Consider a word-addressable logical address space of 8 pages, with 1024 words…

Concept

In paging, every page and every frame have the same size.

For word-addressable memory, an address contains a page or frame number together with a word offset. Its width is the selector width plus the offset width.

Application

1. A page contains 1024 = 210 words, so the word offset needs 10 bits.

2. There are 8 = 23 logical pages, so the logical address needs 3 + 10 = 13 bits.

3. There are 32 = 25 physical frames, so the physical address needs 5 + 10 = 15 bits.

Cross-check

The logical space contains 8 × 1024 = 8192 = 213 addressable words, while physical memory contains 32 × 1024 = 32768 = 215 addressable words. No conversion from words to bytes is used because the question explicitly treats each word as one addressable unit; equal page and frame sizes are part of paging.

Contrast

• The pair (5, 3) contains only the frame-number and page-number widths, without the word-offset width.

• The pair (10, 10) contains only the word-offset width for both addresses.

• The pair (15, 13) combines the appropriate selector width with the common 10-bit word offset.

• The pair (15, 15) uses the physical frame-number width in both addresses, although the logical space has only eight pages.

Result

Therefore, in the requested order (physical, logical), the address widths are (15, 13) bits.