A given application written in Java runs 20 seconds on a des

A given application written in Java runs 20 seconds on a desktop processor A new Java compiler is released that requires only 0.6 as many instructions as the old compiler Bu. It increases CPI by 1.2. How fast can we expect the application to run using this new compiler? Assume that a processors P has a 3 GHz clock rate and a CPI of 1.5. and executes a program in 10 seconds Find the number of cycles and the number of instructions. We are trying to reduce the execution time by 20% but this leads to an increase of 14% in the CPI What clock rate should we have to get this time reduction?

Solution

Answer

1.

20 seconds = Ins_A * CPI_A * ClockRate

ClockRate = 20 seconds / (Ins_A * CPI_A)

Time_B = (0.6 * Ins_A) * (1.2 * CPI_A) * 20 seconds / (Ins_A * CPI_A)

= 0.6 * 1.2 * 20 seconds = 14.4 seconds (Answer)

2.

1.

clock rate = 3 GHz

CPI = 1.5

Execution time = 10 seconds = clock cycles / clock rate

clock cycles = execution time * clock rate

= 10 seconds * 3 * 10⁹ cycles/seconds

= 30 * 10⁹ cycles

Number of instructions = clock cycles / CPI

= 30 * 10⁹ / 1.5

= 20 * 10⁹ instructions

2.

Reduced execution time = 10(1 - 0.2) = 10 * .8 = 8 seconds

Increased CPI = 1.5 * (1 + .14)

= 1.5 * 1.14

= 1.71

Time = instruction count * CPI * execution time

= 20 * 10⁹ instructions * 1.71 * 8 seconds

= 273.6 * 10⁹ seconds

Therefore, clock rate = 1/time

= 1/(273.6*10⁹) seconds

= 0.00365 * 10^-9

  = 0.00365 GHz