2kr Factorial r Factorial Designs - cse.wustl.edujain/cse567-08/ftp/k_182kr.pdf · 2kr Factorial r...

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18-1 ©2006 Raj Jain CSE567M Washington University in St. Louis 2 2 k k r Factorial r Factorial Designs Designs Raj Jain Washington University in Saint Louis Saint Louis, MO 63130 [email protected] These slides are available on-line at: http://www.cse.wustl.edu/~jain/cse567-06/

Transcript of 2kr Factorial r Factorial Designs - cse.wustl.edujain/cse567-08/ftp/k_182kr.pdf · 2kr Factorial r...

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18-1©2006 Raj JainCSE567MWashington University in St. Louis

22kkr Factorial r Factorial DesignsDesigns

Raj Jain Washington University in Saint Louis

Saint Louis, MO [email protected]

These slides are available on-line at:http://www.cse.wustl.edu/~jain/cse567-06/

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18-2©2006 Raj JainCSE567MWashington University in St. Louis

OverviewOverview

! Computation of Effects! Estimation of Experimental Errors! Allocation of Variation! Confidence Intervals for Effects! Confidence Intervals for Predicted Responses! Visual Tests for Verifying the assumptions! Multiplicative Models

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18-3©2006 Raj JainCSE567MWashington University in St. Louis

22kkr Factorial Designsr Factorial Designs

! r replications of 2k Experiments⇒ 2kr observations.⇒ Allows estimation of experimental errors.

! Model:

! e = Experimental error

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18-4©2006 Raj JainCSE567MWashington University in St. Louis

Computation of EffectsComputation of Effects

Simply use means of r measurements

! Effects: q0= 41, qA= 21.5, qB= 9.5, qAB= 5.

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18-5©2006 Raj JainCSE567MWashington University in St. Louis

Estimation of Experimental ErrorsEstimation of Experimental Errors

! Estimated Response:

Experimental Error = Estimated - Measured

∑i,j eij = 0

Sum of Squared Errors:

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18-6©2006 Raj JainCSE567MWashington University in St. Louis

Experimental Errors: ExampleExperimental Errors: Example! Estimated Response:

! Experimental errors:

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18-7©2006 Raj JainCSE567MWashington University in St. Louis

Allocation of VariationAllocation of Variation

! Total variation or total sum of squares:

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18-8©2006 Raj JainCSE567MWashington University in St. Louis

DerivationDerivation! Model:

Since x's, their products, and all errors add to zero

Mean response:

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18-9©2006 Raj JainCSE567MWashington University in St. Louis

Derivation (Cont)Derivation (Cont)

Squaring both sides of the model and ignoring cross product terms:

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18-10©2006 Raj JainCSE567MWashington University in St. Louis

Derivation (Cont)Derivation (Cont)

Total variation:

One way to compute SSE:

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18-11©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.3: MemoryExample 18.3: Memory--Cache StudyCache Study

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18-12©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.3 (Cont)Example 18.3 (Cont)

Factor A explains 5547/7032 or 78.88%Factor B explains 15.40%Interaction AB explains 4.27%1.45% is unexplained and is attributed to errors.

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18-13©2006 Raj JainCSE567MWashington University in St. Louis

Confidence Intervals For EffectsConfidence Intervals For Effects! Effects are random variables.! Errors ∼ N(0,σe) ⇒ y ∼ N( , σe)

! q0 = Linear combination of normal variates⇒ q0 is normal with variance σe

2/(22r)Variance of errors:

! Denominator = 22(r-1) = # of independent terms in SSE⇒ SSE has 22(r-1) degrees of freedom.

Estimated variance of q0: sq02=se

2/(22r)

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18-14©2006 Raj JainCSE567MWashington University in St. Louis

Confidence Intervals For Effects (Cont)Confidence Intervals For Effects (Cont)

! Similarly,

! Confidence intervals (CI) for the effects:

! CI does not include a zero ⇒ significant

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18-15©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.4Example 18.4! For Memory-cache study: Standard deviation of errors:

! Standard deviation of effects:

! For 90% Confidence: t[0.95,8]= 1.86

! Confidence intervals: qi ∓ (1.86)(1.03) = qi ∓ 1.92q0= (39.08, 42.91)qA=(19.58, 23.41)qB=(7.58, 11.41)qAB= (3.08, 6.91)! No zero crossing ⇒ All effects are significant.

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18-16©2006 Raj JainCSE567MWashington University in St. Louis

Confidence Intervals for ContrastsConfidence Intervals for Contrasts! Contrast M Linear combination with ∑ coefficients =

0! Variance of ∑ h

! For 100(1-α)% confidence interval, use t[1-α/2; 22(r-1)].

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18-17©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.5Example 18.5

Memory-cache studyu = qA+ qB - 2qAB

Coefficients= 0, 1, 1, and -2 ⇒ Contrast

t[0.95;8]=1.8690% Confidence interval for u:

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18-18©2006 Raj JainCSE567MWashington University in St. Louis

Conf. Interval For Predicted ResponsesConf. Interval For Predicted Responses! Mean response :

! The standard deviation of the mean of m responses:

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18-19©2006 Raj JainCSE567MWashington University in St. Louis

Conf. Interval for Predicted Responses (Cont)Conf. Interval for Predicted Responses (Cont)

100(1-α)% confidence interval:

! A single run (m=1):

! Population mean (m=∞

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18-20©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.6: MemoryExample 18.6: Memory--cache Studycache Study! For xA= -1 and xB = -1:! A single confirmation experiment:

! Standard deviation of the prediction:

! Using t[0.95;8] = 1.86, the 90% confidence interval is:

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18-21©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.6 (Cont)Example 18.6 (Cont)! Mean response for 5 experiments in future:

! The 90% confidence interval is:

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18-22©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.6 (Cont)Example 18.6 (Cont)! Mean response for a large number of experiments in future:

! The 90% confidence interval is:

! Current mean response: Not for future. Use contrasts formula.

! 90% confidence interval:

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18-23©2006 Raj JainCSE567MWashington University in St. Louis

AssumptionsAssumptions

1. Errors are statistically independent.2. Errors are additive. 3. Errors are normally distributed.4. Errors have a constant standard deviation σe.5. Effects of factors are additive

⇒ observations are independent and normally distributed with constant variance.

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18-24©2006 Raj JainCSE567MWashington University in St. Louis

Visual TestsVisual Tests1. Independent Errors:

! Scatter plot of residuals versus the predicted response ! Magnitude of residuals < Magnitude of responses/10

⇒ Ignore trends ! Plot the residuals as a function of the experiment number! Trend up or down ⇒ other factors or side effects

2. Normally distributed errors: Normal quantile-quantile plot of errors

3. Constant Standard Deviation of Errors: Scatter plot of y for various levels of the factor Spread at one level significantly different than that at other⇒ Need transformation

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18-25©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.7: MemoryExample 18.7: Memory--cachecache

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18-26©2006 Raj JainCSE567MWashington University in St. Louis

Multiplicative ModelsMultiplicative Models! Additive model:

! Not valid if effects do not add. E.g., execution time of workloads.ith processor speed= vi instructions/second.jth workload Size= wj instructions

! The two effects multiply. Logarithm ⇒ additive model:

! Correct Model:

Where, y'ij=log(yij)

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18-27©2006 Raj JainCSE567MWashington University in St. Louis

Multiplicative Model (Cont)Multiplicative Model (Cont)! Taking an antilog of effects:

uA = 10qA, uB=10qB, and uAB=10qAB

! uA= ratio of MIPS rating of the two processors! uB= ratio of the size of the two workloads.! Antilog of additive mean q0 ⇒ geometric mean

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18-28©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.8: Execution TimesExample 18.8: Execution Times

Additive model is not valid because:! Physical consideration ⇒ effects of workload and processors do

not add. They multiply.! Large range for y. ymax/ymin= 147.90/0.0118 or 12,534

⇒ log transformation! Taking an arithmetic mean of 114.17 and 0.013 is inappropriate.

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18-29©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.8 (Cont)Example 18.8 (Cont)! The residuals are not small as compared to the response.

! The spread of residuals is large at larger value of the response.⇒ log transformation

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18-30©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.8 (Cont)Example 18.8 (Cont)

! Residual distribution has a longer tail than normal

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18-31©2006 Raj JainCSE567MWashington University in St. Louis

Analysis Using Multiplicative ModelAnalysis Using Multiplicative Model

!

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18-32©2006 Raj JainCSE567MWashington University in St. Louis

Variation Explained by the Two ModelsVariation Explained by the Two Models

! With multiplicative model:" Interaction is almost zero." Unexplained variation is only 0.2%

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18-33©2006 Raj JainCSE567MWashington University in St. Louis

Visual TestsVisual Tests

! Conclusion: Multiplicative model is better than the additive model.

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18-34©2006 Raj JainCSE567MWashington University in St. Louis

Interpretation of ResultsInterpretation of Results

! The time for an average processor on an average benchmark is 1.07.

! The time on processor A1 is nine times (0.107-1) that on an average processor. The time on A2 is one ninth (0.1071) of that on an average processor.

! MIPS rate for A2 is 81 times that of A1.! Benchmark B1 executes 81 times more instructions than B2.! The interaction is negligible.

⇒ Results apply to all benchmarks and processors.

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18-35©2006 Raj JainCSE567MWashington University in St. Louis

Transformation ConsiderationsTransformation Considerations! ymax/ymin small ⇒ Multiplicative model results similar to

additive model.! Many other transformations possible. ! Box-Cox family of transformations:

! Where g is the geometric mean of the responses:

! w has the same units as y.! a can have any real value, positive, negative, or zero.! Plot SSE as a function of a ⇒ optimal a! Knowledge about the system behavior should always take

precedence over statistical considerations.

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18-36©2006 Raj JainCSE567MWashington University in St. Louis

General 2General 2kkr Factorial Designr Factorial Design! Model:

! Parameter estimation:

Sij = (i,j)th entry in the sign table.! Sum of squares:

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18-37©2006 Raj JainCSE567MWashington University in St. Louis

General 2General 2kkr Factorial Design (Cont)r Factorial Design (Cont)! Percentage of y's variation explained by jth effect =

! Standard deviation of errors:

! Standard deviation of effects:

! Variance of contrast ∑ hi qi, where ∑ hi=0 is:

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18-38©2006 Raj JainCSE567MWashington University in St. Louis

General 2General 2kkr Factorial Design (Cont)r Factorial Design (Cont)! Standard deviation of the mean of m future responses:

! Confidence intervals are calculated using t[1-α/2;2k(r-1)].! Modeling assumptions:

" Errors are IID normal variates with zero mean." Errors have the same variance for all values of the

predictors." Effects and errors are additive.

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18-39©2006 Raj JainCSE567MWashington University in St. Louis

Visual Tests for 2Visual Tests for 2kkr Designsr Designs

! The scatter plot of errors versus predicted responses should not have any trend.

! The normal quantile-quantile plot of errors should be linear.

! Spread of y values in all experiments should be comparable.

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18-40©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.9: A 2Example 18.9: A 2333 Design3 Design

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18-41©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.9 (Cont)Example 18.9 (Cont)! Sum of Squares:

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18-42©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.9 (Cont)Example 18.9 (Cont)! The errors have 23(3-1) or 16 degrees of freedom. Standard

deviation of errors:

! Standard deviation of effects:

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18-43©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.9 (Cont)Example 18.9 (Cont)

! % Variation:

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18-44©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.9 (Cont)Example 18.9 (Cont)! t[0.95,16]=1.337! 90% confidence intervals for parameters: qi ∓ (1.337)(0.654)

= qi ∓ 0.874

! All effects except qABC are significant.

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18-45©2006 Raj JainCSE567MWashington University in St. Louis

Example 18.9 (Cont)Example 18.9 (Cont)! For a single confirmation experiment (m = 1)

With A = B = C = -1:

! 90% confidence interval:

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18-46©2006 Raj JainCSE567MWashington University in St. Louis

Case Study 18.1: Garbage collectionCase Study 18.1: Garbage collection

!

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18-47©2006 Raj JainCSE567MWashington University in St. Louis

Case Study 18.1 (Cont)Case Study 18.1 (Cont)

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18-48©2006 Raj JainCSE567MWashington University in St. Louis

Case Study 18.1 (Cont)Case Study 18.1 (Cont)

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18-49©2006 Raj JainCSE567MWashington University in St. Louis

Case Study 18.1: ConclusionsCase Study 18.1: Conclusions

! Most of the variation is explained by factors A (Workload), D (Chunk size), and the interaction A D between the two.

! The variation due to experimental error is small⇒ Several effects that explain less than 0.05% of variation (listed as 0.0%) are statistically significant.

! Only effects A, D, and AD are both practically significant and statistically significant.

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18-50©2006 Raj JainCSE567MWashington University in St. Louis

SummarySummary

! Replications allow estimation of measurement errors⇒ Confidence Intervals of parameters⇒ Confidence Intervals of predicted responses

! Allocation of variation is proportional to square of effects! Multiplicative models are appropriate if the factors multiply! Visual tests for independence normal errors

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18-51©2006 Raj JainCSE567MWashington University in St. Louis

Exercise 18.1Exercise 18.1

Table 18.11 lists measured CPU times for two processors on two workloads. Each experiment was repeated three times. Analyze the design.

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18-52©2006 Raj JainCSE567MWashington University in St. Louis

HomeworkHomework

Updated Exercise 18.1: The following table lists measured CPU times for two processors on two workloads. Each experiment was repeated three times. Analyze the design. Determine percentage of variation explained, find confidence intervals of the effects, and conduct visual tests.