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A processor takes $12$ cycles to complete an instruction I. 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?

  1. $1.83$
  2. $2$
  3. $3$
  4. $6$
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6 Answers

Best answer
73 votes
73 votes
For non pipeline processor we have n instruction and each instruction take
$12$ cycle so total $12n$ instruction.

For pipeline processor we have each stage strict to $6ns$ so time to complete
the $n$ instruction is $6\times 6+ (n-1)\times 6$.

$\lim_{n \to \infty }\dfrac{12n}{36 + (n-1)\times 6}=\dfrac{12}{6} =2$.

Correct Answer: $B$
edited by
30 votes
30 votes
Speed Up= Time without Pipeline / Time with Pipeline

Time without Pipeline =12 cycle

Time with Pipeline = 6 cycle    take maximun one

Speed up =12/6=2
22 votes
22 votes

Speed Up= Time without Pipeline / Time with Pipeline

                 =n*tn/[k+(n-1)]tp

where, Time without Pipeline= n*tn

Time with Pipeline= [k+(n-1)]tp

but in the question it is given that "assuming that a very large number of instructions are to be executed" ie. n>>k

therefore ,Time with Pipeline= n*tp   (as n is very large so k becomes zero)

now, speedup= n*tn/n*tp

                      =tn/tp

                       =12/6 = 2 

Answer:

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