Avg stall cycles per instruction

Question

Suppose that in $500$ memory references there are $50$ misses in the first level cache and $20$ misses in second level cache. Assume miss penalty from the $L_{2}$ cache to memory is $100$ cycles. The hit time of $L_{2}$ cache is $20$ cycle. The hit time of $L_{1}$ cache is $10$ cycles. If there are $2.5$ memory references per instruction. How many average stall cycle per instruction?

Comments

$\frac{30*20+20*120}{500}*2.5 = 15$

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for 500 instructions = hit in L2*time + miss in L2*(L2+memory time) = 30*20 + 20*120 = 3000

for 1 instruction= 3000/500 = 6cycles.

Hence for 2.5 memory ref per instruction = 6 cycles/instruction * 2.5 = 15 cycles/instruction

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How is this formula derived?

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bro please give the formula

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Joshi nitish... While solving this you have assumed it to be simultaneous by default isn't?? Otherwise miss rates would be local instead of global

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@ashwin Miss rate in L2 is 20 cycles = 20/500 = 0.04 right? but u took 0.4

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Among 500 references 50 are missed, penalty for these is access time of L2

50*20=1000

among 40 missed references from L1,L2 cannot find 20 hence penalty is 20*100=2000

total stalls=1000+2000=3000

Memory references/Instrn =2.5 , 500 references from 200 instructions

3000/200=15 stall cycles/instruction

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https://gateoverflow.in/101983/co-cache-stall-cycles

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Why you are assuming no stalls for L1 hit? When we access L1 then at time CPU is ideal, this should also be consider as CPU stall.

Answer 1

To calculate stall cycles:there are total 500 references to L1.50 are miss in L1. it means that if we would have been lucky we would have got them( 50 references) in L1 only.but they were a miss so we have to go to L2 .Hence this is the stall time or extra time.

so for 30 references out of 50 we need to go till L2. So extra time is 20 clock cycles.

for 20 references out of 50 we need to go till main mem . so extra time is (20cc+100cc)=120 cc.

Rest u can understand from the snap :)

Hope it helps :)

Comments

time to search in cache and not found in that level cache

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thanks a lot

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@srestha @bikram  sir

what is the meaning of stall cycle / instruction of 

stall cycle per memory reference  ?

 

Answer 2

No of stalls for 500 memory references=$50*20+20*100=3000$

$2.5$ memory references = $1$ instruction

$\therefore500$ memory references= $200$ instructions

$\therefore$ Average Stalls/Instruction=$\frac{3000}{200}=15$

Comments

@Sukannya

No of stalls=miss rate of L1*miss rate of L2 +miss rate of L2* miss penalty of L2

how r u getting this?

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Srestha Di, No of stalls/memory access = Miss rate of L1* miss penalty of L1=miss rate of L1(hit time of L2+miss penalty from L2)

For 1 memory reference=$\frac{50}{500}(20+\frac{20}{50}*100)=\frac{50}{500}*20+\frac{20}{500}*100$

$\rightarrow$ For 500 references=$50*20+20*100$

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@Sukannya

clear explaination here

https://gateoverflow.in/91103/me-flt-4-q-58

right?

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Yes, @srestha Di

Answer 3

Avg Access time= (10 +.1*10+.1*.4*100 )*2.5 Cycles. Stall is when reference doesn't  hit in cache 1 and it occurs (.1*10+.1*.4*100) times so stall cycles will be (.1*10+.1*.4*100)*2.5 = 15 cycles.

Answer 4

Stall cycle created due to  some miss happen in cache and CPU need to access memory

In this question we can see ,

there are $50$ misses in $L_{1}$ cache for which we need to access $L_{2}.$

In $L_{2}$ there are $30$ hits, but again for $200$ misses we need to access Main Memory.

So, total stall cycles are $30*20+20*(20+100)=3000$ stall cycles

Now,

for $500$ memory reference  , there are $3000$ stall cycles.

for $1$ memory reference  , there are $3000/500$ stall cycles.

Here, each instruction needs $2.5$ memory reference.

So, for $2.5$ memory reference  , there are $(3000/500)*2.5=15$ stall cycles or $15 stall/instruction$