Consider an M/M/1/loss (i.e., if an arrival sees a busy server, the arrival is lost) queue with the arrival intensity λ and the service intensity µ. At time 0 this queueing system is empty. Find the probability that the queue is empty at time t.
What I have attempted so far:
Let P0(t) and P1(t) be the be the probability that there is 0 and 1 customer in the system.
I know that if at t=0 there is a customer in the system, then
P0(t) = µ/(λ+ µ)(1 − e^−((λ+µ)t))
P1(t) = 1/(λ+µ)(µe^−((λ+µ)t)+λ)
and the steady state probabilities are:
P0=lim t→∞ P0(t) = µ /(λ+µ) P1=lim t→∞ P1(t) = λ/(λ+µ)
How can I approach this if initially there are NO customers in the system (as the question states)?
This is a $M/M/1/1$ queue (finite buffer of size $2$). Let $X(t)$ be the number of customers in the system at time $t$, then $\{X(t):t\geqslant 0\}$ is a continuous-time Markov chain on $\{0,1\}$ with generator $$ Q = \pmatrix{-\lambda&\lambda\\\mu&-\mu}. $$ From $\det Q= 0$ and $\mathsf{Tr}(Q)=-(\lambda+\mu)$ we see that $Q$ has eigenvalues $0$) and $-(\lambda+\mu)$ (since the product of the eigenvalues is equal to the determinant and the sum equal to the trace). From $$ Q\cdot\pmatrix{1\\1} = \pmatrix{0\\0} = 0\cdot\pmatrix{1\\1} $$ and $$ Q\cdot\pmatrix{\lambda\\-\mu} = \pmatrix{-(\lambda+\mu)\lambda\\-(\lambda+\mu)(-\mu)} $$ we see that $(1,1)$ and $(\lambda,-\mu)$ are eigenvectors of $Q$. It follows that $Q = ADA^{-1}$ where $$ A = \pmatrix{1&\lambda\\1&-\mu},\quad D=\pmatrix{0&0\\0&-(\lambda+\mu)}., $$ and so for each nonnegative integer $n$, $Q^n = AD^nA^{-1}$. For $t\geqslant0$ we compute \begin{align} \exp(tQ) &= \sum_{n=0}^\infty \frac{(tQ)^n}{n!}\\ &= \left( \begin{array}{cc} \frac{\lambda }{\lambda +\mu }e^{ -(\lambda +\mu )t}+\frac{\mu }{\lambda +\mu } & \frac{\lambda }{\lambda +\mu }-\frac{ \lambda }{\lambda +\mu }e^{ -(\lambda +\mu )t} \\ \frac{\mu }{\lambda +\mu }-\frac{ \mu }{\lambda +\mu }e^{ -(\lambda +\mu )t} & \frac{\lambda }{\lambda +\mu }+\frac{ \mu }{\lambda +\mu }e^{ -(\lambda +\mu )t} \\ \end{array} \right). \end{align} Let $P_{ij}(t) = \mathbb P(X(t) = j\mid X(0)=i)$. Then $P(t)=e^{tQ}$ satisfies Kolmogorov's forward equation $P'(t) = P(t)Q$, so $e^{tQ}$ gives the transient behavior of the process. The probability that the queue is empty at time $t$ is thus $$ [e^{tQ}]_{00}\mathbb P(X(0)=0) + [e^{tQ}]_{10}\mathbb P(X(0)=1), $$ and in particular if we assume $P(X(0)=0)=1$ then we have $$ \mathbb P(X(t)=0) = \frac{\lambda }{\lambda +\mu }e^{ -(\lambda +\mu )t}+\frac{\mu }{\lambda +\mu }. $$