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Consider a computer having L1- data cache and main memory. The latencies of different kind accesses are as follows; cache hit 1-cycle; cache miss, 105 cycles; main memory access with cache disabled, 100 cycles. When you run a program with an overall miss rate of 5%, what will be the average memory access time(in CPU cycles)
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6.2 ???
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yes , ans is 6.2
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0.95*1 + 0.05*105 = 6.2
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main memory access with cache disabled, 100 cycles

Why it is written here.

Since miss in cache implicit means accessing main memory only.

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In case of miss why you are not adding 100 cycles for accessing main memory
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because it is already included in 105 cycles

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We have given that cache hit then 1 cycle

which we can say (not always) that cache access time is 1 cycle

Now if there is cache hit or cache miss 1 cycle is required to search cache.

cache miss =105 cycle means 1 cycle for cache access + 100 cycles for main memory access + 4 cycle other overhead

If cache miss cycle was not given then T avg = 0.95*1+0.05*101 = 6 cycles

But we are given cache miss time T avg = 0.95*1+0.05*105= 6.2 cycles

This is my interpretation of this question
by Boss (18.4k points)
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