CO-Pipelining-Q9

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A 7 stage pipeline with following stage delays 100, 150,190,200,400,250,350 is changed to 5 stage pipeline with 100, X, 150, 140, 200 to increase the speed up percentage to 100 percent.
The main reason to move to lesser no of stages is the efficiency of the 7 stage pipeline was only 40 %. What is maximum value X can take?

asked 6 days ago in Computer Organization by gbmentor (6,810 points)
reshown 6 days ago by gbmentor

3 Answers

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Best answer

Ans is 357.14  

answered 5 days ago by getgatebook (22,080 points)
Sir how 7 instructions will be executed in 400 units of time ? And plz help me in understanding the statement throughput= 7instructions/400units.
Sir I personally think 1 instruction should be executed in 400 units.plz clear my doubt sir
can someone explain throughput concept as well how can we infer that 5 stage throughput will be twice that of 7 stage
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X is not more than 200 so X so 200 is correct one
answered 5 days ago by tsnikhilsharmagate2018 (5,940 points)
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Speed up= \frac{time -non pipeline- execution}{ time- taken -in- pipeline}=\frac{n*t_n}{(n+k-1)*t_p}= \eta *k

where n= efficiency and k=no of stages

Case 1 - Speed up=\frac{n*t_n}{(7+n-1)*400}=\frac{n*t_n}{(6+n)*400}=0.4*7=2.8

 

 

Case 2-Speed up=\frac{n*t_n}{(5+n-1)*X}=\frac{n*t_n}{(4+n)*X}=1.0*5=5

 

n*t_n=5*(4+n)*X

n*t_n=2.8*(6+n)*400

If we solve the above two equation,we will get X=224 Assuming (any constant+n) ~ n for very large n

 

 

answered 5 days ago by soumyadeeppro20 (1,640 points)
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