
From:  Nicholas Jankowski 
Subject:  Re: Help in parameter estimation 
Date:  Tue, 30 Aug 2016 11:02:51 0400 
On Tue, Aug 30, 2016 at 10:45 AM, Bharath R <address@hidden> wrote:Hi Nicholas,Thanks for the mail. I am sorry that I have used mat earlier, now I changed the function as follows:function F=f(x,p)A=xlsread('OGparameters.xlsx'); x=[A(:,3),A(:,5),A(:,2),A(:,6),A(:,7)]; p(1)=20;p(2)=1;p(3)=100;p(4)=1;p(5)=0.001;p(6)=1;p(7)=0.01;typeinfo(x)pin=[p(1);p(2);p(3);p(4);p(5);p(6);p(7)]; m(1,1)=20;for i=1:length(x)m(i+1,1)=((p(1)m(i,1))/(p(2)*p(3))+((p(4)*x(i,2)*(x(i,3)m( i,1)))+(p(5)*x(i,4))+(p(6)*x(i ,5)))/(p(3))+p(7))+m(i,1); endforF=m(2:end);endfunctionI am generating a matrix of [length(x), 1] as output from the function. My measured value is also a matrix of [length(x), 1].What I would like to do now is to estimate parameters p. I am using the following command to run the optimization program ( I have tried with nonlinear curve fit as well):[L,p,cvg,iter]=leasqr(x,y,pin,f) pin ,x and y are initialized. The problem is when I run the optimization algorithm, it throws me the following error:error: leasqr: subscript indices must be either positive integers less than 2^31 or logicalserror: called fromleasqr at line 329 column 9I look forward to your reply.Regards,BharathOn Tue, Aug 30, 2016 at 4:01 PM, Nicholas Jankowski <address@hidden> wrote:On Tue, Aug 30, 2016 at 6:00 AM, Bharath R <address@hidden> wrote:function m=f(x,p)m(1,:)=20;for i=1:length(x)m(i+1,:)=((p(1)m(i,:))/(p(2)*p(3))+((p(4)*x(i,2)*(x(i,3)m( i,:)))+(p(5)*x(i,4))+(p(6)*x(i ,5)))/(p(3))+p(7))+m(i,:); endformat=m(2:end);endfunctionis your intent to return the variable m or the variable mat from function f? you are currently returning m and never using mat. since mat is a different size than m, this could make a significant difference when you use the results of f later.your declaration for m is odd. it asks to make all of the columns in row1 equal to 20. since m has not yet been initialized, the number of rows is 1. so the output is a 1x1 array, or a scalar, and m = 20. then in the for loop, if the result of the m(i+1,:) equation was 6, you would be appending it as a new row to m, creating a 2x1 array [20 ; 6] . Since you went to the trouble of using the : operator, but have it doing nothing, I suspect this was not your intent?Can you provide a small sample that I can use for values of x?
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