Programming-for-Output-Problems
October 6, 2023ISRO-2018
October 6, 2023UGC NET CS 2014 Dec – paper-3
| Question 1 |
A hierarchical memory system that uses cache memory has cache access time of 50 nano seconds, main memory access time of 300 nanoseconds, 75% of memory requests are for read, hit ratio of 0.8 for read access and the write-through scheme is used. What will be the average access time of the system both for read and write requests ?
| 157.5 n.sec. | |
| 110 n.sec. | |
| 75 n.sec | |
| 82.5 n.sec.
|
Question 1 Explanation:
Avg. time to read the data = 0.8(50) + 0.2(300+50) = 40 + 70 = 110 ns
Avg. memory access time = (Seq. of data read request * time to read data) + (freq. Of data write request * time to write the data)
= (0.75 × 110) + (0.25 × 300)
⤿Since write through is mentioned in question. So write operation will take place in memory and not in cache.
= 82.5 + 75.0
= 157.5 ns
Avg. memory access time = (Seq. of data read request * time to read data) + (freq. Of data write request * time to write the data)
= (0.75 × 110) + (0.25 × 300)
⤿Since write through is mentioned in question. So write operation will take place in memory and not in cache.
= 82.5 + 75.0
= 157.5 ns
Correct Answer: A
Question 1 Explanation:
Avg. time to read the data = 0.8(50) + 0.2(300+50) = 40 + 70 = 110 ns
Avg. memory access time = (Seq. of data read request * time to read data) + (freq. Of data write request * time to write the data)
= (0.75 × 110) + (0.25 × 300)
⤿Since write through is mentioned in question. So write operation will take place in memory and not in cache.
= 82.5 + 75.0
= 157.5 ns
Avg. memory access time = (Seq. of data read request * time to read data) + (freq. Of data write request * time to write the data)
= (0.75 × 110) + (0.25 × 300)
⤿Since write through is mentioned in question. So write operation will take place in memory and not in cache.
= 82.5 + 75.0
= 157.5 ns
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