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Statistics. CSE 807. Experimental Design and Analysis. How to: Design a proper set of experiments for measurement or simulation. Develop a model that best describes the data obtained. Estimate the contribution of each alternative to the performance. Isolate the measurement errors.

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statistics

Statistics

CSE 807

experimental design and analysis
Experimental Design and Analysis

How to:

  • Design a proper set of experiments for measurement or simulation.
  • Develop a model that best describes the data obtained.
  • Estimate the contribution of each alternative to the performance.
  • Isolate the measurement errors.
  • Estimate confidence intervals for model parameters.
  • Check if the alternatives are significantly different.
  • Check if the model is adequate.
example
Example
  • Personal workstation design.
  • Processor:68000, Z80, or 8086.
  • Memory size: 512K, 2M, or 8M bytes.
  • Number of Disks: One, two, three, or four.
  • Workload: Secretarial, managerial, or scientific.
  • User education: High school, college, or Post-graduate level.
terminology
Terminology
  • Response Variable: Outcome.

E.g., throughput, response time.

  • Factors: Variables that affect the response variable.

E.g., CPU type, memory size, number of disk drivers, workload used, and user’s educational level.

Also called predictor variables or predictors.

  • Levels: The value that a factor can assume.

E.g., the CPU type has three levels:

68000, 8080, or Z80.

# of disk drives has four levels.

Also called treatment.

terminology cont d
Terminology (cont’d)
  • Primary Factors: The factors whose effects need to be quantified.

E.g., CPU type, memory size only, and number of disk drives.

  • Secondary Factors: “Factors whose impact need not be quantified.

E.g., the work loads.

  • Replication: Repetition of all or some experiments.
terminology cont d6
Terminology (cont’d)
  • Design: The number of experiments, the factor level and number of replications for each experiment.

E.g., Full Factorial design with 5 replications:

3 X 3 X 4 X 3 X 3 or 324 experiments, each repeated five times.

  • Experimental Unit: Any entity that is used for experiments.

E.g., users. Generally, no interest in comparing the units.

Goal - minimize the impact of variation among the units.

terminology cont d7
Terminology (cont’d)
  • Interaction => Effect of one factor depends upon the level of the other.

Non-interacting Factors

Interacting Factors

common mistakes in experimentation
Common Mistakes in Experimentation

1. The variation due to experimental error is ignored.

2. Important parameters are not controlled.

3. Effects of different factors are not isolated.

4. Simple one-factor-at-a-time designs are used

5. Interactions are ignored.

6. Too many experiments are conducted.

Better: two phases.

types of experimental designs
Types of Experimental Designs
  • Simple Designs: Vary one factor at a time
    • #of Experiments =

Not statistically efficient.

Wrong conclusions if the factors have interaction.

Not recommended.

types of experimental designs cont d
Types of Experimental Designs (cont’d)
  • Full Factorial Design: All combinations.
    • # of Experiments =

Can find the effect of all factors.

Too much time and money.

May try 2k design first

types of experimental designs cont d11
Types of Experimental Designs (cont’d)
  • Fractional Factorial Designs: Save time and expense.

Less information.

May not get all interactions.

Not a problem if negligible interactions.

exercise
Exercise
  • The performance of a System being designed depends upon the following three factors:

a. CPU type: 68000, 8086, 80286

b. Operating System type: CPM, MS-DOS, UNIX

c. Disk drive type: A, B, C

How many experiments are required to analyze the performance if

a. There is significant interaction among factors.

b. There is no interaction among factors

c. The interactions are small compared to main effects.

2 k factorial designs
2k Factorial Designs
  • k factors, each at two levels.
  • Easy to analyze.
  • Helps in sorting out impact of factors.
  • Good at the beginning of study.
  • Valid only if the effect is unidirectional.

E.g., memory size, the number of disk drives

2 2 factorial designs

Cache

Size

Memory size

4M Bytes

16M Bytes

1K

2K

15

25

45

75

22 Factorial Designs
  • Two factors, each at two levels

Performance in MIPS

-1 if 4M bytes memory

1 if 16M bytes memory

-1 if 1M bytes cache

1 if 2M bytes cache

xA=

xB=

model
Model

y = q0 + qAxA + qBxB +qABxAxB

15= q0 - qA - qB + qAB

45= q0 + qA - qB - qAB

25= q0 - qA + qB - qAB

75= q0 + qA + qB + qAB

y = 40 + 20xA + 10xB + 5xAxB

Interpretation: Mean performance = 40 MIPS

Effect of memory = 20 MIPS

Effect cache = 10 MIPS

Interaction between memory and cache = 5 MIPS

computation of effects
Computation of Effects

Model: y = q0 + qAxA + qBxB +qABxAxB

Substitution:

y1 = q0 - qA - qB + qAB

y2 = q0 + qA - qB - qAB

y3 = q0 - qA + qB - qAB

y4 = q0 + qA + qB + qAB

computation of effects cont d
Computation of Effects (cont’d)

Solution:

q0 =1/4 (y1 + y2 + y3 + y4)

qA =1/4 (-y1 + y2 - y3 + y4)

qB =1/4 (-y1 - y2 + y3 + y4)

qAB =1/4 (y1 - y2 - y3 + y4)

Notice that effects are linear combinations of responses.

Sum of the coefficients is zero => contrasts.

Notice: qA = Column A x Column y

qB = Column B x Column y

qAB = Column A x Column B x Column y

allocation of variation
Allocation of Variation
  • Importance of a factor = proportion of the variation explained
  • Sample variance of
  • Variation of y Numerator

= sum of squares total (SST)

allocation of variation cont d
Allocation of Variation (cont’d)

For a 22 design:

Variation due to

Variation due to

Variation due to interaction

SST = SSA + SSB + SSAB

Fraction explained by

Variation  Variance

derivation
Derivation

Model:

yi = q0 + qAxAi + qBxBi +qABxAixBi

Notice

1. The sum of entries in each column is zero:

2. The sum of the squares of entries in each column is 4:

derivation cont d
Derivation (cont’d)
  • 3. The columns are orthogonal (inner product of any two columns is zero):
derivation cont d25
Derivation (cont’d)

Variation of y

Product terms

example26
Example

Memory-cache study:

Total Variation

Total variation = 2100

Variation due to memory = 1600 (76%)

Variation due to cache = 400 (19%)

Variation due to interaction = 100 (5%)

case study interconnection net
Case Study: Interconnection Net

Memory interconnection networks:

Omega and Crossbar.

Memory reference patterns:

random and Matrix

Fixed factors:

1. Number of processors was fixed at 16.

2. Queued requests were not buffered but blocked.

3. Circuit switching instead of packet switching.

4. Random arbitration instead of round robin.

5. Infinite interleaving of memory => no memory back contention.

2 2 design for interconnection networks
22 Design for Interconnection Networks

Factors Used in the Interconnection Network Study

Level

Response

interconnection network study cont d

Para-

meter

Mean Estimate

Variation Explained

T

N

R

T

N

R

q0

qA

qB

qAB

0.5725

0.0595

-0.1257

-0.0346

3.5

-0.5

1.0

0.0

1.871

-0.145

0.413

0.051

17.2%

77.0%

5.8%

20%

80%

0%

10.9%

87.8%

1.3%

Interconnection Network Study (cont’d)
interpretation of results
Interpretation of Results
  • Average throughput = 0.5725
  • Most effective factor = B = reference pattern => The address patterns chosen are very different.
  • Reference pattern explains  0.1257 (77%) of variation
  • Effect of network type = 0.0595

Omega networks = Average + 0.0595

Crossbar networks = Average - 0.0595

Difference between the two = 0.119

  • Slight interaction (0.0346) between reference pattern and network type.
general 2 k factorial designs
General 2k Factorial Designs

k factors at two levels each.

2kexperiments.

2keffects:

k main effects

Two factor interactions

Three factor interactions...

2 k design example
2k Design Example

Three factors in designing a machine:

Cache size

Memory size

Number of processors

2 k design example cont d

Cache

Size

4M Bytes

16M Bytes

1 Proc

2 Proc

1 Proc

2 Proc

1K Byte

2K Byte

14

10

46

50

22

34

58

86

2k Design Example (cont’d)
analysis

=18%+4%+71%+4%+1%+2%+0%

=100%

Number of Processors (C) is the most important factor

Analysis
exercise35

A1

A2

C1

C2

C1

C2

B1

B2

100

40

15

30

120

20

10

50

Exercise

Analyze the 23 design:

a. Quantify main effects and all interactions.

b. Quantify percentages of variation explained.

c. Sort the variables in the order of decreasing importance

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