###### Pipelining

October 19, 2023###### Pipelining

October 19, 2023# Pipelining

Question 13 |

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?

1.83 | |

2 | |

3 | |

6 |

Question 13 Explanation:

Let there be n instructions.

For a non-pipelined processor each instruction takes 12 cycles.

So for n instructions total execution time be 12 × n = 12n

For a pipelined processor each instruction takes

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

So for n instructions total execution time be,

(1×6 + (n-1) × 1) × 6

= (6 + n – 1) × 6

= (5 + n) × 6

= 30 + 6n

∴ Speedup = time without pipeline/time with pipeline = 12n/30+6n

So, if n is very large,

For a non-pipelined processor each instruction takes 12 cycles.

So for n instructions total execution time be 12 × n = 12n

For a pipelined processor each instruction takes

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

So for n instructions total execution time be,

(1×6 + (n-1) × 1) × 6

= (6 + n – 1) × 6

= (5 + n) × 6

= 30 + 6n

∴ Speedup = time without pipeline/time with pipeline = 12n/30+6n

So, if n is very large,

Correct Answer: B

Question 13 Explanation:

Let there be n instructions.

For a non-pipelined processor each instruction takes 12 cycles.

So for n instructions total execution time be 12 × n = 12n

For a pipelined processor each instruction takes

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

So for n instructions total execution time be,

(1×6 + (n-1) × 1) × 6

= (6 + n – 1) × 6

= (5 + n) × 6

= 30 + 6n

∴ Speedup = time without pipeline/time with pipeline = 12n/30+6n

So, if n is very large,

For a non-pipelined processor each instruction takes 12 cycles.

So for n instructions total execution time be 12 × n = 12n

For a pipelined processor each instruction takes

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

So for n instructions total execution time be,

(1×6 + (n-1) × 1) × 6

= (6 + n – 1) × 6

= (5 + n) × 6

= 30 + 6n

∴ Speedup = time without pipeline/time with pipeline = 12n/30+6n

So, if n is very large,

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