error in matlab code

Question

Devapriya 2024-12-18

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2172275-error-in-matlab-code

评论： Walter Roberson 2025-1-13

I have been trying to solve an optimal problem in which a body of mass m have to [be] transfered to y(tf) = 0 from its initial position y(o) = 0 such that its final velocity is zero satisfying the performance index.

, where t = [to, tf].

But I am not getting the optimal trajectories using this code. Could anyone help me to find where the mistake [is]?

clear all

c0 = [0.1; 0.1; 0.1; 0.1; 5]; % Make a starting guess at the solution

[c, fval] = fsolve (@myfun1, c0) % Call optimizer

Solver stopped prematurely. fsolve stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 5.000000e+02.

c = 5×1

0.0046 -0.0198 -0.0881 0.7725 12.1323

fval = 5×1

-0.2275 0.0271 -0.0881 -0.0968 0.0024

c1 = c(1); c2 = c(2); c3 = c(3); c4 = c(4); c5 = c(5);

t=0:0.01:c5;

x1op = -(c1*(t.^3))/6-(c2*(t.^2))/2+(c3.*t)+c4

x1op = 1×1214

0.7725 0.7716 0.7707 0.7698 0.7690 0.7681 0.7672 0.7663 0.7655 0.7646 0.7638 0.7629 0.7620 0.7612 0.7603 0.7595 0.7586 0.7578 0.7569 0.7561 0.7552 0.7544 0.7536 0.7527 0.7519 0.7511 0.7502 0.7494 0.7486 0.7477

<mw-icon class=""></mw-icon>

x2op =-(c1*(t.^2))/2+-(c2.*t)+c3

x2op = 1×1214

-0.0881 -0.0879 -0.0877 -0.0875 -0.0873 -0.0871 -0.0869 -0.0867 -0.0865 -0.0863 -0.0861 -0.0859 -0.0857 -0.0855 -0.0853 -0.0851 -0.0850 -0.0848 -0.0846 -0.0844 -0.0842 -0.0840 -0.0838 -0.0836 -0.0834 -0.0833 -0.0831 -0.0829 -0.0827 -0.0825

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uop = -(c1.*t)-c2

uop = 1×1214

0.0198 0.0198 0.0198 0.0197 0.0197 0.0196 0.0196 0.0195 0.0195 0.0194 0.0194 0.0193 0.0193 0.0193 0.0192 0.0192 0.0191 0.0191 0.0190 0.0190 0.0189 0.0189 0.0188 0.0188 0.0187 0.0187 0.0187 0.0186 0.0186 0.0185

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lam1= c1;

H = 1.5 * uop.^2 + x2op.* lam1;

figure (1)

plot(t, uop)

figure(2)

plot(t,x1op,t,x2op)

figure(3)

plot(t,H)

function F = myfun1(c)

x10 = 1;

x20 = 0;

x1f = 0;

x2f = 0;

u=-(c(1)*c(5))-c(2);

ud = c(1);

% the conditions x1(t = 0), x1(t = tf), x2(t = 0), x2(t = tf)

f1 = c(4)-x10;

f2 =-(c(1)*(c(5)^3))/6 -(c(2)*(c(5)^2))/2 +(c(3)*c(5)) + 1 -x1f;

f3 = c(3)-x20;

f4 = -(c(1)*(c(5)^2))/2 -(c(2)*c(5))-x2f;

f5 = 1.5 *u.^2 -f4.*ud ;% the condition H(tf) = 0

F = [f1;f2;f3;f4;f5]; % the five unknown system F(C) = 0

end

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Walter Roberson 2025-1-13

在 MATLAB Online 中打开

clear all

c0 = [0.1; 0.1; 0.1; 0.1; 5]; % Make a starting guess at the solution

opts = optimoptions('fsolve', 'MaxIterations', 1e6, 'MaxFunEval', 1e7);

[c, fval] = fsolve (@myfun1, c0, opts) % Call optimizer

Equation solved, inaccuracy possible. The vector of function values is near zero, as measured by the value of the function tolerance. However, the last step was ineffective.

c = 5×1

0.0000 -0.0001 -0.0075 0.9828 210.0654

fval = 5×1

-0.0172 0.0000 -0.0075 -0.0026 0.0000

c1 = c(1); c2 = c(2); c3 = c(3); c4 = c(4); c5 = c(5);

t=0:0.01:c5;

x1op = -(c1*(t.^3))/6-(c2*(t.^2))/2+(c3.*t)+c4

x1op = 1×21007

0.9828 0.9827 0.9826 0.9826 0.9825 0.9824 0.9823 0.9823 0.9822 0.9821 0.9820 0.9820 0.9819 0.9818 0.9817 0.9817 0.9816 0.9815 0.9814 0.9814 0.9813 0.9812 0.9811 0.9811 0.9810 0.9809 0.9808 0.9808 0.9807 0.9806

<mw-icon class=""></mw-icon>

x2op =-(c1*(t.^2))/2+-(c2.*t)+c3

x2op = 1×21007

-0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075

<mw-icon class=""></mw-icon>

uop = -(c1.*t)-c2

uop = 1×21007

1.0e-03 * 0.1026 0.1026 0.1026 0.1026 0.1026 0.1026 0.1025 0.1025 0.1025 0.1025 0.1025 0.1025 0.1025 0.1025 0.1025 0.1024 0.1024 0.1024 0.1024 0.1024 0.1024 0.1024 0.1024 0.1024 0.1023 0.1023 0.1023 0.1023 0.1023 0.1023

<mw-icon class=""></mw-icon>

lam1= c1;

H = 1.5 * uop.^2 + x2op.* lam1;

figure (1)

plot(t, uop)

●

figure(2)

plot(t,x1op,t,x2op)

figure(3)

plot(t,H)

function F = myfun1(c)

x10 = 1;

x20 = 0;

x1f = 0;

x2f = 0;

u=-(c(1)*c(5))-c(2);

ud = c(1);

% the conditions x1(t = 0), x1(t = tf), x2(t = 0), x2(t = tf)

f1 = c(4)-x10;

f2 =-(c(1)*(c(5)^3))/6 -(c(2)*(c(5)^2))/2 +(c(3)*c(5)) + 1 -x1f;

f3 = c(3)-x20;

f4 = -(c(1)*(c(5)^2))/2 -(c(2)*c(5))-x2f;

f5 = 1.5 *u.^2 -f4.*ud ;% the condition H(tf) = 0

F = [f1;f2;f3;f4;f5]; % the five unknown system F(C) = 0

end

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Answer 1

Divyanshu 2025-1-13

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2172275-error-in-matlab-code#answer_1557355

Hello @Devapriya,

I think the reason for the incorrect trajectories can be the error 'Solver stopped prematurely'. This MATLAB answer thread addresses a similar issue, just try to follow this and re-run your script.

Hopefully, after modifying the 'MaxIterations' and 'MaxFunctionEvaluations' parameters you would get correct results from the script.

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