I have a bit a naive question about Poisson process.

Suppose I have a station that receives a number of packets $n(i)$ each time slot $i$. If the number of received messages is less than $M$ then the $n(i)$ packages will all be sent.

If $n(i) > M$ then $n(i)-M$ packets will be lost and $M$ packets will be sent. Let assume packets arrive at a Poisson rate $\lambda$. What is the mean of loss rate (total lost packages / total number of received packets)?

I formulated the problem in this way:

  • Let $N(t)$ be the total number of packets at time $t$, $N(t) = \sum_{i=0}^{t} n(i)$
  • The total number of lost packets is $e(t) = \sum_{i=0}^{t} (n(i)-M) I_{(n(i) >M)}$

The loss rate will then be $\eta(t) = \frac{e(t)}{N(t)}$.

The mean loss rate will be:

$$\lim_{t \to \infty} E[\eta(t)]= E\left[\frac{\sum_{i=0}^{t} (n(i)-M) I_{(n(i) >M)}}{\sum_{i=0}^{t} n(i)}\right]$$

But I don't know how to calculate it.

Any leads would be appreciated.


1 Answer 1


The expected number of lost packets can be re-expressed as the expected number of arriving packets minus the expected number of sent packets:

$\lambda - \left(\sum_{n=0}^M np(n;\lambda) + M(1-P(M;\lambda))\right)$

where $p(n;\lambda)$ is the probability of seeing $n$ packets in a time slot given an arrival rate $\lambda$ and $P(M;\lambda)$ is the probability of seeing $M$ or fewer packets ... The first term captures the events where $n \leq M$, when we send all $n$ packets, and the second where $n > M$, when we send $M$ packets.

Writing out the Poisson distribution function in the second term and dropping the term where $n=0$ (as it equals zero, so contributes nothing to the sum) gives:

$\sum_{n=0}^M np(n;\lambda) = e^{-\lambda}\sum_{n=1}^M n\left(\frac{\lambda^n}{n!}\right)$

which can be rearranged as:

$\lambda e^{-\lambda}\sum_{n=0}^{M-1}\frac{\lambda^n}{n!}$

which in turn equals $\lambda P(M-1;\lambda)$. Substituting this into the first expression results in:

$\mathbb{E}[\text{number of lost packets}] = \lambda[1-P(M-1;\lambda)] - M[1-P(M;\lambda)]$

Noting that $P(M;\lambda) = P(M-1;\lambda) + p(M;\lambda)$ allows us some more, albeit slight, simplification:

$\mathbb{E}[\text{number of lost packets}] = (\lambda - M)[1-P(M-1;\lambda)] + Mp(M;\lambda)$

Checking our algebra by comparing a simulation with our result gives:

> M <- 3
> lambda <- 2
> x <- rpois(1000000,lambda)
> xM <- pmin(x,M)
> cat(" Simulation result:   ",lambda-mean(xM), "\n",
+     "Mathematical result: ",(lambda-M)*(1-ppois(M-1,lambda)) + M*dpois(M,lambda))
 Simulation result:    0.218311 
 Mathematical result:  0.2180175
  • $\begingroup$ Hi jbowman, I just want to say waw, thanks a million. I like the way you reformulated the problem, it is more handy. I suspected from the beginning that my problem basically was in the formulation. $\endgroup$
    – sirus
    Dec 22, 2013 at 14:56

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