# Service probability for M/M/1 queue with reneging

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

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). The equivalent formula for multiple heterogeneous servers can be found in Section 3 here.

 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