Find KKT solution to a given problem
显示 更早的评论
I'm trying to solve kkt problem with matlab but I'm having problem with my code, please, I want some one to assist me in correcting errors in my codes
problem:
Maximize R= 208*x1**0.323168*x2**0.172084*x3**0.085998*x4**0.188794*x5**0.229956
subject to:
2.8 * x1 + 22 * x2 + 4 * x3 + 0.272 * x4 + x5 <= 492
22 * x2 + 4 * x3 <= 212
x5 <= 20
Bounds
x1= (lb = 40, ub = 80)
x2= (lb = 5, ub = 12)
x3= (lb = 4, ub = 11)
x4= (lb = 20, ub = 70)
x5= (lb = 7, ub = 20)
X0 = [41,6,5,22,8]
Proposed solution:
fun = @(x)-208*x(1)^0.323168*x(2)^0.172084*x(3)^0.085998*x(4)^0.188794*x(5)^0.229956;
x0=[41,6,5,22,8]; %initial guess for the solution
%express first constraint in the form A*x<=b:
Aeq=[41,6,5,22,8];
beq = 1;
A=[2.8,22,4,0.272]; b=492;
C=[0,22,4,0,0]; d=212;
E=[0,0,0,0,1]; f=20;
%express constraints 2 and 3 by defining vectors lb and ub:
lb=[40,5,4,20,7]; %no lower blound constraints
ub=[80, 12,11,70,20]; %upper bounds on x(4),x(5)
[x,fval,exitflag,output,lambda]=fmincon(fun,x0,A,b,[],[],lb,ub);
disp(x) %value of x at the solution
disp(-fun(x)); %value of the function you wanted to maximize
disp(A*x') %check constraint 1
disp(lambda)
采纳的回答
Hi Ezra,
You can refer to the below code to solve the problem.
% Objective function
fun = @(x)-208*x(1)^0.323168*x(2)^0.172084*x(3)^0.085998*x(4)^0.188794*x(5)^0.229956;
% Initial guess for the solution
x0 = [41, 6, 5, 22, 8];
% Express first constraint in the form A*x <= b
A = [2.8, 22, 4, 0.272, 1]; % Add the missing element for x5
b = 492;
% Combine the second and third constraints into A and b
A = [A; 0, 22, 4, 0, 0; 0, 0, 0, 0, 1]; % Add the second and third constraints
b = [b; 212; 20];
% Lower and upper bounds
lb = [40, 5, 4, 20, 7]; % Lower bounds
ub = [80, 12, 11, 70, 20]; % Upper bounds
% Run fmincon
[x, fval, exitflag, output, lambda] = fmincon(fun, x0, A, b, [], [], lb, ub);
% Display results
disp(x) % value of x at the solution
disp(-fun(x)); % value of the function you wanted to maximize
disp(A*x' <= b) % check constraints
disp(lambda) % display Lagrange multipliers
Hope this helps!
6 个评论
Ezra
2024-2-14
John D'Errico
2024-2-14
编辑：John D'Errico
2024-2-14
This looks reasonable to me. I would probably have maximized the log of the objective though, as that is often a good way to avoid any potential numerical problems. However, in this case, the variables live in a small enough (well bounded) space that not taking the log will not be an issue. Fmincon seems happy with the result.
Sai Teja G
2024-2-14
@Ezra, you can execute the code provided above and then examine the values of the Lagrange multipliers. To do this, enter 'lambda.upper' and 'lambda.lower' in the command window and run these commands to check the corresponding values.
Ezra
2024-2-14
Star Strider
2024-2-14
Try:
disp(lambda.upper)
Ezra
2024-2-14
移动：John D'Errico
2024-2-14
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