Question

consider a machine with 64Mbyte physical memory and a 32 bit virtual address. if the page size is 4 kByte, what is the approximate size of the page table? a)16 Mbyte b)8 Mbyte c) 2 Mbyte d) 24 Mbyte

Answer 1

Virtual Address Space=32 Bit, Page size= 4 kb= 2² x 2¹⁰ = 2¹²

PTE= V.A.S./ Page Size = 2²⁰   ....................eq. 1

Physical memory size= 64MB= 2⁶ x 2¹⁰x 2¹⁰​ = 2²⁶

Then No. of Frames in Physical Memory= 2²⁶ / 2¹² = 2¹⁴

And page table needs to store the address of all these 2¹⁴ page frames. Therefore, each page table entry will contain 14 bits address of the page frame and 1 bit for valid-invalid bit. we suppose 16 bit require it means 2 bytes (approx.)

so Size of Page table = Total PTE (from eq 1) x Size of page table entry= 2²⁰ x 2 = 2MB

Answer 2

32 bit virtual address means 2³²/4K = 2²⁰ entries in the page table as each entry points to a page (the lower 12 bits are not used while looking up in the page table).  

Each PTE must address a memory location of a page which ranges from 0 to 64M/4K = 16K. So, should need 14 bits. 

So, PTE size will be 2²⁰ * 14 bits ≈ 2 MB

Comments

you answer is wrong ...Answer is C

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Yes. It was a mistake. PTE points to a page and not to a physical address. Thanks for pointing out...