Answer:
The answer is below
Explanation:
Suppose that we find a way to double the performance of arithmetic instructions. What is the speedup of our machine? What if we find a way to improve the performance of arithmetic instructions by 10 times?
Solution:
a) number of clock cycles = (240 million * 2) + (70 million * 6) + (100 million * 3) = 1200 * 10⁶
If the performance of arithmetic instructions is doubled, the CPI of arithmetic instructions would be halved. Hence the CPI would be 1 (2 / 2). Therefore:
number of clock cycles = (240 million * 1) + (70 million * 6) + (100 million * 3) = 960 * 10⁶
Hence:
[tex]\frac{CPU\ time_c}{CPU\ time_A} =\frac{960*10^6}{1200*10^6}=0.8 \\\\[/tex]
Therefore they would be an increase of 20%
b) If the performance of arithmetic instructions is multiplied by 10, the CPI of arithmetic instructions would be halved. Hence the CPI would be 0.2 (2 / 10). Therefore:
number of clock cycles = (240 million * 0.2) + (70 million * 6) + (100 million * 3) = 768 * 10⁶
Hence:
[tex]\frac{CPU\ time_c}{CPU\ time_A} =\frac{768*10^6}{1200*10^6}=0.64 \\\\[/tex]
Therefore they would be an increase of 36%
