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 Comments
  Bjorn Gustavsson
      
 on 12 Jul 2019
				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
      
      
 on 12 Jul 2019
				
      Edited: Walter Roberson
      
      
 on 12 Jul 2019
  
			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.
Answers (1)
  Amna Mazen
 on 22 Sep 2021
        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 Comment
  Walter Roberson
      
      
 on 23 Sep 2021
				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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