Pipelining
October 19, 2023Pipelining
October 19, 2023Pipelining
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 nonpipelined 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 + (n1) × 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 nonpipelined 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 + (n1) × 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 nonpipelined 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 + (n1) × 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 nonpipelined 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 + (n1) × 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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