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Consider a non-pipeline processor has clock rate of 25 MHz and CPI of 6, another processor designed with same clock rate and 8 stage instruction pipeline. If program containing 500 instructions is executed on both processors, then the speedup factor is _______.

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$\Rightarrow $ Clock Rate = $25$ MHZ

$\Rightarrow \ 1$ cycle time $= \ 0.04$ usec.

$\Rightarrow $ Cycle Per instruction = $6 \Rightarrow 6 \times 0.04=0.24$ usec [$1$ instruction execution time]

For $500$ instruction Execution time :

$\Rightarrow $ Non Pipeline processor = $500 \times 0.24 = 120$ usec

$\Rightarrow $ Pipelined processor = $8 \times 0.04\times 1 +0.04\  \times 499 = 20.28 $ usec

SpeedUp achieved $=\Large \frac{120}{20.28}$ $=\color{Red}{5.917}$

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