M/M/1 queue

In

• Arrivals occur at rate λ according to a
  Poisson process
  and move the process from state i to i + 1.
• Service times have an exponential distribution with rate parameter μ in the M/M/1 queue, where 1/μ is the mean service time.
• All arrival times and services times are (usually) assumed to be independent of one another.^[2]
• A single server serves customers one at a time from the front of the queue, according to a
  first-come, first-served
  discipline. When the service is complete the customer leaves the queue and the number of customers in the system reduces by one.
• The buffer is of infinite size, so there is no limit on the number of customers it can contain.

The model can be described as a

${\displaystyle Q={\begin{pmatrix}-\lambda &\lambda \\\mu &-(\mu +\lambda )&\lambda \\&\mu &-(\mu +\lambda )&\lambda \\&&\mu &-(\mu +\lambda )&\lambda &\\&&&&\ddots \end{pmatrix}}}$

on the state space {0,1,2,3,...}. This is the same continuous time Markov chain as in a

Stationary analysis

The model is considered stable only if λ < μ. If, on average, arrivals happen faster than service completions the queue will grow indefinitely long and the system will not have a stationary distribution. The stationary distribution is the limiting distribution for large values of t.

Various performance measures can be computed explicitly for the M/M/1 queue. We write ρ = λ/μ for the utilization of the buffer and require ρ < 1 for the queue to be stable. ρ represents the average proportion of time which the server is occupied.

The probability that the stationary process is in state i (contains i customers, including those in service) is^{[3]: 172–173}

${\displaystyle \pi _{i}=(1-\rho )\rho ^{i}.\,}$

Average number of customers in the system

We see that the number of customers in the system is geometrically distributed with parameter 1 − ρ. Thus the average number of customers in the system is ρ/(1 − ρ) and the variance of number of customers in the system is ρ/(1 − ρ)². This result holds for any work conserving service regime, such as processor sharing.^[4]

Busy period of server

The busy period is the time period measured between the instant a customer arrives to an empty system until the instant a customer departs leaving behind an empty system. The busy period has probability density function^[5][6][7][8]

${\displaystyle f(t)={\begin{cases}{\frac {1}{t{\sqrt {\rho }}}}e^{-(\lambda +\mu )t}I_{1}(2t{\sqrt {\lambda \mu }})&t>0\\0&{\text{otherwise}}\end{cases}}}$

where I₁ is a

The Laplace transform of the M/M/1 busy period is given by[11]^{[12][13]: 215}

${\displaystyle \mathbb {E} (e^{-sF})={\frac {1}{2\lambda }}(\lambda +\mu +s-{\sqrt {(\lambda +\mu +s)^{2}-4\lambda \mu }})}$

which gives the moments of the busy period, in particular the mean is 1/(μ − λ) and variance is given by

${\displaystyle {\frac {1+{\frac {\lambda }{\mu }}}{\mu ^{2}(1-{\frac {\lambda }{\mu }})^{3}}}.}$

Response time

The average response time or sojourn time (total time a customer spends in the system) does not depend on scheduling discipline and can be computed using Little's law as 1/(μ − λ). The average time spent waiting is 1/(μ − λ) − 1/μ = ρ/(μ − λ). The distribution of response times experienced does depend on scheduling discipline.

First-come, first-served discipline

For customers who arrive and find the queue as a stationary process, the response time they experience (the sum of both waiting time and service time) has Laplace transform (μ − λ)/(s + μ − λ)^[14] and therefore probability density function^[15]

${\displaystyle f(t)={\begin{cases}(\mu -\lambda )e^{-(\mu -\lambda )t}&t>0\\0&{\text{otherwise.}}\end{cases}}}$

Processor sharing discipline

In an M/M/1-PS queue there is no waiting line and all jobs receive an equal proportion of the service capacity.^[16] Suppose the single server serves at rate 16 and there are 4 jobs in the system, each job will experience service at rate 4. The rate at which jobs receive service changes each time a job arrives at or departs from the system.^[16]

For customers who arrive to find the queue as a stationary process, the Laplace transform of the distribution of response times experienced by customers was published in 1970,^[16] for which an integral representation is known.^[17] The waiting time distribution (response time less service time) for a customer requiring x amount of service has transform^[3]: 356

${\displaystyle W^{\ast }(s|x)={\frac {(1-\rho )(1-\rho r^{2})e^{-[\lambda (1-r)+s]x}}{(1-\rho r^{2})-\rho (1-r)^{2}e^{-(\mu /r-\lambda r)x}}}}$

where r is the smaller root of the equation

${\displaystyle \lambda r^{2}-(\lambda +\mu +s)r+\mu =0.}$

The mean response time for a job arriving and requiring amount x of service can therefore be computed as x μ/(μ − λ). An alternative approach computes the same results using a spectral expansion method.^[4]

Transient solution

We can write a probability mass function dependent on t to describe the probability that the M/M/1 queue is in a particular state at a given time. We assume that the queue is initially in state i and write p_k(t) for the probability of being in state k at time t. Then^[2][18]

${\displaystyle p_{k}(t)=e^{-(\lambda +\mu )t}\left[\rho ^{\frac {k-i}{2}}I_{k-i}(at)+\rho ^{\frac {k-i-1}{2}}I_{k+i+1}(at)+(1-\rho )\rho ^{k}\sum _{j=k+i+2}^{\infty }\rho ^{-j/2}I_{j}(at)\right]}$

where ${\displaystyle i}$ is the initial number of customers in the station at time ${\displaystyle t=0}$,${\displaystyle \rho =\lambda /\mu }$, ${\displaystyle a=2{\sqrt {\lambda \mu }}}$ and ${\displaystyle I_{k}}$ is the

When the utilization ρ is close to 1 the process can be approximated by a reflected Brownian motion with drift parameter λ – μ and variance parameter λ + μ. This heavy traffic limit was first introduced by John Kingman.^[20]

References