Pipeline Speedup Calculator

A pipeline is clocked by its slowest stage: these five stages give a 170 ns clock and a speedup of 3.959276018 for 100 instructions.

Stage delays (ns):

One delay per stage.

Latch (buffer) delay per stage (ns):

Number of instructions n:

Average stall cycles per instruction:

Speedup =

3.95927601809955

Precision

Full answer, then cut to fewer decimals (not rounded):

3.95927601809955
3.9592760180995
3.959276018099
3.95927601809
3.9592760180
3.959276018
3.95927601
3.9592760
3.959276
3.95927
3.9592
3.959
3.95
3.9

Pipeline clock = max stage + latch: 170 ns
Non-pipelined time per instruction: 700 ns
Pipelined time for 100 instructions: 17680 ns
Non-pipelined time for 100 instructions: 70000 ns
Efficiency = speedup ÷ k: 0.7918552036
Throughput: 5.656108597 instructions/µs
Ideal speedup as n → ∞: 4.117647059

Step-by-step solution

Clock = 160 + 10 = 170 ns
Pipelined time = (k + n − 1 + stalls) × clock = (5 + 100 − 1 + 0) × 170 = 17680 ns
Speedup = 70000 ÷ 17680 = 3.959276018

Pipeline Speedup Formula

T = (k + n − 1) × clock, clock = max stage + latch

speedup = n × Σ stages ÷ T

Worked Example

Five stages of 150, 120, 160, 140 and 130 ns with 10 ns latches, 100 instructions:

Clock = 160 + 10 = 170 ns
Pipelined time = (k + n − 1 + stalls) × clock = (5 + 100 − 1 + 0) × 170 = 17680 ns
Speedup = 70000 ÷ 17680 = 3.959276018

Non-pipelined time per instruction: 700 ns · Pipelined time for 100 instructions: 17680 ns · Non-pipelined time for 100 instructions: 70000 ns · Efficiency = speedup ÷ k: 0.7918552036 · Throughput: 5.656108597 instructions/µs · Ideal speedup as n → ∞: 4.117647059

GATE tip

For large n, speedup → (non-pipelined time per instruction) ÷ (pipeline clock); GATE often asks for this limit rather than the exact value.

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S = 1 ÷ [(1 − p) + p/s]

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Example: A 1 ns cache with a 95 % hit rate in front of 100 ns memory: Average memory access time = 6 ns.

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Frequently asked questions

Why is pipeline speedup less than the number of stages?

Unequal stage delays, latch overhead, the fill time (k − 1 cycles) and stalls all reduce it below k.

How do stalls affect speedup?

With an average of s stall cycles per instruction, CPI becomes 1 + s and the ideal speedup divides by 1 + s.

Sources

Last reviewed: 2026-10-15. Factors are stored exactly and computed with 50-digit decimal arithmetic. See how we verify conversion factors.

Cite this page

APA

Scientific Converters. (2026, October 15). Pipeline Speedup Calculator. Retrieved October 25, 2026, from https://scientificconverters.com/calculators/pipeline-speedup-calculator/

IEEE

Scientific Converters, "Pipeline Speedup Calculator," scientificconverters.com. Accessed: Oct. 25, 2026. [Online]. Available: https://scientificconverters.com/calculators/pipeline-speedup-calculator/