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I am considering simple M/M/1 queue with customer impatience. Customer waits and leaves the system if the delay before service is more than an exponential wait time. Arrival rate is $\lambda$, service rate is $\mu$, and abandonment rate due to delay is $\lambda_W$. I was able to find distribution of state probabilities as;

$p_0 \lambda= p_1 \mu $

$p_1 (\mu +\lambda)= p_0 \lambda+ p_2 (\mu +\lambda_W)$

$p_2 (\mu +\lambda+\lambda_W)= p_1 \lambda+ p_3 (\mu +2\lambda_W)$

$p_3 (\mu +\lambda+2\lambda_W)= p_2 \lambda+ p_4 (\mu +3\lambda_W)$ ...

From the equilibrium conditions, I found

$p_1 = p_0 \frac{\lambda}{\mu }$

$p_2 = p_1 \frac{\lambda}{\mu +\lambda_W} =p_0 \frac{\lambda^2}{\mu (\mu + \lambda_W)}$. ..

Considering $\sum_{n=0}^{\infty}p_n = 1$, $p_0 = \Big(1+ \sum_{n=1}^{\infty} \lambda^{n}\prod_{k=1}^{n}\frac{1}{\mu +(k-1)\lambda_{W}} \Big)^{-1}$ I want to find

  • the fraction of customers which leave the system due to delay?

  • Distribution of waiting time given that customer receives service?

Any idea is appreciated thank you

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    Each customer in the system leaves due to delay with rate $\lambda_W$. When the system state is $n$, there are $n$ such customers. So you can write an expression for the total rate of leaving due to delay. This is related to computations of the rate of dropping due to buffer overflow in an M/M/1/m system (i.e., with finite buffer size $m$) which is just $\lambda p_m$. – 2017-02-11
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    Thank you. How do you relate the rate of dropping with the probability of customers leaving the system? – 2017-02-11
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    That was an analogy. I gave an example answer for a different system because I didn't want to give the answer away completely for your particular problem. For the M/M/1/m system, The total rate that something happens is the sum (over all states) of the instantaneous rate it happens in that state multiplied by the probability being in that state. Since dropping only occurs in one state, and happens whenever we get an arrival in that state, we get $$\mbox{drop rate} = p_m [\mbox{rate of dropping while in state $m$}] = p_m\lambda$$ Can you use this method for your problem? – 2017-02-11
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    One way to see it (not necessarily the best way) is to consider the Markov chain $X_t$ to be in "steady state" at time $t$ and for a small value $\delta>0$ use the law of total expectation to write: $$ \frac{1}{\delta}E[\mbox{drops during $[t,t+\delta]$}] = \frac{1}{\delta}\sum_{s \in \mathcal{S}}E[\mbox{drops during $[t,t+\delta]$}|X_t=s]p_s$$ and you can compute an approx for $E[\mbox{drops during $[t,t+\delta]$}|X_t=s]$ that is proportional to $\delta$ and that is accurate as $\delta\rightarrow 0$. – 2017-02-11
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    Thanks a lot. Considering your approach, we can derive the drop ratio, $P{\rm drop}= \sum_{n=2}^{\infty} p_{n} \frac{(n-1)\lambda_{W}}{\mu + (n-1)\lambda_{W}} $. This is fraction of customers that leave the system before service is finished. I also want to derive the distribution of waiting time given that arrival receives service? Do you have any idea on that? – 2017-02-12
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    Two comments: (i) Your equation $p_0\lambda = p_1(\mu+\lambda_W)$ suggests that even a customer currently in service can abandon due to delay, (ii) The fraction of drops is $$\lim_{t\rightarrow\infty} \frac{\mbox{Num drops during $[0,t]$}}{\mbox{Num arrivals during $[0,t]$}} = \frac{\overbrace{\lim_{t\rightarrow\infty}\frac{1}{t}\mbox{Num drops during $[0,t]$}}^{\mbox{drop rate}}}{\underbrace{\lim_{t\rightarrow\infty}\frac{1}{t}\mbox{Num arrivals during $[0,t]$}}_{\lambda}} $$ – 2017-02-12
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    Example: Bob plays baseball for five years. In his first year he bats 3 times and is successful twice. In years 2-5 he gets up a total of 30 times and is successful 4 times. What is his overall batting average? Is it $(1/5)\frac{2}{3} + (4/5)\frac{4}{30} = 0.24$? Or is it $\frac{6}{33} \approx 0.1818$? – 2017-02-12
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    I edited and updated the flow equations. In the equations, customer leaving the system even in service was wrong. Again, from the first comment, $P{\rm drop}= \frac{ \sum_{n=2}^{\infty} p_{n} \lambda_{W} }{\lambda} $. To answer your question about baseball play, I guess it is about whether the system is ergodic or not? If the system is ergodic I guess time average of batting rate must be equal to its ensemble average ? Based on your approach I compute total number of arrivals during [0,t] period. So, the answer is 0.1818? – 2017-02-13
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    The “ergodic” concept is not relevant to the baseball question, which considers an empirical average over 33 tries. That case shows the true ratio is not a weighted sum of ratios. – 2017-02-13
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    @Michael I want to derive distribution of waiting time given that customer receives service. Assuming little`s law applies average queue length is ${L}{\rm q} = \sum_{n=1}^{\infty} (n-1)p_n$, and average delay $ W_{q} = \frac{L_{q}}{\lambda}$. Do you have any idea for the distribution of delay? – 2017-02-15

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