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.
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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
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