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M/M/1 Arrival and Departure Time Occupancy Distributions

Module by: Bart Sinclair

Summary: (Blank Abstract)

The continuous-time chain for the M/M/1 queue has the following state transition graph.

Figure 1
Figure 1 (MM1_probabilities_1.png)

The state probabilities are given by π k =ρk1-ρ π k ρ k 1 ρ .

Looking at the M/M/1 system only at transistions gives rise to the following discrete-time chain.

Figure 2
Figure 2 (MM1_probabilities_2.png)

The single-step probability matrix for the discrete-time chain is

P=01000…μλ+μ0λλ+μ00…0μλ+μ0λλ+μ0…00μλ+μ0λλ+μ…⋮⋮⋮⋮⋮ P 0 1 0 0 0 … μ λ μ 0 λ λ μ 0 0 … 0 μ λ μ 0 λ λ μ 0 … 0 0 μ λ μ 0 λ λ μ … ⋮ ⋮ ⋮ ⋮ ⋮ (1)
Solving π′P=π′ π ′ P π ′ yields π ′ 1μλ+μ= π ′ 0⇒ π ′ 1=1+ρ π ′ 0 π ′ 1 μ λ μ π ′ 0 π ′ 1 1 ρ π ′ 0 π ′ 01+ π ′ 2μλ+μ= π ′ 1⇒ π ′ 2=1+ρρ π ′ 0 π ′ 0 1 π ′ 2 μ λ μ π ′ 1 π ′ 2 1 ρ ρ π ′ 0 π ′ 1λλ+μ+ π ′ 3μλ+μ= π ′ 2⇒ π ′ 3=1+ρρ2 π ′ 0 π ′ 1 λ λ μ π ′ 3 μ λ μ π ′ 2 π ′ 3 1 ρ ρ 2 π ′ 0 π ′ 2λλ+μ+ π ′ 4μλ+μ= π ′ 3⇒ π ′ 4=1+ρρ3 π ′ 0 π ′ 2 λ λ μ π ′ 4 μ λ μ π ′ 3 π ′ 4 1 ρ ρ 3 π ′ 0 and in general, ∀k,k≥1: π ′ k=1+ρρk-1 π ′ 0 k k 1 π ′ k 1 ρ ρ k 1 π ′ 0 .

Normalization: ∑k=0∞ π ′ k= π ′ 0+∑k=1∞1+ρρk-1 π ′ 0= π ′ 01+1+ρ∑i=0∞ρi= π ′ 01+1+ρ1-ρ=1 k 0 π ′ k π ′ 0 k 1 1 ρ ρ k 1 π ′ 0 π ′ 0 1 1 ρ i 0 ρ i π ′ 0 1 1 ρ 1 ρ 1 π ′ 0=1-ρ2 π ′ 0 1 ρ 2 π ′ k≥1=ρk-11-ρ22 π ′ k 1 ρ k 1 1 ρ 2 2

Prarrival sees k jobs ahead of it=Prarrival in state k|arrival=Prarrival in state k & arrivalPrarrival=Prarrival in state kPrarrival arrival sees k jobs ahead of it arrival arrival in state k arrival in state k & arrival arrival arrival in state k arrival (2)
Prarrival=∑k=0∞ π ′ kPrarrival|state k= π ′ 01+∑k=1∞ρk-11-ρ22λλ+μ=1-ρ2+∑k=1∞ρk-11-ρ2λ+μλλλ+μ=1-ρ2+1-ρ2∑k=1∞ρk=1-ρ2+1-ρ211-ρ-1=12 arrival k 0 π ′ k state k arrival π ′ 0 1 k 1 ρ k 1 1 ρ 2 2 λ λ μ 1 ρ 2 k 1 ρ k 1 1 ρ 2 λ μ λ λ λ μ 1 ρ 2 1 ρ 2 k 1 ρ k 1 ρ 2 1 ρ 2 1 1 ρ 1 1 2 (3)
which could have been inferred directly since, for the system to be stable, half of the transistions must be arrivals and half must be departures (in the long run). Prarrival sees k jobs ahead of it= π ′ kλλ+μ12=ρk1-ρ arrival sees k jobs ahead of it π ′ k λ λ μ 1 2 ρ k 1 ρ What about departing jobs?
Prdeparting job leaves k jobs=Prdeparture in state k+1|departure=Prdeparture in state k+1Prdeparture=ρk1-ρ22μλ+μ12=ρk1-ρ2λ+μμμλ+μ12=ρk1-ρ departing job leaves k jobs departure departure in state k+1 departure in state k+1 departure ρ k 1 ρ 2 2 μ λ μ 1 2 ρ k 1 ρ 2 λ μ μ μ λ μ 1 2 ρ k 1 ρ (4)
Hence, the probability that an arrival sees k jobs ahead of it is equal to the probability that a departure leaves k jobs behind it, which is equal to the probability that the system is in state k in steady-state.

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