Question

Consider a pipeline system. Let the time it takes to process a sub operation in each segment be equal to \( t_p = 20 \, \text{ns} \). Assume that the pipeline has \( k = 4 \) segments and executes \( n = 100 \) tasks in sequence. Consider a non-pipeline system, assume that \( t_n = k t_p \) (a non-pipeline system to perform the operation takes a time equal to \( t_n \) to complete each task), where \( t_p = 20 \, \text{ns} \), \( k = 4 \). Find the speedup of a pipeline processing over an equivalent non-pipeline processing to execute 100 tasks.

• A. 3.88
• B. 0.08
• C. 0.88
• D. 1.88

Hint:

In pipeline processing, the throughput improves as multiple tasks are processed in parallel, leading to a speedup factor.

Answer 1

The Correct Option is A

Step 1: Problem Definition.
A pipeline system comprises 4 segments, each with a 20 ns processing time. This system handles 100 sequential tasks. In contrast, a non-pipeline system processes each task in 4 times the segment duration, equating to 80 ns per task. The objective is to determine the speedup achieved by the pipeline system compared to the non-pipeline system.

Step 2: Non-Pipeline System Processing Time.
The total time for 100 tasks in the non-pipeline system is calculated as:\[t_n = k t_p \times n = 4 \times 20 \, \text{ns} \times 100 = 8000 \, \text{ns}.\]

Step 3: Pipeline System Processing Time.
The first task in the pipeline system requires \( 4 \times 20 \, \text{ns} = 80 \, \text{ns} \). Subsequent tasks are processed concurrently, with each taking 20 ns. Therefore, the total time for 100 tasks is:\[t_p = t_p + (n - k) t_p = 80 \, \text{ns} + (100 - 4) \times 20 \, \text{ns} = 80 \, \text{ns} + 1920 \, \text{ns} = 2000 \, \text{ns}.\]

Step 4: Speedup Calculation.
Speedup is computed as the ratio of the non-pipeline time to the pipeline time:\[\text{Speedup} = \frac{t_n}{t_p} = \frac{8000 \, \text{ns}}{2000 \, \text{ns}} = 4.\]

Step 5: Conclusion.
The speedup of pipeline processing over an equivalent non-pipeline processing is 4.