Consider an M/M/1 queue with arrival rate $\lambda$ and service rate $\mu$, where participants renege after a random amount of time which follows an exponential distribution with mean $\tau$. We neglect balking. Is there an expression for the steady-state probability that a participant receives service?


1 Answer 1


The steady-state probability of being served for an M/M/1 queue with exponential reneging times and no balking is $$ p_s=\frac{1+z}{1+r(1+z)} $$ where $r=\lambda/\mu$ is the service intensity, and $$ z=\exp(r\mu\tau)\cdot(r\mu\tau)^{-\mu\tau}\cdot\gamma(\mu\tau+1,r\mu\tau). $$ Here, $\gamma(x,a)$ is the unscaled lower incomplete gamma function: $$ \gamma(s,x)=\int_0^x t^{s-1}\exp(-t)\;\mathrm{d}t. $$ This result was originally given by Ancker and Cafarian in 1963 (paywall)[1]. The equivalent formula for multiple heterogeneous servers can be found in Section 3 here.

[1] Ancker, C.J., & Cafarian, A.V. 1963. Queuing with reneging and multiple heterogeneous servers, Naval Research Logistics Quarterly, Vol 10, No 1, 125–149. DOI: 10.1002/nav.3800100112


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