in CO and Architecture
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How to approach this one?


in CO and Architecture

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Theory :


$n$ be number of instructions,

$k$ is number of stages in pipeline,

$T_P$ is one cycle time of pipelined processor

In pipelined processor only  $1^{st}$ instruction takes $k\times T_P$ time,

all other $(n-1)$ instructions take $1$ cycle time $=(n-1)\ T_P$

Total execution time on pipelined processor assuming no stalls $T_{EP}=k\space T_P+(n-1)T_P=(n-1+k)T_P$

Execution time of each instruction on non-pipelined processor  $T_n=  k\times T_P$

Total execution time on non-pipelined processor  $T_E=n\space k\space T_P$

$Speed Up = \frac {\text{total execution time without pipeline}}{\text{total execution time with pipeline}}=\frac {T_E} {T_{EP}}=\frac{(n\space k\space )T_P}{(n-1+k)T_P}\approx k$

Approximation based on $n>>k,\therefore (n-1+k)\approx n$ 

Now, let $I$ is stalls per instruction in pipelined processor that is $I \times T_P$ time extra per instruction,

Total execution time in pipelined processor,

$T_{EP}=(n-1+k)T_P + (n\times I) T_P =(n+I.n-1+k)T_P=((1+I)n-1+k)T_P$

$Speed Up = \frac {(n\space k)T_P}{((1+I)n-1+k)T_P}\approx \frac k {1+I}... (n>>k)$

Solution :

$k=5$ , $ 20$% instruction cause one stall,

$n$ instruction cause $n\times20$% stall $=0.2n$ stalls.

Stalls per instruction $I=\frac {0.2n}{n}=0.2$

$SpeedUp = \frac {5} {1+0.2} = \frac {5} {1.2} \approx 4.17 $


1 comment

Thanks a ton!

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