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Lets say ,page size = 2^x byte.

so no of bits required for offset =x.

now page table entry = 4 B

now ,max no of entry one page can store is 2^x/4 =2^(x-2).

so each inner and outer level page table fit completely in one page frame . so we can write,

x-2 x-2 x

outer level                                          inner level                                      offset,

now it sums up to virtual address space .2^38.

(x-2)+(x-2)+x=38

x=14.

so page size =16 KB

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