Editing Amdahl's law (section)

=== Serial programs ===
[[File:Optimizing-different-parts.svg|thumb|400px|Assume that a task has two independent parts, ''A'' and ''B''. Part ''B'' takes roughly 25% of the time of the whole computation. By working very hard, one may be able to make this part 5 times faster, but this reduces the time of the whole computation only slightly. In contrast, one may need to perform less work to make part ''A'' perform twice as fast. This will make the computation much faster than by optimizing part ''B'', even though part ''B'''s speedup is greater in terms of the ratio, (5 times versus 2 times).]]

For example, with a serial program in two parts ''A'' and ''B'' for which {{nowrap|1=''T''<sub>''A''</sub> = 3 s}} and {{nowrap|1=''T''<sub>''B''</sub> = 1 s}},

* if part ''B'' is made to run 5 times faster, that is {{nowrap|1=''s'' = 5}} and {{nowrap|1=''p'' = ''T''<sub>''B''</sub>/(''T''<sub>''A''</sub> + ''T''<sub>''B''</sub>) = 0.25}}, then <math display="block">S_\text{latency} = \frac 1 {1 - 0.25 + \frac{0.25}{5}} = 1.25;</math>
*if part ''A'' is made to run 2 times faster, that is {{nowrap|1=''s'' = 2}} and {{nowrap|1=''p'' = ''T''<sub>''A''</sub>/(''T''<sub>''A''</sub> + ''T''<sub>''B''</sub>) = 0.75}}, then <math display="block">S_\text{latency} = \frac 1 {1 - 0.75 + \frac{0.75}{2}} = 1.60.</math>

Therefore, making part ''A'' to run 2 times faster is better than making part ''B'' to run 5 times faster. The percentage improvement in speed can be calculated as
: <math>\text{percentage improvement} = 100 \left(1 - \frac 1 {S_\text{latency}}\right).</math>

* Improving part ''A'' by a factor of 2 will increase overall program speed by a factor of 1.60, which makes it 37.5% faster than the original computation.
* However, improving part ''B'' by a factor of 5, which presumably requires more effort, will achieve an overall speedup factor of 1.25 only, which makes it 20% faster.