Quadratic Optimization for 4D in for Loop
    7 views (last 30 days)
  
       Show older comments
    
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!
1 Comment
  yanqi liu
      
 on 31 Dec 2021
				yes,sir,may be write the equations,and we can use loop to generate cmd string,then use eval to get function handle
Accepted Answer
  yanqi liu
      
 on 31 Dec 2021
        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
2 Comments
  yanqi liu
      
 on 1 Jan 2022
				yes,sir,may be it is non-linear optimization,use fmincon to get compute,may be write your equations in handwriting,we can make some debug 
More Answers (0)
See Also
Categories
				Find more on Systems of Nonlinear Equations in Help Center and File Exchange
			
	Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!