Question
Download Solution PDFConsider a situation in which physical memory contains 62 page frame. How many number of bits will be required in physical and logical address, if a memory management system has 128 pages with 1024 bytes page size ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is option 4) 14 and 16
Key Points
- Page size = 1024 bytes → 210 bytes, so offset requires 10 bits.
- Total number of pages = 128 → 27, so logical page number requires 7 bits.
- Logical Address = Page Number + Offset → 7 + 10 = 17 bits.
- Physical memory contains 62 page frames → Nearest power of 2 ≥ 62 is 64 → 26 page frames.
- Physical Address = Frame Number + Offset → 6 + 10 = 16 bits.
Additional Information
- Logical Address = bits required to uniquely identify each page (Page number) + Offset within page.
- Physical Address = bits to identify the frame + Offset (same as in logical address).
Hence, the correct answer is: option 4) 14 and 16
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