Consider a logical address space of 8 pages of 1024 words each, mapped onto a physical memory of 32 frames. How many bits are there in the physical address and logical address respectively?
- A. 5, 3
- B. 10, 10
- C. 15, 13
- D. 15, 15
Answer: C. 15, 13
Explanation: Logical: 8 pages ($2^3$) * 1024 words ($2^{10}$) = $2^{13}$ (13 bits). Physical: 32 frames ($2^5$) * 1024 words ($2^{10}$) = $2^{15}$ (15 bits).