# Queuing with Exponential impatience times

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

• 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? – Michael Feb 11 '17 at 20:22
• 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 $\delta\rightarrow 0$. – Michael Feb 11 '17 at 20:30
• 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 $$\lim_{t\rightarrow\infty} \frac{\mbox{Num drops during [0,t]}}{\mbox{Num arrivals during [0,t]}} = \frac{\overbrace{\lim_{t\rightarrow\infty}\frac{1}{t}\mbox{Num drops during [0,t]}}^{\mbox{drop rate}}}{\underbrace{\lim_{t\rightarrow\infty}\frac{1}{t}\mbox{Num arrivals during [0,t]}}_{\lambda}}$$ – Michael Feb 12 '17 at 17:11
• 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 $\frac{6}{33} \approx 0.1818$? – Michael Feb 12 '17 at 17:12
• 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. – Michael Feb 13 '17 at 6:49