Constrained linear least squares
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Hi everyone! I have a set of data such as:
y=[0.007696733 0.004758526 0.00547575 0.009628829 0.009749457 0.009073008 0.009647521 0.009795106 0.014071997 0.014208544 0.01758061 0.015072178 0.018425671 0.019217437 0.020314108 0.018246917 0.022694743 0.041827743 0.040799924 0.033812901 0.043516698];
y(t)=y(2:n);
y(t-1)=y(1:n-1);
I want to figure out the two parameters [b,c] of: y(t)=(1+b)*y(t-1)-b/c*(y(t-1))^2
What's more the two parameters need to be over 0;
Thank you so much for reading my question!
回答（1 个）
Star Strider
2014-7-1
编辑：Star Strider
2014-7-1
It is actually nonlinear, because it is nonlinear in the parameters (the partial derivatives of the function with respect to each parameter are functions of themselves or of other parameters).
Using lsqcurvefit (Optimization Toolbox), the code is:
n = length(y);
yd = y(2:n);
yi = y(1:n-1);
f = @(B,y) (1 + B(1)).*y - (B(1)./B(2)).*y.^2; % Objective function
yp = linspace(min(yi), max(yi)); % Plot vector for ‘yi’
bc = lsqcurvefit(f, rand(2,1)*100, yi, yd, [1; 1]*eps, [Inf; Inf])
figure(1)
plot(yd, yi, 'xb')
hold on
plot(yp, f(bc,yp), '-r')
hold off
grid
The fit produces:
b = 2.989459285195542E-01
c = 3.990401160101856E-02
b/c = 7.491625942488537E+00
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2014-7-1
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