A system has a 36-bit physical address and a page size of 4 KB. How many bits are required for the physical frame number?
Options
- A. 22 bits
- B. 24 bits
- C. 28 bits
- D. More than one of the above
- E. None of the above
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Correct Answer (Detailed Explanation is Below)
B. 24 bits
Detailed Explanation
4 KB requires 12 offset bits. Therefore, the physical frame number requires 36 - 12 = 24 bits.
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