To calculate the speedup of a pipeline, we first need to determine the time it takes to execute an instruction without pipelining and the time it takes with pipelining.

1. Without pipelining, the time to execute an instruction is simply the sum of the stage delays. So, we add up 5 nsec, 6 nsec, 11 nsec, and 8 nsec to get a total of 30 nsec.

2. With pipelining, the time to execute an instruction is the maximum stage delay plus the delay of an inter-stage register. The maximum stage delay here is 11 nsec, and the inter-stage register delay is 1 nsec. So, the total is 12 nsec.

3. The speedup is then calculated as the time without pipelining divided by the time with pipelining. So, we divide 30 nsec by 12 nsec to get a speedup of approximately 2.5.

So, the correct answer is d. 2.5.

Question

To calculate the speedup of a pipeline, we first need to determine the time it takes to execute an instruction without pipelining and the time it takes with pipelining.

1. Without pipelining, the time to execute an instruction is simply the sum of the stage delays. So, we add up 5 nsec, 6 nsec, 11 nsec, and 8 nsec to get a total of 30 nsec.

2. With pipelining, the time to execute an instruction is the maximum stage delay plus the delay of an inter-stage register. The maximum stage delay here is 11 nsec, and the inter-stage register delay is 1 nsec. So, the total is 12 nsec.

3. The speedup is then calculated as the time without pipelining divided by the time with pipelining. So, we divide 30 nsec by 12 nsec to get a speedup of approximately 2.5.

So, the correct answer is d. 2.5.

Knowee AI · Accepted Answer

To calculate the speedup of a pipeline, we first need to determine the time it takes to execute an instruction without pipelining and the time it takes with pipelining.

1. Without pipelining, the time to execute an instruction is simply the sum of the stage delays. So, we add up 5 nsec, 6 nsec, 11 nsec, and 8 nsec to get a total of 30 nsec.

2. With pipelining, the time to execute an instruction is the maximum stage delay plus the delay of an inter-stage register. The maximum stage delay here is 11 nsec, and the inter-stage register delay is 1 nsec. So, the total is 12 nsec.

3. The speedup is then calculated as the time without pipelining divided by the time with pipelining. So, we divide 30 nsec by 12 nsec to get a speedup of approximately 2.5.

So, the correct answer is d. 2.5.

Consider an instruction pipeline with four stages with the stage delays 5 nsec, 6 nsec, 11 nsec, and 8 nsec respectively. The delay of an inter-stage register stage of the pipeline is 1 nsec. What is the approximate speedup?Question 7Select one:a.0.25b.250c.25d.2.5

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