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Demand Paging (TestBook Test Series 2)

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EAT = p(page fault service time) +(1-p)(memory acces time)

EAT $=0.05(0.4 \times 7+0.6 \times 15)+0.95(0.0002)\\ =0.05(11.8) +0.00019 \\ =0.59+0.00019\\ =0.59019 \ msec\\ =590.19\ \mu sec$
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