Loss Networks

In Erlang’s telephone line model discussed in the previous post, we considered users competing for a finite set of resources. When insufficient resources are available, the call is lost. A loss network generalises this situation to more complicated resource configurations. We think of links 1, …, J, each with some integer capacity $c_j$. Each incoming call requires 1 units of each link in some subset of the set of links, and lasts for a time distributed as an exponential random variable with parameter 1, independent of everything else in the model. We call this subset the route, and denote by $A_{jr}$ the incidence matrix of links on routes. Calls arrive as PP($\nu_r$)$independently for each route: no queueing occurs – a call is lost if some link required is operating at full capacity. We call the probability of this event $L_r$, the loss probability. Observe that $(n_r)$, the number of calls on each route r, is a Markov chain on the truncated space $\{An\leq c\}$. By checking the DBEs, it is clear that an ED for this Markov chain is proportional to the ED for the MC without the capacity constraint, with state-space restricted to the truncated space. But without capacity constraints, the system is a linear migration process, for which we discovered the form of the ED in the previous section. If we write $H(c)=\mathbb{P}(An\leq c)$ in the linear migration process, we can compute the acceptance probability for the finite capacity system as: $1-L_r=\frac{H(C-Ae_r)}{H(C)}$ Approximating Blocking Probabilities We want to calculate $B_j$, the equilibrium blocking probability, that a given link j is full. We have two methods: firstly, to find the distribution for $(n_r)$ with maximum probability, for which the blocking probabilities appear as shadow prices. And secondly, to make a reasonable approximation about blocking independence, and solve explicitly. We want to show that these methods give the same answers. To maximise the probability $\pi(n)\propto \prod_r \frac{\nu_r^{n_r}}{n_r!}$ on $\{An\leq c\}$, we take logs and maximise using Stirling’s approximation, which is reasonable as we are implicitly working under a regime where the throughput tends to infinity while preserving ratios. The primal problem is $\max\quad \sum_r(x_r\log \nu_r-x_r\log x_r+x_r),\quad\text{s.t. }Ax\leq c$ which has Lagrangian $L(x,y,z)=\sum_r x_r+\sum_r x_r(\log \nu_r-\log x_r-\sum_j y_jA_{jr})+\sum_j y_jc_j-\sum_j y_jc_j$ We observe that complementary slackness here has the form $y.z=0$, and remember that by Strong Duality, which applies here because everything relevant is convex, this equality holds at the primal optimum. Differentiating the Lagrangian at the optimum allows us to specify the optimal x in terms of y: $\bar{x}_r=\nu_r e^{-\sum y_jA_{jr}}$ The dual problem is then to minimise $\min\quad \sum_r \nu_re^{-\sum_jy_jA_{jr}}+\sum_j y_jc_j$ At this point, we make the suggestive substitution $e^{-y_j}=1-B_j$, observing that this gives B non-negative by default since y is non-negative. After further work, we will deduce that these B do indeed have a sensible interpretation as blocking probabilities, but it should be stressed that this is in no way obvious yet. Now complementary slackness asserts: $\sum_rA_{jr}\nu_r\prod_i(1-B_i)^{A_{ir}}\left\{\begin{array}{l l}=c_j& \quad B_j>0\\ \leq c_j & \quad B_j=0\\ \end{array} \right.$ Note that the primal objective function is strictly convex so $\bar{x}$ as discussed is the unique optimum. The dual objective is strictly convex in $yA$, so if A has full rank J, this induces a unique optimum in terms of y. We assume A is full rank (since for example we can perturb slightly) and that there is no degeneracy in the blocking. Now we consider a sequence of networks with proportionally increasing arrival rates and capacities Continue reading Advertisement # Models for Random Access Channels To describe a situation where, for example, mobile phones compete for the use of a satellite for transmitting their messages, we consider a random access network. Here, we assume packets arrive at (necessarily different) stations at Poisson rate $\nu$, and times of transmissions are discrete. A station will always attempt to transmit a packet which has just arrived, then, if that fails, will retransmit after a time given by a independent random variables. Transmissions fail if at least two stations attempt to use the same window. For the ALOHA protocol, we assume these RVs are geometric with mean $f^{-1}$. We hope that the backlog, that is $N_t$ the number of stations with packets to retransmit at time t eventually clears. Say $Z_t$ is the number of attempts in (t,t+1)$, proportional to $Y_t+\text{Bin}(N_t,f)$ where Y is the number of arrivals in (t,t+1), and then $N_{t+1}=N_t+Y_t-1_{Z_t=1}$.

Then the drift in N is positive once N gets large. This is, however, not enough to prove that the system is transient. Instead we compute p(n), the probability that the channel unjams (that is, an instance where no transmissions are attempted) before the backlog increases, given that the backlog is currently n. We can calculate that this probability is $\sim n(1-f)^{n-1}$ and so $\sum p(n)<\infty$. Applying Borel-Cantelli to the record process, that is the times at which the backlog increases, it can be shown that almost surely, the channel unjams only finitely many times.

To maximise throughput, we would take $f=\frac{1}{n}$, but of course the individual stations don’t know what is. However, they could guess. Suppose they view the process and keep a counter:

$S_{t+1}=\max\left(1,S_t+a1_{Z_t=0}+b1_{Z_t=1}+c1_{Z_t=*}\right)$

Then transmission happens with probability $f=\frac{1}{S_t}$, with the idea that S tracks N well, especially when N is large. Observe that

$\mathbb{E}[S_{t+1}-S_t|N_t=n,S_t=s]\rightarrow (a-c)e^{-k}+(b-c)ke^{-k}+c$

where $k=\frac{n}{s}$. So (a,b,c)=(2-e,0,1) for example gives negative S drift when k>1, and positive drift when k<1, which is required to draw S towards N. But S might still diverge asymptotically away from N? We don’t want N to increase too fast when k>1, but

$\mathbb{E}[N_{t+1}-N_t|N_t=n,S_t=s]\rightarrow \nu-ke^{-k}$

so demanding $\nu-ke^{-k} is what we require when $\nu, which is satisfied by the example suggested.

If a station only knows its own transmission attempts, we can use Ethernet, where after r unsuccessful attempts, wait for a time $B_r$ chosen uniformly from $[1,\ldots,2^r]$. This is called binary exponential backoff. When $\nu<\log 2$, get infinitely many successful transmissions almost surely. But for positive recurrence, would have an equilibrium distribution, so by Little’s Law, we can equate arrival and departure rates. Considering the probabilities of 0 and 1 transmissions, obtain

$\nu=\pi_1e^{-\nu}+\pi_0\nu e^{-\nu}\quad\Rightarrow\quad \nu\leq e^{-\nu}\quad\Rightarrow\quad \nu<0.567$

So can be non-recurrent, but have infinitely many successful transmissions!

# Queues and Migration Processes

Simple Queues

A queue generally has the form of a countable state-space Markov Chain. Here we assume that customers are served in the order they arrive (often referred to as: FIFO – First In First Out). The standard Kendall notation is M/M/C/K. Here the Ms stand for Markov (or memoryless) arrivals, and service times respectively, possibly replaced by G if the process admits more general distributions. C is the number of servers, and K the capacity of the system.

The first example is a M/M/C/C queue, motivated by a telephone network. Here, there are c lines, and an arriving call is lost if all lines are busy at the arrival time. We assume arrivals are a PP($\lambda$) process, and the call times are iid Exp($\mu$) RVs. We record the number of busy lines, which is a continuous-time Markov chain on state-space [0,c]. As usual, we assume the system operates in equilibrium, and so in particular we must have $\lambda<\mu c$. It is easy to find the equilibrium distribution. The only non-zero transition probabilities are

$q(i-1,i)=\lambda, \quad q(i,i-1)=\mu i\quad i=1,\ldots,c$

and so can that that $\pi_i=\frac{\nu_i}{i!}\pi_0$ satisfies the DBEs where $\nu:=\frac{\lambda}{\mu}$, sometimes called the traffic intensity. This gives

$\pi_c=\mathbb{P}(c\text{ lines busy})=\frac{\nu^c}{c!}\left(\sum_{k=0}^c \frac{\nu^k}{k!}\right)^{-1}$

and we define this to be $E(\nu,c)$, Erlang’s formula.

Note that if $(X(t),t\in\mathbb{R})$ is a MC in equilibrium, $(X(-t),t\in\mathbb{R})$ is a MC with the same ED, modulo style of discontinuities (ie whether transitions are left-continuous or right-continuous). Therefore, in any queue where all customers get served, eg an M/M/1 queue, for which $\lambda<\mu$ (otherwise the MC is transient, so no ED!), the departure process is the same (in distribution) as the arrivals process.

We can check that this holds for a series of M/M/1 queues, and that in equilibrium, the sizes of the queues are independent. This is merely an extension of the observation that future arrivals for a given queue are independent of the present, and likewise past departures are independent of the future, but the argument is immediately obvious.

Migration Processes

We consider a closed migration process on J colonies, with populations described by a vector $n=(n_1,\ldots,n_J)$. We say that the instantaneous rate of movement of an individual from colony j to colony k is $q(n,T_{jk}n)=\lambda_{jk}\phi_j(n_j)$, for some functions $\phi_j(0)=0$. We can describe the ED of this system in terms of the distribution obtained when a single individual performs a random walk through the state space, with $\phi_j(1)$ taken to be 1 for each j. We call the DBEs for this case the traffic equations, with solutions $(\alpha_j,j\in J)$. Then, by checking the PBEs, it is clear that the equilibrium distribution of the original migration process satisfies

$\pi(n)\propto \prod_{j=1}^J \frac{\alpha_j^{n_j}}{\prod_{r=1}^{n_j}\phi_j(r)}$

This is important, as it would have been computationally difficult to solve the original equations for an ED.

The same result holds for an open migration process, where individuals can enter and leave the system, arriving at colony j at rate $\nu_j$, and leaving at rate $\mu_k\phi_k(n_k)$. Note that this has the same form as if each colony was served by a PP($\alpha_j\lambda_j$), with departures at rate $\lambda_j\phi_j(n_j):=(\mu_j+\sum_k\lambda_{jk})\phi_j(n_j)$. But (obviously) this interpretation is not equivalent to the model

The time reversal is also an OMP. One has to check that the transition rates have the correct form, and so the exit process from each colony (in equilibrium, naturally), is PP($\alpha_j\mu_j$)\$. Most importantly, given a communicating class, we can think of the restriction to this class as an OMP, so in particular, rates of transition between classes are Poisson.

Little’s Law

Informally, the mean number of customers should be equal to the arrival rate multiplied by the mean sojourn time in the system of a customer. This is easiest to formalise by taking an expectation up to a regeneration time. This is T, the first time the system returns to its original state (assumed to be 0 customers), an a.s. finite stopping time.

Set $L:=\mathbb{E}\frac{1}{T}\int_0^Tn(s)ds$, the average number of customers, and $W:=\frac{\mathbb{E}\sum_1^N W_n}{\mathbb{E}N}$ where N is the number of customers arriving in [0,T], and $W_n$ is the waiting time of the n-th customer.

Little’s Law asserts that $L=\lambda W$. Note that for a Markov Chain in equilibrium, can define L more simply as $\mathbb{E}n$ and similarly for W.

It is easiest proved by considering the area between the arrivals process and the departure process in two ways: integrating over height and width. Note that working up to a regeneration time is convenient because at that time the processes are equal.

The migration processes above are said to be linear if $\phi_j(n_j)=n_j$. This process has the form of a Markov chain, and so even in a more general state space, the distribution of points in disjoint subsets of the state-space are independent Poisson processes.

Often though, we start with no individuals in the system, but still the distribution is given by a time-inhomogenous Poisson random measure. The mean is specified by

$M(t,E)=\nu\int_0^t P(u,E)du$

where $\nu$ is the net arrival rate, and $P(u,E)$ is the probability that an individual is in E, a time interval of u after arriving.

As one would suspect, this is easiest to check through generating functions, since independence has a straightforward generating function analogue, and the expression for a Poisson RV is manageable.