= Solution
For the <M-M-1 queue with Bernoulli feedback>, assume $0<\delta\leq1$. The number of service visits before departure is geometric on $1,2,\ldots$, with probabilities $\delta(1-\delta)^{j-1}$, independent of the independent exponential visit lengths. The transform of the total time actually being served is
$$
\sum_{j=1}^\infty\delta(1-\delta)^{j-1}\left(\frac\mu{\mu+s}\right)^j
=\frac{\delta\mu}{\delta\mu+s}.
$$
Thus
$$
\boxed{B_{\rm total}\sim\operatorname{Exp}(\delta\mu).}
$$
This excludes the waiting between visits. A feedback completion changes the customer's position but not the number in the system. By exponential <memoryless property>, the latter process has births of rate $\lambda$ and true deaths of rate $\delta\mu$ while nonempty; unsuccessful completions are self-events. It is therefore an M/M/1 number process with effective service rate $\delta\mu$, giving
$$
\boxed{\lambda<\delta\mu,\qquad\pi_j=\left(1-\frac\lambda{\delta\mu}\right)\left(\frac\lambda{\delta\mu}\right)^j.}
$$
For positive $\lambda$ equality is null recurrent, larger arrival rate is unstable, and $\delta=0$ admits no departure-based equilibrium.
To prove the output claim, state the relevant <time reversal of a continuous-time Markov chain> theorem: an equilibrium continuous-time chain satisfying <detailed balance> has the same reversed path law. In this reversible <birth-death process>, final departures are precisely downward jumps; time reversal turns them into upward jumps. Upward jumps are supplied by a state-independent Poisson clock of rate $\lambda$, so their process is Poisson. Hence \b[the stationary final-departure process is Poisson of rate $\lambda$]. This is the output conclusion of <Burke theorem> for the effective number process; it does not say the process of all service completions has that rate.
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