Question

Assume a program requires the execution of$\mathbf{50}\mathbf{\times }\mathbf{106}$FP instructions, $\mathbf{110}\mathbf{\times }\mathbf{106}$INT instructions,$\mathbf{80}\mathbf{\times }\mathbf{106}$L/S instructions, and$\mathbf{16}\mathbf{\times }\mathbf{106}$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:

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

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

1.14.3 By how much is the execution time of the program improved if the CPI of INT and FP instructions is reduced byand the CPI of L/S and Branch is reduced by?

Short answer

1.14.1

CPI is negative so it’s not possible

1.14.2

CPI of L/S instructions must improve by 5 times

1.14.3

Time improvement times

Step-by-step solution

Determine the formulae

Write the formula to calculate the execution time

$Executiontime=\sum \frac{Clockcycles}{clockrate}$ …… (1)

Write the formula to determine the CPI

$CPI=\frac{Executiontime\times clockrate}{Instructioncount}$ …… (2)

Write the formula to determine the clock cycles

$Clockcycle=Instructioncount\times CPI$ ……. (3)

Now, we want the program to run two times faster, so we have to find a new CPI

1.14.1

Given,

Total FP Instructions =$50\times {10}^6$

INT Instruction =$110\times {10}^6$

L/S Instructions =$80\times {10}^6$

Branch Instructions =$16\times {10}^6$

Each processor clock rate is=2Ghz

Now, calculate Execution time

$Executiontime=\sum \frac{Clockcycles}{clockrate}$

For FP (Floating point) instructions, Initial Execution time

localid="1655307257244" $$\begin{gathered} =\frac{\left(50\times {10}^6\times 1\right)+\left(110\times {10}^6\times 1\right)+\left(80\times {10}^6\times 4\right)+\left(16\times {10}^6\times 2\right)}{2\times {10}^9} \\ =\frac{512\times {10}^6}{2\times {10}^9} \end{gathered}$$

$=256\times {10}^{-3}$

For 16 processors

Clock cycle$=512\times {10}^6$

Execution time=$256\times {10}^{-3}$

Half the number of clock cycles to improve the CPI of FP instructions,

$$\begin{gathered} CP{I}_{FPimproved}=\frac{{\displaystyle \frac{512\times {10}^6}{2}}-\left[\left(110\times {10}^6\times 1\right)+\left(80\times {10}^6\times 4\right)+\left(16\times {10}^6\times 2\right)\right]}{50\times {10}^6} \\ =\frac{-206\times {10}^6}{50\times {10}^6} \\ =-4.12<0 \end{gathered}$$

We cannot improve CPI of FP instructions when we run the program two times faster because it would be negative

Determine the New CPI

Half the number of clock cycles to improve the CPI of LS instructions:

Given No. of L/S instruction =

$$\begin{gathered} CP{I}_{L/Simproved}=\frac{{\displaystyle \frac{512\times {10}^6}{2}-\left[\left(50\times {10}^6\times 1\right)+\left(110\times {10}^6\times 4\right)+\left(16\times {10}^6\times 2\right)\right]}}{80\times {10}^6} \\ =\frac{64\times {10}^6}{80\times {10}^6} \\ =0.8<0 \end{gathered}$$

For the program to run two times faster, we must improve the CPI of L/S instruction as,

Given CPI of L/S instruction= 4

$\frac{4}{0.8}=5$

Therefore, CPI of L/S instructions must improve by 5 times

Determine the execution time of the program if CPI is reduced

Calculate CPI for floating point (FP), Integer (INT), Load/Store (L/S), and Branch instructions.

Reduce 40% on Floating point,

$$\begin{gathered} CP{I}_{FP}=1-1\times 0.4 \\ =0.6 \end{gathered}$$

Reduce 40% on INT,

$$\begin{gathered} CP{I}_{INT}=1-1\times 0.4 \\ =0.6 \end{gathered}$$

Reduce 30% on load Load/Store,

$$\begin{gathered} CP{I}_{L/S}=4-4\times 0.3 \\ =2.8 \end{gathered}$$

Now calculate clock cycles,

$clockcycle=InstructionCount\times CPI$

$$\begin{gathered} Clockcycle=\left(50\times {10}^6\times 0.6\right)+\left(110\times {10}^6\times 0.6\right)+\left(80\times {10}^6\times 2.8\right)+\left(16\times {10}^6\times 1.4\right) \\ =342.4\times {10}^6 \end{gathered}$$

Initial Execution time is,

$$\begin{gathered} =\frac{342.4\times {10}^6}{2\times {10}^9} \\ =171.2\times {10}^{-3} \end{gathered}$$

The performances for 16 processors are,

Old execution time $=256\times {10}^{-3}$sec

Execution time after reducing CPI =$171.2\times {10}^{-3}$sec

Improving execution time of program$$\begin{gathered} =\frac{0.256}{0.171} \\ =1.497times \end{gathered}$$

times

Time improvement 1.497 times