I'm guessing you have nine replicates of a 2^5 factorial design with a couple of missing values. If so, define a variable to designate the replicates and use it as a blocking factor in the ANOVA. If you want to treat the replicates as a random rather than a fixed factor, then look into the nlme or lme4 packages.
HTH, Dennis On Sun, Nov 13, 2011 at 4:33 PM, Giovanni Azua <brave...@gmail.com> wrote: > Hello, > > I have one replication (r=1 of the 2^k*r) of a 2^k experimental design in the > context of performance analysis i.e. my response variables are Throughput and > Response Time. I use the "aov" function and the results look ok: > >> str(throughput) > 'data.frame': 286 obs. of 7 variables: > $ Time : int 6 7 8 9 10 11 12 13 14 15 ... > $ Throughput : int 42 44 33 41 43 40 37 40 42 37 ... > $ No_databases : Factor w/ 2 levels "1","4": 1 1 1 1 1 1 1 1 1 1 ... > $ Partitioning : Factor w/ 2 levels "sharding","replication": 1 1 1 1 1 1 1 > 1 1 1 ... > $ No_middlewares: Factor w/ 2 levels "2","4": 1 1 1 1 1 1 1 1 1 1 ... > $ Queue_size : Factor w/ 2 levels "40","100": 1 1 1 1 1 1 1 1 1 1 ... > $ No_clients : Factor w/ 1 level "128": 1 1 1 1 1 1 1 1 1 1 ... >> head(throughput) > Time Throughput No_databases Partitioning No_middlewares Queue_size > 1 6 42 1 sharding 2 40 > 2 7 44 1 sharding 2 40 > 3 8 33 1 sharding 2 40 > 4 9 41 1 sharding 2 40 > 5 10 43 1 sharding 2 40 > 6 11 40 1 sharding 2 40 >> >> throughput.aov <- >> aov(Throughput~No_databases+Partitioning+No_middlewares+Queue_size,data=throughput) >> summary(throughput.aov) > Df Sum Sq Mean Sq F value Pr(>F) > No_databases 1 28488651 28488651 53.4981 2.713e-12 *** > Partitioning 1 71687 71687 0.1346 0.713966 > No_middlewares 1 5624454 5624454 10.5620 0.001295 ** > Queue_size 1 50892 50892 0.0956 0.757443 > Residuals 281 149637226 532517 > --- > Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 >> > > This is somehow what I expected and I am happy, it is saying that the > Throughput is significatively affected firstly by the number of database > instances and secondly by the number of middleware instances. > > The problem is that I need to integrate multiple replications of this same > 2^k so I can also account for experimental error i.e. the _r_ of 2^k*r but I > can't see how to integrate the _r_ term into the data and into the aov > function parameters. Can anyone advice? > > TIA, > Best regards, > Giovanni > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
