# Amdahl's Law when the speedup of new component = ∞

I was doing a problem about Amdahl's Law when I noticed that $s,(local$ $speed-up)$ was incalculable. The wording:

A computer takes 100 seconds to execute a program. The multiplication operations take the 80% of the time. By how much we have to improve the speed of the multiplier in such a way that the program runs 5 times faster?

So I applied the law:

$5 = \frac{1}{\frac{0.8}{Sm}+0.2}$

$\frac{0.8}{Sm}=0$

What does this mean? Am I doing something wrong?

• Looks good to me. Before any speedup, multiplication takes 80 seconds, leaving 20 seconds for everything else. 5 times as fast as 100 seconds is 20 seconds total. So after the speedup the multiplication would have to take 0 seconds in order to be that fast. – cardobard_box Oct 14 '17 at 18:18
• I understand it now! Thanks you so much – Antonio M. Burgueño Oct 14 '17 at 22:14