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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