Simple optimisation question
in [Matlab]
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From: Frédéric Bergeron on 28 Jun 2010 09:36 Hi all, First, thanks for the answer I got in the past in this forum, hope I'll get some today! :) My situation: I got a set of 1-dimensional data (data) and I want to find the scalar x that respect a previously defined ratio R: R=sum(data(over)-x)/sum(x-data(under)); where: data(over) represent all the data over x: data(over)=(data>=x); data(under) represent all the data under x: data(under)=(data<x); So R is the ratio of the sum of difference between data over x and x, over the sum of difference between data under x and x. My problem is that I need the value of x to define data(over) and data(under). Is there's a function that could help me? I read about lsqlin() that can solve a problem with inequality constraint but I didn't get success at using it. Thanks in advance! Best regards,
From: Alan Weiss on 28 Jun 2010 09:53 On 6/28/2010 9:36 AM, Frédéric Bergeron wrote: > Hi all, > > First, thanks for the answer I got in the past in this forum, hope I'll > get some today! :) > > My situation: > I got a set of 1-dimensional data (data) and I want to find the scalar x > that respect a previously defined ratio R: > > R=sum(data(over)-x)/sum(x-data(under)); > where: > data(over) represent all the data over x: data(over)=(data>=x); > data(under) represent all the data under x: data(under)=(data<x); > > So R is the ratio of the sum of difference between data over x and x, > over the sum of difference between data under x and x. > > My problem is that I need the value of x to define data(over) and > data(under). Is there's a function that could help me? I read about > lsqlin() that can solve a problem with inequality constraint but I > didn't get success at using it. > > Thanks in advance! > Best regards, I think you would do better to use fzero. Alan Weiss MATLAB mathematical toolbox documentation
From: Matt J on 28 Jun 2010 10:56 "Frédéric Bergeron" <frederic.bergeron(a)logiag.com> wrote in message <i0a8gm$kju$1(a)fred.mathworks.com>... > My problem is that I need the value of x to define data(over) and data(under). Is there's a function that could help me? I read about lsqlin() that can solve a problem with inequality constraint but I didn't get success at using it. ======= I could be wrong, but it seems like you can solve this by simple algebra. Let's make the following definitions N=length(data); %known S=sum(data); %known O=sum(data(over)-x) U=sum(x-data(under)) Then you have 2 equations that allow you to solve for O and U: O-R*U=0 O+U=S; Once you know O and U, you have the relationship O-U=S-N*x which can be used to solve explicitly for x x=(S+U-O)/N
From: Matt J on 28 Jun 2010 11:40 "Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <i0ad6n$91i$1(a)fred.mathworks.com>... > > I could be wrong, but it seems like you can solve this by simple algebra. Let's make the following definitions > > N=length(data); %known > S=sum(data); %known > O=sum(data(over)-x) > U=sum(x-data(under)) > > Then you have 2 equations that allow you to solve for O and U: > > O-R*U=0 > O+U=S; ======== Forget it. I pulled the equation O+U=S from my imagination somehow. In any case, there are at most N different possibilities for the sets data(over) and data(under). For each one, x ranges over a known interval separating data(over) from data(under) and in that interval O -R*U is a known linear function of x. You could simply loop over each of these N sets and test whether O(x)-R*U(x) has a root in that interval.
From: Matt J on 28 Jun 2010 12:14
"Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <i0afpa$3is$1(a)fred.mathworks.com>... > In any case, there are at most N different possibilities for the sets > data(over) and data(under). For each one, x ranges over a known interval separating data(over) from data(under) and in that interval O -R*U is a known linear function of x. You could simply loop over each of these N sets and test whether > O(x)-R*U(x) has a root in that interval. =========== The following code implements this, assuming that all data points are unique. I leave the generalization as an exercise. %data R=0.75; data=rand(1000,1); %engine data=sort(data); S=sum(data); N=length(data); Sunder=cumsum(data(1:end-1)); Sover=S-Sunder; Nunder=(1:N-1).'; Nover=N-Nunder; X=(Sover+R*Sunder)./(Nover+R*Nunder); x=X(X>=data(1:end-1) & X<=data(2:end)) %Check over=(data>=x); under=~over; Rcheck=sum(data(over)-x)/sum(x-data(under)) |