Consider 4 stages in pipeline S1,S2,S3,S4 where each stage requires 2ns
here k=no of stages=4, n is no of task =1, and tp=clock cycle time in pipeline=Max Stage Delay=2ns
Then, 1 task execution in pipeline=(k+n-1)tp=4*2=8 ns
Then, 1 task execution in non pipeline=tn=S1+S2+S3+S4=2+2+2+2=8ns
Speed Up=tn/tp=8/2=4
Efficiency=SpeedUp/No of stages in Pipeline=4/4=1 ===>100%
Now,
Consider 4 stages in pipeline S1,S2,S3,S4 where S1=2ns,S2=8ns,S3=6ns,S4=1ns
here k=no of stages=4, n is no of task =1, and Let tp=clock cycle time in pipeline=Max Stage Delay=8ns
Then, 1 task execution in pipeline=(k+n-1)*tp=4*8=32ns
Then, 1 task execution in non pipeline=tn=S1+S2+S3+S4=2+8+6+1=17ns
Speed Up=tn/tp=17/8=2.125
Efficiency=SpeedUp/No of stages in Pipeline=2.125/4=0.53125==>53.125%
So, B require about the same amount of time is correct