A non-pipeline system takes 50ns to process a task. The same task can be processed in six-segment pipeline with a clock cycle of 10ns. Determine approximately the speedup ratio of the pipeline for 500 tasks.

Correct option is B

Speedup factor is defined as the ratio of time required for non-pipelined execution to that of time required for pipelined execution.
To calculate the speedup ratio of the pipeline system, we use the formula for speedup:
​$$\text{Speedup} = \frac{\text{Time without pipeline}}{\text{Time with pipeline}}$$​​
​$$\mathrm{S = \frac{T_{pipelined}}{T_{non-pipelined}}}$$​​
Where:
• $$T_{\text{non-pipelined}}$$​ = Time for non-pipelined execution
• $$T_{\text{pipelined}}$$​ = Time for pipelined execution
Given Data:
• Time for non-pipelined execution per task, $$t_n = 50 \ \text{ns}$$​​
• Time for pipelined execution per task, $$t_p = 10 \ \text{ns}$$​​
• Number of stages in the pipeline, k = 6
• Number of tasks, N = 500
Formulas:
1. Time for non-pipelined execution:
​$$T_{\text{non-pipelined}} = t_n \times N$$​​
2. Time for pipelined execution:
​$$T_{\text{pipelined}} = t_p \times \left(1 + (N - 1) \times k \right)$$​​
Calculation:
1. Time for non-pipelined execution:
​$$T_{\text{non-pipelined}} = 50 \times 500 = 25000 \, \text{ns}$$​​
2. Time for pipelined execution:
​$$T_{\text{pipelined}} = 6 \times 10 + (500 - 1) \times 10 = 60 + 4990 = 5050 \, \text{ns}$$​​
3. Speedup factor:
​$$S = \frac{25000}{5050} \approx 4.95$$​​
Information Booster:
1. The non-pipelined system takes 50 ns per task, so the total time for 500 tasks is 25000 ns.
2. The pipelined system, after the initial latency of 6 stages (or 60 ns), processes each remaining task in 10 ns, resulting in a total pipelined time of 5050 ns.
3. The speedup factor is 4.95, which shows that the pipelined system is significantly faster than the non-pipelined system.
4. The calculated speedup factor helps understand how efficient a pipeline can be compared to a non-pipelined approach.