Hello,

I currently run aov in the following way:

> throughput.aov <- aov(log(Throughput)~No_databases+Partitioning+No_middlewares+Queue_size,data=throughput)
> summary(throughput.aov)

                Df Sum Sq Mean Sq  F value    Pr(>F)    
No_databases     1 184.68 184.675 136.6945 < 2.2e-16 ***
Partitioning     1  70.16  70.161  51.9321 2.516e-12 ***
No_middlewares   2  44.22  22.110  16.3654 1.395e-07 ***
Queue_size       1   0.40   0.395   0.2926    0.5888    
Residuals      440 594.44   1.351                       
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

In order to compute the fraction of variation I need to know the total Sum Sq. and I assume it is like this:

SST = SS-No_databases + SS-Partitioning + SS-No_middlewares + SS-Queue_size 
        = 184.68 + 70.16 + 44.22 + 0.40
        = 299.46

So the fraction of variation explained by the No_databases would be:

SST/SS-No_databases = 184.68/299.46 = 0.6167101 ... and finally I can say that the No_databases explains 61.6% of the variation in Throughput.

Is this correct? if so, how can I do the same calculations using R? I haven't found the way to extract the Sum Sq out of the throughput.aov Object. Is there a function to get this 0.6167101 and 61.6% results without having to do it manually? even better if I get a table containing all these fraction of variations?

Since this is a 2^k experiment, I cant see how the Residuals fit in the formula. When I introduce replications (blocking factor) then I can also include a SSE term into the SST calculation. 

TIA,
Best regards,
Giovanni


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