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Q. consider a system with 2k pages ,512 frames ,pa=22 bits. calculate the page table size in bytes?
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I know here page table entry size is not giving.but if not giving then we consider with the help of frame numbers in bytes= one page table entry size ?

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Given,

Number of pages $= 2k = 2*2^{10} = 2^{11}$

Number of frames $= 512 = 2^9$

Number of bits to represent physical address $= 22$

Page table size = Number of pages * Page table entry size

If no additional information is given then,

Page table entry size = Number of bits to represent frames.

Number of bits to represent frames $=9$

Page table size $= 2^{11} * 9$ bits $= \cfrac{2^{11} * 9}{8}$ bytes
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