A memory management system has 64 pages with 512 bytes page size. Physical…

Solution summary: Logical address = 15 bits, Physical address = 14 bits.

• Logical address calculation:

  • Page size = 512 bytes = 2^9, so offset bits = 9.

  • Number of pages = 64 = 2^6, so page-number bits = 6.

  • Total logical address bits = page-number bits + offset bits = 6 + 9 = 15.

• Physical address calculation:

  • Number of physical frames = 32 = 2^5, so frame-number bits = 5.

  • Offset bits are the same (9), so physical address bits = frame-number bits + offset bits = 5 + 9 = 14.

Final answer: logical address requires 15 bits; physical address requires 14 bits.

Common mistake: confusing counts (like 32 frames or 64 pages) with bit widths; always convert counts to powers of two to find the number of bits.