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suppose that execution time for a program is directly proportional to instruction access time and that...

suppose that execution time for a program is directly proportional to instruction access time and that access to an instruction in the cache is 20 times faster than access to an instruction in the main memory. assume that there is 96% chance to find the requested instruction in the cache (probability to find the requested instruction in the cache is 0.96), and also assume that if an instruction is not found in the cache, it must be first fetched from the main memory to the cache and then fetched from the cache to be executed. compute the ratio of the program execution time without the cache to the program execution time with cache. this ratio is usually defined as the speedup factor resulting from the presence of the cache

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Answer #1

`Hey,

Note: Brother if you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

a)

Speedup = ( execution time of without cache)/(execution time with cache)

(20+20)/(0.96*(1÷20)+(1-0.96)*((1÷20 ) +(1÷20)+20))=46.94

b)

For part b make (1-0.96)÷2 to make it half

(20+20)/(0.98*(1/20)+(0.02)*((1/20 ) +(1/20)+20))=88.69

Kindly revert for any queries

Thanks.

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