Wednesday 1 July 2015

Q1,p3,d14. 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 B.
Let total read operations are 75.and write operations are 25.
Hit ratio of read =.8
Fitst it check cache, if fails then go for memory  access.
Total read access time 75x50x.8+75x350x.2=3000+5250=8250ns.
Total write access time=25x50x.8+25x350x.2=1000+1750ns
Average access time=(8250+2750)/100=110ns.
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1 comment:

  1. Songita21 December 2015 at 09:27

    The answer is wrong ... It will be A.. 157.5 ns.. You are right up to read access time but for write access time no need to consider cache memory it'll be in main memory ... so Total write access time = 25*300=7500.. Average access time= (8250+7500)/100= 157.5 ns ..

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