Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 30 / 1 / ii / Solution

An open migration process has finitely many colonies and state $n\in {\mathbb{N}}^J$. External arrivals to colony $j$ form independent Poisson processes with rates ${\nu }_j$. When $n_j=k>0$, individuals leave that colony at total rate $g_j(k)>0$, with $g_j(0)=0$. A departing individual moves to $k$ with probability $p_{jk}$ or exits with probability $p_{j0}=1-{\sum }_k{p}_{jk}$. The routing choices and all arrival and service randomness are independent. The routing matrix $P=(p_{jk})$ is substochastic and transient: every individual eventually exits, equivalently its spectral radius is less than one. Self-routing is allowed, but creates no state change.

The traffic equations of an open migration process, using row vectors, areThe inverse matrix exists because ${\sum }_{m\ge 0}{P}^m$ converges; its $(k,j)$ entry counts expected visits to $j$ by an individual entering at $k$. Thus ${\alpha }_j$ includes both external arrivals and internal migrations. Restrict first to the accessible colonies with ${\alpha }_j>0$.

SetIf each $Z_j<\infty$, the candidate product-form stationary distribution of an open migration process isHere is a global balance for a continuous-time Markov chain proof, which does not incorrectly assume that the routing is reversible. The weight ratio is $\pi (n+e_j)/\pi (n)={\alpha }_j/g_j(n_j+1)$. After dividing the incoming balance equation at $n$ by $\pi (n)$, external arrivals contribute ${\sum }_{j:n_j>0}{\nu }_j{g}_j(n_j)/{\alpha }_j$, exits contribute ${\sum }_j{\alpha }_j{p}_{j0}$, and migrations from $k$ to $j\ne k$ contribute ${\sum }_{j:n_j>0}{\sum }_{k\ne j}{g}_j(n_j){\alpha }_k{p}_{kj}/{\alpha }_j$. Combining the first and third contributions using the traffic equations of an open migration process yields ${\sum }_j{g}_j(n_j)(1-p_{jj})$. Summing those same traffic equations of an open migration process gives ${\sum }_j{\alpha }_j{p}_{j0}={\sum }_j{\nu }_j$. The incoming rate is therefore exactly the outgoing rateThis proves stationarity. With finite arrival rates, finite service rates in each state and transient routing, the process is nonexplosive: only finitely many individuals enter in a finite interval, and each makes finitely many routing transitions before exit, almost surely. On the accessible irreducible Markov chain, a finite normalizer gives the unique stationary distribution and positive recurrence. Colonies with zero effective arrivals are empty in this equilibrium. For a nonempty accessible colony, a divergent $Z_j$ prevents normalization of the equilibrium weights.

For the single-server specialization, $g_j(k)={\varphi }_j$ for $k>0$. Each colony has the stationary distribution of an M-M-1 queue:requiring ${\varphi }_j>{\alpha }_j$ on every colony with positive traffic. The service rates do not change the traffic equations of an open migration process.

For positive ${\alpha }_j$, write $B=F-{\sum }_j{\alpha }_j$ and $s_j={\varphi }_j-{\alpha }_j$. A stable allocation requires $B>0$ and $s_j>0$, with ${\sum }_j{s}_j=B$. The Cauchy-Schwarz inequality givesEquality holds precisely when $s_j$ is proportional to $\sqrt{{\alpha }_j}$. Hence the unique optimal service-capacity allocation in an open migration process and its mean population areThere is no stable allocation when $F\le {\sum }_j{\alpha }_j$ and some traffic is positive. If some ${\alpha }_j=0$, those empty colonies contribute no population; allocate the spare capacity to the positive-traffic colonies by the same rule. Allowing zero capacity at an unused colony attains this optimum. If strictly positive ${\varphi }_j$ are required even there, the displayed value is only an infimum, approached as their capacities tend to zero. If every ${\alpha }_j=0$, the equilibrium is empty and any admissible allocation minimizes its mean population at zero.