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Considered a 32 bit virtual address is used for paging with a page size of 1024 bytes.Two level paging is implemented with an equal number of entries in every page table of the system.if Page table entry sizes 2 bytes .what is the maximum size of main memory supported by the above system?

a.16 MB

b.32 MB

c. 64 MB

d. None
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How you gets bits for frames = 16 bits?
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yes ,for maximum memory size we assume PTE only have frame bits i.e 16 bits .

ANS. 64MB
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got it.. thanks
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1 Answer

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PAS = Number of frames * Frame size

Bits for number of frames = Bits in PTE (Assuming that all bits give frame number)  = 16 bits

thus answer  = $2^{16}$ * $2^{10}$ = 64MB
edited by

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