The calculation the MAD (Mean Absolute Deviation)/Mean ratio is this, according to the title: $$ \frac{\overline{\left | Forecast - Demand \right |}}{\overline{Demand}}$$ However, the calculation is often shown as this, because of the derivation from wMAPE (weighted Mean Absolute Percentage Error), with sum not average: $$ \frac{1}{\sum Demand}\sum_{}^{}Demand\frac{\left | Forecast - Demand \right |}{Demand}=\frac{\sum_{}^{}\left | Forecast - Demand \right |}{\sum_{}^{}Demand} $$ I understand why it's shown as that, because of the derivation, but that means that the title (mean) and the calculation (sum) don't match up. Are there pros and cons to using mean or sum? The key ones that I can think for using average are:

  • Fair comparison between numerators and denominators of different populations, because sum is affected by number of observations
  • Average is less affected by missing data than sum is, when you apply that aggregation across a column
  • $\begingroup$ I can't follow this post because the notation and the language are so ambiguous. You seem to express a ratio of means as a weighted mean relative difference, but where you're going with that and what you are trying to ask are unclear. I wish I could suggest what to do, but I truly can't figure out what your question is. $\endgroup$
    – whuber
    Commented Apr 13, 2023 at 20:39

1 Answer 1


If you are a little more careful and explicit in your formulas, it might well get a lot clearer.

Specifically, one typically calculates means or sums over forecasts and actuals for the same periods (as opposed, say, to calculating the MAD over the forecast horizon, but the mean of actuals over the training sample). And then both the numerator and the denominator have the same number of summands, say $N$. But then the sums you have yield the exact same ratio as dividing the means instead - because going from the sums to averages only means that we divide both the numerator and the denominator by $N$, and this cancels.

Some more on this can be found in Kolassa & Schütz (2007).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.