Question

Assume your computer clock runs at the speed of 2.5 GHz (G=10^9).

Answer 1

Answer:

a) $T=4*10^{-4}μs$

b) $N=2.5*10^6 cycles$

c) 10000 programs.

Explanation:

a) We know that the frequency is the inverse of the period, so:

$f=\frac{1}{T}\\\\T=\frac{1}{f}\\T=4*10^{-10}s$

1μs is equal to $1*10^{-6}s$

so $T=4*10^{-4}us$

b) If in a second there are 2.5*10^9 cycles:

$N=2.5*10^9*(1*10^{-3})=2.5*10^6 cycles$

c) we have to make a conversion, we know that a program takes 100*10^(-3) milliseconds, that is, 1*10^(-4) seconds so in 1 second we can execute:

$P=\frac{1s}{1*10^{-4}s}=10000$

10000 programs.