Question

Compilers can have a profound impact on the performance of an application. Assume that for a program., compiler A results in a dynamic instruction count of 1.0E9 and has an execution time of 1.1 s, while compiler B results in a dynamic instruction count of 1.2E9 and an execution time of 1.5 s.

a) Find the average CPI for each program given that the processor has a clock cycle time of 1 ns.

b) Assume the compiled programs run on two different processors. If the execution times on the two processors are the same, how much faster is the clock of the processor running compiler A’s code versus the clock of the processor running compiler B’s code?

c) A new compiler is developed that uses only 6.0E8 instructions and has an average CPI of 1.1. What is the speedup of using this new compiler versus using compiler A or B on the original processor?

Short answer

a) CPI of A=1.1

CPI of B=1.25

b) The clock of the processor running compiler A’s code is 1.37 times faster.

c) The speedup of using this new compiler versus using compiler A or B on the original processor A and B are 1.67 and 2.27 respectively

Step-by-step solution

Determine the formulae

Write the formula to calculate CPI

$CPI=\frac{ExecutionTime\times Clockcycle}{Instructioncount}$

…..(1)

Write the formula to determine the Clock cycle

$Clockcycle=Instructioncount\times CPI$

…..(2)

Determine the average CPI

a. To find the average CPI for each program, consider that the processor has a 1ns clock cycle time.

$CPI=\frac{ExecutionTime\times Clockcycle}{Instructioncount}$

$CPIofA=\frac{1.1\times 1}{1.0\times E9}$

$CPIofB=\frac{1.5\times 1}{1.2E9}$

Determine how much faster compiler A is

b.To find how much faster A is, let us consider the compiled programs run on different processors.

Now find the ratio between clock rates of B and A as follows:

$$\begin{gathered} Clockcycles=InstructionCount\times CPI \\ \frac{ClockCycl{e}_B}{ClockCycl{e}_A}=\frac{1.2e^9\times 1.25}{1.0e^9\times 1.1} \\ =1.3636 \\ =1.37 \end{gathered}$$

Compiler A is 1.37 times faster than Compiler B.

Determine the speed up of the new compiler versus compile A and B

c.

New compiler is developed with the instruction count 6.0E8 and has an average of 1.1.

The ratio of execution between the A and the new execution time is calculated as follows:

$$\begin{gathered} \frac{ExecutiontimeofA}{Executiontimeofnewcompiler}=\frac{1.1}{0.66} \\ =1.666666667 \\ =1.67 \end{gathered}$$

$$\begin{gathered} \frac{ExecutiontimeofB}{Executiontimeofnewcompiler}=\frac{1.5}{0.66} \\ =2.27272727 \\ =2.27 \end{gathered}$$