Calculate number of needed simulations First I have to apologize for any uncorrect naming or categorisation of my question, as I am an electrical engineer rather than a mathematican.
I have a simulation that outputs for a given number of input parameters one output value. As the simulation is pretty complicated, I don't know the statistical connection between input and output. I can run the simulation with different seeds to produce different outputs for the same input (it depends on some random numbers).
How do I choose the number of simulations I have to run with the same input, but different seeds to make sure I get significant results?
Perhaps make 10 simulations, calculate the deviation and with this deviation I can calculate the real number of needed simulations?
Edit1: Some more pieces of information: the simulation is a traffic simulation. I have input parameters (like 70% of all cars are minivans or 60% of all junctions are controlled by traffic lights). And I have output values (fuel consumption). Some internal decissions depend on random numbers (when a traffic light turns green, what route to take, ...).
I want to know how many different seeds for the PRNG I have to choose, to make the output at least a bit "sure".
Edit2: Some more general pieces of information: You know my setup, and the output (fuel consumption). Now I want to check, how the fuel consumption depends on the number of minivans in the city. So I change this number from 0% to 100%. Because there are some random parts in the simulation, you don't get a nice curve, but one with a few outliers. So I thought: Ok, run it 5 times with different seeds and take the mean of all 5. And voila, I get a smooth curve. 
This number 5 was just guessed by me. It looks good in the graph, but has no mathematical background. But perhaps some of you could help me with this. Thanks!
 A: I try to summarize all answers by you in order to have a single place for everything important.
Steps to calculate the needed number of simulations:


*

*Run the simulation with a default number of runs $R_0$. I've seen $R_0 = 1000$ most of the times. Now you should have a vector with the results $x_0$ where $length(x_0) = R_0$.

*Calculate the mean value $\overline{x}_0$ and the standard deviation $s_0$.

*Specify the allowed level of error $\epsilon$ and the uncertainty $\alpha$ you are willing to accept. Normally you choose $\epsilon = \alpha = 0.05\%$.

*Use this equation to get the required number of simulations:
$R \geq (\frac{Z_{1-(\alpha / 2)} \cdot s_0}{\epsilon \cdot \overline{x}_0})^2$, where $Z_{1-(\alpha / 2)}$ is the $1 − (\alpha /2)$ quantile of the standard normal distribution.

*[Use the student-t-distribution rather than the normal distribution for small $R_0$]
I hope this will help everybody who will look for an answer.
A: This question is more difficult to answer than you imagine.  It depends on the input, the output and degree of precision required on the output.  One thing to do is add say $100$ simulations to the current number and if the results seem not to have changed much there could be sufficient convergence.  If not keep going until you converge.  This assumes that as the input distributions become representative the output distribution will be well represented or if the output is a estimate it will have come close to converging to its expected value.  The simplest case is when the output is a single proportion that is a binomial proportion. Then the variance for the output has variance bounded by $\frac{1}{4n}$ where $n$ is the number of simulations.  Then you can take n large enough so that the variance of the estimator is as small as you would require.  This may seem like an unusual situation.  But it comes up a lot when comparing estimation technique.  For example I have done simulations to compare bootstrap confidence interval methods.  To see if the actual confidence level close to what it is suppose to be we simulate sampling from a particular population distribution and compute the proportion of times the interval includes the true parameter.  We might want the standard deviation of the estimate to be less than say $0.001$. This can be achieved since the standard deviation is the square root of the variance and is less than $\frac{1}{2 \sqrt{n}}$ So this will be achieved if we take $n> \frac{4}{(0.001)^2}$.
A: Powerbar, this answer is to address your last comment.  Perhaps you can think of your simulation as representing a model for fuel consumption as a function of % minivans on the road plus a random noise component.  So every time you run the simulation you get a slightly distorted picture of the curve because of the additive noise component. Each simulation is giving you at various % minivans simulated the value of the function + a random component. If the simulations are independent and the random components are independent (a very reasonable assumption) then averaging the values at each % minivan will improve the estimate of the "actual" fuel consumption value because the average reduces the variance of the noise by a factor of 1/n at each point.  In your case choosing 5 reduces it by a factor of 1/5.  The standard deviation is reduced by a factor of 1/sqrt(5) or 1/2.236.  So the variation from the curve is 2.236 times smaller for the averaged data.  Had you used 3 or 4 it would have helped also but not as much.  For n larger than 5 would help even more but may not be necessary since your eye saw satisfactory smoothness.
