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Operating System :: Memory Management

Home > Operating System > Memory Management > General Questions

1. Which of the following programming technique/structures are not good for demand paged environment?

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Answer: Option C

Explanation:

Here is no explanation for this answer

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2. For 1 MB memory, the number of address lines required,

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Answer: Option C

Explanation:

Here is no explanation for this answer

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3. Virtual memory size depends on

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Answer: Option A

Explanation:

Here is no explanation for this answer

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Answer: Option B

Explanation:

Here is no explanation for this answer

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5. In round robin scheduling, if time quantum is too large then it degenerates to

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Answer: Option D

Explanation:

Here is no explanation for this answer

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6. A byte addressable computer has memory capacity of 2 power m Kbytes and can perform 2 power n operations an instruction involving three operands and one operator needs maximum of___________ bits

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Answer: Option C

Explanation:

Here is no explanation for this answer

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7. A 12 address lines maps to the memory of

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Answer: Option B

Explanation:

Here is no explanation for this answer

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8. In which of the following page replacement policies, Balady's anomaly occurs?

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Answer: Option A

Explanation:

Here is no explanation for this answer

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9. For 1 MB memory, the number of address lines required,

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Answer: Option B

Explanation:

The answer should be 20.Bcz in 1MB=2^20 and number of address lines is in power of two only.

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10. Consider a machine with 64 MB physical memory and a 32-bit virtual address space. If the page size s 4KB, what is the approximate size of the page table?

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Answer: Option C

Explanation:

Number of pages = 232 / 4KB = 220 as we need to map every possible virtual address.

So, we need 220 entries in the page table. Physical memory being 64 MB, a physical address must be 26 bits and a page (of size 4KB) address needs 26-12 = 14 address bits. So, each page table entry must be at least 14 bits.

So, total size of page table = 220 * 14 bits ~ 2 MB

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