[Scilab-users] non linear optimization why produce Nan

karoli Karoli at greenae.org
Wed Mar 27 09:20:33 CET 2019


hello team
why my optimization produce an Nan result, any help


[fd,SST,Sheetnames,Sheetpos]=xls_open('D:\MAG2.xls')
[Value,TextInd]=xls_read(fd,Sheetpos(1))

y=45
i=33
p0=[53.8;1527;43.7;55.7;3424.7]

function T= dyke(p,Value, i, y)  
    h =1.0 - cosd( y )^2*sind( i )^2
    I= atand( tand( i )./sind( y ) )
    O=2*I - p(1)    //P(1) = dip of the dyke(β)  O is degree
    q1 = atand ( (( Value(:,1)-p(2)) - p(3) )./p(4) )
    q2 = atand ( (( Value(:,1)-p(2)) + p(3) )./p(4) )  //p(2) = horizontal
location of the centre of top of dyke model, p(3)= the half width of dyke,
p(4) = the depth to top of the model ,q1 and q2 are degrees
    Q = q1 - q2
    A=%pi*Q./180 // convertion to radisn
    r1 = sqrt ( p(4)^2 + ((Value(:,1)-p(2))-p(3))^2 )
    r2 = sqrt ( p(4)^2 + ((Value(:,1)-p(2))+p(3))^2 )
    R=r1./r2
    T = 2*h*p(5)*sind(p(1))*(  sind(O)*A - cosd(O)*log(R))  //P(5) the
intesity of magnetisation
endfunction

function e=dyke_error(p,Value, i, y)
    e= dyke(p,Value, i, y)-Value(:,2)
endfunction.

->[f,p,g]=leastsq(dyke_error,p0)
 g  =
 
    Nan  
    Nan  
    Nan  
    Nan  
    Nan  
 p  =
 
    53.8    
    1527.   
    43.7    
    55.7    
    3424.7  
 f  =
 
    Nan  
 



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