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Consider a system with byte-addressable memory, 32-bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is _______.
    Correct answer is '4'. Can you explain this answer?
    Most Upvoted Answer
    Consider a system with byte-addressable memory, 32-bit logical address...
    Given information:
    - Byte-addressable memory
    - 32-bit logical addresses
    - 4 kilobyte page size
    - Page table entries of 4 bytes each

    Calculating the number of pages:
    - 32-bit logical addresses can address 2^32 bytes of memory
    - Dividing this by the page size of 4 kilobytes (or 2^12 bytes) gives 2^20 pages

    Calculating the size of the page table:
    - Each page table entry is 4 bytes
    - Multiplying this by the number of pages gives 4 * 2^20 bytes
    - Converting this to megabytes gives 4 MB

    Therefore, the size of the page table in the system is 4 megabytes.
    Free Test
    Community Answer
    Consider a system with byte-addressable memory, 32-bit logical address...
    Data
    1 word (W) = 1 byte (B)
    Logical address = 232 B
    page size = 4 KB = 212 B
    page table entry (PTE) = y B
    Formula:
    Page table size (PTS) = number of pages × y
    Calculation:
    number of pages = 232/212  = 220   
    Page table size (PTS) = number of pages × y
    PTS = 220 × 4B
    ∴ PTS = 4 MB
    Important points
    1 MB = 220 B
    where B stands for byte
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    Consider a system with byte-addressable memory, 32-bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is _______.Correct answer is '4'. Can you explain this answer?
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