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If an instruction takes $i$ microseconds and a page fault takes an additional $j$ microseconds, the effective instruction time if on the average a page fault occurs every $k$ instruction is:

  1. $i + \dfrac{j}{k}$

  2. $i +(j\times k)$

  3. $\dfrac{i+j}{k}$

  4. $({i+j})\times {k}$

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9 Answers

2 votes
2 votes
1 instruction takes i ms.

k instructions will take k*i ms.

1 page fault in every k instructions. Page Fault Service Time is j ms.

So,

Total time required to execute k instructions is (k*i+j) ms.

 

so here ,

Effective instruction time = (k*i+j)/k

i.e, i + j/k
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In this question, a page fault occurring every k instructions. So that means we need to manage a page fault after every k instructions. Now we write this expression by this information,

                                                        k* EIT = k*i + j

                                                           EIT = i + j/k                                    , where EIT = Effective instruction time

 

Correct Answer : Option A

Answer:

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