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/
Quiz of this Question
Attention reader! Don’t stop learning now. Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.
Learn all GATE CS concepts with Free Live Classes on our youtube channel.