Saving vector result from each loop if solution is correct.

Let's say DMP.m uses fsolve and output a solution vector, named xx. It can be `Equation solved', 'No solution found', or 'Equation solved but inaccuracy possible.'

I want to store the vector if it tells `Equation solved'.

So for example, my code is

for r=1:10
    DMP
    if exitflag==1
        xxx(r,:)=xx 
    end
end


Unrecognized function or variable 'r'.

But Matlab gives me the error message: Unrecognized function or variable 'r'.

Could you help me the coding?

Answer 1

[Star Strider]

Use the exitflag output from fsolve to determine whether a solution was found.

.

[Deokjae Jeong]

Thank you!

[Star Strider]

My pleasure!

.

[Deokjae Jeong]

Why Matlab says "Unrecognized function or variable 'r'."?

[Star Strider]

I am not certain what is throwing that error. I created my own version of ‘DMP’ to test it, and it appears to work correctly as originally written (with my ‘DMP’ code, at least) —

for r=1:10
    [xx,exitflag] = DMP(r);
    if exitflag==1
        xxx(r,:)=xx;
    end
end

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 0.6180

fv = 3.2817e-08

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 1.0000

fv = 1.0658e-13

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 1.3028

fv = 1.7319e-13

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 1.5616

fv = 2.2162e-08

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 1.7913

fv = 2.7710e-10

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 2.0000

fv = 7.5939e-13

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 2.1926

fv = 3.4635e-11

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 2.3723

fv = 5.3424e-08

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 2.5414

fv = 7.0818e-09

exitflag = 1

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xx = 2.7016

fv = 6.7687e-08

exitflag = 1

xxx

xxx = 10×1

0.6180 1.0000 1.3028 1.5616 1.7913 2.0000 2.1926 2.3723 2.5414 2.7016

function [xx,exitflag] = DMP(r)
[xx,fv,exitflag] = fsolve(@(x)x.^2+x - r, rand)
end

I only see the posted code and nothing else, so what is not posted may be causing the problem.

.

[Deokjae Jeong]

Oh.. great thank you so much.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

[Deokjae Jeong]

I changed r to i then the problem disappeard.

There is another problem, however. Matlab says "Index in position 1 is invalid. Array indices must be positive integers or logical values."

clear
for i=1:10
    DMP;
    if exitflag==1
        xxx(i,:)=xx;
    end
end

[Star Strider]

I am not certain what MATLAB version/release is being used, however in older releases, ‘i’ and ‘j’ when used alone were considered the imaginary operator , so here ‘i’ is not a real number. Either substitute it for another variable name or use ‘ii’. Then it will not be confused with the imaginary operator.

.

[Deokjae Jeong]

When I change i to k, it gives me the previous error message: "Unrecognized function or variable 'k'."

By the way, xx is a vector: xx = 0.0411 0.1280 0.0153 24.3166 0.0323 0.0067

for k=0:10
    DMP;
    if exitflag==1
        xxx(k,:)=xx;
    end
end

[Star Strider]

I suspect that ‘k’ may be being used int the ‘DMP’ function, however is not being passed to it as an argument, and that throws the error. Since I have never seen the ‘DMP’ function (except the one I wrote to test the code that worked correctly) or the reference to it in the posted code (note that the ‘DMP’ call alone as presented in the posted code will run it, however will not pass anything to it or return anything from it), that is only a guess.

My test code sends ‘r’ to it as an argument and returns the necessary information from it in the function call to it. It shows the correct way to call ‘DMP’ with its arguments, and return values from it to the calling script, so they can be used approporiately.

.

[Deokjae Jeong]

Thank you for your detailed answer. Here is the link for my entire code in case you can see it. Anyway, you helped me so much already.

https://www.dropbox.com/sh/x2m8pzq34gq00dz/AABiMpkxivVl3_hmHY44DzSfa?dl=0

[Star Strider]

Note that ‘DPM’ is a script, not a function, so it will not return anything. I changed it to be a function here.

Try this version (it takes too long to run it here, so I ran it offline on my computer) —

for r=1:10
    [xx,exitflag] = DMP;
    if exitflag==1
        xxx(r,:)=xx;
    end
end

xxx

function F = equations(x,beta,w,p,z,b,d)
u=x(1);
v=x(2);
theta=x(3);
eta=x(4);
a=x(5);
lambda=x(6);
c=x(7);
F(1)=((abs(lambda)+b)*(1-u)*u)/((abs(lambda)+b)*(u-1)+100*a*u)-v;
F(2)=p-(p*c)/(beta*(100*a/(1+theta)))*(1-beta*(1-d)*(1-abs(lambda)))-w;
F(3)=abs(eta)*p+(1-abs(eta))*z+abs(eta)*theta*p*c*(1-d)-w;
F(4)=v/u-theta;
F(5)=-.3697544+.0341093/abs(eta)+7.533068/c-theta;
F(6)=-.0410965+3.417642*abs(d)-abs(lambda);
F(7)=66.2881*abs(eta)-1061.607*lambda+693.8852*a+3.338711*p-4.223466*w-c;
F(8)=-.0162202/c/eta+.0009784*c+.0016295/eta+1.396799*lambda-.0047*p+.0056439*w-a;
%F(9)=.0523244+2.903846*v-.6319993/c-.0003297/eta/eta+.0000121/eta/eta/eta-u
F(10)=(beta*100*a*(p-eta*p+eta*z-z))/(beta*100*a*eta*theta*p*(1-d)+p*(1+theta)*(1-beta*(1-d)*(1-lambda)))-c;
end

function [xx,exitflag] = DMP
beta=0.998418435031256; %monthly (yearly=0.981185442086183)
w=21.42592; %Period 4
p=28.1777866666667; %Period 4
z=0.0113878916464317;  %Period 4
b=0.0119785793333333; %Period 3
d=0.0169109966666667; %Period 4

optimoptions('fsolve');
options = optimoptions('fsolve');
%options.Algorithm = 'trust-region'
options.Algorithm = 'Levenberg-Marquardt';
options.MaxFunctionEvaluations = 1000000;
options.MaxIterations = 100000;
options.StepTolerance = 1e-20;
options.FunctionTolerance = 1e-5;
options.OptimalityTolerance: 1e-5;

st = 0.01;
ed = 0.04;
u_ini = (ed-st).*rand(1,1) + st;
st = 0.005;
ed = 0.017;
v_ini = (ed-st).*rand(1,1) + st;
theta_ini=v_ini/u_ini;
st = 0.02;
ed = 0.06;
a_ini = (ed-st).*rand(1,1) + st;
st = 0.06;
ed = 0.20;
eta_ini = (ed-st).*rand(1,1) + st;
st = 0.01;
ed = 0.03;
lambda_ini = (ed-st).*rand(1,1) + st;
st = 12;
ed = 30;
c_ini = (ed-st).*rand(1,1) + st;

x0=[u_ini,v_ini,theta_ini,eta_ini,a_ini,lambda_ini,c_ini];

[x fval exitflag]=fsolve(@equations,x0,options,beta,w,p,z,b,d);

u=x(1);
v=x(2);
theta=x(3);
eta=x(4);
a=x(5);
lambda=x(6);
c=x(7);
xx=[a,eta,lambda,c,u,v];
end

It ran without error, and produced (in one run) —

xxx =

0.0399 0.0820 0.0167 18.9749 0.0229 0.0101

0 0 0 0 0 0

0.0413 0.1399 0.0165 24.0175 0.0418 0.0078

0.0399 0.0820 0.0167 18.9749 0.0229 0.0101

0.0399 0.0820 0.0167 18.9749 0.0229 0.0101

0.0399 0.0820 0.0167 18.9749 0.0229 0.0101

0.0399 0.0820 0.0167 18.9749 0.0229 0.0101

0.0399 0.0820 0.0167 18.9749 0.0229 0.0101

The zeros indicate a run in which the solution was not found. The random values are the reason that it will not always converge on a solution. This varies between runs.

.

[Deokjae Jeong]

Oh.... It works. I am so thankful to you. I learned a lot.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.