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If instruction takes i microsecond and page fault takes j mi

If instruction takes i microsecond and page fault takes j micro second  then the effective instruction time if the average page fault occurs every k instruction is:

1. i+ j/k

2.i+j*k

3.(i+j)*k

4.(i+j)/k

3Comments
Vivek Vikram Singh @vivek14
24 Sep 2014 06:37 pm

The rate of page fault occurring is " in every K inst" so, rate would be  1/K.

Now each page fault takes J microsec , so this will be resulting in J/K.

Now each instruction takes I microsec. 

so effective time will be = I+ J/K 

Please tell the answer and let me know if i am wrong or unclear.

Aparajita Mehta @aparajita
25 Sep 2014 04:32 pm

I actually don't know the correct answer but it seems right.

Vikash @vikasch
7 Oct 2014 03:35 pm

for every page fault k instr. so rate will be "1/k "

now as we formulae

effective m/m access time = P*S + (1-P)*M

                                            = (1/k)*(i+j) + (1-(1/k))*i          // here j is added bcz as given que. page fault takes j micro sec

                                                        = i/k+j/k+i-i/k                

                                                         = i+j/k .....................ANS (B)