Question

我约束了要使用Matlab解决的ODE优化问题，我首先使用ode15s解决了ODE，该工作将使这个初始值有效，而且我使用fminsearch也获得了很好的结果，没有约束，当我使用fmincon时，问题开始了，它给了我以下警告：

Warning: Length of lower bounds is < length(x); filling in missing lower bounds with -Inf. 
> In checkbounds (line 33)
  In fmincon (line 318)
  In Optimization (line 64) 
Warning: Length of upper bounds is < length(x); filling in missing upper bounds with +Inf. 
> In checkbounds (line 47)
  In fmincon (line 318)
  In Optimization (line 64) 
Warning: Failure at t=2.340250e+01.  Unable to meet integration tolerances without reducing the step size below the smallest value
allowed (5.684342e-14) at time t. 
> In ode15s (line 668)
  In ObjFun (line 6)
  In barrier
  In fmincon (line 813)
  In Optimization (line 64)

我试图消除限制，但没有改变...

y1_max = 5; y2_max = 2.3; 
y01 = 1;    y02 = 1.6;

%Control function parameter
Umin = -0.3; Umax = -0.1;

%Creation the structure of model parameters
params.T0=T0; params.Tf=Tf;
params.y1d=y1d; params.y2d=y2d;
params.y01=y01; params.y02=y02;
params.y2_max=y2_max;

n=2;
U0=-0.2*ones(1,n);
params.n=n;
params.U=U0; params.Umax=Umax; params.Umin=Umin;
params.dt=(Tf-T0)/n;

%for initial value of optimization parameters
options = odeset('RelTol',1e-7,'AbsTol',1e-7);
U=U0; 

[t,y]=ode15s(@ODEsolver,[T0,Tf],[y01 y02],options,params);

%adding the ode solution as input prameters
params.t=t; params.y1= y(:,1); params.y2= y(:,2);

U0=-0.2*ones(1,n);
params.n=n; params.U=U0;
params.dt=(Tf-T0)/n;

A = [];
B = [];
Aq = []; Bq = [];
options1 = optimset('MaxIter',5000);
[x,fval,exitflag,output] = fmincon(@ObjFun,U0,A,B,Aq,Bq,Umin,Umax,IneqConst(U0,params),options1,params)

function y = ODEsolver(t,y,params)
    dt = params.dt;
    nu = floor(t/dt)+1;
    nu = min(nu,params.n-1);

    t1 = (nu-1)*dt;
    Ut = params.U(nu) + ((t-t1)/dt)*(params.U(nu+1) - params.U(nu));

    dy1a = ((0.63*y(1))/(1 + y(1)));
    dy1b = (Ut*y(1)*y(2))/(1+0.6*y(1));
    dy1 = dy1a + dy1b;

    dy2a = (0.15*y(2))/(0.05-0.2*y(2));
    dy2 = -0.2*y(2) + dy1b * dy2a;

    y = [dy1; dy2];
end

function ObjFun1 = ObjFun(u,params)
   % Calculating the value of the optimization criteria 
   params.U=u;

   options = odeset('RelTol',1e-8,'AbsTol',1e-8);
   [t,y]=ode15s(@ODEsolver,[0,100],[params.y01, 
   params.y02],options,params);

   ObjFun1 =trapz(t,(y(:,1)-params.y1d).^2);
end

function [c,Const] = IneqConst(u, params)
    params.U=u;
    options = odeset('RelTol',1e-8,'AbsTol',1e-8);
    [t,y]=ode15s(@ODEsolver,[0,100],[params.y01, params.y02],options,params);

    c =[];
    yCon = y(:,2)-params.y2_max;
    Const = trapz(t,(abs(yCon)+yCon).^2);
end

Answer 1

此代码中的错误是边界的维数（Umin，Umax）应为与控制函数U大小相同的数组。

因此，如果控制功能参数为n = 2，则Umin = [-0.3 -0.3]和Umax = [-0.1 -0.1]

无法满足fmincon中的积分公差来解决ODE优化问题

1 个答案: