Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 35 / 1 / Solution

Assume $\nu ,\lambda ,\mu >0$ and regard each line as an individual circulating through three colonies. Then $n$ is a closed migration process on $\{n\in {\mathbb{Z}}_{\ge 0}^3:n_1+n_2+n_3=N\}$. With $e_j$ the coordinate vectors, its nonzero transition intensities areThe memoryless property of the exponential distributions and the independent Poisson process make these counts a continuous-time Markov chain. In particular, the accepted-arrival rate is $\nu$, not $\nu n_1$: free lines are availability tokens, not independent callers. An attempted arrival when $n_1=0$ is lost and causes no state change, so it does not enter the off-diagonal generator.

For the cyclic routing $1\to 2\to 3\to 1$, take equal traffic weights at all colonies. The product-form stationary distribution of a closed migration process gives weightsHere is a direct global balance for a continuous-time Markov chain verification. Set ${\varphi }_1(k)=\nu 1_{\{k>0\}}$, ${\varphi }_2(k)=\lambda 1_{\{k>0\}}$, ${\varphi }_3(k)=\mu k$, with cyclic subscripts. The predecessor for a $j\to j+1$ transition is $m=n+e_j-e_{j+1}$, allowed when $n_{j+1}>0$. The product weights obeySumming these incoming weights over the three edges gives $w(n){\sum }_j{\varphi }_j(n_j)$, precisely the outgoing weight. Thus $\pi (n)=w(n)/G_N$ is stationary. The finite state space is irreducible, so this stationary distribution is unique.

Let $k=n_2+n_3$ be the occupied-line count. Summing over its automated-service count $i=n_3$ givesso $G_N={\sum }_{k=0}^{N}H(k)$. The external Poisson process sees the stationary state probabilities by Poisson arrivals see time averages. A call is lost exactly when $n_1=0$, equivalently $k=N$. Therefore

For two independent identical operators, the middle transition rate becomes $\lambda \min (n_2,2)$: one call receives one operator's service, while two or more calls keep both operators busy. Define $D_2(0)=1$ and $D_2(m)={\prod }_{h=1}^m\min (h,2)=2^{m-1}$ for $m\ge 1$. The same global balance for a continuous-time Markov chain calculation applies toConsequently the blocking probability in a finite-line sequential-service network isThe subtraction corrects the $n_2=0$ term, for which $D_2(0)=1$ rather than $2^{-1}$. In particular $H_2(0)=1$, and when $N\le 1$ replacing one operator by two does not change blocking.