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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이전 댓글 표시
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);
댓글 수: 3
Bjorn Gustavsson
2019년 7월 12일
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
2019년 7월 12일
편집: Walter Roberson
2019년 7월 12일
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.
답변 (1개)
Amna Mazen
2021년 9월 22일
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.
댓글 수: 1
Walter Roberson
2021년 9월 23일
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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