A non-pipelined system takes 30 ns to process a task. The same task can be…

CONCEPT: For N tasks in an ideal k-stage pipeline with clock period T, the elapsed time is (k + N - 1)T because the pipeline must first fill and then completes one task per cycle. Speedup is the non-pipelined elapsed time divided by the pipelined elapsed time.

APPLICATION

1. Define N = 100 tasks, k = 4 stages, non-pipelined task time = 30 ns, and pipeline clock period T = 10 ns.

2. The non-pipelined elapsed time is N × 30 ns = 100 × 30 ns = 3000 ns.

3. The pipelined elapsed time is (k + N - 1)T = (4 + 100 - 1) × 10 ns = 1030 ns.

4. Therefore, speedup = 3000 ÷ 1030 = 2.9126…, which rounds to 2.91.

CROSS-CHECK: For a very large batch, the speedup approaches 30 ÷ 10 = 3. A finite batch has pipeline-fill overhead, so a value slightly below 3 is consistent.

Result: The pipeline speedup for 100 tasks is 2.91.