# CO-Pipelining-Q9

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?

reshown Jun 13

Ans is 357.14

answered Jun 13 by (31,090 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
throughput is the number of instruction execution per unit time so,
400 -------------------> 1 instruction
1 unit time ----------> 1/400
Tp = 1/400

in the second case,
200 -------------------> 1 instruction
1 unit time ----------> 1/200
Tp' = 1/200

so Tp' = 2XTp
X is not more than 200 so X so 200 is correct one
answered Jun 13 by (19,690 points)

$Speed up= \frac{time -non pipeline- execution}{ time- taken -in- pipeline}=\frac{n*t_n}{(n+k-1)*t_p}= \eta *k$

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 Jun 13 by (1,720 points)