# How to optimize the value of x(2)

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Raj Arora on 2 May 2022
Edited: Torsten on 2 May 2022
CODE FOR COMPUTING MEDIAN AND MAXIMUM LIKELIHOOD VALUE. I WANT TO CALCULATE OPTIMUM VALUE OF STANDARD DEVIATION BASED ON INOUT DATA FILE
function likeli
da = load("C:\Users\Admin\OneDrive - IIT Bombay\Desktop\standard deviation\fragility1.txt")%%Damage state in terms of 0 and 1
n = length(da)
xdamage = da(:,1)
x0 = [2]%x valueis a median values%
options = optimset('LargeScale','off','Display','off','TolX',0.001,'TolFun',0.001)
[x,fval] = fminsearch(@myfun,x0,options,n,WH,xdamage)%fval is the likeli hood function%
function f = myfun(x,n,WH,xdamage);
options = optimset('LargeScale','on','Display','off','TolX',0.001,'TolFun',0.001)
p1=0.0;
for i=1:n
x(2) = 0.5;
yx=(log(WH(i)/x(1)))/x(2)%%x(2) is standard deviation
if yx >= 5.0;
y1=5.0;%%maximum value normcdf can take taken as 5%%
elseif yx<=-5.0;
y1=-5.0;
else
y1 = yx;
end
y2=normcdf(y1)
p1=p1+log(((y2)^xdamage(i))*((1.0-y2)^(1.0-xdamage(i))))%%Maximum likeli hood%
end
f=-p1;
return
Hello all I have a querry, I am solving one problem in which I have to compute optimum value of x(2), In this code I have taken x(2) value as constant 0.5. Here da is a file having 0 and 1 (100 values; defined as damage state) and the other WH is file having values between 1.25-2.0 (100 values). Basically this value is computed using some formula 0 means no failure and 1 mean failure.

Torsten on 2 May 2022
Edited: Torsten on 2 May 2022
da = load("C:\Users\Admin\OneDrive - IIT Bombay\Desktop\standard deviation\fragility1.txt")%%Damage state in terms of 0 and 1
n = length(da);
xdamage = da(:,1);
wh = WH(:,1);
x0 = [2 0.5];%x valueis a median values%
%options = optimset('LargeScale','off','Display','off','TolX',0.001,'TolFun',0.001)
%[x,fval] = fminsearch(@myfun,x0,options,n,wh,xdamage)%fval is the likeli hood function%
[x,fval] = fminsearch(@(x)myfun(x,n,wh,xdamage),x0);%fval is the likeli hood function%
function f = myfun(x,n,wh,xdamage);
p1=0.0;
for i=1:n
yx=(log(wh(i)/x(1)))/x(2);%%x(2) is standard deviation
if yx >= 5.0;
y1=5.0;%%maximum value normcdf can take taken as 5%%
elseif yx<=-5.0;
y1=-5.0;
else
y1 = yx;
end
y2=normcdf(y1);
p1=p1+log(((y2)^xdamage(i))*((1.0-y2)^(1.0-xdamage(i))));%%Maximum likeli hood%
end
f=-p1;
end
Raj Arora on 2 May 2022
Thanks torsten for your prompt reply. But this is still not giving appropriate answer the standard deviation and maximum likelihood value coming are very small which doesnot seems correct to me.
Torsten on 2 May 2022
Edited: Torsten on 2 May 2022
I didn't check your equations for correctness - I only gave you the way how to work with two inputs for fminseach.
If you describe in more detail what you are doing in your code, maybe someone can also help in terms of content.