Question:

Consider a demand paging memory management system with 32-bit logical address, 20-bit physical address, and page size of 2048 bytes. Assuming that the memory is byte addressable, what is the maximum number of entries in the page table?

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To calculate the number of entries in the page table, divide the total address space by the page size.
Updated On: Apr 4, 2025
  • \( 2^{21} \)
  • \( 2^{20} \)
  • \( 2^{22} \)
  • \( 2^{24} \)
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The Correct Option is A

Solution and Explanation

The logical address is 32 bits, and the page size is 2048 bytes (\( 2^{11} \) bytes).
The number of pages in the system can be calculated by dividing the total address space by the page size: \[ \frac{2^{32}}{2^{11}} = 2^{21}. \] The page table must have an entry for each page, so the number of entries in the page table is \( 2^{21} \). Thus, the correct answer is \( \boxed{A} \).
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