Facing problem in solving simultaneous nonlinear equation with 3 unknowns

x = optimvar('x',3); 
eq1 = 0.7133/(1-0.7133)== x(1).*(1-exp(-(1-x(1)).*127./x(2))).*exp(-127/x(3))/(1-x(1));
eq2 = 0.8058/(1-0.8058)== x(1).*(1-exp(-(1-x(1)).*229./x(2))).*exp(-229/x(3))/(1-x(1));
eq3 = 0.7133/(1-0.8708)== x(1).*(1-exp(-(1-x(1)).*421./x(2))).*exp(-421/x(3))/(1-x(1));
prob = eqnproblem;
prob.Equations.eq1 = eq1;
prob.Equations.eq2 = eq2;
prob.Equations.eq3 = eq3;
show(prob)

  EquationProblem : 

	Solve for:
       x


 eq1:
       2.488 == arg_RHS

       where:

             arg5 = exp(((-127) ./ x(3)));
             arg_RHS = (((x(1) .* (1 - exp((((-(1 - x(1))) .* 127) ./ x(2))))) .* arg5) ./ (1 - x(1)));

 eq2:
       4.1493 == arg_RHS

       where:

             arg5 = exp(((-229) ./ x(3)));
             arg_RHS = (((x(1) .* (1 - exp((((-(1 - x(1))) .* 229) ./ x(2))))) .* arg5) ./ (1 - x(1)));

 eq3:
       5.5209 == arg_RHS

       where:

             arg5 = exp(((-421) ./ x(3)));
             arg_RHS = (((x(1) .* (1 - exp((((-(1 - x(1))) .* 421) ./ x(2))))) .* arg5) ./ (1 - x(1)));

x0.x = [0.9 410 8000];
[sol,fval,exitflag] = solve(prob,x0);

Solving problem using fsolve.

Solver stopped prematurely.

fsolve stopped because it exceeded the function evaluation limit,
options.MaxFunctionEvaluations = 3.000000e+02.

disp(sol.x)

   1.0e+03 *

    0.0010
    0.0525
    1.1031

Here is my codes and error popout. I cant get the correct ans

Hope other can help me :) thanks in advance

[Dyuman Joshi]

That's not an error. That's a message from fsolve displaying information regarding the output.

Why not just use fsolve() directly?

Also, the system of equations you have does not seem to have a solution -

syms x [1 3]
eq1 = 0.7133/(1-0.7133)== x(1).*(1-exp(-(1-x(1)).*127./x(2))).*exp(-127/x(3))/(1-x(1));
eq2 = 0.8058/(1-0.8058)== x(1).*(1-exp(-(1-x(1)).*229./x(2))).*exp(-229/x(3))/(1-x(1));
eq3 = 0.7133/(1-0.8708)== x(1).*(1-exp(-(1-x(1)).*421./x(2))).*exp(-421/x(3))/(1-x(1));
sol = vpasolve([eq1 eq2 eq3], x)

sol = struct with fields:
    x1: [0×1 sym]
    x2: [0×1 sym]
    x3: [0×1 sym]

[John D'Errico]

Note that the presence of x1, x2, and x3 both inside and out of exponentials makes this almost certainly one where solve would never have succeeded anyway. With one unknown, the Lambert W and its close cousin, the Wright-Omega function will sometimes succeed. But not with 3 unknowns. So fsolve or lsqnonlin are the only real choices.

[Chu Yan]

yeaa but it seems cannot get the ans also, as shown by @Matt J

Answer 1

[Matt J]

What @Dyuman Joshi says is true. However, your equations don't seem to make sense without upper and lower bounds on x. For example, since you are dividing by x(2) and x(3) in certain places, you are clearly assuming them to be bounded away from zero somehow... When I impose bounds, a solution (least squares only) is found without hitting the iteration limit.

x = optimvar('x',3,'Lower',[0,0,0],'Upper',[1,inf,inf]); 

eq1 = 0.7133/(1-0.7133)== x(1).*(1-exp(-(1-x(1)).*127./x(2))).*exp(-127/x(3))/(1-x(1));
eq2 = 0.8058/(1-0.8058)== x(1).*(1-exp(-(1-x(1)).*229./x(2))).*exp(-229/x(3))/(1-x(1));
eq3 = 0.7133/(1-0.8708)== x(1).*(1-exp(-(1-x(1)).*421./x(2))).*exp(-421/x(3))/(1-x(1));
prob = eqnproblem;
prob.Equations.eq1 = eq1;
prob.Equations.eq2 = eq2;
prob.Equations.eq3 = eq3;
x0.x = [0.9 410 8000];
[sol,fval,exitflag,output] = solve(prob,x0);

Equation problem has bound constraints. Reformulating as a least squares problem.

Solving problem using lsqnonlin.

No solution found.

lsqnonlin stopped because the last step was ineffective. However, the vector of function
values is not near zero, as measured by the value of the function tolerance. 

disp(sol.x)

    0.9969
   40.1046
  681.4420

exitflag,output

exitflag = 
    NoFeasiblePointFound

output = struct with fields:
         firstorderopt: 1.5721e-05
            iterations: 163
             funcCount: 164
          cgiterations: 0
             algorithm: 'trust-region-reflective'
              stepsize: 0.4884
          bestfeasible: []
       constrviolation: []
    equationderivative: "forward-AD"
                solver: 'lsqnonlin'
               message: 'No solution found.↵↵lsqnonlin stopped because the last step was ineffective. However, the vector of function↵values is not near zero, as measured by the value of the function tolerance. ↵↵<stopping criteria details>↵↵lsqnonlin stopped because the sum of squared function values, r, is changing by less ↵ than options.FunctionTolerance = 1.000000e-06 relative to its initial value.↵However, r = 2.894672e-02, exceeds sqrt(options.FunctionTolerance) = 1.000000e-03.'

[Chu Yan]

alritee, thank you very much, will check back my equations