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Consider a system with 1 K pages and 512 frames and each page is of size 2 KB. How many bits are required to represent the virtual address space Memory is
  • a)
    20 bits
  • b)
    21 bits
  • c)
    11 bits
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider a system with 1 K pages and 512 frames and each page is of si...
Virtual address space consists of pages.
Given that,
Number of pages = 1 K = 210
 Page size = 2KB = 211B
Hence virtual address space
= 211 x210
= 221 B
∴  21 bits are required to represent the virtual address space.
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Most Upvoted Answer
Consider a system with 1 K pages and 512 frames and each page is of si...
Explanation:

  • Total number of pages in the system = 1K = 1024 pages

  • Page size = 2 KB = 2^11 bytes

  • Number of frames = 512

  • Frame size = Page size = 2^11 bytes

  • Virtual address space = Total number of pages * Page size = 1024 * 2^11 bytes = 2^21 bytes

  • Number of bits required to represent the virtual address space = log2(Virtual address space) = log2(2^21 bytes) = 21 bits


Hence, the correct answer is option B.
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Consider a system with 1 K pages and 512 frames and each page is of size 2 KB. How many bits are required to represent the virtual address space Memory isa)20 bitsb)21 bitsc)11 bitsd)None of theseCorrect answer is option 'B'. Can you explain this answer?
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