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rstatisticsexponential-distribution

Two exponential distributions

I am trying to simulate two exponential distributions. For example two CPUs processing jobs e.g. one having average service time 10 min (lambda = 0.1) and another one 20 min (lambda = 0.05) and they work independently. Both of them are busy when a new job arrives.

I would like to simulate the waiting time of a new job

Here is what I did so far.

cpu1 = rexp(n = 10000, rate = .1)
cpu2 = rexp(n = 10000, rate = .05)

I generate 10K data points based on exponential distribution. For each of them new job has to wait min(cpu1[i], cpu2[i]) I store all of them in a data frame and compute the mean.

for (i in seq(1, 10000)) {
    if (i == 1) {
       df1 <- data.frame(waiting_time=min(cpu1[i], cpu2[i])) 
    } else {
       df1 <- rbind(df1, data.frame(waiting_time=data.frame(waiting_time=min(cpu1[i], cpu2[i])))
    }
}
mean(df1$waiting_time)

Is this the right way to do the simulation? or am I doing something wrong?

Solution

  • As has been pointed out, mean(pmin(cpu1,cpu2)) is equivalent to the for loop and mean(df1$waiting_time), but much, much faster.

    Or you could skip the simulation altogether since the minimum of two independent random exponential variables is also exponentially distributed with a rate equal to the sum of the two rates. Furthermore, the sum of n iid exponential random variables is gamma-distributed with the same rate parameter and a shape parameter equal to n.

    So we can simply do rgamma(1, 1e4, 0.15)/1e4 or, equivalently, rgamma(1, 1e4, 0.15*1e4) instead of mean(pmin(cpu1,cpu2)), and the results will have identical distributions.