Busy Periods: Definition, Mean, and Transform

What a Busy Period Measures

Queueing performance is not described only by the number of customers present at a typical time or by an individual customer's waiting time. A busy period measures the duration of a whole episode of continuous work: it begins when an arrival finds the system empty and ends the next time the system becomes empty. Its length depends on the service time of the initiating customer and on all arrivals that occur before the system empties. First-step analysis turns that dependence into an equation for the busy-period transform.

Consider an M/M/1 queue with arrival rate \(\lambda>0\) and service rate \(\mu>0\). Arrivals form a Poisson process, service times are independent exponential random variables with rate \(\mu\), and arrivals and service times are independent. Let \((N_t)_{t\ge0}\) be the number in the system on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(S=\{0,1,2,\ldots\}\) and time index \(t\ge0\). Use the natural filtration \((\mathcal{F}_t)_{t\ge0}\), representing the queue history through time \(t\). From state \(n\ge1\), the transition rates to \(n+1\) and \(n-1\) are \(\lambda\) and \(\mu\); from state 0 the rate to 1 is \(\lambda\). Thus the generator has \(q_{n,n+1}=\lambda\), \(q_{n,n-1}=\mu\) for \(n\ge1\), and \(q_{n,n}=-(\lambda+\mu)\); at 0, \(q_{0,1}=\lambda\) and \(q_{0,0}=-\lambda\). Each generator row sums to zero.

The indicator in the transform matters when the queue can remain busy forever with positive probability: an infinite busy period contributes zero to \(f_n(s)\) for \(s>0\). This definition does not assume that the queue has a stationary distribution. In particular, the transform calculation will also cover the critical and overloaded cases.

First-Step Analysis and the Transform

While the system is nonempty, the next event is either an arrival or a service completion. The time to that event is exponential with rate \(\lambda+\mu\); its type is an arrival with probability \(\lambda/(\lambda+\mu)\) and a completion with probability \(\mu/(\lambda+\mu)\). Discounting by \(e^{-s t}\) gives the common factor \((\lambda+\mu)/(\lambda+\mu+s)\). From \(n\ge1\), an arrival changes the state to \(n+1\), while a completion changes it to \(n-1\). Since \(f_0(s)=1\), first-step conditioning gives

Proof. Fix \(s>0\). The recurrence for \(f_n(s)\), after multiplication by \(\lambda+\mu+s\), has characteristic equation \(\lambda r^2-(\lambda+\mu+s)r+\mu=0\). The polynomial is positive at \(r=0\) and equals \(-s<0\) at \(r=1\). Its leading coefficient is positive, so its two roots satisfy \(0<r_-<1<r_+\). Every solution of this second-order recurrence has the form \(f_n=c_-r_-^n+c_+r_+^n\): both powers satisfy the recurrence, and their distinct roots give the two independent solutions. Since \(0\le f_n(s)\le1\) for every \(n\), the term with \(r_+^n\) cannot have a nonzero coefficient; otherwise the sequence would be unbounded. The boundary value \(f_0(s)=1\) then gives \(c_-=1\), so \(f_n(s)=r_-^n\), and in particular \(\beta(s)=r_-\). The quadratic formula gives the displayed expression.

For the probability of a finite busy period, take \(s\downarrow0\). On \(\{B<\infty\}\), \(e^{-sB}\) increases to 1, and off that event the indicator in the definition is zero. Monotone convergence therefore gives \(\lim_{s\downarrow0}\beta(s)=\mathbb{P}_1(B<\infty)\). The smaller root is continuous in the coefficients, including when the roots coincide. At \(s=0\), the equation factors as \((r-1)(\lambda r-\mu)=0\). If \(\lambda<\mu\), the smaller root is 1; if \(\lambda=\mu\), the roots coincide at 1; if \(\lambda>\mu\), the smaller root is \(\mu/\lambda\). This proves the stated probabilities.

To identify the mean, set \(e^{-sB}=0\) on \(\{B=\infty\}\). As \(s\downarrow0\), \((1-e^{-sB})/s\) increases to \(B\), interpreted as infinity if \(B=\infty\). Consequently, monotone convergence yields $$ \mathbb{E}_1[B]=\lim_{s\downarrow0}\frac{1-\beta(s)}{s}. $$ When \(\lambda<\mu\), the root at zero is \(\beta(0)=1\). Differentiating its quadratic equation for \(s>0\) and taking the limit at zero gives $$ \bigl(2\lambda\beta(0)-(\lambda+\mu)\bigr)\beta'(0)=\beta(0), \qquad (\lambda-\mu)\beta'(0)=1. $$ Thus the limit of \((1-\beta(s))/s\) is \(-\beta'(0)=1/(\mu-\lambda)\). When \(\lambda=\mu\), the quadratic equation rearranges to \(\mu(1-\beta(s))^2=s\beta(s)\). Hence $$ \frac{1-\beta(s)}{s}=\frac{\beta(s)}{\mu(1-\beta(s))}\longrightarrow\infty, $$ because \(\beta(s)\to1\). When \(\lambda>\mu\), the event \(B=\infty\) has probability \(1-\mu/\lambda>0\), so \(\mathbb{E}_1[B]=\infty\). This proves all the claims. \(\square\)

Worked Transform Calculations

Why Root Selection Matters

A common mistake is to set \(s=0\) in the quadratic, observe the roots 1 and \(\mu/\lambda\), and select a root using only the fact that a probability is at most 1. That does not settle the overloaded case: when \(\lambda>\mu\), both roots are at most 1, so this bound cannot distinguish them. The correct argument first proves that for every \(s>0\), the transform is the smaller root. Then it takes the limit \(s\downarrow0\). In the overloaded case, that smaller root tends to \(\mu/\lambda\), giving the finite-busy-period probability. Root selection at \(s=0\) alone is insufficient.

The mean formula also has a specific regime of validity. The expression \(1/(\mu-\lambda)\) applies only when \(\lambda<\mu\). At critical load, the busy period ends almost surely but has infinite mean; above critical load, there is positive probability it never ends, which also forces an infinite mean. These distinctions matter when interpreting a busy-period calculation: a transform at positive \(s\), a probability of eventual emptying, and an expected duration answer different questions.