On one processor we have

$$ T_1 = (1-f)W + fW = W $$

On \( p \) processors we have

$$ T_p = (1-f)W + \frac{fW}{p}, $$

resulting in a speedup of

$$ \frac{T_1}{T_p} = \frac{W}{(1-f)W+fW/p} $$

As \( p \) goes to infinity, \( fW/p \) goes to zero, and the maximum speedup is

$$ \frac{1}{1-f}, $$

meaning that if if \( f = 0.99 \) (all but \( 1\% \) parallelizable), the maximum speedup is \( 1/(1-.99)=100 \)!