[R] 2^k*r (with replications) experimental design question

[Dennis Murphy]

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 <bravegag at 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
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