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Traffic engineering objectives; Engset formula

Short answer

The objectives of traffic engineering: dimensioning minimum trunks for busy-hour traffic, meeting grade of service, minimizing cost, traffic measurement and forecasting, efficient routing, overload control. The Engset formula gives the call congestion (loss probability) for finite sources S sharing N servers: B = binom(S-1,N)α^N / Σ_{k=0}^N binom(S-1,k)α^k, where α is offered traffic per idle source.

What are the objectives of traffic engineering in telecommunication? Explain the Engset formula used in traffic source.

Objectives of traffic engineering

  1. Dimensioning: find the minimum number of trunks, switch paths, registers or channels that carry the expected busy-hour traffic.
  2. Meet the grade of service: keep blocking (loss systems) or delay (delay systems) within the target, e.g. GoS = 0.01.
  3. Minimise cost: balance equipment cost against service quality; avoid both over- and under-provision.
  4. Traffic measurement and forecasting: measure busy-hour traffic and predict growth for planning.
  5. Efficient routing: design direct, tandem and alternate (overflow) routes for high utilisation.
  6. Overload control and reliability: keep the network stable during peaks and failures.

Engset formula (finite traffic sources)

The Engset formula gives the loss probability when a finite number of sources S share N servers (S > N) and blocked calls are cleared, e.g. a PBX or small rural exchange. Since busy sources cannot make new calls, the call rate falls as more sources are busy, so Erlang B (which assumes infinite sources) over-estimates loss.

Assumptions: S identical sources; each idle source calls at rate γ\gamma; mean holding time h=1/μh = 1/\mu; α=γh\alpha = \gamma h = offered traffic per idle source; full availability; lost calls cleared; equilibrium.

Birth–death model: in state n (n servers busy) arrival rate =(S−n)γ=(S-n)\gamma, departure rate =nμ= n\mu. Balance:

(S−n+1)γPn−1=nμPn  ⇒  Pn=(Sn)αnP0(S-n+1)\gamma P_{n-1} = n\mu P_n \;\Rightarrow\; P_n = \binom{S}{n}\alpha^n P_0

Normalising over n=0..Nn = 0..N:

Pn=(Sn)αn∑k=0N(Sk)αkP_n = \frac{\binom{S}{n}\alpha^n}{\sum_{k=0}^{N}\binom{S}{k}\alpha^k}

Time congestion (all N busy):

E=(SN)αN∑k=0N(Sk)αkE = \frac{\binom{S}{N}\alpha^N}{\sum_{k=0}^{N}\binom{S}{k}\alpha^k}

Call congestion (probability an arriving call is lost) – the Engset loss formula:

B=(S−1N)αN∑k=0N(S−1k)αkB = \frac{\binom{S-1}{N}\alpha^N}{\sum_{k=0}^{N}\binom{S-1}{k}\alpha^k}

Notes

  • B<EB < E for finite S.
  • As S→∞S \to \infty with Sα=AS\alpha = A, Engset → Erlang B.
  • If S≤NS \le N, no calls are lost (Bernoulli/binomial distribution).

✓ Answer checked by Claude Opus 5.5 (high effort)

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