1

2

Table 4.1 Key characteristics of six passenger aircraft: all figures are approximate; some relate to a specific model/configuration of the aircraft or are averages of cited range of values.

4.1 Performance and Cost/performance

3

Figure 4.1 Performance improvement as a function of cost.

4

4.2 Defining Computer Performance

Performance = 1 / Execution time

Throughput: amount of work performed per unit time. It can be measured as the number of processes per unit time.

Tournaround time: the average time from the moment that a job is submitted until the moment it is completed. It measures how long the average user has to wait for output.

Response time: In an interactive systems, the time from when a user press an Enter or clicks a mouse until the system delivers a final response.

To filter out variable factor (e.g., scheduling, interrupts, I/O delay)

Performance = 1 / CPU Execution time

5

Figure 4.2 Pipeline analogy shows that imbalance between processing power and I/O capabilities leads to a performance bottleneck.

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CPU execution time

= Instructions × (Cycles per instruction ) × (seconds per cycle)

= Instructions × CPI / (Clock rate)

(CPI: cycles per instruction)

Performance comparison

(Performance of M1) / (Performance of M2) =

(Execution time of M2) / (Execution time of M1)

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4.3 Performance Enhancement and Amdahl’s Law

Amdahl’s law

s = 1 / (f+(1-f) / p) ≤ min (p, 1/f)

f: time for instructions that cannot be parallelized.

p: speed-up (by parallel computer or redesign CPU or algorithm)

s: overall speedup

Study Example 4.1

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Figure 4.4 Amdahl’s law: speedup achieved if a fraction f of a task is unaffected and the remaining 1 – f part runs p times as fast.

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Figure 4.5 Running times of six programs on three machines.

4.4 Performance Measurement vs Modeling

10

Table 4.2 Summary of SPEC CPU2000 benchmark suite characteristics.

Benchmarks: real or synthetic programs that are selected for comparative evaluation of machine performance.

SPEC: Standard Performance Evaluation Corporation

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Figure 4.6 Example graphical depiction of SPEC benchmark results.

Study Example 4.3

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Table 4.3 Usage frequency, in percentage, for various instruction classes in four representative applications.

Performance Estimation

System’s peak performance: expressed in instructions per second. (MIPS, MFLOPS)

)(clockrate / CPIAverage nsInstructio time execution CPU

CPI) i-(classfraction) i-(class CPIAverage classes ninctructio All

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Example 4.4 CPI and IPS calculation

Solution

a. For M1, assume all instructions are class I instructions,

Peak performance of M1 = 1 / (Avg. CPI × Clock time)

= 600 / 2.0 = 300MIPS

Notice: Units for Average CPI and clock time are second.

For M2, assume all instructions are class N instructions,

Peak performance of M2 = 1 / (Avg. CPI × Clock time)= 500 / 2.0 = 250MIPS

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b. Average CPI for M1=5.0×0.25+2.0×0.25+2.4×0.5=2.95 Average CPI for M2=4.0×0.25+3.8×0.25+2.4×0.5=2.95c. 1. Average CPI=2.5×0.25+2.0×0.25+2.4×0.5=2.325

MIPS for option 1 = 600/2.325 = 2582. Average CPI=5.0×0.25+1.2×0.25+2.4×0.5=2.75

MIPS for option 1 = 600/2.75 = 2183. MIPS for option 3 = 750/2.95=254. Conclusion: Option 1 has the greatest impact

d. With larger cache, cache miss rate is reduced 2% (from 5% to 3%), that is all CPIs are reduced 10×2%=0.2ns (cache miss imposes 10 cycle penalty)Average CPI M1=(5.0-0.2)×0.25+(2.0-0.2)×0.25+(2.4-0.2)×0.5=2.75This option is comparable to option 2 in c.

e. Average CPI for M1= 5.0×x+2.0×y+2.4×(1-x-y)=2.6x-0.4y+2.4Average CPI for M2= 4.0×x+3.8×y+2.0×(1-x-y)=2x+1.8y+2

We need 600/(2.6x-0.4y+2.4) > 500/(2x+1.8y+2) => 2.56y > 0.2xThat is, x/y < 12.8, M1 runs faster than M2 for the given task.

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Example 4.5 MIPS rating can be misleading

a. Runtime for the output of compiler 1= (600M+400M)/109= 1.4s

Runtime for the output of compiler 2 = (400M+400M)/109= 1.2s

Compiler 2 is faster.

b. Code produced by compiler 2 is 1.4/1.2= 1.17 times as faster as that of compiler 1.

c. Average CPI for compiler 1 = (600M×1+400M×2)/1000M=1.4

Average CPI for compiler 2 = (400M×1+400M×2)/800M=1.5

MIPS rating of compiler 1=1000/1.4=714

MIPS rating of compiler 2=1000/1.5=667

Compiler 1 is faster

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Table 4.4 Measured or estimated execution times for three programs.

4.5 Reporting Computer Performance

Wrong method (arithmetic mean)

Speedup of Y over X=(0.1+10.0+10.0)/3=6.7 (1)

Speedup of X over y=(10.0+0.1+0.10)/3=3.3 (contradictory with (1))

Total time comparison: correct if they are run the same number of times.

Geometric mean:

Speedup of Y over X=(0.1×10.0×10.0)1/3=2.15 (2)

Speedup of X over y=(10.0×0.1×0.10)1/3=0.46 (consistent with (2))

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Example 4.6

Table 4.3 Usage frequency, in percentage, for various instruction classes in four representative applications.

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Answer:

a. CPI for data compression application on M1=0.25×4.0+0.32×1.5+0.16×1.2+0×6.0+0.19×2.5+0.08×2.0=2.31

CPI for data compression application on M2=2.54

CPI for nuclear reactor simulation application on M1=3.94

CPI for nuclear reactor simulation application on M2=2.89

b. Because the programs and clock rates are the same, speedup ratios is given by the ratio of CPIs.

Data compression performance speed up (M2/M1)= 2.31/2.54 = 0.91

nuclear reactor simulation performance speed up (M2/M1)= 3.94/2.89 = 1.36.

c. Overall performance advantage of M2 over M1 is (0.91×1.36)1/2=1.11

19Figure 4.7 Exponential growth of supercomputer performance [Bell92].

4.6 The Quest for High Performance

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Figure 4.8 Milestones in the Accelerated Strategic Computing Initiative (ASCI) program, sponsored by the U.S. Department of Energy, with extrapolation up to the PFLOPS level.

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Problem 4.5

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Problem 4.12