Failure in initial objective function evaluation. FSOLVE cannot continue.

조회 수: 4 (최근 30일)
ma xueyi
ma xueyi 2022년 4월 25일
답변: ma xueyi 2022년 4월 30일
i set two files as main.m and root.m.
Here are my codes.
When I runs main.m,it keeps telling me that "Failure in initial objective function evaluation. FSOLVE cannot continue."
Hope someone can help me with that.Thanks!
[main.m]
clear all; clc
fun = @root;
x0 = [2e-5,0.04,0.1524,1.0194,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
x = fsolve(fun,x0)
[root.m]
global Tinlet Treact T0
Tinlet = 273+25 ;%K
Treact = 273+150 ;%K
T0 = 273+25;%K
global ki1 ki2 kp11 kp12 kp21 kp22 ktal kd ki10 kp110 kp120 kp210 kp220 ktal0 kd0 Eki1 Ekp11 Ekp12 Ekp21 Ekp22 Ektal Ekd R
R = 8.314;
ki10 = 13.0;
Eki1 = 10000;
ki1 = ki10 * exp(-Eki1*(1/Treact-1/T0));
ki2 = 0.0;
kp110 = 3.70e+05;
Ekp11 = 10000;
kp11 = kp110 * exp(-Ekp11*(1/Treact-1/T0));
kp120 = 673;
Ekp12 = 10000;
kp12 = kp120 * exp(-Ekp12*(1/Treact-1/T0));
kp210 = 1.36e+03;
Ekp21 = 10000;
kp21 = kp210 * exp(-Ekp21*(1/Treact-1/T0));
kp220 = 46 ;
Ekp22 = 10000;
kp22 = kp220 * exp(-Ekp22*(1/Treact-1/T0));
ktal0 = 2.5;
Ektal = 10000;
ktal = ktal0 * exp(-Ektal*(1/Treact-1/T0));
kd0 = 1.0e-04;
Ekd = 10000;
kd = kd0 * exp(-Ekd*(1/Treact-1/T0));
function F = fun(x)
F(1) =ktx(2)*x(5)+x(6)*x(2)-ki1*x(1)*x(3)-ki2*x(1)*x(4)-x(14);
F(2) =UDSI0 - x(14)*stoptime-x(1);
F(3) =ktx(2)*x(2)*(x(6)+x(7))-x(15);
F(4) =UDSI1 - x(15)*stoptime-x(2);
F(5) =(ki1*x(1)+kp11*x(5)+kp21*x(6))*x(3)-x(16);
F(6) =UDSI2 - x(16)*stoptime-x(3);
F(7) =(ki2*x(1)+kp12*x(5)+kp22*x(6))*x(4)-x(17);
F(8) =UDSI3 - x(17)*stoptime-x(4);
F(9) =ki1*x(1)*x(3)+kp21*x(6)*x(3)-kp12*x(5)*x(4)-(ktx(2)*x(2)+kd)*x(5)-x(18);
F(10) =UDSI4 - x(18)*stoptime-x(5);
F(11) =ki2*x(1)*x(4)+kp12*x(5)*x(4)-kp21*x(6)*x(3)-(ktx(2)*x(2)+kd)*x(6)-x(19);
F(12) =UDSI5 - x(19)*stoptime-x(6);
F(13) =ki1*x(1)*x(3)+kp11*x(5)*x(3)+kp21*(x(7)+x(6))*x(3)-kp12*x(7)*x(4)-(ktx(2)*x(2)+kd)*x(7)-x(20);
F(14) =UDSI6 - x(20)*stoptime-x(7);
F(15) =ki2*x(1)*x(4)+kp22*x(6)*x(4)+kp12*(first_lamada2+x(5))*x(4)-kp21*x(7)*x(3)-(ktx(2)*x(2)+kd)*first_lamada2-x(21) ;
F(16) =UDSI7 - x(21)*stoptime-x(8);
F(17) =ki1*x(1)*x(3)+kp11*(2*x(7)+x(5))*x(3)+kp21*(x(10)+2*x(8)+x(6))*x(3)-kp12*x(9)*x(4)-(ktx(2)*x(2)+kd)*x(9)-x(22);
F(18) =UDSI8 - x(22)*stoptime-x(9);
F(19) =ki2*x(1)*x(4)+kp22*(2*x(8)+x(6))*x(4)+kp12*(x(9)+2*x(7)+x(5))*x(4)-kp21*x(10)*x(3)-(ktx(2)*x(2)+kd)*x(10)-x(23);
F(20) =UDSI9 - x(23)*stoptime-x(10);
F(21) =(ktx(2)*x(2)+kd)*(x(5)+x(6))-x(24);
F(22) =UDSI10 - x(24)*stoptime-x(11);
F(23) =(ktx(2)*x(2)+kd)*(x(7)+x(8))-x(25);
F(24) =UDSI11 - x(25)*stoptime-x(12);
F(25) =(ktx(2)*x(2)+kd)*(x(9)+x(10))-x(26);
F(26) =UDSI12 - x(25)*stoptime-x(13);
F = [F(1),F(2),F(3),F(4),F(5),F(6),F(7),F(8),F(9),F(10),F(11),F(12),F(13),F(14),F(15),F(16),F(17),F(18),F(19),F(20),F(21),F(22),F(23),F(24),F(25),F(26)];
end
  댓글 수: 5
Alex Sha
Alex Sha 2022년 4월 27일
You haven't given those data yet.
Rik
Rik 2022년 4월 30일
Comment posted as flag by ma xueyi:
thanks, i provided the data otherplace

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채택된 답변

Jan
Jan 2022년 4월 27일
You provide a handle to a script. This cannot work, because fsolve expects a function handle. Most likely you want:
fsolve(@fcn, x0)
instead. Currently the constants are not visible in the function fcn. Maybe you want to declare the variables as globals there also? What is stoptime?
Whenever you mention an error in the forum, paste a copy of the complete error message, not just some parts of it.

추가 답변 (1개)

ma xueyi
ma xueyi 2022년 4월 30일
Actully my codes are now running successfully.
Howerver its answer is not reliable(exitflag = 0)
No matter how big i change the 'MaxFunEvals' and 'MaxIterations' to be, the first-order optimality never changed.

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