a register. This minimizes memory traffic and is favored in RISC designs, enabling single-cycle ALU operations when operands are ready in the register file. Step-by-Step Solution: Identify operand locations: both sources are registers.Execution path: register file → ALU → register file.Result placement: destination is a register (e.g., ADD R1, R2 → R1).No direct memory access happens during the ALU operation itself. Verification / Alternative check: In contrast, RM instructions use one register and one memory operand, while MM instructions use two memory operands (rare in modern CPUs due to pipeline and bandwidth costs). Why Other Options Are Wrong: Option A: Describes memory-to-memory instructions.Option B: Describes register-to-memory (or memory-to-register) class.Option D: Cannot be correct because A and B contradict the RR definition.Option E: Not applicable because Option C is correct. Common Pitfalls: Confusing addressing mode with instruction class—RR defines operand location, not how addresses are computed. Assuming RR always takes one cycle; pipeline hazards can introduce stalls. Final Answer: Operate on two registers and leave the result in a register.

Question 16

Working-Set Model of Process Memory Behavior  According to the working-set theory in operating systems, what does a process’s working set represent?

Accepted Answer

Introduction / Context: The working-set model explains locality of reference in process memory usage and guides policies for page allocation and load control to minimize thrashing. Given Data / Assumptions: Virtual memory with paging is in use. A time window (Delta) defines “recently used”. Locality implies clustered page references. Concept / Approach: The working set W(t, Delta) is the set of distinct pages referenced by a process during the interval of length Delta ending at time t. Ensuring enough frames to hold the working set reduces page faults and prevents thrashing. Step-by-Step Solution: Define Delta: a tunable window representing recent activity.Track page references within this window.Compute W as the unique set of pages touched in the window.Allocate frames ≥ |W| to meet demand and maintain performance. Verification / Alternative check: Empirical observations of program phases show stable working sets during phases and changes at phase boundaries, aligning with locality principles (temporal and spatial). Why Other Options Are Wrong: Option B: Disk scheduling concerns I/O queue order, not working-set content.Option C: Coalescing free holes is a heap/allocator concern, not paging.Option D: CPU scheduling is orthogonal to working-set memory behavior.Option E: Not applicable because Option A is correct. Common Pitfalls: Confusing working set with resident set; working set is a target size, resident set is actual frames held. Choosing Delta too small or too large, leading to under- or over-provisioning. Final Answer: The collection of pages the process actively accesses over a recent window of time.

Question 17

Swapping and Memory Sharing  In classical swapping systems, what does swapping allow regarding use of main memory?

Accepted Answer

Introduction / Context: Swapping is an early memory-management technique where entire processes are moved between main memory and secondary storage to share limited RAM among multiple programs. Given Data / Assumptions: Memory is insufficient to hold all processes at once. CPU scheduling determines which process runs. Whole-process images can be written to disk and restored. Concept / Approach: Swapping enables time-sharing by loading a process into memory to run and writing it back to disk when it is not scheduled, freeing memory for others. This differs from paging, which moves fixed-size pages instead of entire processes. Step-by-Step Solution: Scheduler selects a ready process.If not resident, the OS swaps it in (reads its image into RAM).The process executes for a quantum.When descheduled and memory is needed, the OS may swap it out. Verification / Alternative check: Historically, systems with limited RAM relied on swapping to provide multi-user interactivity; performance improved with paging and demand-paged virtual memory, but the principle of using memory “in turn” remains accurate for swapping. Why Other Options Are Wrong: Option A: Partition count alone does not make swapping “best”.Option B: “Simultaneously” is misleading; swapped-out programs are not in RAM.Option D: Overlaying (manually managing code segments) can be used with or without swapping; they are not mutually exclusive.Option E: Not applicable because Option C is correct. Common Pitfalls: Equating swapping with paging—they are distinct mechanisms. Assuming swapping implies concurrent residency; it does not. Final Answer: Each program to use memory in turn by being swapped in when scheduled.

Question 18

Macro Processors in Assemblers/Compilers  Which tasks must a macro processor perform while handling macro definitions and invocations?

Accepted Answer

Introduction / Context: Macro processors extend assembly or high-level languages by allowing parameterized text substitution. They improve code reuse and readability without runtime overhead. Given Data / Assumptions: Source files may contain macro definitions and macro invocations. The macro processor executes at translation time. Concept / Approach: A macro definition names a template. The macro processor must parse and store definitions, detect invocations, and on each call, expand the template with actual parameters, producing plain source for the subsequent assembler/compiler stage. Step-by-Step Solution: Identify macro boundaries (e.g., MACRO/ENDM directives).Build a macro name table and definition table.When a call is encountered, bind formal parameters to actuals.Emit expanded text in place of the call for further translation. Verification / Alternative check: Tools like m4 or assembler macro facilities follow this pipeline, and listings will show expanded code replacing macro calls. Why Other Options Are Wrong: Options A, B, and C each cover part of the required pipeline; only “All of the above” is complete.Option E is incorrect because macro processing does involve these steps. Common Pitfalls: Confusing macro expansion (compile-time) with function calls (run-time). Mishandling nested macros or recursive expansions leading to infinite loops. Incorrect argument substitution causing subtle bugs in generated code. Final Answer: All of the above.

Question 19

Choosing Application Software  Which information sources are most helpful when evaluating which application software best meets your needs?

Accepted Answer

Introduction / Context: Selecting application software benefits from evidence-based evaluation. Multiple independent sources reduce bias and reveal strengths and weaknesses across real-world use cases. Given Data / Assumptions: You need to compare functionality, performance, cost, and support. Sources vary in perspective: peer users vs. professional reviewers. Concept / Approach: Triangulation—using user recommendations, professional articles, and objective reviews—provides a balanced picture. Check version numbers, test environments, and methodologies to ensure relevance. Step-by-Step Solution: Gather peer feedback matching your domain.Consult magazines or trusted sites for comparative benchmarks.Review formal evaluations with clear criteria (features, TCO, usability).Synthesize findings against your requirements and constraints. Verification / Alternative check: Trial versions or sandbox pilots validate conclusions from secondhand sources before full deployment. Why Other Options Are Wrong: Options A, B, C each contribute meaningfully; selecting only one risks bias.Option E is incorrect because these sources are useful. Common Pitfalls: Relying on marketing claims without independent verification. Ignoring version updates that change features or licensing. Failing to test integration with existing systems. Final Answer: All of the above.

Question 20

Addressing Modes in Instruction Sets  Which statement best characterizes addressing modes in computer architecture?

Accepted Answer

Introduction / Context: Addressing modes describe how an instruction identifies its operands. They are central to ISA design, affecting code density, performance, and compiler simplicity. Given Data / Assumptions: Examples include immediate, register, direct, indirect, indexed, base+offset, displacement, and PC-relative. We distinguish instruction class (e.g., register-to-register) from addressing mode. Concept / Approach: Addressing modes are best seen as a set of variations that manipulate base addresses, indices, constants, and indirection to compute an effective address. This aligns with option B. Option A describes addressing in general; option C confuses addressing modes with instruction classes. Step-by-Step Solution: Identify the ISA’s base addressing structures (registers, memory, immediates).Define how modes compute effective address: EA = base + index * scale + displacement.Recognize special actions: auto-increment/decrement, PC-relative adjustments.Classify instructions separately (RR, RM) from addressing modes they use. Verification / Alternative check: Consider a load with indexed addressing: EA = base + index. Changing to displacement mode uses EA = base + disp. These are different modes using similar primitives. Why Other Options Are Wrong: Option A: Too generic; it describes addressing broadly, not the essence of “modes” as variations.Option C: Describes register-to-register instruction class, not addressing modes.Option D: Cannot be correct because C is unrelated; therefore “all” is false.Option E: Not applicable because B captures the concept accurately. Common Pitfalls: Equating addressing modes with operand location (register vs. memory). Overlooking side effects like auto-increment that are part of the addressing mode semantics. Final Answer: They are variations in how fundamental addressing structures are used and related actions associated with addressing.

Operating Systems Concepts Questions