ISRO CS-2023
January 20, 2024Access-Control-Methods
January 20, 2024UGC NET CS 2008 Dec-Paper-2
| Question 9 |
Suppose it takes 100 ns to access page table and 20 ns to access associative memory. If the average access time is 28 ns, the corresponding hit rate is :
| 100 percent | |
| 90 percent | |
| 80 percent | |
| 70 percent |
Question 9 Explanation:
Given data,
— Access page table time=100 ns
— Associate memory=20 ns
— hit ratio=X
— Miss ratio=1-X
— Average Access Time=28
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= X*20 + (1-X*100) [Note: If X=0.9 then we are getting exact AAT]
= 28 ns
(or)
28=X*20+(1-X*100)
X=0.9
— Access page table time=100 ns
— Associate memory=20 ns
— hit ratio=X
— Miss ratio=1-X
— Average Access Time=28
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= X*20 + (1-X*100) [Note: If X=0.9 then we are getting exact AAT]
= 28 ns
(or)
28=X*20+(1-X*100)
X=0.9
Correct Answer: B
Question 9 Explanation:
Given data,
— Access page table time=100 ns
— Associate memory=20 ns
— hit ratio=X
— Miss ratio=1-X
— Average Access Time=28
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= X*20 + (1-X*100) [Note: If X=0.9 then we are getting exact AAT]
= 28 ns
(or)
28=X*20+(1-X*100)
X=0.9
— Access page table time=100 ns
— Associate memory=20 ns
— hit ratio=X
— Miss ratio=1-X
— Average Access Time=28
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= X*20 + (1-X*100) [Note: If X=0.9 then we are getting exact AAT]
= 28 ns
(or)
28=X*20+(1-X*100)
X=0.9
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