In a demand paging memory system, page table is held in registers. The time…
2014
In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. if an empty frame is available or if the replaced page is not modified, and it takes 20 m.secs., if the replaced page is modified. What is the average access time to service a page fault assuming that the page to be replaced is modified 70% of the time ?
- A.
11.6 m.sec.
- B.
16.4 m.sec.
- C.
28 m.sec.
- D.
14 m.sec.
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Correct answer: B
Answer: 16.4 m.sec.
Compute the weighted average using the probability that the replaced page is modified (70%) and not modified (30%).
Step 1: Contribution when replaced page is modified: 0.7 × 20 m.sec. = 14 m.sec.
Step 2: Contribution when replaced page is not modified/empty frame: 0.3 × 8 m.sec. = 2.4 m.sec.
Step 3: Total average time = 14 + 2.4 = 16.4 m.sec.
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