Answer:
a) [tex]T=4*10^{-4}μs[/tex]
b) [tex]N=2.5*10^6 cycles[/tex]
c) 10000 programs.
Explanation:
a) We know that the frequency is the inverse of the period, so:
[tex]f=\frac{1}{T}\\\\T=\frac{1}{f}\\T=4*10^{-10}s[/tex]
1μs is equal to [tex]1*10^{-6}s[/tex]
so [tex]T=4*10^{-4}us[/tex]
b) If in a second there are 2.5*10^9 cycles:
[tex]N=2.5*10^9*(1*10^{-3})=2.5*10^6 cycles[/tex]
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:
[tex]P=\frac{1s}{1*10^{-4}s}=10000[/tex]
10000 programs.