Memory Management Questions

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171.

If working set window is too large ____________

A

it will not encompass entire locality

B

it may overlap several localities

C

it will cause memory problems

D

none of the mentioned

Correct Answer: B

None.

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172.

If the sum of the working-set sizes increases, exceeding the total number of available frames ____________

A

then the process crashes

B

the memory overflows

C

the system crashes

D

the operating system selects a process to suspend

Correct Answer: D

None.

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173.

Consider the following page reference string. 1 2 3 4 2 1 5 6 2 1 2 3 7 6 3 2 1 2 3 6 For LRU page replacement algorithm with 4 frames, the number of page faults is?

A

10

B

14

C

8

D

11

Correct Answer: A

None.

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174.

Consider the following page reference string. 1 2 3 4 2 1 5 6 2 1 2 3 7 6 3 2 1 2 3 6 For LRU page replacement algorithm with 5 frames, the number of page faults is?

A

10

B

14

C

8

D

11

Correct Answer: C

None.

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175.

Consider the following page reference string. 1 2 3 4 2 1 5 6 2 1 2 3 7 6 3 2 1 2 3 6 For FIFO page replacement algorithms with 3 frames, the number of page faults is?

A

16

B

15

C

14

D

11

Correct Answer: A

None.

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176.

Consider the following page reference string. 1 2 3 4 2 1 5 6 2 1 2 3 7 6 3 2 1 2 3 6 For FIFO page replacement algorithms with 4 frames, the number of page faults is?

A

16

B

15

C

14

D

11

Correct Answer: C

None.

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177.

Consider the following page reference string. 1 2 3 4 2 1 5 6 2 1 2 3 7 6 3 2 1 2 3 6 For Optimal page replacement algorithms with 3 frames, the number of page faults is?

A

16

B

15

C

14

D

11

Correct Answer: D

None.

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178.

A system uses demand paging with a page size of 4 KB. The memory access time is 100 ns, and the page fault service time (including swap in and restart) is 10 ms. If the effective access time (EAT) must not exceed 200 ns, what is the maximum acceptable page fault rate (in terms of number of faults per memory access, rounded to the nearest integer power of 10)?

Correct Answer: 0.00001

Effective Access Time (EAT) = (1 - p) × memory access time + p × page fault service time. Substituting values: EAT = (1 - p) × 100 ns + p × 10,000,000 ns. For EAT ≤ 200 ns, solving gives p ≤ 1 × 10^-5. Thus, the maximum acceptable page fault rate is 10^-5.

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179.

Consider a system with 32-bit virtual addresses and a page size of 8 KB. The system uses a two-level paging scheme where the first-level page table has 1024 entries. Each page table entry is 4 bytes. What is the size (in KB) of the second-level page table?

Correct Answer: 2

Virtual address space = 2^32 bytes. Page size = 8 KB = 2^13 bytes. Thus, total pages = 2^32 / 2^13 = 2^19. First-level page table has 1024 = 2^10 entries. Each entry points to a second-level table. Total pages covered per second-level table = 2^19 / 2^10 = 2^9 = 512 pages. Each second-level table must have 512 entries. Each entry = 4 bytes, so size = 512 × 4 = 2048 bytes = 2 KB.

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