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edited May 8, 2014 at 15:04

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My question is about the required number of simulations for Monte Carlo analysis method. As far as I see the required number of simulations for any allowed percentage error $E$ (e.g., 5) is $$ n = \left\{\frac{100 \cdot z_c \cdot \text{std}(x)}{E \cdot \text{mean}(x)} \right\}^2 , $$

where $\text{std}(x)$ is the standard deviation of the resulting sampling, and $z_c$ is the confidence level coefficient (e.g., for 95% it is 1.96). So in this way it is possible to check that the resulting mean and standard deviation of $n$ simulations represent actual mean and standard deviation with 95% confidence level.

In my case I run the simualtion 7500 times, and compute moving means and standard deviations for each set of 100 sampling out of the 7500 simulations. The required number of simulation I obtain is always less than 100, but % error of mean and std compare to mean and std of entire results is not always less than 5%. In most cases the % error of mean is less than 5% but the error of std goes up to 30%.

What is the best way to determine number of required simulation without know actual mean and std (in my case subjected outcome of simulation is normally distributed)?

Thanks in advance for any help.

In order to have an idea about what may distribution of simulation results look like when iteration is run infinite number of times: Instead of using resulted mean and variance after n number of simulations, I've decided to find a fit function of resulted distribution, but here n has to fullfill allowed % error. I think on that way I can find more correct results on cumulative distrubution function that is related with e.g. 97.5%. Because when I compare the results of 400 and 7000 simulation, fit functions of distribution for both sampling looks like each other only curve of 2nd one is smoother. Also, hence model in MATLAB/Simulink is nonlinear, although generated input parameters are normal distributed, resulted histogram of simulations are not normal for that reason I used "generalized extreme value distribution", which is named as 'gev' in MATLAB. But still, I am quite not sure about this methodolgy, thanks for any command in advance

My question is about the required number of simulations for Monte Carlo analysis method. As far as I see the required number of simulations for any allowed percentage error $E$ (e.g., 5) is $$ n = \left\{\frac{100 \cdot z_c \cdot \text{std}(x)}{E \cdot \text{mean}(x)} \right\}^2 , $$

where $\text{std}(x)$ is the standard deviation of the resulting sampling, and $z_c$ is the confidence level coefficient (e.g., for 95% it is 1.96). So in this way it is possible to check that the resulting mean and standard deviation of $n$ simulations represent actual mean and standard deviation with 95% confidence level.

In my case I run the simualtion 7500 times, and compute moving means and standard deviations for each set of 100 sampling out of the 7500 simulations. The required number of simulation I obtain is always less than 100, but % error of mean and std compare to mean and std of entire results is not always less than 5%. In most cases the % error of mean is less than 5% but the error of std goes up to 30%.

What is the best way to determine number of required simulation without know actual mean and std (in my case subjected outcome of simulation is normally distributed)?

Thanks in advance for any help.

My question is about the required number of simulations for Monte Carlo analysis method. As far as I see the required number of simulations for any allowed percentage error $E$ (e.g., 5) is $$ n = \left\{\frac{100 \cdot z_c \cdot \text{std}(x)}{E \cdot \text{mean}(x)} \right\}^2 , $$

where $\text{std}(x)$ is the standard deviation of the resulting sampling, and $z_c$ is the confidence level coefficient (e.g., for 95% it is 1.96). So in this way it is possible to check that the resulting mean and standard deviation of $n$ simulations represent actual mean and standard deviation with 95% confidence level.

In my case I run the simualtion 7500 times, and compute moving means and standard deviations for each set of 100 sampling out of the 7500 simulations. The required number of simulation I obtain is always less than 100, but % error of mean and std compare to mean and std of entire results is not always less than 5%. In most cases the % error of mean is less than 5% but the error of std goes up to 30%.

What is the best way to determine number of required simulation without know actual mean and std (in my case subjected outcome of simulation is normally distributed)?

Thanks in advance for any help.

In order to have an idea about what may distribution of simulation results look like when iteration is run infinite number of times: Instead of using resulted mean and variance after n number of simulations, I've decided to find a fit function of resulted distribution, but here n has to fullfill allowed % error. I think on that way I can find more correct results on cumulative distrubution function that is related with e.g. 97.5%. Because when I compare the results of 400 and 7000 simulation, fit functions of distribution for both sampling looks like each other only curve of 2nd one is smoother. Also, hence model in MATLAB/Simulink is nonlinear, although generated input parameters are normal distributed, resulted histogram of simulations are not normal for that reason I used "generalized extreme value distribution", which is named as 'gev' in MATLAB. But still, I am quite not sure about this methodolgy, thanks for any command in advance

Improved formatting and wording

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edited May 2, 2014 at 16:29

QuantIbex

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required Required number of simulationsimulations for monte carloMonte Carlo analysis

My question is about the required number of simulations for Monte-Carlo Carlo analysis method,. asAs far as I see the required number of simulations for any allowed percentage error $E$ e(e.g. 5, required number of simulation5) is given by:

$$n=(100 \cdot z_c \cdot \text{std}(x)/(E \cdot \text{mean}(x)))^2$$ $$ n = \left\{\frac{100 \cdot z_c \cdot \text{std}(x)}{E \cdot \text{mean}(x)} \right\}^2 , $$

where $\text{std}(x)$ is the standard deviation of resultedthe resulting sampling, and $z_c$ is the confidence level coefficient e(e.g., for 95% it is 1.96, so). So in this way it is possible to check resultedthat the resulting mean and stdstandard deviation of n$n$ simulations represent actual mean and stdstandard deviation with 95% confidence level.

In my case I run the simualtion 7500 times, and create different meancompute moving means and stdstandard deviations for each set of 100 sampling fromout of the 7500 simulation results (a moving average and std), evensimulations. The required number of simulation I obtain is always less than 100, but % error of mean and std compare to mean and std of entire results is not always less than 5%. In most cases the % error of mean is less than 5% in most case but the error of std goes up to 30%.

What is the best way to determine number of required simulation without know actual mean and std? also in (in my case subjected outcome of simulation is normalnormally distributed, thanks)?

Thanks in advance for any help.

required number of simulation for monte carlo analysis

My question is about the required number of simulations for Monte-Carlo analysis method, as far as I see for any allowed percentage error $E$ e.g. 5, required number of simulation is given by:

$$n=(100 \cdot z_c \cdot \text{std}(x)/(E \cdot \text{mean}(x)))^2$$

where $\text{std}(x)$ is standard deviation of resulted sampling, $z_c$ is confidence level coefficient e.g. for 95% it is 1.96, so in this way it is possible to check resulted mean and std of n simulations represent actual mean and std with 95% confidence level. In my case I run the simualtion 7500 times and create different mean and std of 100 sampling from 7500 simulation results (a moving average and std), even required number of simulation is always less than 100, but % error of mean and std compare to mean and std of entire results is not always less than 5%. % error of mean is less than 5% in most case but error of std goes up to 30%. What is the best way to determine number of required simulation without know actual mean and std? also in my case subjected outcome of simulation is normal distributed, thanks in advance for any help

Required number of simulations for Monte Carlo analysis

My question is about the required number of simulations for Monte Carlo analysis method. As far as I see the required number of simulations for any allowed percentage error $E$ (e.g., 5) is $$ n = \left\{\frac{100 \cdot z_c \cdot \text{std}(x)}{E \cdot \text{mean}(x)} \right\}^2 , $$

where $\text{std}(x)$ is the standard deviation of the resulting sampling, and $z_c$ is the confidence level coefficient (e.g., for 95% it is 1.96). So in this way it is possible to check that the resulting mean and standard deviation of $n$ simulations represent actual mean and standard deviation with 95% confidence level.

In my case I run the simualtion 7500 times, and compute moving means and standard deviations for each set of 100 sampling out of the 7500 simulations. The required number of simulation I obtain is always less than 100, but % error of mean and std compare to mean and std of entire results is not always less than 5%. In most cases the % error of mean is less than 5% but the error of std goes up to 30%.

What is the best way to determine number of required simulation without know actual mean and std (in my case subjected outcome of simulation is normally distributed)?

Thanks in advance for any help.