Clock Frequency CO


#1

A 1.2GHz clock frequency processor takes 8ms to execute a benchmark program. A new processor design is required to finish the same benchmark program in 3ms. What will be the clock frequency of the new processor, if the number of clock cycles required for finishing the program increases by 1.5 times.

Solution

1/(1.2 x 10^9) seconds -> 1 clock cycle
1 second -> 1.2 x 10^9 clock cycles
8 ms -> 8 x 10^-3 x 1.2 x 10^9 = 9.6 x 10^6 clock cycles

Now it is said that the number of clock cycles increases by 1.5 times

= 1.5 x 9.6 x 10^6 clock cycles

= 14.4 x 10^6 clock cycles

So it is said that the same benchmark program would be finished in 3 ms, i.e 14.4 x 10^6 clock cycles would be finishing in 3 ms.

14.4 x 10^6 clock cycles -> 3 ms
1 clock cycle -> (3 x 10^-3)/(14.4 x 10^6) seconds = 1/(4.8 x 10^9) seconds = 4.8 GHz