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stira [4]
2 years ago
14

Assume that we have an application with a total of 500,000 instructions where 20% of them are the load/store instructions with a

n average CPI of 6 cycles, and the rest instructions are ALU instructions with average CPI of 1 cycle. If we double the clock rate without optimizing the memory latency, the average CPI for load/store instruction will also be doubled to 12 cycles. What is the speedup after this change?
Computers and Technology
1 answer:
balu736 [363]2 years ago
5 0

Answer:

1.25

Explanation:

#instructions = 5*10^{5}  

Average CPI (old) = 0.2*6 + 0.8*1 = 1.2+0.8 = 2.0

Average CPI (new) = 0.2*12 + 0.8*1 = 2.4 + 0.8 = 3.2

Assuming Clock Rate = x

Speedup = Execution Time (old) / Execution time (new)  = (2.0*5*10^{5} /x) / (3.2*5*10^{5} /2x)  = 4/3.2 = 1.25

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Answer:

Check the explanation

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Law Incorporation [45]
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The wrong choices can be defined as follows:

  • In option a, it is wrong because the data warehouse transforms data into information that may be used by an organization's decision-making process.
  • In option b, it is wrong because by using Metrics Collector Groups and logging the data, it was able to display real-time market data as well as collect and store data.
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Learn more:

Memory Diagnostic: brainly.com/question/13606234

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