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UGC-NET | UGC NET CS 2014 Dec – III | Question 1

  • Last Updated : 03 Jul, 2018
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A hierarchical memory system that uses cache memory has cache access time of 50 nano seconds, main memory access time of 300 nano seconds, 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 ?
(A) 157.5 n.sec.
(B) 110 n.sec.
(C) 75 n.sec.
(D) 82.5 n.sec.


Answer: (A)

Explanation: cache access time = 50 ns; main memory access time = 300 ns.
Since the system is hierarchical:
tread = hit ratio * cache access time + (1 – hit ratio) *(main memory access time + cache access time)
= 0.8 * 50 + 0.2 * (300 + 50) = 110 ns
twrite = 1 * max(main memory access time, cache access time)
= 1 * max(300,50) = 1* 300 = 300 ns.
read frequency = 75%:; write frequency = 25%.
Now, average access time for both read and write requests = read frequency * tread + write frequency * twrite
0.75 * 110 + 0.25 * 300
= 157.5 ns
So, option (A) is correct.

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