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

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Akshay Pratap Singh
评论： Akshay Pratap Singh 2022-5-14
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)

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采纳的回答

Matt J
Matt J 2022-5-13
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 个评论
Matt J
Matt J 2022-5-13
Ran in:
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

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