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Amdahl’s & Gustafson’s Law Speedup Calculator

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Understanding the Amdahl's and Gustafson's law speedup calculator — concept and principle

When you spread a program over more processors, how much faster does it get? Amdahl's law answers for a fixed problem size; Gustafson's law answers when the problem grows with the machine. This calculator gives both speedups, the theoretical ceiling with unlimited cores and the parallel efficiency, so you can judge whether adding cores is worthwhile.

Amdahl’s law with 90% parallel workWith 90% parallel work, 16 cores cut the time from 1 to 0.156 (6.4×), and even infinite cores cannot beat the 0.1 serial part (10×).The serial 10% never gets faster1 coretime 1.0parallel 0.90.116 corestime 0.1 + 0.9 ÷ 16 = 0.156→ 6.4× faster∞ corestime 0.1→ 10× at mostGustafson: grow the job with the cores16 − 0.1 × 15 = 14.5×Speedup 6.4× on 16 coresefficiency 6.4 ÷ 16 = 40%
Only the blue part shrinks with more cores; the pink serial part sets the 10× ceiling.

How it is calculated

Let P be the parallel fraction (as a decimal) and N the number of processors.

P is measured on the single-processor run: the share of its time that can run in parallel.

The principle

Amdahl's law (1967) splits the original run time into a serial part (1 − P) that no extra processor can shorten and a parallel part P that divides by N. As N grows, the parallel part shrinks towards zero but the serial part stays, so speedup approaches 1 ÷ (1 − P). Even 5% serial code caps speedup at 20×. Gustafson (1988) observed that people usually use bigger machines to solve bigger problems in the same time. If the serial part stays fixed while the parallel work scales with N, the scaled speedup grows almost linearly. Both laws are correct; they answer different questions (strong scaling versus weak scaling).

Worked example

  1. Inputs: P = 90% = 0.9, N = 16.
  2. Amdahl: 0.9 ÷ 16 = 0.05625; 0.1 + 0.05625 = 0.15625; 1 ÷ 0.15625 = 6.4×.
  3. Ceiling: 1 ÷ 0.1 = 10×.
  4. Gustafson: 16 − 0.1 × 15 = 16 − 1.5 = 14.5×.
  5. Efficiency: 6.4 ÷ 16 = 40%.
How Amdahl speedup changes with Parallel fraction (%)How Amdahl speedup changes with Parallel…-20-15-10-5051022.54567.590112135180Example: 6.4Parallel fraction (%)
Each point is the calculator re-run with only “Parallel fraction (%)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
Parallel fraction (%)Amdahl speedup
22.51.267326733
451.72972973
67.52.723404255
906.4
112-18.28571429
135-3.764705882
180-1.454545455

How to read the result

Efficiency shows how much of the hardware you are actually using. At 40%, more than half of the 16 cores' capacity is lost to the serial part, so cutting serial work helps more than adding cores. Compare the Amdahl speedup with the ceiling: if you are already near it, more cores give little.

Common mistakes

Keep in mind: Both laws are idealised. Memory bandwidth, cache effects, load imbalance and communication overhead can make real speedups lower, or occasionally higher (superlinear cache effects).

Frequently asked questions

What is the difference between strong and weak scaling?

Strong scaling keeps the problem size fixed (Amdahl); weak scaling grows the problem with the processors (Gustafson).

How can I estimate my parallel fraction?

Measure speedups at a few core counts and solve Amdahl's formula for P, which is the idea behind the Karp–Flatt metric.

Does Amdahl's law apply to GPUs?

Yes. Data transfer and serial host code act as the serial fraction and limit overall speedup.

Sources: G. M. Amdahl, "Validity of the single processor approach to achieving large scale computing capabilities", AFIPS Conference Proceedings 30, 1967 · J. L. Gustafson, "Reevaluating Amdahl's Law", Communications of the ACM 31(5), 1988

Checked 8 Nov 2026. Standard formulas and RFC/IEEE definitions, computed locally in your browser — nothing you type is sent anywhere.

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