A nonpipelined system takes 300ns to process a task. The same task can be processed in a 5-segment pipeline with a clock cycle of 60ns. Determine the speedup ratio of the pipeline for 100 tasks. What is the maximum speedup that could be achieved with the pipeline unit over the nonpipelined unit

The maximum speedup that could be achieved is approximately 4.81.

Explanation:

From the question, we have:

Speed up ratio = Time taken by a non-pipelined processor / Time taken by a pipelined processor = (n * T) / ((n + (k – 1)) * Tk) …………………. (1)

Where:

n = number of tasks = 100

T = Number ns for a nonpipelined system takes to process a task = 300

k = Type of segment pipeline = 5

Tk = Number of clock cycle of the type of segment pipeline in ns = 60

Substituting the above into equation (1), we have:

Speed up ratio = (100 * 300) / ((100 + (5 - 1)) * 60) = 4.81

Therefore, the maximum speedup that could be achieved is approximately 4.81.

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

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