0 个投票

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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 采纳的回答

the cyclist
the cyclist 2019-11-25

0 个投票

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 2019-11-25
编辑：Jaydeep Kansara 2019-11-25
Thank you for the reply. but, still its running till the end.
Jaydeep Kansara
Jaydeep Kansara 2019-11-25
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 2019-11-25
n is too big
you need a larger breaking point, maybe 1e-5?
Jaydeep Kansara
Jaydeep Kansara 2019-11-30
Hello Mr. Cyclist,
Thanks for your guidance. I got my answer based on your solution.

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更多回答（1 个）

Ridwan Alam
Ridwan Alam 2019-11-25
编辑：Ridwan Alam 2019-11-25

0 个投票

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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Jaydeep Kansara
Jaydeep Kansara 2019-11-25
编辑：Jaydeep Kansara 2019-11-25
Thanks for the reply.
actually i want to break it at 1.00.
it is breaking before that.
Ridwan Alam
Ridwan Alam 2019-11-25
If you increase the end limit of i to 120 (i=1:120), you will see that the break point is reached at i = 118. Because the display is rounding up the floating points, you thought the condition is met earlier, whereas it is not. Hope you understand.
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 : 120
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 >= 1
disp(['i=' num2str(i)])
break
end
end
Ridwan Alam
Ridwan Alam 2019-11-28
Jaydeep, did you get your answer?
Image Analyst
Image Analyst 2019-11-28
He did mark the cyclist's answer as "Accepted".
Jaydeep Kansara
Jaydeep Kansara 2019-11-30
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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