# Finding the optimal solution for a data with two variables

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AAAAAA on 11 Jun 2023
Commented: Alan Stevens on 12 Jun 2023
I have a data set for two variables (two columns) in which each variable has thousands of solutions (rows). I also have the actual solution.
I want to find the best solution (row) with the lowest error considering both variables (not only one variable).
As a simple example containing only 10 solutions (rows), I have the following:
% actual solution:
X1_actual = 0.4722;
X2_actual = 4.4;
% predicted data:[X1_predicted X2_predicted]
Predicted_data = [0.4742 4.4557
0.4739 4.4553
0.4732 4.4549
0.4730 4.4545
0.4725 4.4540
0.4723 4.4536
0.4715 4.4532
0.4714 4.4528
0.4713 4.4505
0.4701 4.4501]
Where the first and second columns represent the predicted values of X1 and X2, respectively.
The problem is that the minimum errors for both variables are not at the same row. As can be seen in this example, the minimum error for X1 is in the 6th row while for X2 it is in the last row.
Is there a scientific method to find the optimal row that considers the minimum errors of both variables?
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AAAAAA on 11 Jun 2023
OK. Thanks What are the other methods so that I search for these methods and compare the results with this method?
Torsten on 11 Jun 2023
Edited: Torsten on 11 Jun 2023
As I wrote: all norms on IR^2 can be used:

Alan Stevens on 11 Jun 2023
Here's a possible approach (I've used relative errors as the values of the two columns are an order of magnitude different):
% actual solution:
X1_actual = 0.4722;
X2_actual = 4.4;
% predicted data:[X1_predicted X2_predicted]
Predicted_data = [0.4742 4.4557
0.4739 4.4553
0.4732 4.4549
0.4730 4.4545
0.4725 4.4540
0.4723 4.4536
0.4715 4.4532
0.4714 4.4528
0.4713 4.4505
0.4701 4.4501];
Relative_error = [Predicted_data(:,1)/X1_actual, ...
Predicted_data(:,2)/X2_actual] ...
- ones(10,2);
combined_error = sqrt(Relative_error(:,1).^2 + Relative_error(:,2).^2);
minerror = min(combined_error)
minerror = 0.0116
minerror_row = find(combined_error == minerror)
minerror_row = 9
AAAAAA on 12 Jun 2023
Great This worked very well. But is the scaling you did very compulsory? How if we use the data as it is without doing any scaling? Will this affect the results?
Alan Stevens on 12 Jun 2023
If the errors of X2 are much larger than those of X1 they will dominate the combined contribution and you might find that the result is the same as if you only used X2!
However, you could try it and see!