Question

Considering equal processing time for each segment, speed-up S achieved by a K-segment instruction pipeline operating on straight sequence of N instructions is given by:
1. \(S=\frac{K(N-K)}{K+N-1}\)
2. \(S=\frac{K+N}{KN}\)
3. \(S=\frac{K(K+1)}{(K+1)N}\)
4. \(S=\frac{KN}{K+N-1}\)

Answer 1

Correct Answer - Option 4 : \(S=\frac{KN}{K+N-1}\)

Data:

number of instructions = N

number of stage (segment) = K

Non-pipeline

Assume each stage take 1 unit of time

Time taken (Twp)  = number of stage × number of instructions

Twp  = K.N 

For pipeline 

Only 1^st instruction takes K unit time then every (N - 1) instruction takes 1 unit time

Time taken(Tp) = (K + N - 1) 

\(S = \frac{T_{wp}}{T_{p}} = \frac{KN}{K+N-1}\)

Therefore option 4 is correct