Hi All, I would like to use R to do some computation. However, before I proceed to write an R script, I have difficulty to find out a solution for the question below. Please note that the question below is not relating to question in R. I have a table below, 2009 2010 2011 2012 2013 2014 Albert 77.28 77.77 77.83 78.11 78.14 78.45 Edmund 87.20 88.00 88.43 89.05 89.07 91.06 Sandy 92.39 92.68 93.68 93.92 96.25 96.49 David 340.60 343.01 344.11 345.75 348.38 351.81 Kean 115.20 115.57 116.45 117.18 117.41 118.46 Janet 128.76 128.93 130.28 132.09 132.68 133.00 Lily 133.92 135.50 135.60 137.10 137.25 137.80 Total 975.35 981.46 986.38 993.20 999.18 1,007.07 Above table shows the total sales made by each sale person from year 2009-2014. A user can change the values in the table, let say, the total sales for Simon from 2009 to 2014 would like to be change to values as below, David 408.72 411.61 412.93 414.90 418.05 422.17 The total sales number should remain unchanged upon the changes made on David. The variance has to be absorbed proportionately by other sales person. Does anyone know a method to distribute the variance to others? Here was my first thought for the solution, For year 2009, the sales for David is changed to 408.72 from 340.60, so the difference is (408.72 -340.60) = 68.12. I'm thinking to divide the difference 68.12 by 6, which is 11.353. The sales for other persons will be subtracted 11.353. 2009 Albert 65.9243 Edmmund 75.8446 Sandy 81.0398 David 408.724 Kean 103.848 Janet 117.405 Lily 122.566 total 975.352 However, I've found out this doesn't work for case like a person sales is not sufficient for the subtraction. Let say if a person's total sales is 10++, then 10++ minus 11.353 would be a negative value. This is just an example, which may not be make sense to the case above. Does anyone have a better solution to distribute the variance, may be by weight or something? You help is very much appreciated. Again, I would like to emphasize that my question above is not related to R code, I just want to find out a methodology to solve the problem above. Thanks! [[alternative HTML version deleted]]