Question

Consider 2 pipelines A & B where pipeline A is having 7 stages of uniform delay of 2ns. Pipeline B is having 6 stages with respect stage delay 6ns. How much time is saved when 150 tasks are pipelined using A instead B.

Answer 1

Answer: 618ns

Explanation:

Time required i pipeline execution (Tp)={ K + ( N - 1 ) }* tp

N=Number of instructions

K=Number of stages

tp =time period for the clock cycle in pipeline ={ MAX(ti) from i=1 to k + Buffer delay }

For A

Time required in pipeline execution (TpA)={ 7 + (150-1) } *2

=312 ns

For B

Time required i pipeline execution (TpB)={6+(150-1) } *6

=930 ns

time saved =930 - 312 = 618 ns