Question

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

Answer 1

The speedup ratio of a 5-segment pipeline with a clock cycle of 60ns processing 100 tasks compared to a nonpipelined system that takes 300ns per task is approximately 4.81. The theoretical maximum speedup for the pipeline is 5, corresponding to the number of pipeline stages.

Step-by-step explanation:

To determine the speedup ratio of a pipeline, we compare the execution time of tasks in a nonpipelined system to that of a pipelined system. In a nonpipelined system, each task takes 300ns, so for 100 tasks, the total time would be 100 × 300ns = 30,000ns. In a 5-segment pipeline with a 60ns clock cycle, the first task will take 5 × 60ns = 300ns to fill the pipeline, and each subsequent task will take 60ns. Therefore, for 100 tasks, the time taken will be 300ns + (99 × 60ns) = 6,240ns.

The speedup ratio is calculated by dividing the processing time of the nonpipelined system by the processing time of the pipelined system: 30,000ns / 6,240ns ≈ 4.81. This means the pipeline can process tasks about 4.81 times faster than the nonpipelined system for 100 tasks. The maximum speedup theoretically could be equal to the number of segments in the pipeline, which is 5 in this case. However, this maximum value assumes perfect conditions without any pipeline stalls or inefficiencies.

Answer 2

The maximum speedup that could be achieved is approximately 4.81.

Step-by-step 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.