n=5 for number of samples
Kernel Mean CI(95%)
2.5.38 411 344-477
2.5.39-gcc32 371 224-519
2.5.38-mm2 95 84-105
The mean is a simple average of the results, and the CI(95%) are the 95%
confidence intervals the mean lies between those numbers. These numbers seem to
be the most useful for comparison.
Comparing 2.5.38(gcc2.95.3) with 2.5.38(gcc3.2) there is NO significant
difference (p 0.56)
Comparing 2.5.38 with 2.5.38-mm2 there is a significant diffence (p<0.001)
After playing with all these it appears I should do the following to contest:
Add sync,swapoff,swapon before each load
Perform noload and process_load twice to ensure no abnormal results
Perform mem_load 3 times
Perform IO_fullmem 5 times (and rename it just IO_load)
Drop IO_halfmem (adds no more useful information and just adds time).
Do a statistical analysis like the above when posting information.
Comments?
Con
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