One intuitive way to think of why this formula holds is that the fraction of work done sequentially is (1-f) (the fraction that is not parallelizable) plus f/n (each processor's share of the parallelizable work). The inverse of this gives the speedup, 1/((1-f) + f/n).
One intuitive way to think of why this formula holds is that the fraction of work done sequentially is (1-f) (the fraction that is not parallelizable) plus f/n (each processor's share of the parallelizable work). The inverse of this gives the speedup, 1/((1-f) + f/n).
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