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# GATE | GATE-CS-2000 | Question 45

Suppose the time to service a page fault is on the average 10 milliseconds, while a memory access takes 1 microsecond. Then a 99.99% hit ratio results in average memory access time of (GATE CS 2000)

(A) 1.9999 milliseconds
(B) 1 millisecond
(C) 9.999 microseconds
(D) 1.9999 microseconds

Explanation: If any page request comes it will first search into page table, if present, then it will directly fetch the page from memory, thus in this case time requires will be only memory access time.

But if required page will not be found, first we have to bring it out and then go for memory access. This extra time is called page fault service time.

Let hit ratio be p , memory access time be t1 , and page fault service time be t2.

```Hence, average memory access time = p*t1 + (1-p)*t2
=(99.99*1 +  0.01*(10*1000 + 1))/100
=1.9999 *10^-6 sec

```

This explanation is contributed by Abhishek Kumar.

Also, see question 1 of https://www.geeksforgeeks.org/operating-systems-set-3/

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