Performance c

Question

Consider the system have L1 data cache with 50 percent of hit rate and take 2 cycles when hit in L1 cache, L2 cache with 70% of hit rate and take 15 cycles when hit in L2 cache and main memory with 100% of the hit rate and 200 cycles when hit in main memory to access a block. If main memory speed is improved by 15% then the improvement in L1 miss time is ____________(upto 2 decimals)

       1. 2.14

       2. 2.78

       3. 1.48

       4. 1.14

Comments

4) 1.14?

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yes answer is 1.14

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You should not post questions with these many grammatical mistakes.

Answer 1

Here basically we need to calculate the performance enhancement which is being asked in terms of L1 cache miss time

So we need to find the L1 cache miss time in the two cases first.

L1 cache miss time_old    =   L1 miss rate * L1 miss penalty + L1 miss rate * L2 miss rate * L2 miss penalty

                                    =   L1 miss rate  * L2 cache hit time + L1 miss rate * L2 miss rate * M.M.A.T

                                    =   0.5 * 15 + 0.5 * 0.3 * 200

                                    =   7.5 + 30

                                    =   37.5

L1 cache miss time_new   =   L1 miss rate * L1 miss penalty + L1 miss rate * L2 miss rate * L2 miss penalty

                                    =    L1 miss rate  * L2 cache hit time + L1 miss rate * L2 miss rate * M.M.A.T

                                    =   0.5 * 15  +  0.5 * 0.3 * (200 / 1.15)

                                    =   7.5 + 26.09

                                    =   33.59

Hence performance improvement factor(speedup)   =   Old time / New time

                                                                           =   37.5 / 33.59

                                                                           =   1.116

                                                                           =   1.12  (correct to 2 decimal places)

Hence 1.12 should be the correct answer .

Ref : http://web.cs.iastate.edu/~prabhu/Tutorial/CACHE/CompPerf.pdf

Comments

It is regarding l2 miss penalty which is main memory access time which is said to be 15% faster in second case..

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Thank you Habib!!

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shouldn't access be hierarchical.Why not add 15ns to miss part?

Answer 2

Initially

L1 miss time= $\cdot 5*[15+\cdot 3*200] =37.5$

After Improvement

L1 miss time=$\cdot 5*[15+\cdot 3*170] =33$

So performance improved = $37\cdot5/33 =1.1363\cong 1.14$