solution - Operating System : Memory Management (Test 2)

Question : Suppose a 16-bit logical address uses 4 bits for segment number and 12 bits for offset.

Solution :
30820260315010821 Correct Answer : 4096 bytes

Description -

Operating System : Memory Management (Test 2)

Question : Which of the following is a major disadvantage of segmentation?

Solution :
687920260315010721 Correct Answer : External fragmentation

Description -

Operating System : Memory Management (Test 2)

Question : Which information is stored in each entry of the segment table?

Solution :
966220260315010633 Correct Answer : Base address and segment limit

Description -

Operating System : Memory Management (Test 2)

Question : A logical address in segmentation consists of which two components?

Solution :
142820260315010548 Correct Answer : Segment number and offset

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Operating System : Memory Management (Test 2)

Question : Which of the following best describes segmentation in an operating system?

Solution :
239620260315010501 Correct Answer : Memory is divided into variable-size logical units called segments

Description -

Operating System : Memory Management (Test 2)

Question : Consider a demand paging system with:
Page fault rate = 0.02
Memory access time = 100 ns
Page fault service time = 8 ms
What is the Effective Access Time (EAT)?

Solution :
826020260315010309 Correct Answer : 160,098 ns

Description -

Operating System : Memory Management (Test 2)

Question : Which of the following best explains why demand paging improves memory utilization?

Solution :
498720260315010038 Correct Answer : Only required pages are loaded into main memory

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Operating System : Memory Management (Test 2)

Question : Which condition best describes thrashing in a demand paging system?

Solution :
392020260315125956 Correct Answer : When page fault rate becomes extremely high and CPU spends most time swapping pages

Description -

Operating System : Memory Management (Test 2)

Question : Which formula correctly represents the Effective Access Time (EAT) in a demand paging system?
p = page fault rate, s = page fault service time, m = memory access time

Solution :
116620260315125850 Correct Answer : EAT = p × s + (1 − p) × m

Description -

Operating System : Memory Management (Test 2)

Question : In a demand paging system, what happens when a process references a page that is not currently in main memory?

Solution :
462720260315125642 Correct Answer : The OS generates a page fault and loads the page from secondary memory

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Operating System : Memory Management (Test 2)

Question : If the physical address space = 64 KB and word size = 8 bytes, what is the number of bits required for the physical address?

Solution :
517920260315125400 Correct Answer : 13 bits

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Operating System : Memory Management (Test 2)

Question : In modern operating systems, which statement about user access to physical addresses is correct?

Solution :
902520260315125317 Correct Answer : Users access physical memory indirectly through logical addresses.

Description -

Operating System : Memory Management (Test 2)

Question : If the physical address space = N words, the number of bits required to represent a physical address is:

Solution :
711120260315125126 Correct Answer : Log2 N

Description -

Operating System : Memory Management (Test 2)

Question : Which statement correctly describes the logical address space?

Solution :
82220260315124209 Correct Answer : It is the set of addresses generated by the CPU for a process.

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Operating System : Memory Management (Test 2)

Question : Which component is primarily responsible for translating logical addresses into physical addresses in an operating system?

Solution :
223020260315124113 Correct Answer : Memory Management Unit (MMU)

Description -

Operating System : Memory Management (Test 2)

Question : When a process tries to access a page that is not currently present in main memory, the event is called:

Solution :
185620260315123541 Correct Answer : Page fault

Description -

Operating System : Memory Management (Test 2)

Question : Which hardware component is primarily responsible for translating logical addresses into physical addresses in a paging system?

Solution :
807720260315123448 Correct Answer : Memory Management Unit (MMU)

Description -

Operating System : Memory Management (Test 2)

Question : Which problem of memory management is significantly reduced by using paging?

Solution :
452720260315123400 Correct Answer : External fragmentation

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Operating System : Memory Management (Test 2)

Question : In paging, logical memory is divided into pages, while physical memory is divided into:

Solution :
575520260315123315 Correct Answer : Frames

Description -

Operating System : Memory Management (Test 2)

Question : Which of the following is the main reason paging is required in an operating system?

Solution :
487420260315123231 Correct Answer : To allow non-contiguous allocation of physical memory to processes

Description -

Operating System : Memory Management (Test 2)