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Weighted score distribution with fixed maximum

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    Weighted score distribution with fixed maximum

    Probably more of a maths than Excel question but going a little round in circles.

    Boiling the problem down a bit. Imagine a range of binary values (1 & 0, true & false, yes & no...). I need to give each value a score. If the value is (for example) false the score is always zero. If true the scores vary depending on weights and a total potential score.

    Ok, no problem yet.

    Now, some of these values could also be "n/a". In this case there is no score but the max potential score for this value needs to be split amongst the other values proportional to their weight so the max score of all the values wich are not "n/a" remains the same.

    Hope that is enough explanation and someone has a clue to solve the puzzle.
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    Re: Weighted score distribution with fixed maximum

    Hi,

    I'm struggling to visualise this.

    If the values are all binary how can a weighting enter the equation? I can see how if values were say 2,4,6 etc. with a weighting to each respectively of 10,11,12 that you could work out a total weighted score ignoring n/a or whatever, but since the non zero score is by definition always 1, I can't see this.

    Would you upload an example table of your data and show whatever results you expect to see and explain how you get from one to the other.
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    Re: Weighted score distribution with fixed maximum

    Something like this should work.

    This setup applies to any binary system you wish to apply.

    I went with unicorns and carebears, because I'm funny like that.

    Attachment 261756

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    Re: Weighted score distribution with fixed maximum

    Working from my phone at the moment so limited to words. Let me try...

    The binary values are just switches. The weights are external. So, if A1:A4 can be either pass or fail and A5 can be pass, fail or n/a. A pass in all scores 100 points. However, due to the 'external' weighting a pass in A1 scores 10, in A2:A3 scores 20 each and in A4:A5 scores 25 each.

    This would be simple enough but if A5 is n/a then the potential 25 points from that value needs to be spread over the other values proportionally. So the total of all the passes (and one n/a) remains the same.

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    Re: Weighted score distribution with fixed maximum

    Got it. Average overall weight will split up the point values.

    So 4 answers, 1 n/a means each value increased by 25% of the skipped question.

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    Re: Weighted score distribution with fixed maximum

    I updated the attachment here.

    The top has a place to enter your binary values, complete with positive and negative definition. You can define them as T/F, 1/0, bananas/oranges, whatever.

    The formulas first count the total positive values and divides that by the sum of all positive and negative values. This gives the answer's average weight.

    In a 5 question test with T and F, featuring 3 True, 1 False, 1 n/a (in that order) the average weight of each question actually answered would be 25%.

    The score is calculated using SUMPRODUCT to find the score value of all questions with n/a, adding them together, and multiplying that against the average weight. This value is then added to that question's original value if the answer had a positive value.

    Back to my example, this would mean question 5 is split 4 ways.

    Question 1 is now worth 10 + (25/4) points
    Question 2 is now worth 20 + (25/4) points
    Question 3 is now worth 20 + (25/4) points
    Question 4 is now worth 25 + (25/4) points, but was F so it's 0 points.

    We then add it up: 68.75 points

    Attachment 261770

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    Re: Weighted score distribution with fixed maximum

    Huge thanks daffodil, I'll download and try it in the morning.

    Just one thing though... "+ (25/4) points", isnt that ignoring the external score? If so the low score pass has greater proportional value which is what I need to avoid.

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    Re: Weighted score distribution with fixed maximum

    You can't spread the n/a values out proportionally like you require, and have the low score not increase proportionally.

    You said you wanted the values split up for the n/a.

    If you passed the first three, failed the 4th, and skipped the last your scores would be: 16.25, 26.25, 26.25, 0 = 68.75


    Did you mean you want to spread the points out disproportionately? Do you want the 10% of 25 points from the n/a to hit 1 instead, and so on and so forth?
    Last edited by daffodil11; 08-29-2013 at 04:59 PM.

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    Re: Weighted score distribution with fixed maximum

    I'm, ya, I guess. Proportional to the remaining external weights.

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    Re: Weighted score distribution with fixed maximum

    I tested it out from the other direction.

    Using the example above..

    Question 1: (100 ÷ Answered Weights[10+20+20+25] ) x ( Answer Value [10]) = (100/75)*10 = 1.33 * 10 = 13.33
    Question 2: (100 ÷ Answered Weights[10+20+20+25] ) x ( Answer Value [20]) = (100/75)*20 = 1.33 * 20 = 26.67
    Question 3: (100 ÷ Answered Weights[10+20+20+25] ) x ( Answer Value [20]) = (100/75)*20 = 1.33 * 20 = 26.67
    Question 4: (100 ÷ Answered Weights[10+20+20+25] ) x ( Answer Value [0]) = (100/75)*0 = 1.33 * 0 = 0


    This makes the overal score 66.67

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