1. Experiment Design Overview Number of factors 1 2 k levels 2:min/max n - cat num regression models2k2k repl interactions & errors 2 k-p weak interactions n x r pts 2kr2kr est. errors full factorial additive? Analysis to get: Main effects Interactions/errors % of variation explained Significance (CI or ANOVA)

2. © 1998, Geoff Kuenning Two-Factor Full Factorial Design Without Replications Used when you have only two parameters But multiple levels for each Test all combinations of the levels of the two parameters At this point, without replicating any observations For factors A and B with a and b levels, ab experiments required

3. Experimental Design ( l 1,0, l 1,1, …, l 1,n1-1 ) x ( l 2,0, l 2,1, …, l 2,n2-1 ) 2 different factors, each factor with n i levels (and possibly r replications -- defer) Categorical levels Factor 1 Factor 2

4. © 1998, Geoff Kuenning What is This Design Good For? Systems that have two important factors Factors are categorical More than two levels for at least one factor Examples - –Performance of different processors under different workloads –Characteristics of different compilers for different benchmarks –Effects of different reconciliation topologies and workloads on a replicated file system

5. © 1998, Geoff Kuenning What Isn’t This Design Good For? Systems with more than two important factors –Use general factorial design Non-categorical variables –Use regression Only two levels –Use 2 2 designs

6. © 1998, Geoff Kuenning Model For This Design y ij is the observation  is the mean response  j is the effect of factor A at level j  i is the effect of factor B at level i e ij is an error term Sums of  j ’s and  j ’s are both zero

7. © 1998, Geoff Kuenning What Are the Model’s Assumptions? Factors are additive Errors are additive Typical assumptions about errors –Distributed independently of factor levels –Normally distributed Remember to check these assumptions!

8. © 1998, Geoff Kuenning Computing the Effects Need to figure out ,  j, and  j Arrange observations in two- dimensional matrix –With b rows and a columns Compute effects such that error has zero mean –Sum of error terms across all rows and columns is zero

9. © 1998, Geoff Kuenning Two-Factor Full Factorial Example We want to expand the functionality of a file system to allow automatic compression We examine three choices - –Library substitution of file system calls –A new VFS –UCLA stackable layers Using three different benchmarks –With response time as the metric

10. © 1998, Geoff Kuenning Sample Data for Our Example LibraryVFSLayers Compile Benchmark94.3 89.5 96.2 Email Benchmark224.9231.8 247.2 Web Server Benchmark733.5702.1797.4

11. © 1998, Geoff Kuenning Computing  Averaging the j th column, By assumption, the error terms add to zero Also, the  j s add to zero, so Averaging rows produces Averaging everything produces

12. © 1998, Geoff Kuenning So the Parameters Are...

13. © 1998, Geoff Kuenning Sample Data for Our Example LibraryVFSLayers Compile Benchmark94.3 89.5 96.2 93.3 Email Benchmark224.9231.8 247.2 234.6 Web Server Benchmark733.5702.1797.4 744.3 _____________________________________________________ 350.9341.1380.3 357.4

14. © 1998, Geoff Kuenning Calculating Parameters for Our Example  = grand mean = 357.4  j = (-6.5, -16.3, 22.8)  i = (-264.1, -122.8, 386.9) So, for example, the model predicts that the email benchmark using a special- purpose VFS will take 357.4 - 16.3 - 122.8 seconds –Which is 218.3 seconds

15. © 1998, Geoff Kuenning Estimating Experimental Errors Similar to estimation of errors in previous designs Take the difference between the model’s predictions and the observations Calculate a Sum of Squared Errors Then allocate the variation

16. © 1998, Geoff Kuenning Allocating Variation Using the same kind of procedure we’ve used on other models, SSY = SS0 + SSA + SSB + SSE SST = SSY - SS0 We can then divide the total variation between SSA, SSB, and SSE

17. © 1998, Geoff Kuenning Calculating SS0, SSA, and SSB a and b are the number of levels for the factors

18. © 1998, Geoff Kuenning Allocation of Variation For Our Example SSE = 2512 SSY = 1,858,390 SS0 = 1,149,827 SSA = 2489 SSB = 703,561 SST=708,562 Percent variation due to A -.35% Percent variation due to B - 99.3% Percent variation due to errors -.35%

19. © 1998, Geoff Kuenning Analysis of Variation Again, similar to previous models –With slight modifications As before, use an ANOVA procedure –With an extra row for the second factor –And changes in degrees of freedom But the end steps are the same –Compare F-computed to F-table –Compare for each factor

20. © 1998, Geoff Kuenning Analysis of Variation for Our Example MSE = SSE/[(a-1)(b-1)]=2512/[(2)(2)]=628 MSA = SSA/(a-1) = 2489/2 = 1244 MSB = SSB/(b-1) = 703,561/2 = 351,780 F-computed for A = MSA/MSE = 1.98 F-computed for B = MSB/MSE = 560 The 95% F-table value for A & B = 6.94 A is not significant, B is

21. © 1998, Geoff Kuenning Checking Our Results With Visual Tests As always, check if the assumptions made by this analysis are correct Using the residuals vs. predicted and quantile-quantile plots

22. © 1998, Geoff Kuenning Residuals Vs. Predicted Response for Example

23. © 1998, Geoff Kuenning What Does This Chart Tell Us? Do we or don’t we see a trend in the errors? Clearly they’re higher at the highest level of the predictors But is that alone enough to call a trend? –Perhaps not, but we should take a close look at both the factors to see if there’s a reason to look further –And take results with a grain of salt

24. © 1998, Geoff Kuenning Quantile-Quantile Plot for Example

25. © 1998, Geoff Kuenning Confidence Intervals for Effects Need to determine the standard deviation for the data as a whole From which standard deviations for the effects can be derived –Using different degrees of freedom for each Complete table in Jain, pg. 351

26. © 1998, Geoff Kuenning Standard Deviations for Our Example s e = 25 Standard deviation of  - Standard deviation of  j - Standard deviation of  i -

27. © 1998, Geoff Kuenning Calculating Confidence Intervals for Our Example Just the file system alternatives shown here At 95% level, with 4 degrees of freedom CI for library solution - (-39,26) CI for VFS solution - (-49,16) CI for layered solution - (-10,55) So none of these solutions are 95% significantly different than the mean

28. © 1998, Geoff Kuenning Looking a Little Closer Does this mean that none of the alternatives for adding the functionality are different? Use contrasts to check contrasts – any linear combination of effects whose coeff add up to zero (ch. 18.5)

29. © 1998, Geoff Kuenning Looking a Little Closer For example, is the library approach significantly better than layers? –Using contrast of  library -  layers, the confidence interval is (-58,-.5) (at 95%) –So library approach is better, with this confidence

30. Two-Factor Full Factorial Design With Replications Replicating a full factorial design allows separating out the interactions between factors from experimental error Without replicating implies assumption that interactions were negligible and could be viewed as errors. Read Chapter 22

31. ( l 1,0, l 1,1, …, l 1,n1-1 ) x ( l 2,0, l 2,1, …, l 2,n2-1 ) x … x ( l k,0, l k,1, …, l k,nk-1 ) k different factors, each factor with n i levels r replications Informal techniques just to answer which combo is of levels is best – rank responses Factor 1 Factor k Factor 2 Full Factorial Design With k Factors

32. Experiment Design Overview Number of factors 1 2 k levels 2:min/max n - cat num regression models2k2k repl interactions & errors 2 k-p weak interactions n x r pts 2kr2kr est. errors full factorial !additive log transform Jain ch 14,15 Jain ch 17 Jain ch 19 Jain ch 18 Jain ch 20Jain ch 23 Jain ch 21 Jain ch 22

33. For Discussion Project Proposal 1.Statement of hypothesis 2.Workload decisions 3.Metrics to be used 4.Method