In a demand-paging system, it takes 250 ns to satisfy a memory access when the…
2025
In a demand-paging system, it takes 250 ns to satisfy a memory access when the requested page is in the resident set. If it is not in the resident set, then the request takes 10 ms if a free frame is found or the page to be replaced is not modified. Such requests are 3% of all the accesses. Otherwise, if there is no free frame and the page to be replaced is modified, then it takes 20 ms. Such pages are 7% of all the accesses. If the page fault rate (PFR) in the system is 10%, then what will be the Effective Memory Access Time (EAT)?
Show answer & explanation
Concept
Effective memory access time is the probability-weighted average of the latencies of all mutually exclusive memory-access outcomes.
Every latency must first be expressed in the same unit, and the outcome probabilities must sum to 1.
Application
The resident-set hit probability is 1 − PFR = 1 − 0.10 = 0.90. The two page-fault cases have probabilities 0.03 and 0.07, so 0.90 + 0.03 + 0.07 = 1.
Convert the hit time to milliseconds: 250 ns = 250 ÷ 1,000,000 ms = 0.00025 ms.
Form the weighted average: EAT = (0.90 × 0.00025) + (0.03 × 10) + (0.07 × 20) ms.
Therefore, EAT = 0.000225 + 0.30 + 1.40 = 1.700225 ms.
Cross-check
Conditioned on a page fault, the mean fault-service time is (0.03/0.10) × 10 + (0.07/0.10) × 20 = 17 ms.
Hence EAT = 0.90 × 0.00025 + 0.10 × 17 = 1.700225 ms, which agrees with the direct weighted average.
Result: The effective memory access time is 1.700225 ms, approximately 1.70 ms (1,700,225 ns).