Question

The results of the SPEC CPU2006 bzip2 benchmark running on an AMD Barcelona has an instruction count of 2.389E12, an execution time of 750 s, and a reference time of 9650 s.

1.11.1 [5] <§§1.6, 1.9> Find the CPI if the clock cycle time is 0.333 ns.

1.11.2 [5] <§1.9> Find the SPECratio.

1.11.3 [5] <§§1.6, 1.9> Find the increase in CPU time if the number of instructions of the benchmark is increased by 10% without affecting the CPI.

1.11.4 [5] <§§1.6, 1.9> Find the increase in CPU time if the number of instructions of the benchmark is increased by 10% and the CPI is increased by 5%.

1.11.5 [5] <§§1.6, 1.9> Find the change in the SPECratio for this change.

1.11.6 [10] <§1.6> Suppose that we are developing a new version of the AMD Barcelona processor with a 4 GHz clock rate. We have added some additional instructions to the instruction set in such a way that the number of instructions has been reduced by 15%. Th e execution time is reduced to 700 s and the new SPECratio is 13.7. Find the new CPI.

1.11.7 [10] <§1.6> Th is CPI value is larger than obtained in 1.11.1 as the clock rate was increased from 3 GHz to 4 GHz. Determine whether the increase in the CPI is similar to that of the clock rate. If they are dissimilar, why?

1.11.8 [5] <§1.6> By how much has the CPU time been reduced?

58 Chapter 1 Computer Abstractions and Technology

1.11.9 [10] <§1.6> For a second benchmark, libquantum, assume an execution time of 960 ns, CPI of 1.61, and clock rate of 3 GHz. If the execution time is reduced by an additional 10% without aff ecting to the CPI and with a clock rate of 4 GHz, determine the number of instructions.

1.11.10 [10] <§1.6> Determine the clock rate required to give a further 10% reduction in CPU time while maintaining the number of instructions and with the CPI unchanged.

1.11.11 [10] <§1.6> Determine the clock rate if the CPI is reduced by 15% and the CPU time by 20% while the number of instructions is unchanged.

Short answer

1.11.1

The value of the CPI is 0.94.

1.11.2

The value of the SPECratio is 12.86.

1.11.3

The increase in CPU time is 10%.

1.11.4

The increase in CPU time is 20%.

1.11.5

The changes in SPECratio for these changes is 10.7.

1.11.6

The new CPI is 1.38.

1.11.7

They are dissimilar because there are differences between the instruction count.

1.11.8

The CPU time is reduced by 7%.

1.11.9

The number of instructions is .

1.11.10

The clock rate is 4 GHz.

1.11.11

The clock rate is 3.83 GHz.

Step-by-step solution

Define the concept.

1.11.1

Cycle per instruction (CPI) depends on the clock cycle time, instruction count, and clock cycle time. The CPI refers to the number of cycles per instruction for a program. This number can be considered an aspect of the performance of the computer processor.

1.11.2

If the reference time is divided by the execution time. Then the result is referred to as the “SPECratio”.

1.11.3

The CPU time is referred to as the instruction count which is multiplied by CPI and which is again multiplied by the clock cycle time.

1.11.4

AlsoThe CPU time is referred to as the instruction count which is multiplied by CPI and which is divided by the clock rate.

1.11.5

The “SPECratio” time is directly proportional to the reference time and reversely proportional to the execution time.

1.11.6

The CPI is directly proportional to the clock cycles of the CPU and reversely proportional to the number of instruction counts.

1.11.7

$TheCPI=\frac{theclockrate\times theCPUtime}{instructioncount}$

So, the clock rate is directly proportional to the CPI. If the clock rate is increased from 3 GHz to 4 GHz then definitely the CPI will be increased.

1.11.8

$TheCPI=\frac{theclockrate\times theCPUtime}{instructioncount}$

So, the CPU time is directly proportional to the CPU. If the CPU time is reduced then definitely the CPI will be reduced.

1.11.9

$theclockrate=\frac{instructioncount\times TheCPI}{theCPUtime}$

The number of instructions can be referred to as “instruction count”.

1.11.10

The clock rate is indirectly proportional to the CPU time. If CPU time is reduced then the clock rate will be increased.

1.11.11

The clock rate is indirectly proportional to the CPU time. If CPU time is reduced then the clock rate will be increased. But the clock rate is directly proportional to the CPI. If the clock rate is reduced then definitely the CPI will be reduced.

Determine the value.

1.11.1

In the given question, the execution time is 750 s.

The instruction count is 2.389E12

Clock cycle time is 0.333 ns = 0.33 * 10^-9 s

$$\begin{gathered} TheCPIis=\frac{Executiontime}{instructiontime\times clockcycletime} \\ =\frac{750}{2.389E12\times 0.33\times {10}^{-9}} \\ =0.9425 \end{gathered}$$

1.11.2

In the given question, the execution time is 750 s.

And the reference time is 9650 s.

$$\begin{gathered} ThevalueofSPECratiois=\frac{referencetime}{executiontime} \\ =\frac{9650}{750} \\ =12.86 \end{gathered}$$

1.11.3

The instruction count is 2.389E12.

In the given question,the number of instructions is increased by 10%.

So,$intructioncount\times 1.1=2.628$= 2.628.

The CPU time is 823 s.

The execution time is 750 s.

Hence increase in CPU time

localid="1650336474517" $$\begin{gathered} \left(\frac{823}{750}-1\right) \\ =0.1 \\ =10\% \end{gathered}$$

1.11.4

The number of instructions of the benchmark is increased by 10%.

The CPI is 0.94 the CPI is increased by 5%.

So, after increasing, the CPI =$CPI\times 1.1$= 1.03.

The CPU time is 901 s.

The execution time is 750 s.

Hence increase in CPU time

$$\begin{gathered} \left(\frac{901}{750}-1\right) \\ =0.2 \\ =20\% \end{gathered}$$

1.11.5

The execution time is 750 s.

The previous reference time is 9650 s.

For these changes in above section,

Changes in reference time = 8025.

$$\begin{gathered} thechangesinSPECratiois=\frac{changesinreferencetime}{executiontime} \\ =\frac{8025}{750} \\ =10.7 \end{gathered}$$

1.11.6

In the given question, theexecution time is reduced to 700 s and the previous clock rate is 4 GHz.

The number of instructions is reduced by 15%.

$$\begin{gathered} ThevalueofSPECratiois=\frac{referencetime}{executiontime} \\ =\frac{9650}{700} \\ =13.78 \end{gathered}$$

And the new SPECratio is 13.78.

$TheCPIis=\frac{executiontime}{instructiontime\times clockcycletime}$

So, the new CPI = 1.38

So, the new instruction count = 2.031E12

$Theclockrate=\frac{theinstructioncount\times theCPI}{theCPUtime}$

Hence, the clock rate is $4.0\times {10}^9$ .

1.11.7

$$\begin{gathered} Theratioofclockrate=\frac{4GHz}{3GHz} \\ =1.33 \end{gathered}$$

$$\begin{gathered} TheratiooftheCPI=\frac{1.38}{0.94} \\ =1.47 \end{gathered}$$

Hence, the increase in the CPI is dissimilar to the increase in the CPU time.

1.11.8

$CPUtimereduction=\frac{500}{600}=0.93$

The CPU time has been reduced by the (100-93) % = 7%.

1.11.9

In the given question, the execution of time is 960 ns.

The CPI is 1.61

The clock rate is 3 GHz.

The execution time is reduced by 10% without affecting the CPI and with a clock rate of 4 GHz,

$$\begin{gathered} Numberofinstructions=\frac{CPUtime\times clockrate}{CPI} \\ =\frac{560\times 0.9\times 4\times {10}^9}{0.94} \\ =2147\times {10}^9 \end{gathered}$$

Hence, the number of instructions = $2147\times {10}^9$.

1.11.10

The CPU time is 901 s and itis reduced by 10%.

The number of instructions andCPI is unchanged.

$$\begin{gathered} Theclockrate=\frac{2147\times {10}^9\times 1.61}{864}[AsClockrate=\frac{theinstructioncount\times theCPI}{theCPUtime}] \\ =4.0\times {10}^9 \\ =4GHz \end{gathered}$$

Hence the CPU time$\times 0.9=864$

1.11.11

The CPIis reduced by 15%.

The CPU timeis reduced by 20%.

The number of instructionsis unchanged.

$$\begin{gathered} CPI\times 0.85=0.94\times 0.85=1.37 \\ Clockrate=\frac{theinstructioncount\times theCPI}{theCPUtime} \\ =\frac{theinstructioncount\times theCPI}{0.9\times theCPUtime} \\ =\frac{1}{0.9}\times thepreviousvalueoftheclockrate \\ =1\times 3.83GHz \\ Executiontime\times 0.8=\left(960\times 0.8\right)s=768s \end{gathered}$$