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Amdahl's law
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== Definition == In the context of Amdahl's law, speedup can be defined as:<ref name=":1" /> <math>\text{Speedup} = \frac{\text{Performance for the entire task when enhancements are applied}}{\text{Performance for the same task without those enhancements}}</math> or <math>\text{Speedup} = \frac{\text{Execution time for the entire task without enhancements}}{\text{Execution time for the same task when enhancements are applied}}</math> Amdahl's law can be formulated in the following way:<ref name=":0">{{Citation |last=Bakos |first=Jason D. |title=Chapter 2 - Multicore and data-level optimization: OpenMP and SIMD |date=2016-01-01 |work=Embedded Systems |pages=49β103 |editor-last=Bakos |editor-first=Jason D. |url=https://linkinghub.elsevier.com/retrieve/pii/B978012800342800002X |access-date=2024-11-18 |place=Boston |publisher=Morgan Kaufmann |doi=10.1016/b978-0-12-800342-8.00002-x |isbn=978-0-12-800342-8}}</ref> : <math>\text{Speedup}_\text{overall} = \frac{1}{(1 - \text{time}_{\text{optimized}}) + \frac{\text{time}_{\text{optimized}}}{\text{speedup}_{\text{optimized}}}} </math> where * <math>\text{Speedup}_\text{overall}</math> represents the total speedup of a program * <math>\text{Time}_{\text{optimized}}</math> represents the proportion of time spent on the portion of the code where improvements are made * <math>\text{Speedup}_{\text{optimized}}</math> represents the extent of the improvement The <math>\text{Speedup}_\text{overall} </math> is frequently much lower than one might expect. For instance, if a programmer enhances a part of the code that represents 10% of the total execution time (i.e. <math>\text{Time}_{\text{optimized}} </math> of 0.10) and achieves a <math>\text{Speedup}_{\text{optimized}} </math> of 10,000, then <math>\text{Speedup}_\text{overall} </math> becomes 1.11 which means only 11% improvement in total speedup of the program. So, despite a massive improvement in one section, the overall benefit is quite small. In another example, if the programmer optimizes a section that accounts for 99% of the execution time (i.e. <math>\text{Time}_{\text{optimized}}</math> of 0.99) with a speedup factor of 100 (i.e. <math>\text{Speedup}_{\text{optimized}}</math>of 100), the <math>\text{Speedup}_\text{overall}</math> only reaches 50. This indicates that half of the potential performance gain (<math>\text{Speedup}_\text{overall}</math> will reach 100 if 100% of the execution time is covered) is lost due to the remaining 1% of execution time that was not improved.<ref name=":0" />
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