Feedback Loops and Visit Ratios in Jackson Networks

Routing Loops Create Additional Visits

In a network with feedback, a customer may leave a service station only to return to it later, directly or through other stations. The external arrival rate at a station therefore need not equal the rate at which that station must serve customers. Traffic equations account for every visit, including repeated visits caused by routing loops. Their solution gives the effective arrival rates needed to assess station loads and network stability.

Consider an open Jackson network with \(d\) single-server stations. The queue-length process \((X_t)_{t\ge0}\) is a continuous-time Markov chain on \(S=\mathbb{Z}_{\ge0}^d\), on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Customers arrive externally to station \(i\) as independent Poisson processes of rates \(\alpha_i\ge0\). At station \(i\), service times are exponential with rate \(\mu_i>0\). After service, a customer goes to station \(j\) with probability \(r_{ij}\), or leaves the network with probability \(r_{i0}=1-\sum_{j=1}^d r_{ij}\). We assume independent routing choices and the usual Jackson-network conditions. Let \(R=(r_{ij})\) be the routing matrix, using the row convention: row \(i\) records the destinations of a customer departing station \(i\).

The traffic equation says that the rate of arrivals to station \(j\) combines external arrivals there and customers routed there after service elsewhere. In particular, \(\lambda_i\) is not generally \(\alpha_i\). If routing sends customers around a loop, they generate further visits before eventually exiting.

Visit Ratios from Routing Paths

First consider a single external stream: each arriving customer enters according to a row probability vector \(b\), and the stream has rate \(a>0\). Thus the external rate vector is \(\alpha=ab\). Suppose every customer eventually exits; a convenient sufficient condition for a finite network is that the spectral radius of \(R\) is less than \(1\).

Proof. Number the customer’s visits by \(0,1,2,\ldots\), with the entry station counted as visit \(0\). The probability row vector for the station of visit \(0\) is \(b\). If the distribution of visit \(n\) is \(bR^n\), then the probability of the next visit being to each station is obtained by routing once more, so that distribution is \(bR^{n+1}\). Therefore, for each station \(j\), its total number of visits is a sum of indicators, one for each possible visit number. Taking expectations and summing the nonnegative terms gives expected visit vector \(\sum_{n\ge0}bR^n\). The finite-dimensional linear-algebra fact that spectral radius less than \(1\) implies \(R^n\to0\) and convergence of \(\sum_{n\ge0}R^n\) gives \(v=b(I-R)^{-1}\). Multiplying the series by \(R\) shows \(v=b+vR\). If \(w\) is another finite solution, then \(w-v=(w-v)R\), and iteration gives \(w-v=(w-v)R^n\to0\); hence \(w=v\). With external rate \(a\), expected visits per unit time are multiplied by the number of external customers per unit time, giving \(\lambda=av\). \(\square\)

For multiple external streams the same reasoning, applied separately to each stream and added, gives \(\lambda=\alpha+\lambda R\). Under the spectral-radius condition the solution is \(\lambda=\alpha(I-R)^{-1}\). Equivalently, \(\lambda=\alpha+\alpha R+\alpha R^2+\cdots\): the terms count external arrivals, their first routed visits, their second routed visits, and so on. This series also explains why the traffic equations are appropriate even when customers can circulate repeatedly.

Stability: Use the Criterion, Not a Circular Inference

The strict inequality matters. A station with \(\lambda_i=\mu_i\) is at critical load, not strictly below capacity, and does not satisfy the theorem’s positive-recurrence condition. If any \(\lambda_i>\mu_i\), the condition also fails. Conversely, rates below capacity at every station are sufficient under the theorem’s open Jackson assumptions. The criterion concerns effective rates, not just the external rates.

For comparison, if the external rate in this same one-station network were \(1\), the traffic equation would give \(\lambda_1=1+\tfrac14\lambda_1\), hence \(\tfrac34\lambda_1=1\) and \(\lambda_1=\tfrac43\). With service rate \(1\), the utilization would be \(\tfrac43>1\), again failing the criterion. These calculations show how feedback can push an apparently modest external rate to critical or excessive total workload.

Common Pitfall: Ignoring the Routed Traffic

A common incorrect calculation sets \(\lambda=\alpha\), as though a customer were served only at the station of external entry. In the first two-station example, that mistake would use rates \((2,1)\), giving apparent utilizations \(2/4=\tfrac12\) and \(1/3=\tfrac13\). The correction is to include routed arrivals: the actual rates are \((\tfrac{14}{5},\tfrac{12}{5})\), with utilizations \(\tfrac7{10}\) and \(\tfrac45\). The wrong values happen to remain below \(1\) in that example, but they understate both stations’ loads and could give a wrong stability conclusion in a network closer to capacity.

A second logical error is to argue that failure of the product-form probabilities to normalize proves instability. A formula for a stationary distribution, by itself, establishes a sufficient condition when it can be normalized; failure to normalize that particular expression does not exclude every possible stationary distribution. For the converse conclusion, use the independently stated Jackson-network stability theorem with its hypotheses, rather than treating product-form normalization as its proof.

The practical sequence is therefore: check that every routing row plus its exit probability sums to \(1\); solve the traffic equations, including all feedback; verify the solution by substitution; and compare each effective rate with its own service rate. Visit ratios help interpret the solution, while the stability theorem supplies the precise positive-recurrence conclusion.