A 5-stage pipeline has stage delays: 180, 250, 150, 170, 250 ns. Inter-stage latch delay = 10 ns. No stalls. Time to process 1000 instructions (μs)?

Answer: 261.04 μsClock period: T_seg = max(180,250,150,170,250) + latch = 250+10 = 260 ns.Total time for n=1000 instructions, K=5 stages:Total = (K + n − 1) × T_seg = (5+1000−1) × 260 = 1004 × 260 = 261040 ns = 261.04 μs.

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A 5-stage pipeline has stage delays: 180, 250, 150, 170, 250 ns. Inter-stage latch delay = 10 ns. No stalls. Time to process 1000 instructions (μs)?

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