# CO-Pipelining-Q1

A pipeline of 2 stages with a delay of x units each is split to n stages. In the new design, each stage has a delay of x/n units. To get the throughput increase of 1700% what would be the n value?

reshown Jun 13

1700/100=17

17+1=18

n=18
answered Jun 13 by (19,690 points)
answered Jun 13 by (160 points)
+1 vote

let initial throughput be 100. required throughput is 1800.

speedup = 1800/100 = 18

In the first pipeline, each instruction will get completed after x units, and in the second pipeline, each instruction will get completed after x/n units

speedup = x / (x/n)

= n

therefore n = 18

answered Jun 14 by (4,100 points)

Comment if you find any issues.

Pipeline 1

One instruction <----- x units
? <-------------------------1 unit

$= \frac{1}{x} units$

Pipeline 2

One instruction ----> x/n units

? -----------------------> 1 units
$= \frac{n}{x} units$

$percentage \ of \ gain\ in \ throughput = \frac{old value- new value}{old value} * 100$

1700 = (n-1)* 100

17 = n-1

18=n

answered Jun 14 by (31,090 points)
edited Jun 15
In pipeline 1, stage delay= x units. If we consider the CPI(cycles per instruction as 1), which is considered until no stalls are given.

Thus each instruction takes x time units.

Throughput in Pipeline 1 = 1/x

Similarly, every instruction in pipeline 2 takes x/n time units.

Thus, throughput in pipeline 2 = 1/(x/n) = n/x

% increase in throughput = ((new-old)/old)*100

1700       =(((n/x)-(1/x))/(1/x)*100

17           =n-1

18           = n
Yes you are correct. corrected the answer.
In this question (http://thegatebook.in/qa/4507) the throughput is (no. of stages / max stage time) so why in the above question. its only 1/x?