D686 Operating Systems for Computer Scientists - Set 1 - Part 1

Correct Answer: Running to Waiting

The process moves from Running to Waiting state when it requests I/O operations, allowing the scheduler to select another ready process for CPU execution.
Option a is incorrect because Ready to Running transition occurs when the scheduler selects a process from the ready queue for execution, not during I/O requests.
Option c is incorrect because Waiting to Ready transition happens after I/O completion when the process can resume execution but isn't yet running.
Option d is incorrect because Running to Terminated transition occurs when a process completes execution or is forcefully terminated, not during I/O operations.

Correct Answer: Belady's Anomaly with FIFO

Belady's Anomaly occurs when increasing memory frames causes more page faults, commonly seen with FIFO page replacement algorithm, contradicting expected behavior patterns.
Option a is incorrect because thrashing involves excessive paging due to insufficient memory for the working set, not increased faults with more frames.
Option c is incorrect because fragmentation relates to memory allocation waste, not page fault behavior with varying frame counts in virtual memory.
Option d is incorrect because starvation with SSTF occurs in disk scheduling when requests wait indefinitely, unrelated to page faults and memory frames.

Correct Answer: Many-to-Many Model

Many-to-Many model maps multiple user threads to multiple kernel threads, providing both efficiency and true parallelism without overwhelming kernel resources.
Option a is incorrect because Many-to-One model maps all user threads to single kernel thread, preventing true parallel execution on multicore systems.
Option b is incorrect because One-to-One model creates kernel thread per user thread, causing overhead and resource limitations with thousands of connections.
Option d is incorrect because single-threaded model cannot handle concurrent connections, making it unsuitable for high-performance web server requirements and scalability needs.

Correct Answer: User space for stability

Microkernel systems run most services in user space, isolating failures and improving system stability since service crashes don't affect kernel operations.
Option a is incorrect because running all services in kernel space is monolithic kernel design, not microkernel architecture which minimizes kernel components.
Option c is incorrect because hardware layer refers to physical components, not where operating system services execute in microkernel or any architecture.
Option d is incorrect because hypervisor relates to virtualization technology for running multiple operating systems, not microkernel service execution location design.

Correct Answer: Busy waiting

Busy waiting occurs when processes repeatedly check conditions in loops, consuming CPU cycles unproductively instead of blocking and yielding processor time.
Option a is incorrect because deadlock involves processes waiting for resources held by each other in circular dependency, not continuous checking loops.
Option b is incorrect because race condition occurs when multiple threads access shared data simultaneously, causing unpredictable results based on timing.
Option d is incorrect because priority inversion happens when lower priority process blocks higher priority process, not from continuous lock checking.

Correct Answer: Priority-based preemptive scheduling

Priority-based preemptive scheduling assigns importance levels to tasks, allowing critical tasks to interrupt lower-priority ones, ensuring deadline compliance for safety-critical operations.
Option a is incorrect because round-robin gives equal time slices to all processes, unable to guarantee critical task deadlines in real-time systems.
Option c is incorrect because first-come first-served cannot prioritize critical tasks, making it unsuitable for time-sensitive aircraft navigation control systems requirements.
Option d is incorrect because shortest job first optimizes average waiting time but doesn't guarantee meeting specific deadlines for critical real-time tasks.

Correct Answer: Readers-Writers with locks

The Readers-Writers problem allows concurrent read access while ensuring exclusive write access through synchronization mechanisms like read-write locks for consistency.
Option a is incorrect because Producer-Consumer problem involves buffer management between processes that generate and consume data, not concurrent read-write access.
Option c is incorrect because Dining Philosophers problem demonstrates resource allocation and deadlock scenarios with shared resources, not database access patterns.
Option d is incorrect because Bounded-Buffer is another name for Producer-Consumer problem with limited buffer size, not multiple reader single writer scenarios.

Correct Answer: Best-fit allocation

Best-fit allocation searches all holes and chooses the smallest adequate space, minimizing wasted space per allocation but creating many tiny fragments.
Option a is incorrect because first-fit allocation selects the first hole large enough for the request without examining all available options comprehensively.
Option c is incorrect because worst-fit allocation chooses the largest available hole, attempting to leave usable larger fragments rather than small ones.
Option d is incorrect because next-fit allocation continues searching from the last allocation point, not examining all holes for the smallest fit.

Correct Answer: Translation lookaside buffer

TLB caches frequently accessed page table entries in fast hardware memory, avoiding repeated page table lookups and improving virtual address translation performance.
Option a is incorrect because page fault handler manages missing pages in memory by loading them from disk, not speeding up address translation.
Option c is incorrect because MMU performs address translation but TLB specifically provides the caching mechanism that accelerates this translation process significantly.
Option d is incorrect because segment descriptor cache relates to segmentation systems, not the page table entry caching in paging-based virtual memory.

Correct Answer: Different process IDs

Fork returns the child's PID to parent and zero to child, allowing conditional logic based on return value to execute different code.
Option a is incorrect because fork creates exact copy including memory space initially, though copy-on-write may separate them later upon modification.
Option c is incorrect because both processes have same program counter value immediately after fork, continuing from the same instruction point.
Option d is incorrect because stack pointer is duplicated like other process resources, with both processes having identical stack contents initially.

Correct Answer: RAID one

RAID one mirrors data across drives, providing complete redundancy and fault tolerance while doubling read performance through parallel access capabilities.
Option a is incorrect because RAID zero stripes data without redundancy, improving performance but offering no fault tolerance for drive failures.
Option c is incorrect because RAID five uses parity distribution across drives, more complex than mirroring while providing single drive failure tolerance.
Option d is incorrect because RAID six implements double parity for two drive failures, adding complexity compared to simple mirroring in RAID one.

Correct Answer: Response time

Response time measures from request submission to first output, crucial for interactive systems where users expect immediate feedback to maintain usability.
Option a is incorrect because turnaround time measures total time from submission to completion, including all processing and waiting periods combined together.
Option b is incorrect because waiting time only counts time spent in ready queue, not including actual execution or initial response generation.
Option d is incorrect because throughput measures number of processes completed per time unit, not individual process response timing for interactivity.

Correct Answer: Access matrix

Access matrix defines permissions by mapping protection domains to objects with specific rights, forming the foundation for access control implementation.
Option a is incorrect because page table maps virtual pages to physical frames for memory management, not for defining user access permissions.
Option c is incorrect because process table maintains information about running processes like state and resources, not access control permissions between domains.
Option d is incorrect because file allocation table tracks disk block usage for files, not security permissions or access rights between domains.

Correct Answer: Process control block

PCB contains complete process state including registers, program counter, stack pointer, and other context information necessary for proper resumption later.
Option a is incorrect because saving only program counter wouldn't preserve register values, stack state, and other critical execution context information needed.
Option c is incorrect because stack pointer alone doesn't capture complete process state like registers, program counter, and other execution context.
Option d is incorrect because register values alone miss other critical state information like memory management data, open files, and scheduling information.

Correct Answer: Suspended state

Suspended state keeps processes in memory but not actively executing, allowing quick resumption while conserving CPU cycles and battery power effectively.
Option a is incorrect because running state means actively executing on CPU, consuming maximum resources which contradicts battery conservation goals for hidden apps.
Option b is incorrect because ready state indicates readiness for immediate execution, still competing for CPU time rather than conserving battery resources.
Option c is incorrect because waiting state typically involves I/O operations or event waiting, not the memory-resident but inactive app scenario described.

Correct Answer: Clustering technology

Clustering connects multiple computers to work cooperatively, providing high availability, load balancing, and appearing as unified system to users externally.
Option a is incorrect because batch processing executes jobs sequentially without user interaction, not creating unified multi-computer systems for resource sharing.
Option b is incorrect because time sharing allows multiple users on single computer through rapid switching, not connecting multiple computers together cooperatively.
Option d is incorrect because multiprogramming keeps multiple programs in memory on single computer, not distributing work across networked computer systems.

Correct Answer: Trap then load

Page fault causes trap to operating system, which then loads required page from secondary storage into memory before resuming execution.
Option a is incorrect because loading cannot occur before page fault detection; the trap must first signal the missing page to initiate loading.
Option c is incorrect because execution cannot proceed when page is absent; the missing page causes immediate trap before any execution occurs.
Option d is incorrect because allocation happens after trap as part of page fault handling, not before the trap signals the missing page.

Correct Answer: Bitmap approach

Bitmap uses one bit per disk block to indicate allocation status, providing fast lookup and efficient space usage for free space management.
Option a is incorrect because linked list chains free blocks together using pointers, less efficient for random access than bitmap's direct indexing approach.
Option c is incorrect because grouping method stores addresses of free blocks in first free block, different from simple bit-per-block representation.
Option d is incorrect because counting technique stores contiguous free block ranges as address-count pairs, not individual bits per block like bitmap.

Correct Answer: First-come first-served

FCFS scheduling processes requests in arrival order, causing short processes to wait for long ones, creating convoy effect and poor response.
Option a is incorrect because shortest job first minimizes average waiting time by processing short jobs first, preventing convoy effect from occurring.
Option b is incorrect because round robin gives time slices to all processes equally, preventing any single long process from blocking others.
Option d is incorrect because priority scheduling serves high-priority processes first, which may cause starvation but not specifically the convoy effect.

Correct Answer: Core location

NUMA systems have varying memory access times based on which core accesses which memory region, with local memory faster than remote.
Option a is incorrect because cache size affects hit rates and capacity but doesn't determine the fundamental NUMA characteristic of location-based access times.
Option c is incorrect because bus speed affects overall transfer rates but NUMA specifically refers to non-uniform times based on core-memory proximity.
Option d is incorrect because clock frequency determines processing speed but NUMA's non-uniformity stems from physical distance between cores and memory regions.