fmincon, nonlcon - too many input arguments

Question

e_frog 2020-12-17

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/696360-fmincon-nonlcon-too-many-input-arguments

评论： Walter Roberson 2020-12-20

Hi i am trying to minimize some output of a system of 2 ODEs that also are bound by a nonlinear constraint. Unfortunately I cant seem to get my syntax for implementing the nonlcon right.

To clarify what I want to achieve with the nonlnear constraint: The last value of X(2) (a velocity) has to be zero.

I know this is a fairly common question on here, but the other questions and answers couldn't solve my problem. This is my specific code:

x_0 = init_vars();
init_conds_odes = [1 0 5 0];
lb = x_0 - 54;
ub = x_0 + 43;
options = [];
nlcon = @nonlcon;
[H] = fmincon(@objective,x_0,[],[],[],[],lb,ub,@nonlcon,options,init_conds_odes);

These are the functions:

function [x_0,tmax] = init_vars()
m = 2;
n = 3;
o = 4;
tmax = 5;
x_0 = [m;n;o;tmax];
end
function [obj_val,X] = objective(H,init_conds_odes)
tmax = H(4);
tspan_ode = [0 tmax];
[~,X] = ode45(@(t,x) ODEs(t,x,H),tspan_ode,init_conds_odes);
z = max(abs(X(:,2)));
obj_val = z;
end
function [dXdt] = ODEs(~,X,H)
m = H(1);
n = H(2);
o = H(3);
dXdt = [X(2);
       X(1)/m+X(3)/n+1;
       X(4);
       X(3)/o];
end
function [c,ceq] = nonlcon(X)
c = [];
ceq(1) = 0 - X(end,2);
end

This is the error message:

Error using test>nonlcon
Too many input arguments.
Error in fmincon (line 639)
        [ctmp,ceqtmp] = feval(confcn{3},X,varargin{:});
Error in test (line 13)
[H] = fmincon(@objective,x_0,[],[],[],[],lb,ub,@nonlcon,options,init_conds_odes);
Caused by:
    Failure in initial nonlinear constraint function evaluation. FMINCON cannot continue.

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Answer 1

Walter Roberson 2020-12-17

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/696360-fmincon-nonlcon-too-many-input-arguments#answer_578135

在 MATLAB Online 中打开

nonlcon will be passed t and x. You are not required to make use of t, but it will be passed.

function [c,ceq] = nonlcon(~, X)

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e_frog 2020-12-18

在 MATLAB Online 中打开

x_0 = init_vars();
init_conds_odes = [1 5 1 1];
lb = x_0 - 0.1;
ub = x_0 + 43;
options = [];
%nlcon = @(H,X) nonlcontest(H,X,tspan_ode,init_conds_odes);
[H] = fmincon(@objective,x_0,[],[],[],[],lb,ub,@(H,X,tspan_ode,init_conds_odes) nonlcontest(H,X),options,init_conds_odes);
function [x_0,tmax] = init_vars()
m = 2;
n = 3;
o = 4;
tmax = 5;
x_0 = [m;n;o;tmax];
end
function [obj_val,X,tspan_ode,init_conds_odes] = objective(H,init_conds_odes)
tmax = H(4);
tspan_ode = [0 tmax];
[t,X] = ode45(@(t,x) ODEs(t,x,H),tspan_ode,init_conds_odes);
%disp(obj_val)
dXdt = zeros(length(t),4);
for i = 1:length(t)
[X(i,:),dXdt(i,:)] = ODEs(t,X(i,:),H);
end
z = max(abs(dXdt(:,2)));
obj_val = z;
end
   
function [X,dXdt] = ODEs(~,X,H)
m = H(1);
n = H(2);
o = H(3);
dXdt = [X(2);
       X(1)/m+X(3)/n+9;
       X(4);
       X(3)/o];
end
function [c,ceq] = nonlcontest(H,~,tspan_ode,init_conds_odes)
[t,X] = ode45(@(t,x) ODEs(t,x,H),tspan_ode,init_conds_odes);
dXdt = zeros(length(t),4);
for i = 1:length(t)
[X(i,:),dXdt(i,:)] = ODEs(t,X(i,:),H);
end
c = [];
ceq(1) = 0 - X(end,2);
end

Thanks for your help so far! I tried to write my nonlcon a little bit different. Can you help me finding out why the nonlcontest does not get enough inputs?

Not enough input arguments.
Error in test>nonlcontest (line 95)
[t,X] = ode45(@(t,x) ODEs(t,x,H),tspan_ode,init_conds_odes);
Error in test>@(H,X,tspan_ode,init_conds_odes)nonlcontest(H,X) (line
13)
[H] =
fmincon(@objective,x_0,[],[],[],[],lb,ub,@(H,X,tspan_ode,init_conds_odes)
nonlcontest(H,X),options,init_conds_odes);
Error in fmincon (line 639)
        [ctmp,ceqtmp] = feval(confcn{3},X,varargin{:});
Error in test (line 13)
[H] =
fmincon(@objective,x_0,[],[],[],[],lb,ub,@(H,X,tspan_ode,init_conds_odes)
nonlcontest(H,X),options,init_conds_odes);
Caused by:
    Failure in initial nonlinear constraint function evaluation.
    FMINCON cannot continue.

e_frog 2020-12-20

编辑：e_frog 2020-12-20

I cant rewrite it as a boundary value problem, because the initial conditions of the odes [pos(1) vel (1) pos(2) vel(2)] (figuratively) must be met. I tried rewriting my new question from this comment. Perhaps you find the time to help me again. I would really much appreciate it!

Rewritten new question

Walter Roberson 2020-12-20

If those initial conditions must be met, then specify them as part of the boundaries.

Remember that if needed you can add an extra variable to act as an extra layer of differentiation or an extra layer of integration (as needed) in order to be able to put boundary conditions on derivatives or positions.

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