Question

A processor takes 12 cycles to complete an instruction 1. The corresponding pipelined processor uses 6 stages with the execution times of 3, 2, 5, 4, 6 and 2 cycles respectively. What is the asymptotic speedup assuming that a very large number of instructions are to be executed?

• A. 1.83
• B. 2
• C. 6
• D. 3

Answer 1

The correct option is B 2

$Nonpipelineprocessortime\left(NP\right)$

$=12cycle\times Numberofinstruction$

$=12ncycles$


$Pipelineprocessortime\left(P\right)$

$=(K+n-1)\times T_P$

$=(6-1+n)\times T_P$


$T_P=max\left(allstagestimes\right)$

$=max(3,2,5,4,6,2)cycles$

$=6cycles$


$\left(P\right)=(6-1+n)\times T_P$

$=(5+n)\times 6cycles$

$=30+6ncycles$


$So,Speedup={lim}_{n\to \infty }\frac{12n}{30+6n}$

$={lim}_{n\to \infty }\frac{12}{6}$
$(ApplyL-Hospitalrulesince(\infty /\infty \left)\right)$

$=\left(\frac{12}{6}\right)=2$


$So,option\left(b\right)iscorrect.$