Solver stopped prematurely. fmincon stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 3.000000e+03. Unable to rectify this error. Please Help
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F_r and T_r are defined.
[m n ]=size(F_r);
[M N]=size(T_r);
My constraints are lengthy but i have checked it is correct.
My code:
aeq=[];
beq=[];
a=[];
b=[];
lb=zeros(1,42);
for j=1:n+N
if(j<=n)
ub(j)=F_r(1);
else
ub(j)=tmax;
end
end
x0=[F_r T_r];
options.Display = 'iter';
[x,fval]=fmincon(@(x)objective2(x,n),x0,a,b,aeq,beq,lb,ub,@(x)constraint2(x,n,UA_r,F_r,T_r),options);
My objective Function:
function f=objective2(x,n)
f=-x(n+10);
end
My contraints:
%% Constraint
function [c,ceq]=constraint2(x,n,UA_r,F_r,T_r)
ceq(1)=x(2)+x(6)-x(1);
ceq(2)=x(3)-x(2);
ceq(3)=x(15)-x(14);
ceq(4)=x(4)-x(3);
ceq(5)=x(12)-x(11);
ceq(6)=x(5)-x(4);
ceq(7)=x(14)-x(13);
ceq(8)=x(10)-x(5)-x(9);
ceq(9)=x(7)-x(6);
ceq(10)=x(17)-x(16);
ceq(11)=x(8)-x(7);
ceq(12)=x(19)-x(18);
ceq(13)=x(9)-x(8);
ceq(14)=x(21)-x(20);
ceq(15)=x(6)*[0.444*(x(n+7)-x(n+6))+(0.00101/2)*(x(n+7)^2-x(n+6)^2)]-[x(16)*[0.409*(x(n+16)-x(n+17))+(0.00096/2)*(x(n+16)^2-x(n+17)^2)]];
ceq(16)=x(6)*[0.444*(x(n+7)-x(n+6))+(0.00101/2)*(x(n+7)^2-x(n+6)^2)]-[UA_r(4)*[(x(n+16)-x(n+7))-(x(n+17)-x(n+6))]/log((x(n+16)-x(n+7))/(x(n+17)-x(n+6)))];
ceq(17)=x(7)*[0.444*(x(n+8)-x(n+7))+(0.00101/2)*(x(n+8)^2-x(n+7)^2)]-[x(18)*[0.429*(x(n+18)-x(n+19))+(0.00099/2)*(x(n+18)^2-x(n+19)^2)]];
ceq(18)=x(7)*[0.444*(x(n+8)-x(n+7))+(0.00101/2)*(x(n+8)^2-x(n+7)^2)]-[UA_r(5)*[(x(n+18)-x(n+8))-(x(n+19)-x(n+7))]/log((x(n+18)-x(n+8))/(x(n+19)-x(n+7)))];
ceq(19)=x(8)*[0.444*(x(n+9)-x(n+8))+(0.00101/2)*(x(n+9)^2-x(n+8)^2)]-[x(20)*[0.414*(x(n+20)-x(n+21))+(0.00096/2)*(x(n+20)^2-x(n+21)^2)]];
ceq(20)=x(8)*[0.444*(x(n+9)-x(n+8))+(0.00101/2)*(x(n+9)^2-x(n+8)^2)]-[UA_r(5)*[(x(n+20)-x(n+9))-(x(n+21)-x(n+8))]/log((x(n+20)-x(n+9))/(x(n+21)-x(n+8)))];
ceq(21)=x(2)*[0.444*(x(n+3)-x(n+2))+(0.000101/2)*(x(n+3)^2-x(n+2)^2)]-x(14)*[0.414*(x(n+14)-x(n+15))+(0.00096/2)*(x(n+14)^2-x(n+15)^2)];
ceq(22)=x(2)*[0.444*(x(n+3)-x(n+2))+(0.000101/2)*(x(n+3)^2-x(n+2)^2)]-UA_r(1)*[((x(n+14)-x(n+3))-(x(n+15)-x(n+2)))/log((x(n+14)-x(n+3))/(x(n+15)-x(n+2)))];
ceq(23)=x(3)*[0.444*(x(n+4)-x(n+3))+(0.00101/2)*(x(n+4)^2-x(n+3)^2)]-UA_r(2)*[((x(n+11)-x(n+4))-(x(n+12)-x(n+3)))/log((x(n+11)-x(n+4))/(x(n+12)-x(n+3)))];
ceq(24)=x(3)*[0.444*(x(n+4)-x(n+3))+(0.00101/2)*(x(n+4)^2-x(n+3)^2)]-x(11)*[0.446*(x(n+11)-x(n+12))+(0.00101/2)*(x(n+11)^2-x(n+12)^2)];
ceq(25)=x(4)*[0.444*(x(n+5)-x(n+4))+(0.00101/2)*(x(n+5)^2-x(n+4)^2)]-x(13)*[0.414*(x(n+13)-x(n+14))+(0.00096/2)*(x(n+13)^2-x(n+14)^2)];
ceq(26)=x(4)*[0.444*(x(n+5)-x(n+4))+(0.00101/2)*(x(n+5)^2-x(n+4)^2)]-UA_r(3)*[((x(n+13)-x(n+5))-(x(n+14)-x(n+4)))/log((x(n+13)-x(n+5))/(x(n+14)-x(n+4)))];
ceq(27)=x(n+2)-x(n+1);
ceq(28)=x(n+6)-x(n+1);
ceq(29)=x(5)*[0.444*x(n+5)+(0.00101/2)*x(n+5)^2]+x(9)*[0.444*x(n+9)+(0.00101/2)*x(n+9)^2]-x(10)*[0.444*x(n+10)+(0.00101/2)*x(n+10)^2];
c(1)=x(n+6)-x(n+7);
c(2)=x(n+17)-x(n+16);
c(3)=10-x(n+16)+x(n+7);
c(4)=10-x(n+17)+x(n+6);
c(5)=x(n+7)-x(n+8);
c(6)=x(n+19)-x(n+18);
c(7)=10-x(n+18)+x(n+8);
c(8)=10-x(n+19)+x(n+7);
c(9)=x(n+8)-x(n+9);
c(10)=x(n+21)-x(n+20);
c(11)=10-x(n+20)+x(n+9);
c(12)=10-x(n+21)+x(n+8);
c(13)=x(n+2)-x(n+3);
c(14)=x(n+15)-x(n+14);
c(15)=10-x(n+14)+x(n+3);
c(16)=10-x(n+15)+x(n+2);
c(17)=x(n+3)-x(n+4);
c(18)=x(n+12)-x(n+11);
c(19)=10-x(n+11)+x(n+4);
c(20)=10-x(n+12)+x(n+3);
c(21)=x(n+4)-x(n+5);
c(22)=x(n+14)-x(n+13);
c(23)=10-x(n+13)+x(n+5);
c(24)=10-x(n+14)+x(n+4);
if(x(n+10)>x(n+5))
c(25)=x(n+10)-x(n+9);
else
c(26)=x(n+10)-x(n+5);
end
ceq(27)=x(13)-F_r(13);
ceq(28)=x(11)-F_r(11);
ceq(28)=x(16)-F_r(16);
ceq(30)=x(18)-F_r(18);
ceq(31)=x(1)-F_r(1);
ceq(32)=x(20)-F_r(20);
ceq(33)=x(n+1)-T_r(1);
ceq(34)=x(n+2)-T_r(2);
ceq(35)=x(n+6)-T_r(6);
ceq(36)=x(n+11)-T_r(11);
ceq(37)=x(n+13)-T_r(13);
ceq(38)=x(n+16)-T_r(16);
ceq(39)=x(n+18)-T_r(18);
ceq(40)=x(n+20)-T_r(20);
3 个评论
Bjorn Gustavsson
2019-7-12
Couldn't you reduce the number or components in x - at least the first 14 ceq seems to be simple linear equalities. That might help a bit.
...and for any given value of n you only look for the smallest value of component x(n+10), so this might be a problem of only looking at the subspace where your constraints are satisfied, maybe linprog has a better algorithm for this problem.
HTH
Walter Roberson
2019-7-12
编辑:Walter Roberson
2019-7-12
Replaces https://www.mathworks.com/matlabcentral/answers/471209-fmincon-stopped-because-it-exceeded-the-function-evaluation-limit-options-maxfunctionevaluations which had a solution for your difficulty.
回答(1 个)
Amna Mazen
2021-9-22
If you changed the Step tolerance to a smaller number. Return it back to its default number. I think that was "1e-10" and your problem may be solved.
1 个评论
Walter Roberson
2021-9-23
The only option the user passed in was 'Display', 'iter', so the step tolerance was at the default value already. The user needed to increase MaxFunctionEvaluations
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