G/M/1 queue

Let ${\displaystyle (X_{t},t\geq 0)}$ be a ${\displaystyle G/M(\mu )/1}$ queue with arrival times ${\displaystyle (A_{n},n\in \mathbb {N} )}$ that have interarrival distribution A. Define the size of the queue immediately before the nth arrival by the process ${\displaystyle U_{n}=X_{A_{n}-}}$. This is a discrete-time Markov chain with stochastic matrix:

${\displaystyle P={\begin{pmatrix}1-a_{0}&a_{0}&0&0&0&\cdots \\1-(a_{0}+a_{1})&a_{1}&a_{0}&0&0&\cdots \\1-(a_{0}+a_{1}+a_{2})&a_{2}&a_{1}&a_{0}&0&\cdots \\1-(a_{0}+a_{1}+a_{2}+a_{3})&a_{3}&a_{2}&a_{1}&a_{0}&\cdots \\\vdots &\vdots &\vdots &\vdots &\vdots &\ddots \end{pmatrix}}}$

where ${\displaystyle a_{v}=\mathbb {E} \left({\frac {(\mu X)^{v}e^{-\mu A}}{v!}}\right)}$.[3]^{: 427–428}

The Markov chain ${\displaystyle U_{n}}$ has a stationary distribution if and only if the traffic intensity ${\displaystyle \rho =(\mu \mathbb {E} (A))^{-1}}$ is less than 1, in which case the unique such distribution is the geometric distribution with probability ${\displaystyle \eta }$ of failure, where ${\displaystyle \eta }$ is the smallest root of the equation ${\displaystyle \mathbb {E} (\exp(\mu (\eta -1)A))}$.[3]^: 428

In this case, under the assumption that the queue is first-in first-out (FIFO), a customer's waiting time W is distributed by:[3]^: 430

${\displaystyle \mathbb {P} (W\leq x)=1-\eta \exp(-\mu (1-\eta )x)~{\text{ for }}x\geq 0}$

The busy period can be computed by using a duality between the G/M/1 model and M/G/1 queue generated by the Christmas tree transformation.[4]

The response time is the amount of time a job spends in the system from the instant of arrival to the time they leave the system. A consistent and asymptotically normal estimator for the mean response time, can be computed as the fixed point of an empirical Laplace transform.[5]