Question

2 Section 1.10 cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions.

1.12.1 [5] <§§1.6, 1.10> One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2.

1.12.2 [10] <§§1.6, 1.10> Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions.

1.12.3 [10] <§§1.6, 1.10> A common fallacy is to use MIPS (millions of

instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance.

Check if this is true for P1 and P2.

1.12.4 [10] <§1.10> Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as

MFLOPS = No. FP operations / (execution time × 1E6)

but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs

Short answer

1.12.1

This is not true for P1 and P2. P1 has the largest clock rate but P2 has the largest performance.

1.12.2

P2 can execute 90% of instructions at the same time that P1 needs to execute 1.0E9 instructions.

1.12.3

This is not true for P1 and P2. P2 has the largest performance but P1 has the largest MIPS.

1.12.4

The MFLOPS of P2 is greater than the MFLOPS of P1.

Step-by-step solution

Define the concept

1.12.1

The CPU time can be defined to as,

$CPUtime=\frac{CPI\times thenumberofinstruction}{clockrate}$

1.12.2

The CPU time is reversely proportional with clock rate. If the CPI and the number of instructions are constant then, if the clock rate is increased then the CPU time is reduced.

1.12.3

MIPS refers to the millions of instructions per second.

And this can be simplified to as,

$$\begin{gathered} MIPS=\frac{thenumberofinstruction}{executiontime\times {10}^6} \\ =\frac{theclockrate}{CPI\times {10}^6} \end{gathered}$$

1.12.4

MFLOPS refers to the millions of floating-point operations per second.

And this can be defined to as,

$TheMFLOPS=\frac{thenumberofFPoperation}{theexecutiontime\times 1E6}$

Details of the calculation

1.12.1

Given,

For P1,

The clock rate is 4 GHz.

The average CPI is 0.9.

The number of execution is 5.0E9 = $5\times {10}^9$.

$TheCPUtimeofP1=\frac{0.9\times 5\times {10}^9}{4\times {10}^9}=1.125sec$

For P2,

The clock rate is 3 GHz.

The average CPI is 0.75.

The number of execution is 1.0E9 =.$1\times {10}^9$$1\times {10}^9$

Hence, the CPU time of P2 < the CPU time of P1.

So, the performance of P2 > the performance of P1.

1.12.2

For P1,

The clock rate is 4 GHz.

The average CPI is 0.9.

The number of execution is 1.0E9 = .

For P2,

The clock rate is 3 GHz.

The average CPI is 0.75.

The number of execution is 1.0E9 = .

P2 can execute instructions.

The clock rate of P2 < the clock rate of P1.

It is already proved that the performance of P2 > the performance of P1.

Hence, P2 can execute 90% of instructions at the same time that P1 needs to execute 1.0E9 instructions.

1.12.3

Given,

For P1,

The clock rate is 4 GHz.

The average CPI is 0.9.

For P2,

The clock rate is 3 GHz.

The average CPI is 0.75.

It is already proved that the performance of P2 > the performance of P1.

But The MIPS of P2 < the MIPS of P1.

1.12.4

Given that, 40% of the instructions executed on both P1 and P2 are floating-point instructions.

For P1,

The number of execution is 5.0E9 = .

40% of the instructions = .

The execution time is 1.125 sec.

For P2,

The number of execution is 1.0E9 = .

40% of the instructions = .

The execution time is 0.225 sec.

It is already proved that the performance of P2 > the performance of P1.

But the MFLOPS of P2 < the MFLOPS of P1.