Question

Consider two different implementations of the same instruction set architecture. The instructions can be divided into four classes according to their CPI(class A, B, C, and D). P1 with a clock rate of 2.5 GHz and CPIs of 1,2,3, and 3, and P2 with a clock rate of 3 GHz and CPIs of 2,2,2, and 2.

Given a program with a dynamic instruction count of 1.0E6 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D, which is faster?

a)What is the global CPI for each implementation?

b) Find the clock cycles required in both cases.

Short answer

P2 is faster

a) Global CPI of P1=2.6

Global CPI of P2=2.0

b) Clock cycles P1=$26\times {10}^5$

Clock cycles P2=$20\times {10}^5$

Step-by-step solution

Determine the formulae.

Write the formula to calculate the execution time

$Executiontime=\frac{NumberofInstructions\times CPI}{ClockRate}$ ….. (1)

Write the formula to determine the CPI

$CPI=\frac{ExecutionTime\times Clockrate}{Instructioncount}$ ….. (2)

Write the formula to determine the clock cycles

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

Determine the faster implementation

To find which implementation works faster, the Execution time of both the implementations can be compared. To find the execution time of P1 and P2, the following formula will be used.

$Executiontime=\frac{NumberofInstructions\times CPI}{Clockrate}$

Given instruction count for the classes are:

$$\begin{gathered} ClassA={10}^5 \\ ClassB=2\times {10}^5 \\ ClassC=5\times {10}^5 \\ ClassD=2\times {10}^5 \end{gathered}$$

The execution time of P1:

$$\begin{gathered} ExecutiontimeofP1=\frac{\left({10}^5+2\times {10}^5\times 2+5\times {10}^5\times 3+2\times {10}^5\times 3\right)}{2.5\times {10}^9} \\ =10.4\times {10}^{-4}s \end{gathered}$$

The execution time of P2:

$$\begin{gathered} ExecutiontimeofP2=\frac{\left({10}^5\times 2+2\times {10}^5\times 2+5\times {10}^5\times 2+2\times {10}^5\times 2\right)}{3\times {10}^9} \\ =6.66\times {10}^{-4}s \end{gathered}$$

By comparing the execution time of P1 and P2, P2 has the lesser execution time, So P2 is faster.

Determine the global CPI for P1 and P2

a)

The global CPI can be found by using the following formula:

$CPI=\frac{Executiontime\times Clockrate}{InstructionCount}$

The global CPI for P1:

$$\begin{gathered} CPIofP1=\frac{10.4\times {10}^{-4}\times 2.5\times {10}^9}{{10}^6} \\ =2.6 \end{gathered}$$

The global CPI of P2:

$$\begin{gathered} CPIforP2=\frac{6.66\times {10}^{-4}\times 3\times {10}^9}{{10}^6} \\ =2.0 \end{gathered}$$

Determine the clock cycles of P1 and P2

$ClockCycles=Instructioncount\times CPI$

Clock Cycle of P1:

$$\begin{gathered} ClockcyclesofP1={10}^5\times 1+2\times {10}^5\times 2+5\times {10}^5\times 3+2\times {10}^5\times 3 \\ =26\times {10}^5 \end{gathered}$$

Clock Cycle of P2:

$$\begin{gathered} ClockCyclesofP2={10}^5\times 2+2\times {10}^5\times 2+5\times {10}^5\times 2+2\times {10}^5\times 2 \\ =20\times {10}^5 \end{gathered}$$