Solving an equation system with lsqnonlin gives wrong/insufficient solution

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Miriam Weiß
Miriam Weiß 2020-8-28
评论： Mario Malic 2020-9-28
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);
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Miriam Weiß
Miriam Weiß 2020-9-28
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

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回答（1 个）

Matt J
Matt J 2020-8-29
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.

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