Test by Bikram | Computer Organization and Architecture | Test 2 | Question: 27

Question

Consider a memory hierarchy system consisting of two levels. The access time of level $1$ is $2$ ns. The miss penalty (The time to get data from level $2$, in case of miss) is $100$ ns.

If the average memory access time is $5$ ns, then in order to reduce the average access time to $40 \%$,  the probability that valid data found in level $1$ is ___________  $\%$

Answer 1

After reducing average access time to 40% , that new average access time will be  (40/100) * 5 = 2 ns

Let the probability that valid data found in level 1 is p. 

2  = p*2 + (1-p)*(102)  [ by using avg mem access time =  Tavg = h*Tc +(1-h)*M

        h    = hit rate
       (1-h) = miss rate
       Tc    = time to access information from cache
       M     = miss penalty  (time to access main memory) ]

2 = p*2 + 102 -102 p

102 p - 2 p = 102 - 2
100 p = 100
p =1 or 100%

 

Comments

This is not correct! time is not reduced by 40%, while, time is reduced to 40%. there is a difference between both terms. 
average memory access time is reduced to 40% means now average memory access time is 2ns not 3ns as you said.
answer of this question will be 100%

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@Bikram sir,

agreed with what manu00x said..

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@manu00x

if possible please show your calculation.

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There is no calculation required here, cache access time is 2ns, and if average access time is 2ns means cache hit should be 100 %.

or
2ns  = h*2ns + (1-h)*(102ns)
2ns = h*2ns + 102ns -102h
100h = 100
h=1 or 100%

Answer 2

As we know,

         Access time for l1 cache --2ns  ,  and miss penalty -- 100ns  , average memory acess time -5ns

In order to reduce access time to 40%=  (40/100)*50= 2ns

average memory access time= probability(hit) * access time l1 cache +probability(miss) * penalty time

 2 = probability(hit)*2+{ 1-probability(hit) }*100     (  as probability(miss)=1-probability(hit)  )

therefore solving above equation probability (hit)= 1

And its asking the percentage therefore=100% hit