Erlang arrivals joining the shorter queue

Abstract

We consider a system in which customers join upon arrival the shortest of two single-server queues. The interarrival times between customers are Erlang distributed and the service times of both servers are exponentially distributed. Under these assumptions, this system gives rise to a Markov chain on a multi-layered quarter plane. For this Markov chain we derive the equilibrium distribution using the compensation approach. The expression for the equilibrium distribution matches and refines tail asymptotics obtained earlier in the literature.

Appendix

In what follows we will assume that the α’s are in absolute value smaller than or equal to the starting solution σ ², with σ the unique root of Eq. (3.4) in (0,1). However, for 0<ρ<1 one can easily verify that σ<ρ, so we will assume that |α|<ρ ².

Lemma 7.1

For ρ<1 and for every α, with |α|∈(0,ρ ²), the equation det(D(α,β))=0 assumes the form

(7.1)

and has exactly k simple roots in the β-plane with 0<|β|<|α|/2.

Proof

Equation (7.1) is a polynomial in β of degree 2k and we will show that exactly k roots are inside the open circle of radius |α|/2. We divide both sides of Eq. (7.1) by α ^k+1 and set z=β/α which gives

(7.2)

with f ₁(z)=2(kρ+1)z−αz ²−1. Then f ₁(z) has two roots:

For 0<ρ<1, we observe that z ₊ is located outside the open unit circle, while

Furthermore, for |z|=1/2,

Hence, by Rouché’s theorem $f_{1}^{k}(z)-(2k\rho)^{k} z^{k+1}$ has exactly k roots inside |z|=1/2, which completes the proof. □

According to Lemma 7.1 all β roots are located in the open circle with radius ρ ²/2, so we assume that |β|<ρ ²/2.

Lemma 7.2

For ρ<1 and for every β, with |β|∈(0,ρ ²/2), the equation det(D(α,β))=0 assumes the form

(7.3)

and has exactly one root in the α-plane with 0<|α|<3|β|/4.

Proof

Equation (7.3) is a polynomial in α of degree k+1 and we will show that there is only one root inside the open circle of radius 3|α|/4. We divide both sides of Eq. (7.3) by β ^k+1 and set x=α/β, which gives

(7.4)

with f ₂(x)=x(2kρ+2−β−x)^k. Then f ₂(x) has the zero root of multiplicity one and the root 2kρ+2−β of multiplicity k larger than the radius 3/4. Furthermore, for |x|=3/4,

Hence, by Rouché’s theorem f ₂(x)−(2kρ)^k has exactly one root in |x|<3/4, which completes the proof. □