Suppose that a given application is run on a 64-processor machine and that 70…

Concept

Amdahl’s law gives the overall speed-up of a program when only part of it can be run in parallel. If a fraction P of the total work is parallelisable and the remaining (1 − P) must stay serial, then on N processors the overall speed-up is S = 1 / ((1 − P) + P/N). Because the serial share never shrinks, it alone fixes the ceiling on what extra hardware can ever buy.

Applying it here

1. Read off the two given quantities: the parallelisable fraction P = 70% = 0.70, and the processor count N = 64.

2. The part that cannot be parallelised keeps its full cost: 1 − P = 1 − 0.70 = 0.30.

3. The parallelisable part is spread over all 64 processors, so its cost falls to P/N = 0.70 / 64 = 0.0109375.

4. Add the two to get the normalised execution time on 64 processors: (1 − P) + P/N = 0.30 + 0.0109375 = 0.3109375.

5. Invert that total for the speed-up: S = 1 / 0.3109375 = 640/199 ≈ 3.216, which to two decimal places is 3.22.

Cross-check

Two independent checks confirm the arithmetic. First, substitute back: 3.216 × 0.3109375 ≈ 1.000, so the reciprocal was taken correctly. Second, let N grow without bound: P/N → 0 and S → 1 / (1 − P) = 1 / 0.30 ≈ 3.33. That is the absolute ceiling imposed by a 30% serial share, so any 64-processor result must sit just below it — and 3.216 does.

The table also shows the diminishing return: moving from 32 to 64 processors adds only about 0.11 to the speed-up, because the fixed 30% serial share now dominates the run.

So the expected performance improvement on the 64-processor machine is a speed-up of about 3.22.