Consider a word-addressable logical address space of 8 pages, with 1024 words…
2013
Consider a word-addressable logical address space of 8 pages, with 1024 words per page, mapped onto a physical memory of 32 frames. In this paging system, each frame has the same size as a page. How many bits are there in the physical address and logical address, respectively?
Answer: C. 15, 13 — ConceptIn 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…
- A.
5, 3
- B.
10, 10
- C.
15, 13
- D.
15, 15
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Show answer & explanation
Correct answer: C
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
A page contains 1024 = 210 words, so the word offset needs 10 bits.
There are 8 = 23 logical pages, so the logical address needs 3 + 10 = 13 bits.
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.