Compilers can have a profound impact on the performance of an applicat

Step 1
a) Remember that
${\displaystyle \text{CPU time}=\text{instructions}\times \text{CPU}\times \text{cycle time,}}$
thus
${\displaystyle \text{CPI}=\frac{\text{COU time}}{\text{instructions}\times \text{cycle time}}}$
Here cycle time ${\displaystyle =1ns={10}^{-9}s}$. Additionally, execution time in this context ${\displaystyle =\text{CPU time}}$. So, for complier A, we have that
${\displaystyle {\text{CPI}}_A=\frac{{\text{CPU time}}_A}{{\text{instructions}}_A\times \text{cycle time}}=\frac{1.1s}{{10}^9\times {10}^{-9}s}=1.1}$
For compiler B, we have
${\displaystyle {\text{CPI}}_B=\frac{{\text{CPU time}}_B}{{\text{instructions}}_B\times \text{cycle time}}=\frac{1.5s}{1.2\times {10}^9\times {10}^{-9}s}=1.25}$
Step 2
b) Recall that
$\text{execution time}=\text{CPU time}=\frac{\text{instructions}\times \text{CPI}}{\text{clock rate}}$
So, since execution times in both cases are the same,
${\displaystyle {\text{execution time}}_1={\text{execution time}}_2}$
which yields
${\displaystyle \frac{{\text{instructions}}_1\times {\text{CPI}}_1}{{\text{clock rate}}_1}=\frac{{\text{instructions}}_2\times {\text{CPI}}_2}{{\text{clock rate}}_2}}$
Therefore, after rearranging,
${\displaystyle {\text{clock rate}}_1=\frac{{\text{instructions}}_1\times {\text{CPI}}_1}{{\text{instructions}}_2\times {\text{CPI}}_2}\times {\text{clock rate}}_2=\frac{{10}^9\times 1.1}{1.2\times {10}^9\times 1.25}\times {\text{clock rate}}_2}$
So,
${\displaystyle {\text{clock rate}}_1=0.73{\text{clock rate}}_2}$
So, the clock rate of processor 1 is actually approximately ${\displaystyle 27\mathrm{\%}}$ slower than the clock rate of processor 2.
Step 3
c) We use the same formula as in (a):
${\displaystyle {\text{CPU time}}_C={\text{instructions}}_C\times {\text{CPI}}_C\times \text{cycle time}=6\times {10}^8\times 1.1\times {10}^{-9}s=0.66s}$
So,
${\displaystyle \frac{{\text{performance}}_C}{{\text{performance}}_A}=\frac{{\text{CPU time}}_A}{{\text{CPU time}}_C}=\frac{1.1s}{0.66s}=1.67}$
${\displaystyle \frac{{\text{performance}}_C}{{\text{performance}}_A}=\frac{{\text{CPU time}}_A}{{\text{CPU time}}_C}=\frac{1.5s}{0.66s}=2.27}$
So, C is faster than A approximately 1.67 times, and faster than B approximately 2.27 times.

Step 1
Given
Cycle Time ${\displaystyle =1ns={10}^{-9}}$
CPU Time for A ${\displaystyle =1.1s}$
Instruction count ${\displaystyle ={10}^9}$
CPU Time for B ${\displaystyle =1.5s}$
Instruction Count ${\displaystyle =1.2\times {10}^9}$
${\displaystyle \text{CPU Time}=\text{Instructions}\times \text{Cycle Time}\times \text{CPI}}$
So, ${\displaystyle \text{CPI}=\frac{\text{CPU Time}}{\text{Instructions}\times \text{Cycle Time}}}$
CPI for ${\displaystyle A=\frac{1.1}{{10}^9\times {10}^{-9}}}$
${\displaystyle \text{CPI}=1.1}$
CPI for ${\displaystyle B=\frac{1.5}{1.2\times {10}^9\times {10}^{-9}}}$
${\displaystyle \text{CPI}=\frac{1.5}{12}}$
${\displaystyle \text{CPI}=1.25}$
Step 2
Given
${\displaystyle \text{Execution Time on A}=\text{Execution Time on B}}$
${\displaystyle \text{CPU Time for A}=1.1s}$
${\displaystyle \text{Instruction count}={10}^9}$
${\displaystyle \text{CPI}=1.1}$
CPU Time for ${\displaystyle B=1.5s}$
Instruction Count ${\displaystyle =1.2\times {10}^9}$
${\displaystyle \text{CPI}=1.25}$
${\displaystyle \text{Execution Time}=\text{Instructions}\times \frac{\text{CPI}}{\text{Clock Rate}}}$
${\displaystyle \text{Execution Time A}=\text{Execution Time B}}$
${\displaystyle {\text{Instruction}}_A\times \frac{{\text{CPI}}_A}{{\text{Clock Rate}}_A}={\text{Instruction}}_B\times \frac{{\text{CPI}}_B}{{\text{Clock Rate}}_B}}$.
Make Clock Rate A the subject of formula
${\displaystyle {\text{Clock Rate}}_A=\frac{{\text{Instruction}}_A\times {\text{CPI}}_A\times {\text{Clock Rate}}_B}{{\text{Instruction}}_B\times {\text{CPI}}_B}}$
${\displaystyle {\text{Clock Rate}}_A=\frac{{10}^9\times 1.1\times {\text{Clock Rate}}_B}{1.2\times {10}^9\times 1.25}}$
${\displaystyle {\text{Clock Rate}}_A=0.73{\text{Clock Rate}}_B}$
So the clock rate of A is ${\displaystyle 27\mathrm{\%}}$ slower than clock rate of B
Step 3
For compiler A and C,
${\displaystyle \text{CPU Time}=\text{Instructions}\times \text{Cycle Time}\text{CPI}}$
${\displaystyle \text{CPU Time of C}=6\times {10}^8\times 1.1\times {10}^{-9}=0.66s}$
Using ratio of CPU time and Compiler performance i,e.
${\displaystyle \text{CPU Time of A}\times \text{Performance of A}=\text{CPU Time of C}\times \text{Performance of C}}$
Given
${\displaystyle \text{CPU Time of A}=1.1}$
${\displaystyle \text{CPU Time of C}=0.66}$
${\displaystyle 1.1\times \text{Performance of A}=0.66\times \text{Performance of C}}$