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I am writing a code for using "fmincon" but it is having some error, Why?

2 vues (au cours des 30 derniers jours)
I am writing the following code. I am not able to find the error? I am new to MATLAB.
dbstop if error
clear all
clc
format longEng
kv = 0.5; kh = 0.1;
r0 = 5; H = 5;
q = 20; a = 1;
b = 0; gamma = 18.4;
omega = 25; phi = 39*(pi/180);
lambda = kv/kh;
syms x(i)
% fun = @(x)A1 /((lambda*A1)+A2)
L = r0*(sin(x(2))*exp((x(2)-x(1))*tan(phi))-sin(x(1)))
% % % % Rate of work due to soil weight in static condition
f1 = -(1/3)*(1/(9*((tan(phi))^2)+1)*(exp(3*(x(2) - x(1))*tan(phi))*(3*tan(phi)*sin(x(2))-cos(x(2)))-(3*tan(phi)*sin(x(1))-cos(x(1)))))
f2 = -(1/6)*(L/r0)*cos(x(2))*exp(x(2)-x(1))*tan(phi)*(2*sin(x(1))+(L/r0))
f3 = -(1/3)*(H/r0)*(sin(x(1)))^2
W1 = (1-kv)*gamma*omega*((r0)^3)*(f1-f2-f3)
% Rate of work due to strip load
Wqs = -(1-kv)*omega*q*b*(a+(b/2)+r0*sin(x(1)))
% % % % Rate of work due to soil weight with kh
fq1 = -(1/3)*(1/(9*((tan(phi))^2)+1)*(exp(3*(x(2) - x(1))*tan(phi))*(3*tan(phi)*cos(x(2))+sin(x(2)))-(3*tan(phi)*cos(x(1))+sin(x(1)))))
fq2 = -(1/3)*(L/(r0)^2)*cos(x(2))*exp(x(2)-x(1))*tan(phi)*(cos(x(1))-(H/r0))
fq3 = -(1/6)*(H/r0)*sin(x(1))*(2*cos(x(1))-(H/r0))
W1kh = kh*gamma*omega*((r0)^3)*(fq1-fq2-fq3)
% Rate of work due to strip load
Wqskh = -kh*omega*q*b*r0*exp((x(2)-x(1))*tan(phi))*cos(x(2))
% Yield acceleration coefficients (ky)
A1 = gamma*omega*((r0)^3)*(f1-f2-f3) - omega*q*b*(a+(b/2)+r0*sin(x(1)))
A2 = omega*q*b*r0*exp((x(2)-x(1))*tan(phi))*cos(x(2)) - gamma*omega*((r0)^3)*(fq1-fq2-fq3)
fun = @(x)A1 /((lambda*A1)+A2)
lb = [0,0];
ub = [3.14,3.14];
A = [];
b = [];
Aeq = [];
beq = [];
x0 = [0,0];
nonlcon = [];
x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon)

Réponse acceptée

Matt J
Matt J le 13 Mai 2022
Your objective function is not returning numbers. It is returning sym variables, e.g.,
>> fun([0,0])
ans =
(6.7465e+03*sin(x(1)) - 2.7771e+03*cos(x(1)) + 2.7771e+03*exp(2.4294*x(2) - 2.4294*x(1))*(cos(x(2)) - 2.4294*sin(x(2))) + 1.9167e+04*sin(x(1))^2 - 4.6563e+04*exp(x(2) - 1*x(1))*cos(x(2))*(sin(x(1)) + exp(0.8098*x(2) - 0.8098*x(1))*sin(x(2)))*(0.1667*sin(x(1)) - 0.1667*exp(0.8098*x(2) - 0.8098*x(1))*sin(x(2))))/(3.0955e+04*sin(x(1)) - 2.0632e+04*cos(x(1)) + 1.3885e+04*exp(2.4294*x(2) - 2.4294*x(1))*(cos(x(2)) - 2.4294*sin(x(2))) + 2.7771e+03*exp(2.4294*x(2) - 2.4294*x(1))*(2.4294*cos(x(2)) + sin(x(2))) + 9.5833e+04*sin(x(1))^2 - 9.5833e+03*sin(x(1))*(2*cos(x(1)) - 1) - 2.3281e+05*exp(x(2) - 1*x(1))*cos(x(2))*(sin(x(1)) + exp(0.8098*x(2) - 0.8098*x(1))*sin(x(2)))*(0.1667*sin(x(1)) - 0.1667*exp(0.8098*x(2) - 0.8098*x(1))*sin(x(2))) + 4.6563e+04*exp(x(2) - 1*x(1))*cos(x(2))*(cos(x(1)) - 1)*(0.0667*sin(x(1)) - 0.0667*exp(0.8098*x(2) - 0.8098*x(1))*sin(x(2))))
>> whos ans
Name Size Bytes Class Attributes
ans 1x1 8 sym
  2 commentaires
Matt J
Matt J le 13 Mai 2022
lb = [0,0];
ub = [3.14,3.14];
A = [];
b = [];
Aeq = [];
beq = [];
x0 = [0,0];
nonlcon = [];
x = fmincon(@fun,x0,A,b,Aeq,beq,lb,ub,nonlcon)
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x = 1×2
2.2538 3.1342
function fval = fun(x)
kv = 0.5; kh = 0.1;
r0 = 5; H = 5;
q = 20; a = 1;
b = 0; gamma = 18.4;
omega = 25; phi = 39*(pi/180);
lambda = kv/kh;
% fun = @(x)
L = r0*(sin(x(2))*exp((x(2)-x(1))*tan(phi))-sin(x(1)));
% % % % Rate of work due to soil weight in static condition
f1 = -(1/3)*(1/(9*((tan(phi))^2)+1)*(exp(3*(x(2) - x(1))*tan(phi))*(3*tan(phi)*sin(x(2))-cos(x(2)))-(3*tan(phi)*sin(x(1))-cos(x(1)))));
f2 = -(1/6)*(L/r0)*cos(x(2))*exp(x(2)-x(1))*tan(phi)*(2*sin(x(1))+(L/r0));
f3 = -(1/3)*(H/r0)*(sin(x(1)))^2;
W1 = (1-kv)*gamma*omega*((r0)^3)*(f1-f2-f3);
% Rate of work due to strip load
Wqs = -(1-kv)*omega*q*b*(a+(b/2)+r0*sin(x(1)));
% % % % Rate of work due to soil weight with kh
fq1 = -(1/3)*(1/(9*((tan(phi))^2)+1)*(exp(3*(x(2) - x(1))*tan(phi))*(3*tan(phi)*cos(x(2))+sin(x(2)))-(3*tan(phi)*cos(x(1))+sin(x(1)))));
fq2 = -(1/3)*(L/(r0)^2)*cos(x(2))*exp(x(2)-x(1))*tan(phi)*(cos(x(1))-(H/r0));
fq3 = -(1/6)*(H/r0)*sin(x(1))*(2*cos(x(1))-(H/r0));
W1kh = kh*gamma*omega*((r0)^3)*(fq1-fq2-fq3);
% Rate of work due to strip load
Wqskh = -kh*omega*q*b*r0*exp((x(2)-x(1))*tan(phi))*cos(x(2));
% Yield acceleration coefficients (ky)
A1 = gamma*omega*((r0)^3)*(f1-f2-f3) - omega*q*b*(a+(b/2)+r0*sin(x(1)));
A2 = omega*q*b*r0*exp((x(2)-x(1))*tan(phi))*cos(x(2)) - gamma*omega*((r0)^3)*(fq1-fq2-fq3);
fval = A1 /((lambda*A1)+A2);
end
Akshay Pratap Singh
Akshay Pratap Singh le 14 Mai 2022
Thanks Matt J, It solved my problem.
Thank you very much.

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