Quadratic Optimization for 4D in for Loop

Question

MarshallSc 2021-12-31

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1620025-quadratic-optimization-for-4d-in-for-loop

评论： yanqi liu 2022-1-1

I need to find the roots (complex in nature) of an objective function in 4D by using quadratic optimization for the function below:

a = [0.0068 0.0036 0.000299 0.0151]; b = [0.0086 0.00453 0.0016 0.00872]
f = @(xj,xk) a(i) - (x(j)*x(k)) * b(l); %i,j,k,l = 1:4   -    simple eqn: f = @(x1,x2) a(1) - (x(2)*x(3)) * b(4)

The problem that I have is that I don't know how to write it in a for loop or permutation manner that each loop takes a specific value of the (a,b) and (xj,xk) from 1:4. Basically it's a nonlinear coordinate transformation. Since my X(i) * X(j) makes the problem quadratic, I need to perform an approximation using the only equality constraint such (imposing the symmetry of the potential function - (i,j) and (k,l) pair become exchangeable):

(x(j)*x(k)) * b(l) + (x(i)*x(l)) * b(k) + (x(k)*x(j)) * b(j) + (x(l)*x(i)) * b(i) =< a(i) + a(j) + a(k) + a(l)

That's my only constraint for optimization that minimizes the objective function. I tried using fmincon but I don't know how to use it in a loop for the equation and the constraint.

I'd appreciate it if someone can help me! Thank you!

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yanqi liu 2021-12-31

yes，sir，may be write the equations，and we can use loop to generate cmd string，then use eval to get function handle

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Answer 1

yanqi liu 2021-12-31

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1620025-quadratic-optimization-for-4d-in-for-loop#answer_865655

在 MATLAB Online 中打开

clc; clear all; close all;
a = [0.0068 0.0036 0.000299 0.0151]; 
b = [0.0086 0.00453 0.0016 0.00872];
% f = @(xj,xk) a(i) - (x(j)*x(k)) * b(l); %i,j,k,l = 1:4   -    simple eqn: f = @(x1,x2) a(1) - (x(2)*x(3)) * b(4)
for i = 1 : length(a)
    eqi = sprintf('f=@(x1,x2) %f- (x(1)*x(2))*%f;', a(i), b(i));
    disp(eqi)
end
f=@(x1,x2) 0.006800- (x(1)*x(2))*0.008600;
f=@(x1,x2) 0.003600- (x(1)*x(2))*0.004530;
f=@(x1,x2) 0.000299- (x(1)*x(2))*0.001600;
f=@(x1,x2) 0.015100- (x(1)*x(2))*0.008720;

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MarshallSc 2021-12-31

编辑：MarshallSc 2021-12-31

Thanks Yanqi for your answer. Do you know how I can incoporate this pemutated equation to solve for the quadratic optimization to find the minimum values of Xs? a & b are just the coefficient of the second order polynomial.

I'm a little bit lost as to what should be done. I'd appreciate it.

yanqi liu 2022-1-1