A non-pipelined system takes 30ns to process a task. The same task can be…
2019
A non-pipelined system takes 30ns to process a task. The same task can be processed in a four-segment pipeline with a clock cycle of 10ns. Determine the speed up of the pipeline for 100 tasks.
- A.
3
- B.
4
- C.
3.91
- D.
2.91
Attempted by 146 students.
Show answer & explanation
Correct answer: D
Solution: compute total times and the speedup.
Non-pipelined total time = number of tasks × time per task = 100 × 30 ns = 3000 ns.
Pipelined total time = (number of stages + number of tasks − 1) × clock cycle = (4 + 100 − 1) × 10 ns = 103 × 10 ns = 1030 ns.
Speedup = non-pipelined total time ÷ pipelined total time = 3000 ÷ 1030 ≈ 2.9126 ≈ 2.91.
Key formula: Speedup = (N × T_non-pipelined) ÷ ((k + N − 1) × T_cycle), where N = number of tasks, k = pipeline stages.
Plugging values: N = 100, T_non-pipelined = 30 ns, k = 4, T_cycle = 10 ns.
Therefore speedup = (100 × 30) ÷ ((4 + 100 − 1) × 10) = 3000 ÷ 1030 ≈ 2.91.
A video solution is available for this question — log in and enroll to watch it.