Calculation of E(P|w)
Let P represent an exponentially distributed time variable with mean \( 1/\mu \). Consider a queuing system where customers arrive according to a Poisson distribution with constant rate \( \lambda \). In this appendix, we show that E(P|w) , the expectation of a time interval given that during this time interval w customer arrivals occur, is equal to \( (w+1)/(\lambda + \mu ) \). Using the definition of conditional expectation and the property that the number of Poisson arrivals during an exponentially distributed time interval with mean \( 1/\mu \) has a geometric distribution with success probability \( \frac{1}{1 + \frac{\lambda }{\mu }} \), we obtain the following expression for E(P|w) :
$$\begin{aligned} E(P|w)&= \frac{\int \limits _{0}^{\infty } t \quad \mu \mathrm {e}^{-\mu \, t} \quad \frac{(\lambda \, t)^w}{w!} \quad \mathrm {e}^{-\lambda \, t} \quad \mathrm {d}t}{(1 - \frac{1}{1 + \frac{\lambda }{\mu }})^w \quad (\frac{1}{1 + \frac{\lambda }{\mu }})} \end{aligned}$$
(13a)
Rearranging terms, and using that \( \frac{(\lambda + \mu )^{w+2} \quad t^{w+1}}{(w+1)!} \quad \mathrm {e}^{-(\lambda + \mu ) \, t} \) is the probability density function of the Erlang distribution with parameters \( (w + 2) \) and \( (\lambda + \mu ) \), we can rewrite expression (13a) as follows:
$$\begin{aligned} E(P|w)= & {} \frac{\frac{\mu \, \lambda ^w \, (w+1)}{(\lambda + \mu )^{w+2}} \int \limits _{0}^{\infty } \frac{(\lambda + \mu )^{w+2} \quad t^{w+1}}{(w+1)!} \quad \mathrm {e}^{-(\lambda + \mu ) \, t}}{\left( 1 - \frac{1}{1 + \frac{\lambda }{\mu }}\right) ^w \quad \left( \frac{1}{1 + \frac{\lambda }{\mu }}\right) } \end{aligned}$$
(13b)
$$\begin{aligned}= & {} \frac{\frac{\mu \, \lambda ^w \, (w+1)}{(\lambda + \mu )^{w+2}} }{\left( 1 - \frac{1}{1 + \frac{\lambda }{\mu }}\right) ^w \quad \left( \frac{1}{1 + \frac{\lambda }{\mu }}\right) } \end{aligned}$$
(13c)
Finally, we obtain the desired result \( E(P|w) = (w+1)/(\lambda + \mu ) \) from (13c) via straightforward algebraic manipulations.