I simulated a system using discrete-events simulation, more particularly next-event simulation. The system is M/M/1 (infinite FIFO waiting room). Whenever the system is jobless, the server shutdowns. When a job arrives, the server restarts if it was off, but it cannot serve until the restart is finished. Restart time is defined by an exponential distribution.

  • $\lambda$ is the arrival rate (expected inter-arrival is $1/\lambda$)
  • $\mu$ is the service rate (expected service time is $1/\mu$)
  • $\theta$ is the restart rate (expected restart time is $1/\theta$)

The simulation stops at time $\tau$. Jobs that entered before that time are guaranteed to be served. $\mu > \lambda$ is guaranteed.

I want to observe :

  • $\mathbb{E}[B]$, the expected service time
  • $\mathbb{E}[S]$, the expected sojourn time (departure - arrival)
  • $P_{SETUP}$, the probability that the server is restarting when a client arrives
  • $P_{OFF}$, the probability that the server is off when a client arrives

I know the theoretical values: $$\mathbb{E}[S] = \frac{1 / \mu}{1 - \rho} + \frac{1}{\theta}$$ $$P_{OFF} = (1 - \rho) \frac{1/\lambda}{1/\lambda + 1/\theta}$$ $$P_{SETUP} = (1 - \rho) \frac{1/\theta}{1/\lambda + 1/\theta}$$

I've programmed this system. Now, I'm wondering how many simulations is enough so that I can conclude the measures obtained from my simulations are representative of the real system.

I'm wondering if I could draw $n$ samples, then $n_2$ samples, with $n_2 = 2 * n$ and test the null hypothesis $H_0 : v = v_2, H_1 : v \neq v_2$. With $v$ any of the above measures from the first batch of $n$ simulation and $v_2$ the same measure from the second batch of $n_2$ simulations. But I fear that it will only prove that doubling the number of simulations is not significant.

I've also seen that we can do the test with only one sample if we have the population measure (and I do) but this gets me nonsensical results. By example if I do

scipy.stats.ttest_1samp(np.random.exponential(1, 1000000), 1)

I get a statistic and pvalue very different from one run to the other and very rarely less than 0.05 which I think is my purpose.

  • 1
    $\begingroup$ These are exponential variables, right? Then, you can compute the P-value explicitly as a function of n for a statistic to estimate $\mu $ (the sample mean). You can then pick n sufficiently large to ensure your worst case null hypothesis is rejected. $\endgroup$
    – Three Diag
    Jan 9 at 22:11
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    $\begingroup$ OMFG, I found my mistake. The procedure @ThreeDiag suggests is correct and that's what I've been trying to do with scipy.stats.ttest_1samp, but I thought it failed. Except it worked and my measures were wrong. I tested the validity of my measures and didn't even consider they were not valid... I had to discard the first measures because my system wasn't stationary. $\endgroup$
    – Adrien H
    Jan 10 at 13:56

Ok, I found on how to achieve it and in fact, using

scipy.stats.ttest_1samp(samples, samples_expected_mean)

was indeed the way to go. That's what @ThreeDiag was suggesting. This is a bilateral statistical test using student law.

This gives the pvalue and I only have to assert the pvalue of the test and accept the null hypothesis if $pvalue > \alpha$ or reject it otherwise.

So why didn't I see it sooner as I've already tried that ?

It worked fine and my data was wrong... The problem is that the process I'm measuring wasn't stationary. I had to discard the first half of my raw measures (before aggregating into means) to have something meaningful.

  • $\begingroup$ But why are you using a t-test? Sample mean is only approximately normal, but you might be able to do better than that given you variables are exponential and their sum follows a gamma distribution. $\endgroup$
    – Three Diag
    Jan 10 at 17:33
  • $\begingroup$ Mmh, I may not understand what I'm doing. What better test could I have done ? $\endgroup$
    – Adrien H
    Jan 10 at 19:17
  • $\begingroup$ Sum of exponential is gamma distributed, so you can compute the probability of the realization of the sample mean exactly (i. E. The pvalue) $\endgroup$
    – Three Diag
    Jan 11 at 13:02

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