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9 votes
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A cache memory needs an access time of 30 ns and main memory 150 ns, what is average access time of CPU (assume hit ratio = 80%)?

  1. 60 ns
  2. 30 ns
  3. 150 ns
  4. 70 ns
in CO and Architecture
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3 Answers

16 votes
 
Best answer
Effective Memory Access Time = Cache hit * Cache access time + Cache miss ( Cache miss Penalty + memory Access time)

 = 0.8(30) + (1-0.8)(30+150) ns

 = 24 + 0.2(180) ns

 = 24 + 36 ns = 60 ns.

Option A.

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0
in question they not asking Effective Memory Access Time ...they are asking only average access time ?
2
Both are same dude.
0 votes
By default Hierarchial access

Tavg  = cache miss + main memory hit

 Cache miss = cache hit ratio * cache access time = 0.8 * 30ns = 24ns

Main memory hit = (1-cache hit ratio) * (main memory hit ratio) * (main mem access time + cache mem access time)

                              = (1-0.8) * (1) * (150ns+ 30ns)

                              = 36ns

(Main memory is final level of hierarchial access so main mem hit ratio = 1)

 

Therefore,

 

Tavg = 24+36 = 60ns
0 votes
Step-1: Hit ratio 80%=0.8 and Miss ratio=20%=0.2
Access time=30ns
Main memory=150ns
Step-2: CPU access time = (Hit ratio*access time) + (Miss ratio*access time+Main memory)
= (0.8*30) + (0.2*(30+150))
= 60 ns
Answer:

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