A non-pipeline system takes 50ns to process a task. The same task can be…

Solution: Compute the total time for the non-pipelined system and for the pipelined system, then take their ratio.

• Non-pipelined total time = number of tasks × time per task = 500 × 50 ns = 25,000 ns.

• Pipelined total time for N tasks with k stages and clock cycle C is (k + N − 1) × C. Here k = 6, N = 500, C = 10 ns, so pipelined time = (6 + 500 − 1) × 10 ns = 505 × 10 ns = 5,050 ns.

• Speedup = (non-pipelined time) / (pipelined time) = 25,000 / 5,050 ≈ 4.9505 ≈ 4.95.

Therefore the approximate speedup for 500 tasks is 4.95. Note that the maximum ideal speedup equals the number of stages (6), but for a finite number of tasks the startup and drain overhead reduce the actual speedup.