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Gate1998_2.18

 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)  \(i + {j \over k}\)

(B)  \(i + j*k\)

(C)  \({i + j \over k}\)

(D) \((i + j)* k\)

Answer

Answer :-(A)
Exp:-
                avg execution time=\(1 \over k\)(i+j) +(1-\(1 \over k\))i
                                              = \((i+j)\over k\) + \((k-1) \over k\)i
                                              =  \(i + {j \over k}\) 

1Comment
Amit Pal @amitpal101
26 Sep 2016 11:37 am

instruction processing time = i
page fault service = j
page fault occur once at every k instructions .
so take a scenario where execution of k instruction has happenned :
so total taken time = i*k +j (page fault service time which will occur once in k instruction execution)
so effective memory access time =( i*k+j )/ k =  i + (j/k)
so ans is (A)