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 Consider a system with 4 KB Pages and 16 Bit physical and logical addresses. The page table entry Contains besides the other information like a valid bit, a dirty bit, and then how many bits are still available on the page table to store permission information when table entry size is 1 byte?

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PTE stores frame no+valid+dirty+permission.

PTE=1B=8bits.

No of frames=$2^{16}/2^{12}$=16.

No of bits for frames=4bits and

8=4+1+1+x

x=2 bits are for permission

+1 vote