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Consider a single-server queue where arrivals are Poisson with rate $10$ per hour. An arriving customer, upon finding $n$ in the system, departs the system with probability $p_n = \frac{n}{n+1}$ (so, as the system becomes more congested, arriving customers are more likely to go elsewhere without ever entering the line.) Suppose that the time to service each customer is exponential with rate $10$ per hour.

(a) Find the birth and death rates

(b) Find the expected time for the system to reach $3$ customers starting from the empty system.

My attempt: (a) The birth rate is $10q_n = \frac{10}{n+1}$, and the death rate is $\frac{10n}{n+1}$.

(b) My thought is: I tried forming the recusion formula by letting $E_i =$ expected time to reach $i$ customers. Now, $E_{i} = \frac{10}{10*2+10}E_{i-1} + \frac{20}{20+10}E_{i+1}$ with $E(0) = 0$ and $E(1) = 6$ minutes. Thus $E(2) = 9$ and $E(3) = (9 - 2)\frac{3}{2} = \fbox{$10.5$}$ minutes. Is this correct?

My question: I am not sure at all about my solution above. Could anyone give me some thoughts in case my way to solve this one was completely wrong?

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The death rate here, i.e. the rate to go from $n$ to $n-1$ when $n \geq 1$, is just $10$ uniformly. It has nothing to do with the customers that leave the system before properly entering it; they never change the state.

The birth rate here is analogous to what happens in the Metropolis dynamics: you "try" to go from $n$ to $n+1$ with rate $10$ but you only succeed with probability $\frac{1}{n+1}$, which is effectively the same as actually going from $n$ to $n+1$ with rate $\frac{10}{n+1}$ (which is what you said). This is a nontrivial but ultimately elementary fact: the sum of $Geo(p)$-many independent exponential random variables with common rate $\lambda$, where the $Geo(p)$ variable is independent of the exponential variables, is an exponential random variable with rate $p \lambda$.

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  • $\begingroup$ Many thanks for your help Ian. How do you think of my solution to part (b)? Of course I need to use the new birth rate to plug in, but is the method correct? $\endgroup$ – user177196 Oct 25 '16 at 17:28
  • $\begingroup$ You should instead take $E_i $ to be the team to get to 3 customers starting from $i $ customers and then derive a recurrence from there. (Also note that the answer should clearly be at least 10+20+30 because that would be the mean time if no service were offered.) $\endgroup$ – Ian Oct 25 '16 at 18:02
  • $\begingroup$ Thank you so much. Is the recurrent $E_{i} = \frac{10}{10+\frac{10}{i+2}}E_{i+1} + \frac{\frac{10}{i}}{\frac{10}{i} + 10}E_{i-1}}$? The reason is because to get from $E_{i+1}$ to $E_i$, the person who arrives must be BEFORE one person who "dies" from the system, and in the system currently having $n$ people, the arrival time ($T_1$) and death time ($T_2$) are exponential random variable with rate $\frac{10}{n+1}$ and $10$, respectively. It's a bit difficult to compute $P(T_1<T_2)$ as the birth rate is not constant here, but changing based on the number of the people currently in the system. $\endgroup$ – user177196 Oct 25 '16 at 18:24
  • $\begingroup$ In general the probability to go from $i$ to $j$ in one step of a CTMC is $\frac{q_{ij}}{\sum_{k \neq i} q_{ik}}$. (This defines the embedded DTMC in a CTMC, also called the "jump chain".) So $E_i$ is given by the sum over $j$ of $E_j+1$ (since it takes a step to to go to $j$) times the probability to go to $j$ from $i$. $\endgroup$ – Ian Oct 25 '16 at 19:28
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    $\begingroup$ I have $E_0=\frac{1+10E_1}{10}$,$E_1=\frac{1+5E_2+10E_0}{15},E_2=\frac{1+\frac{10}{3}E_3+10E_1}{\frac{10}{3}+10},E_3=0$, which gives $2$ indeed. $\endgroup$ – Ian Oct 26 '16 at 0:45

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