Two level of cache

Question

Consider a two level cache system. For 100 memory references 20 misses in 1st level cache,10 misses in second level cache. Miss penalty from second level to memory is 40 cycles. if Total average = 7.6 cycles, then hit time of second level cache ? Assume hit time of second level cache is two times the first level cache?

Ans. 4

Comments

Here x = 2.57.

Then X will be 3.

For minimum Hit Time.

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We have to take Ceil or Floor here ???

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am getting this (here x= time of L1 cache; and 2x = time of L2 cache)

7.6= x+ 0.2x+ 0.2*0.5*40

x=3;

therefore time of L2 cache = 2x = 6

 

Answer 1

https://gateoverflow.in/?qa=blob&qa_blobid=11423634474127333130

In this from slide 17 gives AMAT for a 2-level cache.

Let L1 cache access time=x cycles

Then L2 cache access time=2x cycles.

100 references to L1 cache, out of which 20 missed, $Missrate_{L1}=\frac{20}{100}=0.2$

These 20 misses of L1 cache are looked into L2 cache of which 10 misses occur in L2.

So $Missrate_{L2}(Local\,Miss\,Rate)=\frac{10}{20}=0.5$

$Global\,Miss\,Rate_{L2}=\frac{10}{100}=0.1$

Assuming Local miss rate of L2, goes below calculation.

$Misspenalty_{L1}=Hit\,Time_{L2}+(Miss\,rate_{L2} \times Miss\,Penalty_{L2})=2x+(0.5 \times 40)=2x+20$

$AMAT=Hit\,Time_{L1}+(Miss\,rate_{L1} \times Miss\,Penalty_{L1})$

$7.6=x+0.2(2x+20)$

$7.6=1.4x+4$

$x=2.57$

Since $x$ is the L1 cache access time in cycles, number of cycles must be integral, so it has taken 2 cycles plus 0.57 of another cycle so appx 3 cycles.

So, Access time L2=6 cycles.

Let me know if there is something wrong here.

Comments

@Gate Fever-Yes I am also getting this

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@Minipanda-Usually in most numericals that I have solved on multi-level caches, they directly have given miss penalties.

In case they are not given and block sizes for L1 and L2 cache are given and main memory access time is given then you logically consider Miss penalty as Access time + Block transfer time.

If, you know a numercial based on such data, please share so that we can discuss more.

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Why do we consider this as hierarchial and not simultaneous access ?? How to know when to consider hierarchial and when to consider simultaneous ?

Answer 2

Let the hit time of first level (L1) be x cycle, then hit time of second level (L2)  will be 2x cycle.

AMAT =  L1 access time + L1 miss rate X L2 access time

+ L1 miss rate  X L2 miss rate X Memory access time.

7.6= x + (20/100) * (2x) + (20/100) * (10/20) *  40

x=2.571

we take floor of (2.571)

then x=3 clock cycle

second level time=6 clock cycle.