Question

Assume a program requires the execution of 50 × 10⁶ FP instructions, 110 × 10⁶ INT instructions, 80 × 10⁶ L/S instructions, and 16 × 10⁶ branch instructions. The CPI for each type of instruction is 1, 1, 4, and 2, respectively. Assume that the processor has a 2 GHz clock rate.

a. Find the execution time?

b. By how much must we improve the CPI of FP instructions if we want the program to run two times faster?

c. By how much must we improve the CPI of L/S instructions if we want the program to run two times faster?

d. By how much is the execution time of the program improved if the CPI of INT and FP instructions is reduced by 40% and the CPI of L/S and Branch is reduced by 30%?

Answer 1

Answer a

Execution time = Total Clock Cycles / Clock Rate

Total Clock cycles = (50 x 10⁶ x 1) + (110 x 10⁶ x 1) + (80 x 10⁶ x 4) + (16 x 10⁶ x 2) = 512 x 10⁶

so, execution time = 512 x 10⁶ / 2 x 10⁹ Hz = 256 ms   

Answer b

Now, we want to execute program as 2 times faster by improving the clock cycle for FP instruction

so, assuming FP instruction takes P CPI

Total clock cycles   (50 x 10⁶ x P) + (110 x 10⁶​​​​​​​ x 1) + (80 x 10⁶​​​​​​​ x 4) + (16 x 10⁶​​​​​​​ x 2) = (462 + 50P) x 10⁶

  Execution time = 256/2 = 128 ms

So, 128 x 10^-3 s = (462 + 50P) x 10⁶  / 2 x 10⁹ Hz

462+50P = (128 x 10^-3) (2 x 10³) = 256

P = - 4.12

New CPI is in negative which means it is not possible to execute program two times faster by Improving the clock cycle of FP instructions

Answer c

Now, we want to execute program 2 times faster by improving the clock cycle for L/S instruction

so, assuming L/S instruction takes P CPI

Total clock cycles   (50 x 10⁶ x 1) + (110 x 10⁶​​​​​​​ x 1) + (80 x 10⁶​​​​​​​ x P) + (16 x 10⁶​​​​​​​ x 2) = (192 + 80P) x 10⁶

  Execution time = 256/2 = 128 ms

So, 128 x 10^-3 s = (192 + 80P) x 10⁶  / 2 x 10⁹ Hz

192+80P = (128 x 10^-3) (2 x 10³) = 256

P = 0.8

Answer d

(Execution time)_new = Total Clock Cycles / Clock Rate

Total Clock cycles = (50 x 10⁶ x 0.6) + (110 x 10⁶ x 0.6) + (80 x 10⁶ x 2.8) + (16 x 10⁶​​​​​​​ x 1.4) = 342.4 x 10⁶

so, (Execution time)_new = 342.4 x 10⁶ / 2 x 10⁹ Hz = 171.2 ms   

Speedup = Execution Time / (Execution time)_new = 256/171.2 = 1.495