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Consider a two-level cache hierarchy with $L1$ and $L2$ caches. An application incurs $1.4$ memory accesses per instruction on average. For this application, the miss rate of $L1$ cache is $0.1$; the $L2$ cache experiences, on average, $7$ misses per $1000$ instructions. The miss rate of $L2$ expressed correct to two decimal places is ________.
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It is given that average references per instruction = 1.4
For 1000 instructions total number of memory references = 1000 * 1.4 = 1400
These 1400 memory references are first accessed in the L1.
Since the miss rate of L1 is 0.1, for 1400 L1 references the number of misses = 0.1 * 1400 = 140
We know when there is a miss in L1 we next access the L2 cache.
So number of memory references to L2 = 140
It is given that there are 7 misses in L2 cache. Out of 140 memory references to L2 cache there are 7 misses.
Hence the miss rate in L2 cache = 7/140 = 0.05
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7*100/(1000*1.4) =.5
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1.4 memory access per inst so for 1000 inst. we have 1400 memory accesses.

Now,all the 1400 memory accesses will be first to L1 then if there is a miss they will go to L2.

number of memory access on L2 for 1000 instructions = 0.1 * 1400 = 140

miss rate = no. of misses/total no. of memory accesses = 7/140 = 0.05
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2 votes

“the L2 cache experiences, on average, 7 misses per 1000 instructions”

 i think this is where everyone is getting the doubt lets approach like how most of u might look at this question…

1.4  memory accesses per instructions   , lets assume the processor is executing 1 instruction right now 

now for L1 cache miss rate is 0.1 given   

         $memory\: accesses \, \times global\:miss\:rate\:of \:L1 \:cache = number\: of \:misses \:in \:L1 \:cache$

1.4 X 0.1 = number of misses in L1 cache

number of misses in L1 cache = 0.14

 

now misses for L1 cache  = memory access of for L2 cache 

Its given 7 misses per 1000 instructions for L2 cache 

so misses per instruction = $\frac{7}{1000}$ 

         $memory\: accesses \:per\:instruction\:for \:L2\:cache \, \times local\:miss\:rate\:of \:L2 \:cache = number\: of \:misses \:in \:L2 \:cache\:per\:instructions$

0.14 X  local miss rate of L2 cache  = $\frac{7}{1000}$

$\therefore$   local miss rate of L2 cache = $\frac{7}{1000\times 0.14}$ = $0.05$

 

 

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