Discussion
A memory management system has 64 pages with page size of 512 bytes. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively :
Answer: C
The logical and physical addresses need 15 and 14 bits.
\[ \begin{aligned} \text{logical space} &= 64 \times 512 = 2^{6} \times 2^{9} = 2^{15} \\ \text{physical space} &= 32 \times 512 = 2^{5} \times 2^{9} = 2^{14} \end{aligned} \]
So a logical address needs 15 bits (6 for the page number and 9 for the offset) and a physical address 14 bits (5 for the frame number and 9 for the offset).
Option A reverses the two. 29 and 32 come from multiplying or adding the wrong quantities rather than taking logarithms.
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