Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 213 / 1

An open migration process is a continuous-time Markov chain recording the populations $n=(n_1,\dots ,n_J)$ of finitely many colonies. Independent external Poisson processes bring individuals to colony $j$ at rate ${\nu }_j$. When that colony contains $k>0$ individuals, a departure opportunity occurs at total rate ${\varphi }_j(k)>0$, with ${\varphi }_j(0)=0$. The departing individual moves to colony $l$ with probability $p_{jl}$, or leaves the system with probability $p_{j0}=1-{\sum }_l{p}_{jl}$. Routing choices and the clocks are independent. A return to the same colony does not change the state.

The word open means that the routing matrix $P=(p_{jl})$ is transient: every individual eventually leaves, or equivalently the spectral radius of $P$ is less than one. The traffic equations of an open migration process, with row-vector convention, give the total arrival rates:These rates include migrations as well as external arrivals. On the active colonies assume ${\alpha }_j>0$ and the usual irreducible Markov chain condition. PutThe product form and its normalization condition areTo establish the product-form stationary distribution of an open migration process, first use its unnormalized weight $b(n)={\prod }_j{b}_j(n_j)$. Its ratios satisfy $b_j(k){\varphi }_j(k)={\alpha }_j{b}_j(k-1)$. The incoming external arrivals into $n$, divided by $b(n)$, contribute ${\sum }_j{\nu }_j{\varphi }_j(n_j)/{\alpha }_j$. Incoming exits contribute ${\sum }_i{\alpha }_i{p}_{i0}$. Incoming migrations from $i$ to $j\ne i$ contribute ${\alpha }_i{p}_{ij}{\varphi }_j(n_j)/{\alpha }_j$. Terms involving an empty destination are zero. Thus the full incoming rate divided by $b(n)$ isHere the traffic equations of an open migration process give the second equality, including ${\sum }_i{\alpha }_i{p}_{i0}={\sum }_j{\nu }_j$. The right side is precisely the total outgoing rate. This proves global balance for a continuous-time Markov chain; pairwise detailed balance is not required for general routing.

For finite rates at each finite population, the continuous-time Markov chain is nonexplosive Markov chain: its total population is bounded over a finite time interval by the initial population plus the finitely many external arrivals, so it visits a finite set on which all jump rates are bounded. Consequently finite $Z_j$ normalize the invariant weight to a stationary distribution. Under the stated irreducible Markov chain assumptions, this is the unique stationary distribution, and it is a positive recurrent Markov chain. Conversely, the unique invariant measure of an irreducible recurrent chain is proportional to $b$, so a positive recurrent Markov chain requires its normalization to be finite. A sufficient condition is ${\mathrm{liminf}}_{k\to \infty }{\varphi }_j(k)>{\alpha }_j$; the series criterion itself is the exact condition. Colonies with ${\alpha }_j=0$ have zero stationary population and can be removed from the active class.

For the single-server M/M/1 queue specialization, ${\varphi }_j(k)={\varphi }_j$ for $k>0$, each stationary colony has the geometric distribution of an M/M/1 queue:The traffic equations of an open migration process do not involve the service rates, so the ${\alpha }_j$ remain fixed during this service-capacity allocation in an open migration process. Write $A={\sum }_j{\alpha }_j$, $B=F-A$ and $b_j={\varphi }_j-{\alpha }_j$. A stationary allocation requires $B>0$, with $b_j>0$ and ${\sum }_j{b}_j=B$. The Cauchy-Schwarz inequality givesEquality holds exactly when $b_j$ is proportional to $\sqrt{{\alpha }_j}$. Hence, for positive arrival rates, the unique optimal service allocation and the minimum expected value of the total population areIf $F\le A$ and some arrival rate is positive, no stable service allocation exists. Zero-arrival colonies need no service in their stationary empty class: allowing zero service there gives the same formula restricted to active colonies. If strictly positive service is demanded even at inactive colonies, the displayed minimum is an infimum approached as that unused allocation tends to zero. If all arrivals vanish, the stationary mean population is zero for every allocation that drains the initial population.

An effective bandwidth measures how much capacity a random traffic source requires when unusually large demands matter. A mean-rate description loses burstiness; a peak-rate description can waste capacity when peaks are rare. The parameter $s>0$ in an effective bandwidth controls the emphasis placed on such bursts through the moment-generating function:Here $X$ is the traffic in a chosen time unit. If $X(t)$ instead measures traffic over length $t$, the corresponding rate is $\alpha (s,t)=\log \mathbb{E}[e^{sX(t)}]/(st)$. With independent stationary increments, this is independent of $t$ whenever the moment-generating function exists. More generally both the time scale and dependence structure matter.

The Chernoff bound follows directly from the Markov inequality, since $e^{sX}$ is nonnegative and $X\ge b$ implies $e^{sX}\ge e^{sb}$:Optimize over the admissible positive values of $s$ to obtainFor independent sources $X_1,\dots ,X_N$, moment-generating functions multiply and their effective bandwidths add. Thus a total capacity $c$ has the one-time overflow bound $\mathbb{P}({\sum }_i{X}_i\ge c)\le \exp \{s({\sum }_i{\alpha }_i(s)-c)\}$. In particular, for independent and identically distributed random variables,This is the positive-parameter part of the Legendre transform of a cumulant-generating function. The size of its exponent describes how a service margin reduces the chance of a large aggregate demand. A buffer-overflow event involves traffic over many intervals; a one-time Chernoff bound alone is not automatically a bound on that event. A time-dependent model and, for example, bounds on the union of relevant interval events are then needed.

Assume a finite moment-generating function in a neighborhood of zero. Differentiating the cumulant-generating function gives $K(0)=0$, $K^{\prime }(0)=\mathbb{E}[X]$ and $K^{\prime \prime }(0)=\mathrm{Var}(X)$. Therefore the small-parameter limit is the mean traffic rate:The variance term shows how an effective bandwidth charges extra capacity for burstiness. Also $K$ is a convex function with $K(0)=0$, so $K(s)/s$ is nondecreasing for $s>0$ in its domain.

For the other limit, let $M=\mathrm{esssup}X$, the essential supremum. If $M<\infty$, $X\le M$ almost surely gives $\alpha (s)\le M$. For every $a<M$, $p_a=\mathbb{P}(X\ge a)>0$, andTaking a lower limit and then $a\uparrow M$ proves the large-parameter limit is peak traffic:If $X$ has no finite upper bound and all positive exponential moments exist, the same lower bound for arbitrarily large $a$ makes the limit infinite. If positive exponential moments cease to exist at a finite parameter, the effective bandwidth is infinite beyond that domain, and the infinite limit is understood in this extended sense. For example, a normal distribution with mean $m$ and variance $v$ has $\alpha (s)=m+sv/2$: its unbounded upper tail has no finite peak. Thus effective bandwidth interpolates between mean and peak demand while permitting independent sources to be aggregated by addition.