Respuesta :
Answer:
Following are the solution to the given question:
Explanation:
In point 1:
I = number of program code completed.
In such a block of the system memory, the number of iterations to read would be 64/W, while W is the length of its band.
Miss penalty (cycling):
Miss penalty is written for the data cache:
[tex]\to I \times \frac{100}{1000}\times 0.02(\frac{64}{W} + 1)[/tex]
Miss penalty for the cache instruction:
[tex]\to I \times 0.003 \times (\frac{64}{W} + 1)[/tex]
The [tex]CPI \leq 2[/tex], Hit time + miss penalty Hit time:
[tex]= I + I \times [0.25 \times 0.02 + 0.1 \times 0.02 + 0.003] \times (\frac{64}{W} + 1) \leq I \times 2.[/tex]
[tex]W \geq \frac{64}{99} \` \ 0.646 \frac{Byte}{cycle}.[/tex]
In point 2:
Miss penalty read data cache:
[tex]\to I \times 0.25 \times 0.02 \times (1 + 0.3) \times (\frac{64}{W} + 1)[/tex]
Miss penalty for the data cache:
[tex]\to I \times 0.1 \times 0.02 \times(1 + 0.3) \times (\frac{64}{W} + 1)[/tex]
Read the penalty for miss:
[tex]\to I \times 0.003 \times (\frac{64}{W} + 1)[/tex]
Total runtime = time hit + penalty mismatch
[tex]= I + I \times [0.25 \times 0.02 \times 1.3 + 0.1 \times 0.02\times 1.3 + 0.003]\times (\frac{64}{W} + 1) \leq 2I[/tex]
[tex]0.0121 \times (\frac{64}{W} + 1) \leq 1 \\\\ W \geq 0.784 \frac{Byte}[cycle}[/tex]
