In a paging scheme, 16-bit addresses are used with a page size of 512 bytes.…
2025
In a paging scheme, 16-bit addresses are used with a page size of 512 bytes. If the logical address is 0000010001111101, how many bits are used for the page number and offset ? Compute the page number and the offset as well. What will be the physical address, if the frame address corresponding to the computed page number is 15 ?
Show answer & explanation
Concept
For a page size of 2p bytes, the p low-order bits of a logical address give the offset within the page.
The remaining high-order bits give the page number. Address translation replaces that page number with the mapped frame number while preserving the offset.
Application
Since 512 = 29, the offset uses 9 bits and the page number uses 16 - 9 = 7 bits.
Split the logical address after 7 bits: 0000010 | 001111101.
The page number is 00000102 = 2, and the offset is 0011111012 = 125 bytes.
The mapped frame number is 15, which is 00011112 in 7 bits.
Preserve the 9-bit offset and replace the page field: 0001111 | 001111101 = 00011110011111012. Thus the physical address is 780510 (0x1E7D).
Cross-check
The original address is 2 × 512 + 125 = 1149, which equals 00000100011111012. The translated address is 15 × 512 + 125 = 7805, and 125 is below the 512-byte page size.