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Discussion/Midterm QuestionPowerPoint Presentation

Discussion/Midterm Question

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- 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 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 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 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 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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