Derive the address-field widths
Pages contain 2^12 bytes, so the offset needs 12 bits. Logical page number: 14 − 12 = 2 bits. Physical frame number: 15 − 12 = 3 bits.
Start with a logical hexadecimal address
The CPU requests 0x1A4A. The prefix 0x means hexadecimal. Each hexadecimal digit represents four binary bits.
Expand every hex digit into four bits
0x1A4A = 0001 1010 0100 1010. The first 2 zeros are display padding; the actual logical address has 14 bits.
Split page number from offset
Keep the rightmost 12 bits as the offset. The remaining 2 bits are page 1. The split is determined by page size, not by spaces between hexadecimal digits.
Index the page table
Use page 1 as the row index. This example’s pages are present. Row 1 maps to physical frame 3; page and frame numbers need not match.
Build the physical address
Write frame 3 as 011 (3 bits), then append the unchanged 12-bit offset. The result is 15 bits long.
Regroup into four-bit chunks
0011 1010 0100 1010. Group from the right; add leading display zeros if needed. This regrouping changes notation, not the address value.
Read the physical address in hex
0011 1010 0100 1010 → 0x3A4A. Check: frame 3 × 4096 bytes + offset 2634 = 14922 (0x3A4A).
Derive the address-field widths
Pages contain 2^12 bytes, so the offset needs 12 bits. Logical page number: 14 − 12 = 2 bits. Physical frame number: 15 − 12 = 3 bits.