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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Consider a machine with $16\ GB$ main memory and $32$-bits virtual address space, with page size as $4KB$. Frame size and page size is same for the given machine.
The number of bits reserved for the frame offset is:
The number of bits reserved for the frame offset is:
Solution
Frame size is same as page size.Given,
Page size $=4KB$
$4KB=4\times1024=4096\ bytes$
$4096=2^{12}$
Offset bits are used to identify a byte within a page/frame.
So,
Number of frame offset bits $=\log_2(2^{12})=12$
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