I am solving 20 nonlinear equation using fsolve to find 20 variables' value. But i previously know one of variable's value. That variable is x(17) in eqn F(17). x(17)=1200. Based on this fixed variable i want to find remaining variables' value.

function F = root16d(x)
Qs_ref=0;
Pw_ref=0.322;
F(1) = -0.5*((x(1)*x(2))-(x(3)*x(4)))-Qs_ref;      %Qr=0.5*((x(5)*x(8))-(x(7)*x(6))
F(2) = 0.5*((x(1)*x(4))+(x(3)*x(2)))+0.5*((x(5)*x(6))+(x(7)*x(8)))+Pw_ref;
%% Algebric Equation 
F(3) = 0.5*((x(1)*x(9))+(x(3)*x(10)))+0.5*((x(1)*x(4))+(x(3)*x(2)))-0.5*((x(11)*x(15))+(x(12)*x(16)));
F(4) = 0.5*((x(1)*x(2))-(x(3)*x(4)))+0.5*((x(1)*x(10))-(x(3)*x(9)))+0.5*((x(11)*x(16))-(x(12)*x(15)));
%% SS of Tr line
delta=0.1049;
                Rl=0.02;                  we=2*pi*60;  
                Xl=0.5;                   wb=2*pi*60;  
                Xc=0.75*0.5;
                EB=1;
                Ebq=EB*cos(delta);                                                                                  
                Ebd=EB*sin(delta);                                                                            
%               
F(5) = -(Rl*wb/Xl)*x(9)-we*x(10)-(wb/Xl)*x(13)+(wb/Xl)*(x(1)-Ebq);
F(6) = we*x(9)-(Rl*wb/Xl)*(x(10))-(wb/Xl)*x(14)+(wb/Xl)*(x(3)-Ebd);
F(7) = (wb*x(9)*Xc)-(we*x(13));
F(8) = (wb*x(10)*Xc)+(we*x(13));
%% two aglebric equation 
            Xtg=0.3;
F(13)= x(1)-x(11)+(Xtg*x(16));
F(14)= x(3)-x(12)-(Xtg*x(15));
F(15)= x(15)-x(4)-x(9);
F(16)= x(16)-x(2)-x(10);
%% Power flow equation for DC capacitor placed between GSC and RSC
            C=50*14000e-6;%C=1;
F(17)= (100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))/(-C*1200));%100e6**x(17)
%% State Space model of Generator Turbine shaft model
            Dt=0;                       Ht=4.29;
            Dtg=1.5;                    Hg=0.9;                                           Ktg=0.15;
            Xm=3.95279;                 Tw=0.322/(0.75);
            Wbase=2*pi*60;                    Te=0.5*Xm*(((x(4)+x(6))*x(8))-((x(2)+x(8))*x(6)));
F(18)= ((-Dt-Dtg)/(2*Ht))*x(18)+((Dtg/(2*Ht))*x(19))-(x(20)/(2*Ht))+(Tw/(2*Ht));
F(19)= ((Dtg/(2*Hg))*x(18))+((-Dt-Dtg)/(2*Hg))*x(19)+(x(20)/(2*Hg))-(Te/(2*Hg));
F(20)=(Ktg*Wbase)*x(18)-(Ktg*Wbase)*x(19);
%% State Space model of DFIG
                Rs = 0.00488;          Wb= 2*pi*60%2*pi*60;          
                Rr = 0.00549;          We= 2*pi*60;%2*pi*60;
                Xls = 0.09231;         Wr= 2*pi*60*x(19);%2*pi*60*x(19)
                Xlr = 0.09955;     
                Xm = 3.95279;
                Xss = Xls+Xm;   Xrr = Xlr+Xm;
F(9)=x(2)*((Xm^2*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (We*(Xlr + Xm)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) - x(8)*((We*Xm*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Xm*(Xlr + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) + (Wb*x(1)*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Wb*Xm*x(5))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Rs*Wb*x(4)*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) + (Rr*Wb*Xm*x(6))/(Xlr*Xls + Xlr*Xm + Xls*Xm)
F(10)= x(6)*((We*Xm*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Xm*(Xlr + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) - x(4)*((Xm^2*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (We*(Xlr + Xm)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) + (Wb*x(3)*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Wb*Xm*x(7))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Rs*Wb*x(2)*(Xlr + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) + (Rr*Wb*Xm*x(8))/(Xlr*Xls + Xlr*Xm + Xls*Xm)
F(11)=x(2)*((We*Xm*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Xm*(Xls + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) + x(8)*((We*Xm^2)/(Xlr*Xls + Xlr*Xm + Xls*Xm) - ((Xlr + Xm)*(Xls + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) + (Wb*x(5)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Wb*Xm*x(1))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Rr*Wb*x(6)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) + (Rs*Wb*Xm*x(4))/(Xlr*Xls + Xlr*Xm + Xls*Xm)
F(12)=(Wb*x(7)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - x(6)*((We*Xm^2)/(Xlr*Xls + Xlr*Xm + Xls*Xm) - ((Xlr + Xm)*(Xls + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) - x(4)*((We*Xm*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Xm*(Xls + Xm)*(We - Wr))/(Xlr*Xls + Xlr*Xm + Xls*Xm)) - (Wb*Xm*x(3))/(Xlr*Xls + Xlr*Xm + Xls*Xm) - (Rr*Wb*x(8)*(Xls + Xm))/(Xlr*Xls + Xlr*Xm + Xls*Xm) + (Rs*Wb*Xm*x(2))/(Xlr*Xls + Xlr*Xm + Xls*Xm)

[Walter Roberson]

Please do not close questions that have an Answer.

I spent a number of hours investigating this for you, and it is frustrating to have my work vanish.

Answer 1

[Walter Roberson]

x(17) does not appear in your function, so setting a particular value for it does not make any difference.

What does make a difference is that you have 20 equations in 19 variables (since x(17) is unused). That is either going to have no solutions or an infinite number of solutions. My tests suggest that this particular set of equations has no consistent solutions.

[chirag rohit]

I already put the value 1200 instead of x(17) to present my question.

See in the line F(17) and compare it with below equation

F(17)= (100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))/(-C*x(17)))

i want to keep value of x(17) as 1200 (because its already known to me) but at the same time i want such a solution that it satisfy all equation from F(1) to F(20)

[Stephan]

The more important point is that your equations dont seem to have a consistent solution.

[chirag rohit]

Yes sir... I appreciate your answer...I am aware of it... Is there any way in which we can use fsolve for n number of equation in which 1 variable is known and value of remaining (n-1) variables are unknown. Condition is i can't remove nth equation because it also has to be satisfy with obtained values of (n-1) variables.

[Walter Roberson]

 fsolve(@(x) root20D([x(1:16), 1200, x(18:end)]), x0([1:16, 18:end]), options)

You are going to need options increasing the maximum function evaluations and maximum interations quite a lot, because fsolve needs to spend a long time searching for a consistent solution to the inconsistent equations.

[chirag rohit]

Yes sir... i am doing the same..

[Walter Roberson]

I continued to analyze the equations you posted. It is feasible to use calculas techniques on sum(F.^2) to reduce down to 12 variables, with the other 7 being closed form solutions based on the 12. The best sequence is to eliminate X6, X5, X12, X9, X3, X4, X2. It is not feasible to find closed form solutions for the remaining variables.

Having reduced to 12 variables, I did sort of semi-intelligent monte carlo simulation of the residue that remained. It did not appear to be possible to reduce the residue to 0. The best I found so far was at

[ X1 == 0.6529759180536074, X15 == 4.144202714116702, X16 == -2.679195707057889, X19 == -0.7587878291303904, X2 == 0.5740191093184646, X3 == 0.4596545081443212, X4 == 0.8911261003071187, X5 == 0.9678333808777941, X6 == -1.027511586190954, X7 == 1.125758319506284, X8 == -0.4233336402438206, X9 == 3.252655602042921]

with a SSE of 0.000396420099420918 .

But when you substitute these back into F, the system you get is definitely inconsistent.

[chirag rohit]

yes it is true.. Thank you sir..

Answer 2

[Alex Sha]

The equation F(17) seems to be meanless:

F(17)= (100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))/(-C*x(17)))

equal to:

(100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))/(-C*x(17)))=0

equal to:

(100e6*((0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16))))=0

equal to:

(0.5*(x(5)*x(6)+x(7)*x(8)))+(0.5*(x(11)*x(15)+x(12)*x(16)))=0

equal to:

x(5)*x(6)+x(7)*x(8)+x(11)*x(15)+x(12)*x(16)=0

so x(17) actually do nothing, it leads to 19 variables but 20 equations, and there is no idea solution.