Optimise a reference that cuts my curve into 2 equal sections

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Hello,
I am collecting data from excel, then basic idea is that i want to determine a constant value line that cuts my curve into 2 eqaual sections (area above the line and the curve = area under).
The objective is to determine the value of the red line.

Accepted Answer

Matt J
Matt J on 12 Feb 2022
Edited: Matt J on 12 Feb 2022
%t= time, X=consumption, x=unknown midline
tc=t-t(1);
Xc=cumtrapz(t,X);
x=optimvar('x');
sol=solve( optimproblem('Objective',x,'Constraints',Xc-x*tc<=90,'ObjectiveSense','minimize') );
lb=sol.x; %lower bound
sol=solve( optimproblem('Objective',x,'Constraints',10<=Xc-x*tc,'ObjectiveSense','maximize') );
ub=sol.x; %upper bound
if lb>ub
disp 'Problem is infeasible'
else
xunc=trapz(t,X)/(t(end)-t(1)); %unconstrained solution
x=min(max(xunc,lb),ub); %constrained solution
end
  3 Comments
Matt J
Matt J on 12 Feb 2022
Edited: Matt J on 12 Feb 2022
You probably need to upgrade your Matlab version.
If you cannot upgrade, you can try my revised version.

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More Answers (2)

William Rose
William Rose on 12 Feb 2022
If you mean that the area between the red line and the blue curve abve the line equals the area between the red line and the blue curve below the line, then the height of the red line is simply the mean value of the blue line data.
  5 Comments
dpb
dpb on 12 Feb 2022
Attach the data as a .mat file.
Just looking at the graph, I'd venture it isn't possible to meet the constraints that the integral of the area be <95 when the magnitudes of the integrand are in the thousands. Just by a very crude approximation of the leftmost area (brown?) as roughly a rectangle of height ~(2800-1800) and a duration of 12, the integral would be roughly 1000*12 --> 12,000. Unless there's a very small scaling factor to be applied, it "just ain't a-gonna' happen".
fadi awar
fadi awar on 12 Feb 2022
Kindly find attached the files.
I am to get the result according to my limits even if in this case of chosen data it wont matter a lot.
But Future wise I will be chaniging the data.
Thank you

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Catalytic
Catalytic on 12 Feb 2022
Edited: Catalytic on 12 Feb 2022
midline = trapz(x,y)/(max(x)-min(x))

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