In a 6 segment pipeline, If the respective delays are 10, 5, 30, 15, 20, and 40 unit. What is the speedup using pipeline.
(A) 100%
(B) 200%
(C) 300%
(D) 400%


Answer: (C)

Explanation: Speedup (S) = t_n / t_p
Where t_n

= time consumed in non-pipelined manner 
= 10 + 5 + 30 + 15 + 20 + 40 
= 120 unit 

And, t_p

= time consumed in pipelined manner
= max (10, 5, 30, 15, 20, and 40)
= 40 unit 

Therefore, speedup (S),

= tn / tp
= 120 / 40 
= 3
= 300% 

So, option (C) is correct.


Quiz of this Question