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Short note: Queuing theory in delay system

Short answer

Delay systems let calls wait instead of being lost. Queueing theory uses models like M/M/1 and M/M/n (Erlang C) to compute wait probability, mean queue length, and delay. For M/M/1, L = ρ/(1-ρ), W_q = ρ/(μ-λ); for M/M/n, C(n,A) gives probability of wait, with mean wait W_q = C·h/(n-A). Little's law: L = λW.

Write a short note on queuing theory in delay system.

A delay system (queueing system or lost-calls-delayed system) is one in which a call that finds all servers busy is not lost but waits in a queue until a server becomes free. Queueing theory gives the probability of waiting, the mean waiting time and the queue length. Examples: calls waiting for a register (dial tone delay), messages in a store-and-forward data network, callers held at a call centre, jobs waiting for an SPC processor.

Elements of a queueing system

 arrivals     queue (buffer)      n servers
 --lambda-->  [ | | | | ]  --->  [S1]
                                 [S2]  ---> departures
                                 [..]
                                 [Sn]
  1. Arrival process: usually Poisson with rate λ\lambda.
  2. Service time distribution: usually exponential with mean h=1/μh = 1/\mu.
  3. Number of servers nn.
  4. Queue capacity: infinite or finite.
  5. Queue discipline: FIFO, LIFO, random, priority.
  6. Number of sources: infinite or finite.

Kendall's notation A/B/n/K/N/D: arrival distribution / service distribution / servers / capacity / sources / discipline. Example: M/M/1, M/M/n, M/D/1 (M = Markov/exponential, D = deterministic, G = general).

Main results

Little's law (holds for any queue):

L=λW,Lq=λWqL = \lambda W, \qquad L_q = \lambda W_q

M/M/1 queue (ρ=λ/μ<1\rho = \lambda/\mu < 1):

P(k)=(1−ρ)ρkL=ρ1−ρ,Lq=ρ21−ρW=1μ−λ,Wq=ρμ−λ\begin{aligned} P(k) &= (1-\rho)\rho^k \\ L &= \frac{\rho}{1-\rho}, \quad L_q = \frac{\rho^2}{1-\rho} \\ W &= \frac{1}{\mu-\lambda}, \quad W_q = \frac{\rho}{\mu-\lambda} \end{aligned}

M/M/n queue (Erlang C formula). With A=λhA = \lambda h erlang offered to nn servers (A<nA < n), the probability that a call has to wait is

C(n,A)=Ann!nn−A∑k=0n−1Akk!+Ann!nn−AC(n,A) = \frac{\frac{A^n}{n!}\frac{n}{n-A}}{\sum_{k=0}^{n-1}\frac{A^k}{k!} + \frac{A^n}{n!}\frac{n}{n-A}}

Mean waiting time of all calls and of delayed calls:

Wq=C(n,A) hn−A,Wd=hn−AW_q = \frac{C(n,A)\,h}{n-A}, \qquad W_{d} = \frac{h}{n-A}

Probability of waiting longer than tt:

P(W>t)=C(n,A) e−(n−A)t/hP(W > t) = C(n,A)\,e^{-(n-A)t/h}

Grade of service in delay systems

The GoS is stated as the probability of delay, or the probability of waiting more than a given time, e.g. "dial tone within 3 s for 99% of calls". Delay systems give higher server utilisation than loss systems, but need A<nA < n for stability; as ρ→1\rho \to 1 the waiting time grows very rapidly.

Example

An M/M/1 server with λ=8\lambda = 8/min and μ=10\mu = 10/min has ρ=0.8\rho = 0.8, L=4L = 4, W=1/(10−8)=0.5W = 1/(10-8) = 0.5 min and Wq=0.4W_q = 0.4 min.

✓ Answer checked by Claude Opus 5.5 (high effort)

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