La cancelación del poder via inscripción de la sucesión intestada

We have already mentioned in Section 6.1 the fact that the output process of an M/M/1 is Poisson. This is one of the results of the so-called Burke’s theorem [17]. In steady-state, the departure process of a stable M/M/1, where ρ <1, is a Poisson process with parameterλ and is independent of the number in the queue after the departures occur. If we already know that the output process is Poisson, given that the arrival rate isλand given that there are no losses, all the traffic that enters must depart. Therefore the rate of the output process must also be equal to λ.

We have shown in Section 6.1 the reversibility of M/M/1 queue-size process showing that the detailed balance equations and the normalizing equation yield the steady-state distribution of the queue-size process.

By reversibility, in steady-state, the arrival process of the reversed process must also follow a Poisson process with parameter λ and this process is the departure process of the forward process. Therefore the departures follow a Poisson process and the inter-departure times are independent of the number in the queue after the departures occur in the same way that inter-arrival times are independent of a queue size before the arrivals.

Now that we know that in steady-state the departure process of a sable M/M/1 queue is Poisson with parameter λ, we also know that, in steady-state, the inter-departure times are also exponentially distributed with parameterλ. We will now show this fact without using the fact that the departure process is Poisson directly. Instead, we will use it indirectly to induce PASTA for the reversed arrival process to obtain that, following a departure, in steady-state, the queue is empty with probability 1−ρ and non-empty with probability ρ. If the queue is non-empty, the time until the next departure is exponentially distributed with parameter µ – this is the service-time of the next customer. If the queue is empty, we have to wait until the next customer arrival which is exponentially distributed with parameter λ and then we will have to wait until the next departure which will take additional time which is exponentially distributed. All together, if the queue is empty, the time until the next departure is a sum of two exponential random variables, one with parameter λ and the other with parameter µ. Let U1 and U2 be two independent exponential random variables with parameters λ and µ,

respectively. Define U =U1+U2, notice that U is the convolution of U1 andU2, and note that U has hypoexponential distribution. Having the density fU(u), the density fD(t) of a random variable D representing the inter-departure time will be given by

fD(t) =ρµe−µt+ (1−ρ)fU(t). (299) Knowing thatfU(u) is a convolution of two exponentials, we obtain

fU(t) =

Z t

u=0

λe−λuµe−µ(t−u)du

= λµ

µ−λ e

−λt₋

Then by the latter and (299), we obtain fD(t) = ρµe−µt+ (1−ρ) λµ µ−λ e −λt_−e−µt (300) which after some algebra gives

fD(t) =λe−λt. (301)

This result is consistent with Burke’s theorem.

Homework 6.8

Complete all the algebraic details in the derivation of equations (299) – (301).

Another way to show consistency with Burke’s theorem is the following. Consider a stable (ρ <1) M/M/1 queue. Let d be the unconditional number of departures, in steady state, that leave the M/M/1 queue during a small interval of time of size , and let d(i) be the number of departures that leave the M/M/1 queue during a small interval of time of size if there are

i packets in our M/M/1 queue at the beginning of the interval. Then, P(d(i) > 0) = o() if

i= 0, and P(d(i) = 1) =µ+o() if i >0. Therefore, in steady-state,

P(d = 1) = (1−ρ)0 + (ρ)µ+o() =λ+o(),

which is a property consistent with the assertion of Poisson output process with parameter λ

in steady-state.

Homework 6.9

So far we have discussed the behaviour of the M/M/1 departure process in steady-state. You are now asked to demonstrate that the M/M/1 departure process may not be Poisson with parameter λ if we do not assume steady-state condition. Consider an M/M/1 system with arrival rate λ and service rateµ, assume that ρ=λ/µ <1 and that there are no customers in the system at time 0. Derive the distribution of the number of customers that leave the system during the time interval (0, t). Argue that this distribution is, in most cases, not Poisson with parameter λtand find a special case when it is.

Guide

Let D(t) be a random variable representing the number of customers that leave the system during the time interval (0, t). LetXp(λt) be a Poisson random variable with parameterλtand consider two cases: (a) the system is empty at timet, and (b) the system is not empty at time

t. In case (a), D(t) = Xp(λt) (why?) and in case (b) D(t) = Xp(λt)−Q(t) (why?) and use the notation used in Section 2.5 P00(t) to denote the probability that in time t the system is

empty, so the probability that the system is not empty at time t is 1−P00(t). Derive P00(t)

using Eqs. (230) and (231). Then notice that

Consider the limit

Dk(t) = lim

∆t→0P[Q(t) =k |a departure occurs within(t−∆t, t)].

Considering the fact that the reversed process is Poisson and independence between departures before timet and Q(t), we obtain that

Dk(t) =P[Q(t) =k]. (302)

Then, by taking the limit of both sides of (302), we show that the queue size seen by a leaving customer is statistically identical to the queue size seen by an independent observer.

Homework 6.10

Write a simulation of the M/M/1 queue by measuring queue size values in two ways: (1) just before arrivals and (2) just after departures. Verify that the results obtained for the mean queue size in steady-state are consistent. Use confidence intervals. Verify that the results are also consistent with analytical results. Repeat your simulations and computation for a wide range of parameters values (different ρ values). Plot all the results in a graph including the confidence intervals (bars).