1
$\begingroup$

I have a markov chain from https://ieeexplore.ieee.org/document/9766097 as follows

enter image description here

The state transition matrix is as below

enter image description here

The analysis pertains to packet drop rate for ONE device with certain arrival($\lambda$) and transmission success probability($\mu_1$) where $\bar\lambda$ stands for no arrivals ($1-\lambda$) and $\bar\mu_1$ pertains to transmission failure probability ($1-\mu_1$).

The drop rate for ONE device is given by $\pi_d\bar\mu_1$ where $\pi_d$ is the steady state probability for the markov chain with delay bound $d$ (after which a packet will be dropped) in state $d$

I need to calculate the combined drop rate for N devices which are identical and symmetrical (i.e. have the same probability for successful transmission). What is the correct approach to extend this analysis for N devices? May be it is trivial, but I am missing the point.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

I think you mean expected drop rate (since actual drop rate is stochastic).

Let $D_i$ be the drop rate of device $i$. You are asking about the following quantity:

$$E\left[ \sum_{i=1}^N D_i \right]$$

Where we are assuming the $D_i$ are iid.

You pointed out that $E[D_i] = \pi_d\bar \mu_1$ therefore, by linearity of expectation, we should expect $N\pi_d\bar \mu_1$ combined drop rate.

The steady state distribution of combined drop rate would be the discrete convolution of $N$ copies of $\pi_d$

$\endgroup$
9
  • $\begingroup$ minor input, $\pi_d\bar \mu_1$ is the drop probability derived from Markov analysis. Is it ok to understand it as an expected value? $\endgroup$
    – Yukti Kaura
    Aug 5 at 8:59
  • 1
    $\begingroup$ @YuktiKaura -- I interpreted it as an average drop rate because $\pi_d$ is a steady-state probability (over many packets) for one device. Over a large number of packets we find the device spends $\pi_d$ percent of its time in the final state, and of the times its in the final state, it will fail again $\bar \mu_1$ percent of the time, so the packet will be dropped. So, if I have several devices, each getting the same packet rate and having same behavior, then each on, on average will also drop at rate $\pi_d \bar \mu_1$ $\endgroup$
    – Annika
    Aug 5 at 19:30
  • $\begingroup$ Thank you... Got it, but when we multiply the expected value with $N$ the drop rate may no longer stay within the range [0,1] $\endgroup$
    – Yukti Kaura
    Aug 6 at 13:11
  • $\begingroup$ @YuktiKaura since it's a rate I wasn't expecting that to be an issue -- if you have 1000 devices they will drop $1000\pi_d\bar \mu_1$ packets per unit time. Or...are you asking for the probability that ANY of the devices will drop something (like in reliability engineering) in which case it would be something like $1-(\pi_d\mu_1)^N$ $\endgroup$
    – Annika
    Aug 6 at 13:33
  • $\begingroup$ Precisely, the formulation $N\pi_d\mu_1$ would give us the expected number of packets dropped across total $N$ devices. Drop rate would then be the ratio of number of packets dropped (from the formulation above) to the number of packets transmitted ain't it? $\endgroup$
    – Yukti Kaura
    Aug 6 at 14:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .