Difficulty in extracting the particular variable value (a_mean, a_std) from the code
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I want to fit an quadratic plateau equation and find the statistics of the parameters.
The form of equation are given as:
y=a*(x)^2+b*(x)+c, x < xc(Ymax)
y=yp, x >= xc(Ymax)
x=[0,40,80,100,120,150,170,200],
y=[1865,2855,3608,4057,4343,4389,4415,4478]
The code which I have tried for this purpose is given as:
function Fiting_ex()
global a_iter b_iter c_iter
a_iter = 0;
b_iter = 0;
c_iter = 0;
x=[0; 40; 80; 100; 120; 150; 170; 200];
y=[1865; 2855; 3608; 4057; 4343; 4389; 4415; 4478];
X = [x.^2 x ones(numel(x),1)];
A = y\X;
a0=A(1);
b0=A(2);
c0=A(3);
B0 = [a0; b0; c0];
[Bm,normresm] = fminsearch(@(b) myfun(b,x,y),B0);
a=Bm(1);
b=Bm(2);
c=Bm(3);
xc=-b/(2*a);
p=c-(b^2/(4*a));
yfit = zeros(numel(x),1);
for i=1:numel(x)
if (x(i) < xc)
yfit(i) = a.*x(i).^2+ b*x(i)+c;
else
yfit(i) = p;
end
end
plot(x,yfit,'*')
hold on;
plot(x,y);
hold off
% Statistic on optimization process
a_mean = mean(a_iter(2:end)); % mean value
a_var = var(a_iter(2:end)); % variance
a_std = std(a_iter(2:end)); % standard deviation
function f = myfun(Bm, x, y)
global a_iter b_iter c_iter
a_iter = [a_iter Bm(1)];
b_iter = [b_iter Bm(2)];
c_iter = [c_iter Bm(3)];
yf = Bm(1)*(x).^2+Bm(2)*(x)+Bm(3);
a=Bm(1);
b=Bm(2);
c=Bm(3);
xc=-b/(2*a);
p=c-(b^2/(4*a));
yfit = zeros(numel(x),1);
for i=1:numel(x)
if (x(i) < xc)
yfit(i) = a.*x(i).^2+ b*x(i)+c;
else
yfit(i) = p;
end
end
f = norm(y - yfit);
When I run this script, no error have shown in command window, but I am unable to get the values of the variables (a_mean, a_std, a_var, yfit), I mean it did not display the values of these variables. How can I extrat these value from this code.
I have another query related to the equation that if I want to repalce 'x' with 'x+n', where n is also a unknown parameter. Also I need to find the mean, standard deviation of this parameter (n) too. Then what will be the changes in above code.
Kindly help me for this...
2 Comments
KALYAN ACHARJYA
on 3 Jan 2019
particular variable value
Which variable?
madhuri dubey
on 3 Jan 2019
Accepted Answer
More Answers (0)
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