Excel How to take a weighted average where blank values cause proportional weighting shifts

0 votes

I have five categories. There is a weight for each category in row 2 and a value in row 3. I want a formula that computes the weighted average of each category to go in cell G3.

The main qualification is that some categories won't have values. If a category is empty, the other categories with values should get an equal amount of its weighting before calculating the weighted average. Here are two examples:

enter image description here

In this photo, cell C3 is missing a value. There are 4 categories remaining that do have values. So its weighting of 0.2 will be redistributed four ways, so each of the remaining categories will have a (02./4 =0.05) increase in their weightings in the weighted average. The highlighted cell is the desired result.

I will further illustrate with one additional scenario:

enter image description here

Two categories in the aforementioned case contain missing data and a weighting of 0.15 each. Three categories each have values. Therefore, the three remaining categories will receive a new weighting of 0.15/3 = 0.05 for each missing category. The 2 missing categories will result in an enhanced weighting for each category of 2*0.05 = 0.1. The appropriate weighted average can be seen in the yellow box.

One more restriction: It would be ideal if the formula were flexible enough to accommodate ranges larger than 5 categories. Any suggestions on how to put something similar into practice?

Sep 26, 2022 in Others by Kithuzzz
• 27,740 points
102 views

1 answer to this question.

0 votes

This seems to work for me, though I think it may be made simpler by following the steps specified in the question:

=SUMPRODUCT(B2:F2*B3:F3)+SUMPRODUCT(SUMPRODUCT((B3:F3="")*B2:F2)*B3:F3)/COUNT(B3:F3)

enter image description here

If the blank cells are not actually blank but contain "" returned as the result of a formula, you could try

=SUMPRODUCT(B2:F2*N(+B3:F3))+
SUMPRODUCT(SUMPRODUCT((B3:F3="")*B2:F2)*N(+B3:F3))/COUNT(B3:F3)

As suggested by the brilliant @XOR LX in this answer

answered Sep 27, 2022 by narikkadan
• 51,240 points

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