GATE CSE 1998 | Question: 2.18, UGCNET-June2012-III: 48

Question

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:

• A. $i + \dfrac{j}{k}$

• B. $i +(j\times k)$

• C. $\dfrac{i+j}{k}$

• D. $({i+j})\times {k}$

Comments

T = [ (1)(i+j) + (k-1)(i) ] / k = (ik + j)/k

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i will always take

now every kth of a second, j time is taking place

i.e. $i + \frac{1}{j}k$

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Effective $m_{acc}$ time=  $(1-p) *m_{acc} + (p)*P_{faults}$ Here $P_{faults}$ means time needed to service Page fault.

Answer 1

Consider k instructions: 

instruction #	1	2	3	...	k
time	i	i	i	...	i+j

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

Answer 2

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

Answer 3

One more way we can do is as follows

 

EMAT= m+p(ps)...{y=mx+c}

i+j/k

 

A)

Answer 4

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