optimize four functions together

2023 5 24
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Hello,
I have four functions and I want to optimize them together by ga. I know that I can solve each function alone and I already got an answer about that, but if I have all of them. The values I want to get is F1=0.405, F2=24.736 ,F3=0.525, F4=14.97. I approciate any help.
F1=@(x) 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;
F2 = @(x) 30.07+71.68*x(1)-21.83*x(2)-306.55*x(1)*x(2)+179*x(2)^2;
F3 = @(x) 0.54-18.32*x(3)-10.6*x(1)-3.22*x(2)+0.3*x(4)+273.71*x(3)*x(1)+60.28*x(1)*x(2)-19.81*x(2).^2;
F4 = @(x) 17.39+1246.36*x(3)+348.83*x(1)-88.27*x(2)-43.72*x(4)-24455.25*x(3)*x(1)-1164.66*x(1)*x(2)+347.38*x(2)*x(4);
FitnessFunction=[F1;F2;F3;F4];
% [ fn, fc, f0, ff] ; % the range like this
lb = [0.001,0.01,0.0002,0.1];
ub = [0.045,0.1,0.0045,0.2];
numberOfVariables = 4;
A = []; b = [];
Aeq = []; beq = [];
[x,fval] = ga(FitnessFunction, numberOfVariables, A, b, Aeq, beq, lb, ub)
Many thanks

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Walter Roberson
Walter Roberson 2023-5-24
移动：Matt J 2023-5-24

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nadia nadi
nadia nadi 2023-5-24
移动：Matt J 2023-5-24
Thank you very much
Walter Roberson
Walter Roberson 2023-5-25
Note that gamultiobj tries to find places that locally optimize all of the functions together -- places in which changing any of the model parameters a little, increases the function value. Places, in other words, where the curl of the set of functions is positive for all of the functions simultaneously.
This is quite a different requirement than optimizing each of the functions individually.
Imagine for example that one of the functions is (x-10)^2+1 and another of the functions is (x-20)^2+1 -- then if you optimize them individually you can easily get x==10 and x==20, but if you try to optimize them together then the better you fit the first function the worse the second one fits, and the compromise point is x==15 where the two functions are equally bad.
nadia nadi
nadia nadi 2023-5-25
Thanks, I understand that. I tried but couldn’t get the same results. I may need to look at it more. Best
Walter Roberson
Walter Roberson 2023-5-25
gamultiobj finds Pareto points, rather than optimizing each function individually. To use gamultiobj to optimize each function independently but with a single optimization call, you would need to use independent variables for each. For example if each function was over two variables you could use 4*2=8 variables to allow the four functions to work independently.
Walter Roberson
Walter Roberson 2023-5-25
Ran in:
Ran in:
Option 1: functions are independent, but for some reason you want to call an optimizer only once instead of making four separate optimization calls. Note that this approach will always be less efficient than making separate optimization calls:
F1=@(x) 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;
F2 = @(x) 30.07+71.68*x(1)-21.83*x(2)-306.55*x(1)*x(2)+179*x(2)^2;
F3 = @(x) 0.54-18.32*x(3)-10.6*x(1)-3.22*x(2)+0.3*x(4)+273.71*x(3)*x(1)+60.28*x(1)*x(2)-19.81*x(2).^2;
F4 = @(x) 17.39+1246.36*x(3)+348.83*x(1)-88.27*x(2)-43.72*x(4)-24455.25*x(3)*x(1)-1164.66*x(1)*x(2)+347.38*x(2)*x(4);
FitnessFunction = @(x)[F1(x(1:2));F2(x(3:4));F3(x(5:8));F4(x(9:12))];
lb = [0.001,0.01,0.001,0.01,0.001,0.01,0.0002,0.1,0.001,0.01,0.0002,0.1]
lb = 1×12
0.0010 0.0100 0.0010 0.0100 0.0010 0.0100 0.0002 0.1000 0.0010 0.0100 0.0002 0.1000
ub = [0.045,0.1,0.045,0.1,0.045,0.1,0.0045,0.2,0.045,0.1,0.0045,0.2]
ub = 1×12
0.0450 0.1000 0.0450 0.1000 0.0450 0.1000 0.0045 0.2000 0.0450 0.1000 0.0045 0.2000
numberOfVariables = length(lb);
A = []; b = [];
Aeq = []; beq = [];
[x,fval] = gamultiobj(FitnessFunction, numberOfVariables, A, b, Aeq, beq, lb, ub)
Optimization terminated: average change in the spread of Pareto solutions less than options.FunctionTolerance.
x = 70×12
0.0069 0.0874 0.0029 0.0699 0.0449 0.0998 0.0044 0.1174 0.0228 0.0671 0.0045 0.1266 0.0197 0.0478 0.0209 0.0822 0.0237 0.0424 0.0014 0.1303 0.0023 0.0850 0.0003 0.1963 0.0428 0.0262 0.0249 0.0678 0.0434 0.0967 0.0043 0.1480 0.0424 0.0103 0.0036 0.1437 0.0450 0.0133 0.0194 0.0335 0.0375 0.0753 0.0036 0.1438 0.0368 0.0509 0.0022 0.1538 0.0233 0.0527 0.0104 0.0886 0.0072 0.0194 0.0009 0.1574 0.0022 0.0842 0.0009 0.1781 0.0283 0.0451 0.0159 0.0898 0.0034 0.0173 0.0005 0.1603 0.0018 0.0845 0.0008 0.1790 0.0295 0.0369 0.0066 0.0876 0.0050 0.0162 0.0014 0.1669 0.0107 0.0921 0.0010 0.1376 0.0097 0.0709 0.0010 0.0641 0.0448 0.0706 0.0045 0.1184 0.0090 0.0996 0.0045 0.1625 0.0277 0.0369 0.0066 0.0872 0.0031 0.0162 0.0003 0.1669 0.0105 0.0910 0.0010 0.1376 0.0281 0.0523 0.0101 0.0892 0.0037 0.0188 0.0009 0.1593 0.0021 0.0844 0.0009 0.1784
fval = 70×4
0.2461 29.5665 -0.1761 18.1384 0.1969 30.4569 0.2000 8.0086 0.0949 30.6783 -0.1471 25.7319 0.0731 30.7326 -0.0284 20.2926 0.1861 30.0021 0.4339 9.0363 0.1641 30.2558 0.4850 8.7437 0.1546 29.8277 0.4598 11.2062 0.2375 29.4582 -0.0617 13.7948 0.1615 29.8246 0.4979 11.2112 0.1686 29.9956 0.4701 8.9272
Walter Roberson
Walter Roberson 2023-5-25
Ran in:
Ran in:
Option 2: variables are shared, x(1) is the same variable for each, x(2) is the same for each, x(3) is the same for each that uses it, etc.
F1=@(x) 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;
F2 = @(x) 30.07+71.68*x(1)-21.83*x(2)-306.55*x(1)*x(2)+179*x(2)^2;
F3 = @(x) 0.54-18.32*x(3)-10.6*x(1)-3.22*x(2)+0.3*x(4)+273.71*x(3)*x(1)+60.28*x(1)*x(2)-19.81*x(2).^2;
F4 = @(x) 17.39+1246.36*x(3)+348.83*x(1)-88.27*x(2)-43.72*x(4)-24455.25*x(3)*x(1)-1164.66*x(1)*x(2)+347.38*x(2)*x(4);
FitnessFunction = @(x)[F1(x(1:2));F2(x(1:2));F3(x(1:4));F4(x(1:4))];
lb = [0.001,0.01,0.0002,0.1];
ub = [0.045,0.1,0.0045,0.2];
numberOfVariables = length(lb);
A = []; b = [];
Aeq = []; beq = [];
[x,fval] = gamultiobj(FitnessFunction, numberOfVariables, A, b, Aeq, beq, lb, ub)
Optimization terminated: average change in the spread of Pareto solutions less than options.FunctionTolerance.
x = 20×4
0.0450 0.0100 0.0045 0.2000 0.0435 0.0991 0.0023 0.1445 0.0018 0.0867 0.0011 0.1504 0.0015 0.0672 0.0020 0.1515 0.0015 0.0627 0.0021 0.1536 0.0027 0.0203 0.0025 0.1472 0.0353 0.0101 0.0043 0.1983 0.0021 0.0138 0.0023 0.1839 0.0409 0.0872 0.0031 0.1794 0.0404 0.0106 0.0044 0.1578
fval = 20×4
0.0695 32.9573 0.0889 24.2877 0.1360 31.4632 -0.1465 17.8785 0.2620 29.6038 0.1286 9.4199 0.2646 29.4904 0.2332 11.2367 0.2645 29.4862 0.2587 11.4209 0.2497 29.8765 0.4412 14.0291 0.1098 32.2882 0.1753 22.0645 0.2491 29.9413 0.4849 12.5083 0.1403 31.3677 -0.0786 18.1736 0.0891 32.6241 0.1160 24.8790

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nadia nadi
nadia nadi 2023-5-25
编辑：nadia nadi 2023-5-25

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Hello Walter,

I have two variables in First and second equations and four variables in the third and fourth. So you mean I need to write 12 different variables, I set their range and run the code ??

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Torsten
Torsten 2023-5-25
编辑：Torsten 2023-5-25
If these 12 variables are not interconnected, you have 4 independent problems to solve. Thus call "fmincon" 4 times, with F_i as objective function for the i-th call.

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