+1 vote
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Suppose: TLB lookup time = 20 ns

TLB hit ratio = 80%

Memory access time = 75 ns

Swap page time = 500,000 ns

50% of the pages are dirty

OS uses a single level page table

What is the effective access time (EAT) if we assume the page fault rate is 10%? Assume the cost to update the TLB, the page table, and the frame table (if needed) is negligible.

A
15106 ns
B
76000 ns
C
15200 ns

D
75110 ns
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0
$$\frac{80}{100}[20 + 75] + \frac{20}{100}[\frac{90}{100}(20+75+75) + \frac{10}{100}(\frac{50}{100}\times500000 + \frac{50}{100}(500000 + 500000))]$$

I’m not getting the correct answer too, was any solution given? I want to know how they calculated that.
+1

EMAT = h ( T + M)  +  (1 – h) [T + (1 - p) * 2M + p( (1-d) * (D+M) + d * (2D + M) )]  = 0.8( 20 + 75)  +  0.2[ ( 20 + 0.9 * 2(75) ) +  0.10( (0.5) * (500,000) + 0.5( 2 * 500,000)]  = 76 + 0.2 [ 155 + 0.10( 2,50,000 + 0.5( 10,00,000)]  = 76 + 0.2 [ 155 + 0.10( 7,50,000)]  = 76 + 0.2 [ 155 + 75,000 ]    // I think, they made 76 + (0.2 * 155) + 75,000  = 76 + 15,031                  //  76 + 75031  = 15,107                      //   75,107