Decision tree; finite source blocking formulas

Decision tree of traffic formulas

The right traffic formula depends on three questions: is the number of sources large (infinite) or small (finite)? Is the system a loss system or a delay system? And what happens to a blocked call? A blocked call can be cleared (LCC: it is lost and the user goes away), held (LCH: the user keeps trying for the holding time) or delayed (LCD: it waits in a queue).

                  Traffic formula
                        |
         +--------------+--------------+
    Infinite sources            Finite sources
   (Poisson arrivals)          (N users, N small)
         |                             |
  +------+------+               +------+------+
 LCC   LCH    LCD              LCC   LCH    LCD
  |     |      |                |     |      |
Erlang Poisson Erlang         Engset Binomial Finite
  B            C                            queue

Main formulas:

• Erlang B (A erlang offered to n trunks): $E_{1,n}(A) = \dfrac{A^n/n!}{\sum_{k=0}^{n} A^k/k!}$
• Erlang C (probability of delay): $E_{2,n}(A) = \dfrac{\frac{A^n}{n!}\frac{n}{n-A}}{\sum_{k=0}^{n-1}\frac{A^k}{k!} + \frac{A^n}{n!}\frac{n}{n-A}}$
• Poisson (LCH): $P = e^{-A}\sum_{k=n}^{\infty} \dfrac{A^k}{k!}$

Blocking formulas for finite sources

When the number of sources $N$ is not much larger than the number of servers $n$, the arrival rate falls as more sources become busy. Erlang B then overestimates the blocking, so finite-source formulas are used.

1. Engset formula (lost calls cleared, N > n). Let $\beta$ be the offered traffic per idle source. The probability that $k$ servers are busy is

$$P(k) = \frac{\binom{N}{k}\beta^k}{\sum_{i=0}^{n}\binom{N}{i}\beta^i}, \quad k = 0,1,\dots,n$$

• Time congestion (fraction of time all n servers are busy): $E = P(n)$.
• Call congestion (fraction of calls lost): an arriving call comes from an idle source, so it sees only the other $N-1$ sources:

$$B = \frac{\binom{N-1}{n}\beta^n}{\sum_{i=0}^{n}\binom{N-1}{i}\beta^i}$$

For finite sources $B < E$, because fewer idle sources exist to make calls when the group is busy.

2. Binomial formula (lost calls held, or N ≤ n). Each source is busy with probability $a$, independently of the others, so the number of busy sources follows

$$P(k) = \binom{N}{k} a^k (1-a)^{N-k}$$

A call from one source is blocked when $n$ or more of the other $N-1$ sources are already busy, so the blocking probability is

$$P(\text{blocking}) = \sum_{k=n}^{N-1}\binom{N-1}{k}a^k(1-a)^{N-1-k}$$

If $N \le n$ the sum is empty and there is no blocking at all.

As $N \to \infty$ with total traffic fixed, Engset tends to Erlang B and the binomial formula tends to the Poisson formula.