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How to terminate the loop

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I want to terminate the loop. Condition is satisfying at i = 96, but loop is still running till the end.
after the termination, i want to know the value of i.
point_load(1) = 0;
for i = 1 : 100
point_load(i + 1) = point_load(i) + 1;
PL_BM(i) = point_load(i) * beam_length / 4;
PL_BM = repelem(PL_BM(i),n);
failure_domain_BM = ( RV_BM - PL_BM(i) ) < 0;
p_f_BM = sum(failure_domain_BM (:) == 1) / n;
if p_f_BM == 1
break
disp(point_load(i+1))
end
end

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Accepted Answer

the cyclist
the cyclist on 25 Nov 2019
Due to floating-point error, it may be that the condition
if p_f_BM == 1
is not exactly met. With floating-point numbers, it's better to check for equality within some tolerance. Try something like
if abs(p_f_BM-1) < tol
for some appropriately small value of tol.

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Jaydeep Kansara
Jaydeep Kansara on 25 Nov 2019
This is the entire code, if it helps.
b18 = 3+0+0+7+6+0+4+9; % b18 = 29
b58 = 6+0+4+9; % b58 = 19
n = 100000000; % No of samples = 10,000,000
rng default
beam_length = 5; % Beam length = 5 m
% Bending Moment Capacity follows a log-normal distribution
mean_BM = 2 * b18; % mean of BM = 58 kNm
COV_BM = 0.01 * b58; % COV of BM = 0.19
variance_BM = (mean_BM * COV_BM)^2; % Variance of BM = 121.4404
mu_BM = log(mean_BM^2 / sqrt(mean_BM^2 + variance_BM)); % Location parameter of BM = 4.0427
sigma_BM = sqrt(log(1 + (variance_BM / mean_BM^2))); % Scale paramaeter of BM = 0.1883
RV_BM = lognrnd(mu_BM, sigma_BM, 1, n); % Random variables of BM
% Load Test
% Apply 50 kN load gradually in 1 kN load increment
point_load(1) = 0;
breaking_point = 10^-10;
for i = 1 : 100
point_load(i + 1) = point_load(i) + 1;
PL_BM(i) = point_load(i) * beam_length / 4;
PL_BM = repelem(PL_BM(i),n);
failure_domain_BM = ( RV_BM - PL_BM(i) ) < 0;
p_f_BM = sum(failure_domain_BM (:) == 1) / n;
if abs(p_f_BM - 1) < breaking_point
break
end
end
Ridwan Alam
Ridwan Alam on 25 Nov 2019
n is too big
you need a larger breaking point, maybe 1e-5?
Jaydeep Kansara
Jaydeep Kansara on 30 Nov 2019
Hello Mr. Cyclist,
Thanks for your guidance. I got my answer based on your solution.

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More Answers (1)

Ridwan Alam
Ridwan Alam on 25 Nov 2019
Edited: Ridwan Alam on 25 Nov 2019
b18 = 3+0+0+7+6+0+4+9; % b18 = 29
b58 = 6+0+4+9; % b58 = 19
n = 10000000; % No of samples = 10,000,000
rng default
beam_length = 5; % Beam length = 5 m
% Bending Moment Capacity follows a log-normal distribution
mean_BM = 2 * b18; % mean of BM = 58 kNm
COV_BM = 0.01 * b58; % COV of BM = 0.19
variance_BM = (mean_BM * COV_BM)^2; % Variance of BM = 121.4404
mu_BM = log(mean_BM^2 / sqrt(mean_BM^2 + variance_BM)); % Location parameter of BM = 4.0427
sigma_BM = sqrt(log(1 + (variance_BM / mean_BM^2))); % Scale paramaeter of BM = 0.1883
RV_BM = lognrnd(mu_BM, sigma_BM, 1, n);
point_load(1) = 0;
for i = 1 : 100
point_load(i + 1) = point_load(i) + 1;
PL_BM(i) = point_load(i) * beam_length / 4;
%PL_BM = repelem(PL_BM(i),n);
failure_domain_BM = sum( RV_BM < PL_BM(i) );
p_f_BM = failure_domain_BM / n
if p_f_BM >= 0.9999
disp(['i=' num2str(i)])
break
end
end
this gave me i=93

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Ridwan Alam
Ridwan Alam on 28 Nov 2019
Jaydeep, did you get your answer?
Image Analyst
Image Analyst on 28 Nov 2019
He did mark the cyclist's answer as "Accepted".
Jaydeep Kansara
Jaydeep Kansara on 30 Nov 2019
Hello Ridwan,
Thabks for the reply. I got the answer. It was a floating point error. When I kept the difference too small around 5e-4, i got the answer. Thanks again.

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