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

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The state transition matrix is as below

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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.

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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 $\pi_d$

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  • $\begingroup$ 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? $\endgroup$ Aug 5 at 8:59
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    $\begingroup$ @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] $\endgroup$ 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? $\endgroup$ Aug 6 at 14:05

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