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This is one of a question in a past paper:

On a simple paging system with $4GB$ of physical memory,$16GB$ of virtual memory,and a page size of $8KB$

asked in Theory of Computation by (7 points)
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1 Answer

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I hope you are trying to find the number of bits for all the mentioned terms. Assuming word size to be $1$ $Byte$, as nothing is mentioned about it.

Page size is $8KB$ = $2^{3+10}$ = $2^{13}Bytes$. So in page offset $13$ $bits$ are there.

Virtual memory(or logical memory) is $16 GB$ = $2^{4+30}$ = $2^{34}Bytes$. So total bits in virtual address is $34 bits$.

This means that page number has $34 -13 = 21 bits $.

Physical memory is $4GB$ = $2^{2+30}$ = $2^{32}Bytes$. This means that physical address is $32 bits$. 

Answer would change is word size is NOT 1 Byte.

answered by Junior (879 points)

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