CPE 619 Two-Factor Full Factorial Design With ReplicationsOverviewModelModel (cont’d)Computation of EffectsExample 22.1: Code SizeExample 22.1: Log TransformationExample 22.1: Computation of EffectsExample 22.1: InteractionsComputation of ErrorsAllocation of VariationAnalysis of VarianceANOVA for Two Factors w ReplicationsExample 22.4: Code Size StudyConfidence Intervals For EffectsExample 22.5: Code Size StudyExample 22.5: Conf. Intervals (cont’d)Example 22.5: CI for InteractionsExample 22.5: Visual TestsSummaryCPE 619 General Full Factorial Designs With k FactorsSlide 22General Full Factorial Designs With k FactorsModel ParametersCase Study 23.1: Paging ProcessCase Study 23.1 (cont’d)Slide 27Slide 28Case Study 23.1: ANOVA TableCase Study 23.1: Simplified modelCase Study 23.1: Simplified Model (cont’d)Case Study 23.1: Error ComputationCase Study 23.1: Visual TestCase Study 23.1: Final ModelObservation MethodExample 23.1: Measured ThroughputsExample 23.1: ConclusionsRanking MethodExample 23.2: ConclusionsRange MethodSlide 41CPE 619Two-Factor Full Factorial DesignWith ReplicationsAleksandar MilenkovićThe LaCASA LaboratoryElectrical and Computer Engineering DepartmentThe University of Alabama in Huntsvillehttp://www.ece.uah.edu/~milenkahttp://www.ece.uah.edu/~lacasa2OverviewModelComputation of Effects Estimating Experimental ErrorsAllocation of VariationANOVA Table and F-TestConfidence Intervals For Effects3ModelReplications allow separating out the interactions from experimental errorsModel: With r replicationsWhere4Model (cont’d)The effects are computed so that their sum is zero:The interactions are computed so that their row as well as column sums are zero:The errors in each experiment add up to zero:5Computation of EffectsAveraging the observations in each cell:Similarly,  Use cell means to compute row and column effects6Example 22.1: Code Size7Example 22.1: Log Transformation8Example 22.1: Computation of EffectsAn average workload on an average processor requires a code size of 103.94 (8710 instructions)Proc. W requires 100.23 (=1.69) less code than avg processorProcessor X requires 100.02 (=1.05) less than an average processor …The ratio of code sizes of an average workload on processor W and X is 100.21 (= 1.62).9Example 22.1: InteractionsCheck: The row as well column sums of interactions are zero Interpretation: Workload I on processor W requires 0.02 less log code size than an average workload on processor W or equivalently 0.02 less log code size than I on an average processor10Computation of ErrorsEstimated Response:Error in the kth replication:Example 22.2: Cell mean for (1,1) = 3.8427Errors in the observations in this cell are:3.8455-3.8427 = 0.00283.8191-3.8427 = -0.0236, and3.8634-3.8427 = 0.0208Check: Sum of the three errors is zero11Allocation of VariationInteractions explain less than 5% of variation  may be ignored12Analysis of VarianceDegrees of freedoms:13ANOVA for Two Factors w Replications14Example 22.4: Code Size StudyAll three effects are statistically significant at a significance level of 0.1015Confidence Intervals For EffectsUse t values at ab(r-1) degrees of freedom for confidence intervals16Example 22.5: Code Size StudyFrom ANOVA table: se=0.03. The standard deviation of processor effects:The error degrees of freedom:ab(r-1) = 40  use Normal tablesFor 90% confidence, z0.95 = 1.64590% confidence interval for the effect of processor W is:1 ¨ t s1 = -0.2304 ¨ 1.645 £ 0.0060 = -0.2304 ¨ 0.00987 = (-0.2406, -0.2203) The effect is significant17Example 22.5: Conf. Intervals (cont’d)The intervals are very narrow.18Example 22.5: CI for Interactions19Example 22.5: Visual TestsNo visible trend.Approximately linear ) normality is valid20SummaryReplications allow interactions to be estimatedSSE has ab(r-1) degrees of freedomNeed to conduct F-tests for MSA/MSE, MSB/MSE, MSAB/MSECPE 619General Full Factorial Designs With k FactorsAleksandar MilenkovićThe LaCASA LaboratoryElectrical and Computer Engineering DepartmentThe University of Alabama in Huntsvillehttp://www.ece.uah.edu/~milenkahttp://www.ece.uah.edu/~lacasa22OverviewModelAnalysis of a General DesignInformal MethodsObservation MethodRanking MethodRange Method23General Full Factorial Designs With k FactorsModel: k factors ) 2k-1 effectsk main effectstwo factor interactions,three factor interactions, and so on. Example: 3 factors A, B, C:24Model ParametersAnalysis: Similar to that with two factors The sums of squares, degrees of freedom, and F-test also extend as expected25Case Study 23.1: Paging ProcessTotal 81 experiments26Case Study 23.1 (cont’d)Total Number of Page Swapsymax/ymin = 23134/32 = 723  log transformation27Case Study 23.1 (cont’d)Transformed Data For the Paging Study28Case Study 23.1 (cont’d)Effects:AlsoSix two-factor interactions,Four three-factor interactions, andOne four-factor interaction.29Case Study 23.1: ANOVA Table30Case Study 23.1: Simplified modelMost interactions except DM are small.Where,31Case Study 23.1: Simplified Model (cont’d)Interactions Between Deck Arrangement and Memory Pages32Case Study 23.1: Error Computation33Case Study 23.1: Visual TestAlmost a straight lineOutlier was verified34Case Study 23.1: Final ModelStandard Error= Stdv of sample mean= Stdv of Error35Observation MethodTo find the best combinationExample: Scheduler DesignThree Classes of Jobs:Word processingInteractive data processingBackground data processingFive Factors 25-1 design36Example 23.1: Measured Throughputs37Example 23.1: ConclusionsTo get high throughput for word processing jobs:1. There should not be any preemption (A=-1)2. The time slice should be large (B=1)3. The fairness should be on (E=1)4. The settings for queue assignment and re-queueing do not matter38Ranking MethodSort the experiments.39Example 23.2: Conclusions1. A=-1 (no preemption) is good for word processing jobs and also that A=1 is bad2. B=1 (large time slice) is good for such jobs. No strong negative comment can be made about B=-13. Given a choice C should be chosen at 1, that is, there should be two queues4. The effect of E is not clear5. If top rows chosen, then E=1 is a good choice40Range MethodRange = Maximum-MinimumFactors with large range are important