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Queueing Theory Lecture Notes
ENEE 426, Spring 2008
March 4, 2008
1 Little’s Theorem
N (t) = Number of customers in queue at time t α(t) = Number of customers who arrived in [0, t] Ti = Time spent in system by ith customer
Let Nt =
t
∫ (^) t
0
N (τ )dτ (1)
be the time average of N (t). Let
N = lim t→∞ Nt (2)
be the steady-state time average. Let
λt =
α(t) t
be the time average arrival rate over [0, t]. Let
λ = lim t→∞ λt (4)
be the steady-state arrival rate. Let
Tt =
α(t)
α ∑(t)
i=
Ti (5)
be the time average customer delay over [0, t]. Let
T = lim t→∞ Tt (6)
be the steady-state delay. Little’s theorem:
N = λT (7)
2 Probabalistic Version
Define random variable
pn(t) = P r[n customers in queue at time t] (8)
Then let
N¯ (t) = E[pn(t)] =
∑^ ∞
n=
npn(t) (9)
In steady state, pn = limt→∞ pn(t), ∀n, then
N¯ = E[pn] =
∑^ ∞
n=
npn (10)
Note that this is also
N¯ = lim t→∞ N^ ¯ (t) (11)
For delay, assume knowledge of T¯k, average delay for customer k. Then
T¯ = lim k→∞ T^ ¯k (12)
In ergodic systems, the time average equals the stead- state average, thus N = N¯ and T = T¯. Lastly, define
λ = lim t→∞
E[probabalistic α(t)] t
Then Little’s Theorem still holds:
N = λT (14)
3 Examples
Example 1: Packets arrive at n nodes in a network with rates λ 1 , ..., λn. N is the average number of packets inside the network. Then the average delay per packet is T =
N
∑n i=1 λi
Also, Ni = λiTi. Example 2: k packets per second arrive with the first arrival at t = 0. Each requires αk time to trans- mit. Processing and propagation time equal P. In- terarrival times constant.
T = αk + P (16)
λ =
k
N = λT = α +
P
k
which holds only with time averages.
4 Poisson Processes
Used to measure the number arrivals between time 0 and time t. If A(t) is a Poisson Process, then:
- A(t + τ ) ≥ A(t) for τ ≥ 0
- A(t) ∈ I+
- For arrival rate λ, A(t) ∼ P oi(τ ; λ), then
P [A(t + τ ) − A(t) = n] = e−λt^
(λτ )n n!
and the average is
E[P oi(τ ; λ)] = τ λ (20)
- If tn is the time of the nth arrival, and τn = tn+1 − tn is the interarrival time, then
P [τn = τ ] = λe−λτ^ (21)
which is an exponential probability density func- tion, with mean E[τn] = (^1) λ and variance V ar[τn] = (^) λ^12.
- If Ai(t) are independent Poisson processes and Ai(t) ∼ P oi(t; λi) then if
Aˆ(t) =
∑^ n
i=
Ai(t) (22)
then the sum process has distribution Aˆ(t) ∼ P oi(t; ˆλ), where
λˆ =
∑^ n
i=
λi (23)
5 Discrete-Time Markov
Chains
M [n] is a Markov process at time n. Markov means that the value of M [n + 1] depends only on the value of M [n], and is independent of M [0], ..., M [n−1]. Let
P r[M [n] = j|M [n − 1] = i] = qij (24)
Let Q be a matrix of values qij for all i, j. Q is the transition probability matrix. Let
vi[n] = P r[M [n] = i] (25)
and v[n] be a vector of values vi[n]. Then
v[n + 1] = v[n]Q (26)
which can be inductively expanded to
v[n] = v[0]Qn^ (27)
These are called the Chapman-Kolmogrov Equations. To find the steady-state probability, compute
v = lim n→∞ v[n] = lim n→∞ v[0]Qn^ (28)
6 Continuous-Time Markov
Chains (CTMC)
M (t) is a Markov process at time t. The time spent in state i is distributed exp(λi), where λi is the exit rate from state i. The probability of transitioning to state j from state i is, as before, qij. Let λ be a vector of λi. Then the rate of transition between states is λQ. Let pj = P [in state j at steady-state] and Tj (t) be the time in state j from [0, t], then
pj = lim t→∞
Tj (t) t = lim t→∞ P r[M (t) = j] (29)
7 CTMC Queues
M (t) is the number of customers in the system. The rate at which M (t) increases by one is λ and the rate at which it decreases is μ, which are the arrival and departure rates from the system. Our goal is to compute pj = lim t→∞ P r[M (t) = j] (30)
Look at the transitions from state n to state n + 1. They must be within 1 of each other. Thus
pnλ = pn+1μ (31)
If we let ρ = λ/μ, then
pn+1 = ρpn (32)
Inductively, we can compute that
pn = ρnp 0 (33)
For ρ < 1 we have
∑^ ∞
n=
pn =
∑^ ∞
n=
ρnp 0 = p 0 1 − ρ
Solving, we get p 0 = 1 − ρ and pn = ρn(1 − rho).
