Function lsqnonlin() has no optimization effect

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Yuwen Zhu
Yuwen Zhu 2020-5-17
回答： Matt J 2020-5-18
It is a teaching example, but I can't get the right answer in my software. How can I solve this problem?
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Yuwen Zhu
Yuwen Zhu 2020-5-18
@Matt J@Ang Feng Thanks for your answers. I try this code again, and it works right.
Yuwen Zhu
Yuwen Zhu 2020-5-18
And I have another question. My code has no optimization effection.
function f=objfunc(x)
PNO=[60.8,60.8,60.8,60.8,60.8,60.8,15.2,30.4,60.8,121.59,151.99,202.65,60.8,60.8,60.8,60.8,60.8];
PO2=[253.3,506.6,1013.3,2026.5,5066.3,10132.5,2026.5,2026.5,2026.5,2026.5,2026.5,2026.5,2026.5,2026.5,2026.5,2026.5,2026.5];
rNO=[7.12e-05,9.66e-05,1.31e-04,1.77e-04,2.66e-04,3.60e-04,5.32e-05,9.72e-06,1.77e-04,3.25e-04,3.948e-04,5.07e-04,1.79e-04,1.77e-04,1.79e-04,1.76e-04,1.79e-04];
a=log(PNO);
b=log(PO2);
y=log(rNO);
f=x(1)+a*x(2)+b*x(3)-y;
x0=[-16.800207,1.2008,0.4151];
[x,resnorm]=lsqnonlin(@objfunc,x0)
Local minimum found.
Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
<stopping criteria details>
x =
-16.8802 1.2008 0.4151
resnorm =
4.4249

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回答（1 个）

Matt J
Matt J 2020-5-18
And I have another question. My code has no optimization effect
In this case, your initial guess already is optimal. You can see this by solving for the optimum x analytically, which is possible in this case because your model is linear, and because you have no bound constraints:
a=log(PNO(:));
b=log(PO2(:));
y=log(rNO(:));
x=[ones(size(a)),a,b]\y(:)
x =
-16.8802
1.2008
0.4151

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