cache

Question

consider single level  cache woth access time 5ns  line size of 128 bytes and hit ratio is 0.97.Main memory uses block transfer capability that has first 8 bytes access time 50 ns and for remaining words 10ns What is access time when there is cache miss(Assume cache waits until line is fetched from memory  and then reexecutes for hit)

Comments

looks like 205ns for me.

Answer 1

Access Time on miss = Cache access + Memory fetch + Cache access (re-access cache as mentioned in question)

$= 5 + (50 + (128-8)/8 \times 10) + 5 \\= 210 ns$

Comments

128 bytes is the cache line size. First 8 byte takes 50 ns to reach cache from main memory and remaining words (128-8 = 120)/8 takes 10ns. Yes, here word size is taken as 8 bytes- because otherwise there is no point in saying first 8 byte takes one time and remaining takes another time- this shows the memory bandwidth and is of 8 bytes.

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@arjun sir why in memory fetch you have not multiplied the factor 0.03(for miss ratio)

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Since Miss is certain... Given in the question...

Answer 2

Is it 62 ns ?