Solving an equation system with lsqnonlin gives wrong/insufficient solution

I want so solve a system of 6 equations with lsqnonlin. The solution is not satisfactory, as estimated values are too far from expected and sufficient values.

I have tried to improve the solving by changing the initial values, Algorithm and Step Tolerance - without any success.

Can anyone help?

        A = 3.6855e03;
        B = 7.6168e03;
        C = 1.0747e08;
        D = 0.0013;
        E = 5.4871e-04;
        F = 2.2370e-05;
        p_sys = 3e05;
        
        fun=@(y)[D - y(1) - y(4);                                            % 1
            E - y(2) - y(5);                                                 % 2
            F - y(3) - y(6);                                                 % 3
            1 - (A*(y(1)/(y(1)+y(2)+y(3))))/((y(4)/(y(4)+y(5)+y(6)))*p_sys);    % 4
            1 - (B*(y(2)/(y(1)+y(2)+y(3))))/((y(5)/(y(4)+y(5)+y(6)))*p_sys);    % 5
            1 - (C*(y(3)/(y(1)+y(2)+y(3))))/((y(6)/(y(4)+y(5)+y(6)))*p_sys)];   % 6

        y0 = [D, E, F, D, E, F];
        lb = [0 0 0 0 0 0];
        ub = [D, E, F, D, E, F];
        lsg = lsqnonlin(fun,y0,lb,ub);
        
        check_1 = D - lsg(1,1) - lsg(1,4);
        check_2 = E - lsg(1,2) - lsg(1,5);
        check_3 = F - lsg(1,3) - lsg(1,6);
        check_4 = 1-(A*lsg(1,1)/(lsg(1,1)+lsg(1,2)+lsg(1,3)))/((lsg(1,4)/(lsg(1,4)+lsg(1,5)+lsg(1,6)))*p_sys);
        check_5 = 1-(B*(lsg(1,2)/(lsg(1,1)+lsg(1,2)+lsg(1,3))))/((lsg(1,5)/(lsg(1,4)+lsg(1,5)+lsg(1,6)))*p_sys);
        check_6 = 1-(C*(lsg(1,3)/(lsg(1,1)+lsg(1,2)+lsg(1,3))))/((lsg(1,6)/(lsg(1,4)+lsg(1,5)+lsg(1,6)))*p_sys);

[Mario Malic]

Hey Miriam,

There are a couple of things that I could see from the code:

• Your values for A - F are both, large and small. If some of them represent same physical property, make sure you have the same units (i.e. not to use pressure values both in Pa and MPa).
• Set your starting guess to middle of your bounds
• If you tweaked optimisation options, make sure you call them as well in your optimisation function

lsg = lsqnonlin(fun,x0,lb,ub,options)

• If your values are as they are, you can also try an option to scale a problem, see ScaleProblem in documentation.
• If you can provide objective function gradients to optimisation algorithm, it would be useful

[Miriam Weiß]

Hello Mario,

please excuse my late reply and thank you for your answer.

A good solution was to change the first three equations:

        fun=@(y)[(D - y(1) - y(4))*10000;                       % 1
            (E - y(2) - y(5))*10000;                            % 2
            (F - y(3) - y(6))*10000;                            % 3
            1 - (A*(y(1)/(y(1)+y(2)+y(3))))/((y(4)/(y(4)+y(5)+y(6)))*p_sys);    % 4
            1 - (B*(y(2)/(y(1)+y(2)+y(3))))/((y(5)/(y(4)+y(5)+y(6)))*p_sys);    % 5
            1 - (C*(y(3)/(y(1)+y(2)+y(3))))/((y(6)/(y(4)+y(5)+y(6)))*p_sys)];   % 6

[Mario Malic]

Great to hear it worked out well.

Answer 1

[Matt J]

In addition to what Mario said, the first 3 equations can be used to eliminate y(4:6). With only 3 unknowns it would be computationally feasible to sample the objective function on say a 500x500x500 search grid of values for y(1:3). This would give you a decent idea of where to choose your initial guess.