I have been comparing two solutions and used weights for individual areas based on their importance. I was interested which solution is better, i.e. which one has the higher score.

The thing is I simply multiplied each score with individual weight (Security with 0,8, UI with 0,2 etc) and then summed them all up and compared. Does it make sense or I need to use weighted average?

In the following example, why the authors used weighted average and not simply multiplied and sum up results using 0.5 etc instead of 5?

E.g.: enter image description here


If the sum of the weights is in both cases the same (e.g. 1), you can do it like you did, otherwise you should use the weighted average.

  • $\begingroup$ Thanks, I updated my question. Not sure when I can use what you mentioned and why. $\endgroup$ – Spietro May 15 '15 at 6:54
  • $\begingroup$ Okay, let me explain through you example. In all cases, we have that the sum of the weights is 8. Therefore you could take the sum instead of the average. However, suppose that in the setup, the size was more important, and was given weight 2, with the others still having weight 1. Then you have to take the average, because the sum of the weights of the parts you are comparing is not the same anymore. However, I don't think it is such a big deal to calculate the weighted average every time, so why wouldn't you do that anyway? $\endgroup$ – wythagoras May 15 '15 at 6:59
  • 1
    $\begingroup$ You mean weighted average, right. No it is okay for me to calculate it, I just was not sure why, what is the difference. And in the example, the results differ with sum and weighted average. $\endgroup$ – Spietro May 15 '15 at 7:00
  • $\begingroup$ Yes, I meant that. Do you understand it now? $\endgroup$ – wythagoras May 15 '15 at 7:01
  • $\begingroup$ Not entirely but getting to it :) You say I can use the sum but in that case, I get different results (in the first case it would be 4*01 + 2*0,5 + 1*0,2 which gives 1,6) $\endgroup$ – Spietro May 15 '15 at 7:10

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