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Assume we have a computer where the cycles per instruction (CPI) is 1.0 when all memory accesses hit in the cache.The only data accesses are loads and stores, and these total 50% of the instructions. If the miss penalty is 25 clock cycles and the miss rate is 2%, how much faster would the computer be if all instructions were cache hits?

  1.   1.5
  2.   0.7
  3.   2.75
  4.   1.75
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CPI = 1 with 100% cache hits.

Average Instruction Time = Fraction of Non-memory Instructions * 1 + Fraction of Load/Store Instructions * 1 * hit% in cache + Fraction of Load/Store Instructions * (1+miss penalty) * (1-hit%)

= 0.5 * 1 + 0.5*.98*1 + 0.5*.02*26

= .99 + 0.26 = 1.25.

So, speed up due to ideal cache will be 1.25. 

To get the answer in choice, you can modify question as below:

http://www.inf.ed.ac.uk/teaching/courses/car/Notes/2015-16/solution04.pdf

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