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Consider a paging system with 48 bit virtual address space. Each address defers to a byte in memory. Suppose the size of the page is 16 kB and the main memory size is 16 GB. Each page table entry contains frame number and 2 protection bits, 1 valid bit and 1 dirty bit. The minimum size of page table _____________ (in GB).
    Correct answer is '48'. Can you explain this answer?
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    Consider a paging system with 48 bit virtual address space. Each addre...
    Given,
    Page table size = 16 kB = 2
    14
    Main memory size = 2
    34
    As we know,
    #Frame 
    = 2
    34
    /2
    14
    = 2
    20
    Each page table entry has 20 + 4 = 24 bits.
    Minimum page table size = # page × page table entry size
    = 2
    48
    /2
    14
    × 3 B
    = 2
    34
    × 3 B
    = 48 GB
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    Community Answer
    Consider a paging system with 48 bit virtual address space. Each addre...
    The minimum size of the page table in this scenario is 48 GB. Let's break down the calculations to understand how we arrive at this answer.

    1. Calculating the number of pages:
    The virtual address space is 48 bits, which means there are 2^48 possible addresses. Since each address refers to a byte in memory, the total number of bytes in the virtual address space is 2^48 bytes.
    Since the page size is 16 kB (16 * 1024 bytes), the number of pages can be calculated as:
    Number of Pages = Total Bytes / Page Size = (2^48 bytes) / (16 * 1024 bytes) = 2^(48-14) = 2^34

    2. Calculating the number of page table entries:
    Each page table entry contains the frame number (to identify the physical memory location where the page is stored) and 2 protection bits, 1 valid bit, and 1 dirty bit. Let's calculate the size of each entry:
    Size of each entry = Size of frame number + 2 protection bits + 1 valid bit + 1 dirty bit
    We know that the main memory size is 16 GB, and the page size is 16 kB. Therefore, the number of frames can be calculated as:
    Number of Frames = Main Memory Size / Page Size = (16 * 1024 MB) / (16 * 1024 bytes) = 2^20
    Since the frame number is used to identify each frame, it requires log2(Number of Frames) bits. In this case, it is 20 bits.

    Hence, the size of each entry can be calculated as:
    Size of each entry = 20 bits + 2 bits + 1 bit + 1 bit = 24 bits = 3 bytes

    3. Calculating the size of the page table:
    Now that we know the number of pages and the size of each entry, we can calculate the total size of the page table:
    Size of page table = Number of Pages * Size of each entry = (2^34) * (3 bytes) = 3 * 2^34 bytes

    To convert the size from bytes to GB, we divide it by (1024^3):
    Size of page table = (3 * 2^34 bytes) / (1024^3 bytes/GB) ≈ 3 * 2^(34-30) GB ≈ 3 * 2^4 GB ≈ 48 GB

    Therefore, the minimum size of the page table in this scenario is 48 GB.
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    Consider a paging system with 48 bit virtual address space. Each address defers to a byte in memory. Suppose the size of the page is 16 kB and the main memory size is 16 GB. Each page table entry contains frame number and 2 protection bits, 1 valid bit and 1 dirty bit. The minimum size of page table _____________ (in GB).Correct answer is '48'. Can you explain this answer?
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    Consider a paging system with 48 bit virtual address space. Each address defers to a byte in memory. Suppose the size of the page is 16 kB and the main memory size is 16 GB. Each page table entry contains frame number and 2 protection bits, 1 valid bit and 1 dirty bit. The minimum size of page table _____________ (in GB).Correct answer is '48'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider a paging system with 48 bit virtual address space. Each address defers to a byte in memory. Suppose the size of the page is 16 kB and the main memory size is 16 GB. Each page table entry contains frame number and 2 protection bits, 1 valid bit and 1 dirty bit. The minimum size of page table _____________ (in GB).Correct answer is '48'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a paging system with 48 bit virtual address space. Each address defers to a byte in memory. Suppose the size of the page is 16 kB and the main memory size is 16 GB. Each page table entry contains frame number and 2 protection bits, 1 valid bit and 1 dirty bit. The minimum size of page table _____________ (in GB).Correct answer is '48'. Can you explain this answer?.
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