GATE CSE 2025 | Set 2 | Question: 45

Question

Given a computing system with two levels of cache (L1 and L2) and a main memory. The first level (L1) cache access time is $1$ nanosecond (ns) and the "hit rate" for L1 cache is $90 \%$ while the processor is accessing the data from L1 cache. Whereas, for the second level (L2) cache, the "hit rate" is $80 \%$ and the "miss penalty" for transferring data from L2 cache to L1 cache is $10$ ns . The "miss penalty" for the data to be transferred from main memory to L2 cache is $100$ ns .

Then the average memory access time in this system in nanoseconds is __________ . (rounded off to one decimal place)

Comments

'Miss Penalty' sounds like its talking about the AMAT for L2 cache.

But on taking 10ns as AMAT for L2, we get negative L2 access time.

Considering it access time of L2 gives the right answer.

Question doesnt seem well worded.

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1 +  (0.1) [ 10 + (0.2)*100 ]  => 4

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OR
0.9* 1  +  (0.1)* [0.8 * 11  +  0.2 * 111)  => 4

Answer 1

.

Comments

apart from it I have studied coa from a teacher who have said 
if question is mentioning hirarchy or level word then u should use hirarchical access 
https://gateoverflow.in/460037/gate-cse-2025-set-1-question-43

i have followed same concept in above question and got wrong answer in Gate 2025
so don't believe any trick of anyone blindly 

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the miss penalty is defined differently in the que. bcuz when the miss in L1 happens the L2 will require 10ns to transfer the data to L1 to which they refer miss penalty, but also remember we already accessed L1 cache so its access time should be included when calc  the AMAT.

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Is this \(1+10+100\) right? Because in the given answer you have mentioned \(1+10+101\), but the calculation doesn't come up right if that's the case.

\[ \boxed{ 0.9(1)+0.1(0.8)(1+10)+0.1(0.2)\underbrace{(1+10+100)}_{\text{L1 access + L2 miss penalty + Main-memory miss penalty}} } \]
On an L1 miss + L2 miss, the request goes through all three stages sequentially:

\[ \boxed{1\text{ ns} + 10\text{ ns} + 100\text{ ns}=111\text{ ns}} \]

Answer 2

Answer - 4ns

Comments

If explicitly mentioned then only we use simultaneous cache :

H*T(L1) + M(L1)*[ H(L2)*T(L2)+M(L2)*T(MM)]

Else we always consider hirarcheial cache (By default):

T(L1) + M(L1)[ H(L2)*T(L2)+M(L2)*T(MM)]

Am I correct..!?

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No, there is no concept like simultaneous Cache. Many standard resources does not even mention this concept.

If any such concept should be used, GATE must mention that & explain how does it actually works...

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Ok understood thank you

Answer 3

So the question is asking for average memory access time.

AMAT in this question can be found by using a simple logic. Access the memory one by one and calculate average memory access time.

In the given question:

AMAT = 0.9*1 + 0.1*(0.8*(1+10)) + 0.1*0.2*(1+10+100) = 4 ns

This means first find hit in L1, then find miss in L1 and hit in L2 and teh find miss in L1 and miss in L2 and hit in mainmemory.

This will give answer as 4 ns.

 

Comments

they have given miss penalty why we are inclucing the L1 cache access time here ?
L2 - L1 penalty is 10 ns, and MEM to L2 is 100
the equation should be 0.9(1)+0.1(0.8(10)+0.2(100)) right ?

 

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Here, we have assumed that 90% of the time, when there will be a hit, it will take 1ns. Now for the remaining 10% of the time, if we get a hit in L2, then it will take:.1*0.8*(1 + 10) ns. If we just take 0.1*0.8*10, then we are not including the L1 access time. But here according the the answer, we first access L1, which is a miss, then access L2, which is a hit. Hence time will be L1 miss * L2 hit * (L1 access time + L2 access time)

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is there any way to tell, based on the Question, when to do sequential additition and when to do addition only of the given miss penalty, usually i get confused.

Answer 4

 

So, 

AMAT = 1 +  0.1 ( 10 + 0.2 (100) )

           = 1 + 0.1 ( 10 + 20 )

           = 1 + 0.1 (30)

           = 1 + 3

           = 4 ns

Answer 5

Average memory access time recursion formula = L1 access time + L1 Miss rate * L1 miss penalty.

L1 Miss penalty = L2 access time + L2 Miss rate * L2 miss penalty

and given in the question that L2 miss penalty which is main memory access time as 100ns

So, AMAT = 1 + 0.1(10+0.2*100)

AMAT = 1+0.1(10+20)

AMAT => 1+ 3 =4...

Answer 6

Category : Recall (Formula Substitution)

Check : Computer Architecture: A Quantitative Approach -Hennessy (For the Formula)