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Discussion/Midterm Question. Assume a system with a two level page table. The virtual memory address space is 32 bits and the physical memory address space is 16 bits. . Factors in determining size of page. Page table fits on one page Equal sized L1 and L2 page tables

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Factors in determining size of page
Factors in determining size of page memory address space is 32 bits and the physical memory address space is 16 bits.

  • Page table fits on one page

  • Equal sized L1 and L2 page tables

  • Size of a page table entry

    • size of physical memory (16 bit address space)

    • bookkeeping bits


Fitting a page table on one page
Fitting a page table on one page memory address space is 32 bits and the physical memory address space is 16 bits.

(size of page) (size of page table)

  • Size of page

  • # of PTE

  • Size of PTE

(size of page table) =

(# of PTEs) * (size of PTE)

  • (2offset)

  • (2(32-offset)/2)

(16 bits - offset + bookkeeping ≈ 1-4 bytes)

2offset ≥ 2(32-offset)/2 (assuming 1 byte PTE)

offset ≥ 10.667


Fitting a page table on one page1
Fitting a page table on one page memory address space is 32 bits and the physical memory address space is 16 bits.

(size of page) (size of page table)

  • Size of page

  • # of PTE

  • Size of PTE

(size of page table) =

(# of PTEs) * (size of PTE)

  • (2offset)

  • (2(32-offset)/2)

(16 bits - offset + bookkeeping ≈ 1-4 bytes)

2offset ≥ 2(32-offset)/2*22 (assuming 4 byte PTE)

offset ≥ 12

12 bit offset works under either assumption


Solutions
Solutions memory address space is 32 bits and the physical memory address space is 16 bits.

  • Offset = 12

    • 4KB pages

  • VPN1 = VPN2 = 10

    • 210 entries per page table


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