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 Suppose that we are considering an enhancement that runs 10 times faster than the original machine but is usable only 40% of the time. What is the overall speedup gained by incorporating the enhancement.
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@LRU

 

Se = 10 ….

F = 40/100

= 0.4 …

 

So = 1 / (( 1 - F) + F/Se)

= 1/ (0.6 + 0.4/10)

= 1/0.64

= 1.56…

 

 

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here question telling about overall  speed up

so here you take nonpipeline instruction   taking  100 unit time then in this question the speedup only on the 40 percent of the time means here pipelined  instruction 60 unit  time same as   nonpipeline and 40 unit time speedup so 40 time speed up by 10 then it is taking only 4 unit time  so overall time in pipeline is 4+60=64

so overall speedup=ET old means nonpipeline/pipelined =100/64=1.5625