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Consider a single server queuing system, where customers arrive according to a Poisson proccess with rate $\lambda$, service times are exponential with rate $\alpha$. When an individual comes into the system, there is a person on the waiting line and one other being attended. The individual will wait a maximum time of $\delta$, if he is not being attended in that time, he will go away from the system.

What is the probability that the individual is going to be attended?

I'm trying to do is, I condition that the first arrival is lower than $\omega$, son I use the acumulative function, and then I multiply that for the $e^(\delta-\omega)$. So the limit goes to $1$.

This is what I did

http://www.wolframalpha.com/input/?i=int+(a*e%5E-(a(s-y))*(1+-+e%5E(-a+y)))dy+from+y%3D0+to+s)

The thing is that I just had a test with this questions and the answers were:

$$a.\ (\alpha\cdot\delta)\cdot e^{-(\alpha\cdot\delta)}$$ $$b.\ (\alpha\cdot\delta)^2\cdot e^{-(\alpha\cdot\delta)} $$ $$c.\ (\alpha^2\cdot\delta)\cdot e^{-(\alpha\cdot\delta)}$$ $$d.\ (\alpha\cdot\delta)^2\cdot e^{-(\alpha\cdot\delta)}$$

Am I wrong or the tests was wrong? Because here when the rate and time are big, in the limit is goes to $0$

JPC
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1 Answers1

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The probability that a customer arriving to the system with a customer in service is attended is simply $$\frac\mu{\mu+\delta}. $$ As for the limiting probability that a customer arriving to the system is attended:

This is a $M/M/1/2$ queue with reneging. Let $\{X(t):t\geqslant 0\}$ be the number of customers in the system at time $t$, then $X(t)$ is a continuous-time Markov chain on state space $\{0,1,2\}$ with transition rates $$ q_{ij} = \begin{cases} \lambda,& j=i+1\\ \mu,& (i,j) = (1,0)\\ \mu+\delta,& (i,j) = (2,1)\\ 0,& \text{otherwise}. \end{cases} $$ The detailed balance equations are given by \begin{align} \lambda\pi_0 &= \mu\pi_1\\ \lambda\pi_1 &= (\mu+\delta)\pi_2, \end{align} and yield $$ \pi_1 = \frac\lambda\mu \pi_0,\quad \pi_2 = \frac\lambda\mu\left(\frac\lambda{\mu+\delta}\right)\pi_0. $$ From $\sum_{i=0}^2\pi_i=1$ we find $$ \pi_0 = \left(1+\frac\lambda\mu\left(1+\frac\lambda{\mu+\delta}\right)\right)^{-1}, $$ so $$ \pi_1 = \lambda \left(\mu +\lambda\left(1+\frac\lambda{\mu+\delta}\right)\right)^{-1} $$ and $$ \pi_2 = \lambda^2 \left((\lambda+\mu)(\mu+\delta)+\lambda^2\right)^{-1}. $$ It follows that the limiting probability that a customer arriving to the system is attended is given by $$ \frac{\pi_0}{\pi_0+\pi_1} + \frac{\pi_1}{\pi_0+\pi_1}\left(\frac\mu{\mu+\delta}\right) = \frac{\mu (\lambda +\mu+\delta )}{(\lambda+\mu)(\mu+\delta)}. $$

Math1000
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