# How do I use the random function in Fsolve Matlab?

2 views (last 30 days)
Bowen Yang on 17 Mar 2022
Commented: Torsten on 26 Mar 2022
Hello everyone,
In the following codes, I am trying to determine the variable x that lets beta equal target beta. This does not work for me when I use fsolve. Could anyone please take a look at it and help me with that?
Thank you so much!
% example
clear
clc
% define values
n = 1e4;
target_beta = 2;
x = 2; % predefine x
% parameters
Mean_R = 2000 ; %mean
CoV_R = 0.1;%coefficient of variation
Std_R = Mean_R * CoV_R;
% Monte Carlo Simulation
Ri = Mean_R + Std_R .* norminv(rand(n, 1));
Mean_Q = x * 1100; % Determine which variable, x, will let the beta = target beta
CoV_Q = 0.18;
Std_Q = Mean_Q * CoV_Q;
Qi = Mean_Q + Std_Q .* norminv(rand(n, 1)); % simuate the Q term
g = Ri - Qi; % limit state function g = R - Q
m = find(g < 0); % count the number of failure cases
f = length(m); % find the failure cases number in the g = R-Q vector
pr_failure = f / n; % probability of failure
beta = norminv(1 - pr_failure); %Calculate the reliability index beta
% fsolve
alpha = fsolve(@(x) (beta-target_beta),2)

Torsten on 22 Mar 2022
This might help to get beta as a function of x:
X = (0*2000/1100:0.1:2000/1100+10).';
n = 1e4;
trials = 1e3;
beta = zeros(numel(X),1);
Pr_failure = zeros(numel(X),1);
for j=1:numel(X)
pr_failure = zeros(trials,1);
x = X(j);
for i = 1:trials
RAND = rand(n,2);
% parameters
Mean_R = 2000 ; %mean
CoV_R = 0.1;%coefficient of variation
Std_R = Mean_R * CoV_R;
% Monte Carlo Simulation
Ri = Mean_R + Std_R .* norminv(RAND(:,1));
Mean_Q = x * 1100; % Determine which variable, x, will let the beta = target beta
CoV_Q = 0.18;
Std_Q = Mean_Q * CoV_Q;
Qi = Mean_Q + Std_Q .* norminv(RAND(:,2)); % simuate the Q term
g = Ri - Qi; % limit state function g = R - Q
m = find(g < 0); % count the number of failure cases
f = length(m); % find the failure cases number in the g = R-Q vector
pr_failure(i) = f / n; % probability of failure
end
pr_failure = mean(pr_failure);
pr_failure = max(eps,pr_failure);
beta(j) = norminv(1 - pr_failure); %Calculate the reliability index beta
Pr_failure(j) = pr_failure;
end
figure(1)
plot(X,beta)
figure(2)
plot(X,Pr_failure)
Torsten on 26 Mar 2022
I was hoping fsolve could be used if the number of trials was big enough to stabilize the distribution for pr_failure (and thus could return stable values to fsolve), but I was not successful. The precision needed to calculate the derivatives in fsolve is too high.

Priyanka Kondapalli on 22 Mar 2022
Edited: Priyanka Kondapalli on 24 Mar 2022
Hi,
I do not see any issue with the code provided by you.However, recheck the equation. Please refer to the link below which provides more details on how to use Fsolve.