Question

When a program is run on Computer X, 50% of the execution time is CPU time . A better Computer Y reduces the execution time by 10%. It is know that Computer Y has a clock rate of 2.3 GHz, and it takes Computer Y 20% more clock cycles to execute the program. In addition, Computer Y can only reduce CPU time. What is the clock rate in GHz of Computer X?

Answer 1

First let the total execution time taken by computer x be 2t seconds

Therefore, the cpu time for computer x = 2t/2 = t seconds

Also total time of execution on computer y is 10% less than of x, so it is = 90% of 2t = 1.8t seconds

Now the reduction in total time = 2t - 1.8t = 0.2t seconds

Since reduction is only possible in cpu time, therefore cpu time for computer y = t - 0.2t = 0.8t

As computer y took 20% more cycles than x, therefore, cycles of y = 120% of cycles of x

Assuming clock speed of computer x as 'Cx' we have from above relation : 2.3 * 0.8t = 120% * Cx * t

This gives, Cx = 1.53 GHz