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    Page table, physical and logical addresses

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    Consider a simple paging system with the following parameters: 2^32 bytes of physical memory, page size of 2^10 bytes, and 2^16 pages of logical address space.

    a. How many bits are in a logical address?
    b. How many bytes in a frame?
    c. How many bits in the physical address specify the frame?
    d. How many entries in the page table?
    e. How many bits in each page table entry? Assume that each page table entry includes a valid/invalid bit.

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    Solution Preview

    a. Logical address space = 2^16 pages = 2^16 pages * 2^10 bytes/page = 2^26 bytes
    Hence, number of bits required to specify a logical address = 26

    b. A frame has same size as a page, ...

    Solution Summary

    Solution gives the computations along with brief explanation.