Skip to main content

Infinite and pure loss systems; Engset model

Short answer

An infinite-source system has constant arrival rate as sources are many. A pure loss system clears blocked calls with no queue. The Engset model is a pure loss system with finite sources S>N; arrival rate falls as more sources are busy. State probabilities use binomial coefficients; time congestion is P_N, and call congestion B uses (S-1 choose N), giving B < E.

What is infinite system and pure loss system? Describe the tele-traffic Engset model.

Infinite (source) system

An infinite-source system is one in which the number of traffic sources (subscribers) is very large compared with the number of servers. The call arrival rate therefore does not depend on how many sources are already busy: arrivals are Poisson with a constant mean rate λ\lambda. Example: a junction route between two large exchanges. Erlang B, Erlang C and Poisson formulas assume infinite sources.

Pure loss system

A pure loss (lost-calls-cleared) system has N servers and no waiting room. A call that finds all N servers busy is rejected (lost) and disappears; it is not queued and is assumed not to retry. Its quality is measured by the probability of loss (GoS). In Kendall notation, M/M/N/N. Example: a group of trunks with no queue; Erlang B formula.

Engset model

The Engset model is a pure loss system with a finite number of sources S (S > N), such as a PBX with 40 extensions sharing 8 outgoing lines. As more sources become busy, fewer idle sources remain to make calls, so the arrival rate falls.

Assumptions

  1. S identical sources, N servers, full availability, S>NS > N.
  2. Each idle source makes calls at rate γ\gamma; mean holding time h=1/μh = 1/\mu. Offered traffic per idle source α=γ/μ=γh\alpha = \gamma/\mu = \gamma h.
  3. Blocked calls are cleared; statistical equilibrium.

State diagram (birth–death)

   S*g     (S-1)*g          (S-N+1)*g
 0 ----> 1 ------> 2 ... N-1 ---------> N
   <----   <------     <---------------
    mu      2*mu              N*mu

Arrival rate in state n is (S−n)γ(S-n)\gamma; departure rate is nμn\mu.

State probabilities (from cut equations (S−n+1)γPn−1=nμPn(S-n+1)\gamma P_{n-1} = n\mu P_n):

Pn=(Sn)αn∑k=0N(Sk)αk,n=0,1,…,NP_n = \frac{\binom{S}{n}\alpha^n}{\sum_{k=0}^{N}\binom{S}{k}\alpha^k},\quad n = 0,1,\dots,N

Time congestion (fraction of time all servers busy):

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

Call congestion (Engset loss formula, probability that a call attempt is lost), which is what the caller experiences:

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}

Here B<EB < E, because when all servers are busy fewer sources are free to call.

Limits: if S→∞S \to \infty with Sα=AS\alpha = A constant, both formulas tend to Erlang B. If S≤NS \le N there is no loss and the busy-server distribution is binomial (Bernoulli).

Use: dimensioning PBX trunks, small rural exchanges, concentrators and cell sectors with few users, where Erlang B would over-estimate blocking.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Telecommunications chapter 6 questions →