Telephone Traffic

Queuing system

A queuing (delay) system in telecommunication is one in which a call or packet that finds all servers (trunks, registers, processors, links) busy is not lost but waits in a queue until a server becomes free. Examples: calls waiting for an operator in a call centre, digits waiting for a common-control register, packets waiting in a router buffer. Its performance is measured by probability of delay, mean waiting time and queue length, rather than by lost-call probability.

 arrivals    +----------+     +-----------+   departures
 ---lambda-->|  queue   |---->| N servers |--------->
             | (buffer) |     |  (mu each)|
             +----------+     +-----------+

Characteristics in Kendall's notation

Kendall's notation describes a simple queue as A / B / C / K / N / D (often shortened to A/B/C):

1. Arrival pattern: usually Poisson with mean rate $\lambda$ calls per unit time, so inter-arrival times are exponential.
2. Service pattern: usually exponential holding time with mean $h = 1/\mu$.
3. Number of servers working in parallel.
4. System capacity: limited buffers make the system partly a loss system.
5. Source population: large (infinite) or small (finite) number of subscribers.
6. Discipline: order in which waiting calls are served.

Examples

• M/M/1: Poisson arrivals, exponential service, one server — e.g. a single public phone; utilisation $\rho = \lambda/\mu$, mean number waiting $L_q = \rho^2/(1-\rho)$.
• M/M/N (Erlang C): N trunks with queuing, used for call centres.
• M/M/N/N: no waiting room — the Erlang B loss system.
• M/D/1: fixed packet length on a link.

Kendall–Lee notation is a compact six-part code that describes the main features of a queuing system. Kendall gave the first three symbols; Lee added the last three.

$$(a\,/\,b\,/\,c) : (d\,/\,e\,/\,f)$$

When the last three are FCFS/∞/∞ they are often left out, so M/M/1 means $(M/M/1):(FCFS/\infty/\infty)$.

Meaning of the codes

• M (Markovian): Poisson arrivals with rate $\lambda$, i.e. exponential inter-arrival times; or exponential service with mean $1/\mu$. Memoryless.
• D: every inter-arrival or service time is the same.
• $E_k$: sum of k exponential phases; less random than M.
• G/GI: any distribution.

Examples

1. (M/M/1):(FCFS/∞/∞) – single public call office with Poisson callers and exponential call durations; callers wait in line, infinite population.
   • Utilisation $\rho = \lambda/\mu$, mean number in system $L = \rho/(1-\rho)$, mean time in queue $W_q = \rho/(\mu-\lambda)$.
   • If $\lambda = 10$ calls/h and mean call = 3 min ($\mu = 20$/h): $\rho = 0.5$, $L = 1$, $W_q = 0.5/10$ h $= 3$ min.
2. (M/M/N):(FCFS/∞/∞) – Erlang delay (Erlang C) model: N operators in a call centre with queuing.
3. (M/M/N):(GD/N/∞) – Erlang loss (Erlang B) model: N trunks, no waiting space, blocked calls lost.
4. (M/M/N):(GD/N/S) – Engset model: finite number S of sources, no waiting.
5. (M/D/1):(FCFS/K/∞) – router output link with fixed-size packets and a buffer of K packets.

The notation lets an engineer pick the correct formula (Erlang B, Erlang C, Engset, M/M/1) by simply reading the system description.

Traffic (Erlang) definition: traffic intensity $A = C\,h / T$, where C = number of calls in period T and h = mean holding time. So the number of calls is $C = A\,T/h$.

Given: N = 30 servers, A = 15 E, h = 3 min, T = 1 h = 60 min.

Calls by the group as a whole

$$\begin{aligned} C_{group} &= \frac{A\,T}{h} \\ &= \frac{15 \times 60}{3} \\ &= 300\ \text{calls per hour} \end{aligned}$$

Calls by a single server

Assuming traffic is shared equally, each server carries

$$\begin{aligned} A_1 &= \frac{15}{30} = 0.5\ \text{E} \end{aligned}$$

i.e. each server is busy 50% of the time = 30 min in the hour.

$$\begin{aligned} C_1 &= \frac{A_1\,T}{h} = \frac{0.5 \times 60}{3} = 10\ \text{calls per hour} \end{aligned}$$

Check: $30 \times 10 = 300$ calls.

Answer: a single server handles 10 calls/hour; the group handles 300 calls/hour.

Telephone traffic engineering is the study and design of switching and transmission resources (trunks, switch paths, registers, processors) so that a network gives an acceptable grade of service at minimum cost. It uses probability and queuing theory to relate traffic load, number of servers and service quality.

Basic quantities

• Traffic intensity (Erlang): average number of simultaneously busy servers, $A = C\,h/T$ (C calls of mean holding time h in period T). 1 E = one circuit busy continuously. 1 E = 36 CCS (hundred call-seconds per hour).
• Holding time (h): mean duration a server is occupied per call; usually taken as exponential.
• Busy hour: the continuous 60-min period of the day with the highest traffic; networks are dimensioned for it. Busy hour calling rate and day-to-busy-hour ratio are used to forecast it.
• Grade of Service (GoS): probability that a call is lost (loss system) or delayed beyond a time (delay system), e.g. GoS = 0.01 means 1 call in 100 lost in the busy hour.

Types of systems

Traffic assumptions

1. Pure chance traffic: Poisson call arrivals (large number of independent sources).
2. Exponential holding times.
3. Statistical equilibrium: traffic does not change during the period.
4. Full availability: any free server can be used.

Main formulas

• Erlang B: $B = \dfrac{A^N/N!}{\sum_{k=0}^{N} A^k/k!}$ – probability of loss with N trunks.
• Erlang C: probability of delay in an N-server delay system; mean delay $= P_D\,h/(N-A)$.
• Engset: loss with a finite number of sources (PBX).

Uses

• Dimensioning number of trunks between exchanges and of switching stages.
• Choosing between direct and tandem routes (overflow and alternate routing).
• Planning call centres, cellular channels per cell, and buffers in data networks.
• Measuring and forecasting load for network growth.

Example: with A = 5 E offered and GoS 0.01, the Erlang B table gives N = 11 trunks.

Traffic engineering therefore balances cost (fewer circuits) against service (low blocking or delay).

Grade of Service (GoS) is a measure of the quality of service of a telephone network in the busy hour. In a loss system it is the proportion of calls that are lost because all servers are busy:

$$GoS = \frac{\text{number of calls lost}}{\text{number of calls offered}} = \frac{A_{lost}}{A_{offered}}$$

In a delay system it is the probability that a call waits longer than a stated time.

Key points

• GoS lies between 0 and 1; a smaller value means better service. Typical design values: 0.002 for local exchange paths, 0.01 for trunks, 0.02–0.05 for cellular channels.
• It is fixed for the busy hour, so service is better at other times.
• For Poisson traffic offered to N trunks, GoS equals the Erlang B blocking probability $B = E_{1,N}(A)$.
• For a connection passing through several stages or links in series, overall GoS ≈ sum of the individual GoS values (when each is small): $B \approx B_1 + B_2 + \dots$

Example

In the busy hour 1000 calls are offered and 10 are lost: GoS = 10/1000 = 0.01 (1%).

Use

GoS is the design target in traffic engineering: given the offered traffic and GoS, the number of trunks is found from Erlang B (or Engset/Erlang C) tables. A tighter GoS needs more circuits and so costs more.

The Extended Erlang B (EEB) model is a version of Erlang B that allows for blocked callers who try again. Plain Erlang B assumes blocked calls are cleared and never return; in practice many callers redial, which raises the actual offered load. EEB therefore gives a more realistic, slightly larger number of trunks.

Parameters

• $A_0$ – fresh (first-attempt) traffic in Erlangs
• N – number of trunks
• R – recall factor: fraction of blocked calls that are re-attempted (0 to 1)

Iterative method

1. Start with offered traffic $A = A_0$.
2. Find blocking $B = E_{1,N}(A)$ from Erlang B:

$$B = \frac{A^N/N!}{\sum_{k=0}^{N} A^k/k!}$$

3. Blocked traffic $= B\,A$; retried traffic $= R\,B\,A$.
4. New offered traffic $A = A_0 + R\,B\,A$.
5. Repeat steps 2–4 until A stops changing. Final B is the blocking; lost traffic $= (1-R)\,B\,A$.

 A0 -->(+)--> A --> N trunks --> carried
        ^           |
        |    blocked B*A
        +--R*B*A----+--(1-R)B*A--> lost

Notes

• With R = 0 it reduces to ordinary Erlang B.
• With R = 1 every blocked call returns (lost calls held behaviour).
• It is used in call-centre and trunk dimensioning where customers retry.

Lost calls and grade of service

• Lost call: a call attempt that cannot be connected because all servers (trunks/switch paths) are busy; in a loss system it is cleared and not served.
• Grade of Service (GoS): the proportion of calls lost in the busy hour, $GoS = \dfrac{\text{calls lost}}{\text{calls offered}} = \dfrac{A_l}{A_o}$. A lower GoS means better service (e.g. 0.01 = 1 call in 100 lost).

Given: calls offered $C_o = 600$, calls lost $C_l = 40$, mean holding time $h = 120\ \text{s} = \tfrac{1}{30}$ h, period T = 1 h.

Traffic in Erlangs $= \dfrac{\text{calls} \times h}{T}$.

(a) Traffic offered

$$\begin{aligned} A_o &= \frac{C_o\,h}{T} = 600 \times \tfrac{1}{30} \\ &= 20\ \text{E} \end{aligned}$$

(b) Traffic carried

Calls carried $= 600 - 40 = 560$

$$\begin{aligned} A_c &= 560 \times \tfrac{1}{30} = 18.67\ \text{E} \end{aligned}$$

(c) Traffic lost

$$\begin{aligned} A_l &= A_o - A_c = 40 \times \tfrac{1}{30} = 1.33\ \text{E} \end{aligned}$$

(d) Grade of service

$$\begin{aligned} GoS &= \frac{C_l}{C_o} = \frac{40}{600} = 0.0667 \end{aligned}$$

(e) Duration of congestion

For pure-chance (Poisson) traffic the proportion of time all trunks are busy (time congestion) equals the proportion of calls lost (call congestion), so congestion time in the busy hour is

$$\begin{aligned} T_c &= GoS \times 3600\ \text{s} = \frac{40}{600} \times 3600 \\ &= 240\ \text{s} \end{aligned}$$

Answer: offered traffic = 20 E, carried traffic = 18.67 E (= 18 2/3 E), lost traffic = 1.33 E, GoS = 0.0667 (about 1 in 15), congestion duration = 240 s = 4 min.

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 $\lambda$. 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, $S > N$.
2. Each idle source makes calls at rate $\gamma$; mean holding time $h = 1/\mu$. Offered traffic per idle source $\alpha = \gamma/\mu = \gamma h$.
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 $(S-n)\gamma$; departure rate is $n\mu$.

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

$$P_n = \frac{\binom{S}{n}\alpha^n}{\sum_{k=0}^{N}\binom{S}{k}\alpha^k},\quad n = 0,1,\dots,N$$

Time congestion (fraction of time all servers busy):

$$E = P_N = \frac{\binom{S}{N}\alpha^N}{\sum_{k=0}^{N}\binom{S}{k}\alpha^k}$$

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

$$B = \frac{\binom{S-1}{N}\alpha^N}{\sum_{k=0}^{N}\binom{S-1}{k}\alpha^k}$$

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

Limits: if $S \to \infty$ with $S\alpha = A$ constant, both formulas tend to Erlang B. If $S \le N$ 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.

Kendall's notation

Kendall's notation A/B/C/K/N/D describes a queue: A = arrival process, B = service-time distribution, C = number of servers, K = system capacity, N = source population, D = queue discipline. M = Markovian (Poisson arrivals / exponential service), D = deterministic, G = general. Defaults are K = ∞, N = ∞, FCFS, so M/M/1 means Poisson arrivals, exponential holding time, one server, unlimited queue, first come first served.

PCO analysis with M/M/1

Assumption: the PCO (public call office) is open 24 hours and the 150 calls arrive at random (Poisson) over the day; holding time is exponential. (A 12-hour working day is checked afterwards.)

Parameters

$$\begin{aligned} \lambda &= \frac{150}{24} = 6.25\ \text{calls/h} \\ \mu &= \frac{60}{1.5} = 40\ \text{calls/h} \\ \rho &= \frac{\lambda}{\mu} = \frac{6.25}{40} = 0.15625 \end{aligned}$$

M/M/1 results

$$\begin{aligned} P(\text{wait}) &= \rho = 0.156 \\ L &= \frac{\rho}{1-\rho} = 0.185\ \text{persons in system} \\ L_q &= \frac{\rho^2}{1-\rho} = 0.0289\ \text{persons waiting} \\ W_q &= \frac{\rho}{\mu-\lambda} = \frac{0.15625}{33.75}\ \text{h} = 0.278\ \text{min} \approx 16.7\ \text{s} \\ W &= \frac{1}{\mu-\lambda} = \frac{1}{33.75}\ \text{h} = 1.78\ \text{min} \end{aligned}$$

Check for a 12-hour day ($\lambda = 12.5$/h):

Conclusion

The booth is busy only 16–31% of the time and the average wait is under one minute, with on average far less than one person in the queue. So, on the basis of average daily traffic, a second PCO is not justified; the complaint arises from short busy-hour peaks when many people arrive together. A second booth would be needed only if busy-hour measurements show $\rho$ approaching about 0.7–0.8 (for example, if more than about 28–32 calls arrive in one hour, where $W_q$ rises to several minutes).

Formulas in the traffic engineering decision tree

The choice of formula depends on (1) whether the number of sources is infinite or finite, and (2) how blocked calls are treated: cleared (LCC), held (LCH) or delayed (LCD).

                 Traffic sources
           +-------------+-------------+
       Infinite                     Finite
   +-------+-------+          +-------+-------+
  LCC     LCH     LCD        LCC     LCH     LCD
Erlang B Poisson Erlang C   Engset  Binomial Finite
                                            queue

Numerical

Given: N = 20 servers, A = 10 E, h = 3 min, T = 60 min. Using $A = C\,h/T$, so $C = A\,T/h$.

Group as a whole:

$$\begin{aligned} C_{group} &= \frac{10 \times 60}{3} = 200\ \text{calls/h} \end{aligned}$$

Single server (traffic shared equally):

$$\begin{aligned} A_1 &= \frac{10}{20} = 0.5\ \text{E} \\ C_1 &= \frac{0.5 \times 60}{3} = 10\ \text{calls/h} \end{aligned}$$

Check: $20 \times 10 = 200$.

Answer: each server handles 10 calls/hour; the group handles 200 calls/hour.

Queuing theory is the mathematical study of waiting lines. In a delay system, a call or packet that finds all servers busy waits in a queue instead of being lost. Queuing theory predicts how long it waits and how many wait, given the arrival rate, service rate and number of servers.

Principles of queuing theory

 arrivals   +-------------+    +-----------+
 --lambda-->| queue/buffer|--->| N servers |--> out
            +-------------+    | rate mu   |
                               +-----------+

1. Arrival process: usually Poisson with mean rate $\lambda$ (exponential inter-arrival times).
2. Service process: holding/transmission time, often exponential with mean $1/\mu$.
3. Servers: N trunks, operators, processors or output links.
4. Queue capacity and discipline: finite or infinite buffer; FIFO, priority, etc.
5. Kendall notation A/B/C/K/N/D summarises the system (e.g. M/M/1, M/M/N).
6. Traffic intensity $A = \lambda/\mu$ Erlangs; utilisation per server $\rho = A/N$ must be below 1, otherwise the queue grows without limit.
7. Little's law: $L = \lambda W$ and $L_q = \lambda W_q$ (mean number = arrival rate × mean time).

Key delay-system results

• M/M/1: $W_q = \dfrac{\rho}{\mu(1-\rho)}$, $L_q = \dfrac{\rho^2}{1-\rho}$, $P(\text{wait}) = \rho$.
• M/M/N (Erlang C): probability of delay $C(N,A)$; mean delay of all calls $W = \dfrac{C(N,A)\,h}{N-A}$; probability of waiting longer than t: $C(N,A)\,e^{-(N-A)t/h}$.
• M/D/1 (fixed packet length): $W_q = \dfrac{\rho}{2\mu(1-\rho)}$, half that of M/M/1.

Applications in telecommunication networks

• Common-control exchanges: callers wait for dial tone (register/processor queue); designed so that, e.g., not more than 1.5% wait over 3 s.
• Call centres and operator services: number of agents chosen with Erlang C for a target service level.
• Packet networks (routers, switches, ATM): packets queue in output buffers; queuing delay, jitter and buffer overflow (loss) are predicted with M/M/1, M/D/1 or M/M/1/K.
• Signaling links (SS7) and processors: message delay versus link load.

How queuing theory helps manage delay and improve service

1. Dimensioning: finds the minimum number of servers or link capacity for a target mean delay or probability of delay.
2. Load limits: shows that delay rises sharply as $\rho \to 1$ (e.g. M/M/1 $W/h$ = 2 at $\rho = 0.5$ but 10 at $\rho = 0.9$), so links are kept at moderate utilisation.
3. Buffer sizing: chooses queue length K to balance packet loss and delay.
4. Scheduling and priority: priority queues give voice/video low delay while data waits (QoS).
5. Trunking efficiency: shows one large pooled server group gives less delay than several small separate groups.
6. Capacity planning and SLAs: predicts performance as traffic grows, guiding upgrades before service degrades.

Grade of Service (GoS)

In a loss system (blocked calls cleared, no queue), GoS is the proportion of calls offered in the busy hour that are lost:

$$GoS = \frac{\text{calls lost}}{\text{calls offered}} = \frac{A_o - A_c}{A_o}$$

It is a measured or design target set by the operator, e.g. 0.01 for trunk routes, and it describes the service seen by users over the whole network or a route.

Blocking probability

Blocking probability $P_B$ is the probability, calculated from a traffic model, that all N servers are busy when a call arrives. For Poisson traffic A offered to N trunks (Erlang B):

$$P_B = E_{1,N}(A) = \frac{A^N/N!}{\sum_{k=0}^{N}A^k/k!}$$

Two forms: time congestion (fraction of time all servers are busy) and call congestion (fraction of calls lost). With infinite Poisson sources they are equal, so $GoS = P_B$; with finite sources (Engset) call congestion is less than time congestion.

Use in design and evaluation

1. Dimensioning: forecast busy-hour traffic A, choose GoS (e.g. 0.01), then find N from Erlang B tables. Example: A = 5 E, GoS 0.01 gives N = 11 trunks.
2. End-to-end budget: a call through several links in tandem has total GoS ≈ $B_1 + B_2 + \dots$, so the overall target is split among links (e.g. 0.002 per switch, 0.005 per trunk).
3. Cost trade-off: lower GoS needs more circuits; designers pick the GoS that balances cost and customer satisfaction.
4. Routing design: high-usage direct routes are dimensioned with a high GoS and overflow to a final route with low GoS.
5. Evaluation: measured lost calls in the busy hour are compared with the target; if exceeded, trunks are added or traffic re-routed.
6. Cellular planning: number of channels per cell chosen for 2% blocking.

Grade of Service (GoS)

GoS is the proportion of calls lost (or delayed) in the busy hour because all servers are busy: $GoS = \dfrac{\text{calls lost}}{\text{calls offered}}$. Smaller value = better service.

Busy hour

The busy hour is the continuous 60-minute period of the day during which the traffic is maximum. Exchanges and trunks are dimensioned for busy-hour traffic so that GoS is met even at peak.

Numerical

Given: calls offered = 1400, calls lost = 14, h = 3 min = 1/20 h, T = 1 h.

(i) Traffic offered

$$\begin{aligned} A_o &= \frac{C_o\,h}{T} = \frac{1400 \times 3}{60} = 70\ \text{E} \end{aligned}$$

(ii) Grade of service

$$\begin{aligned} GoS &= \frac{14}{1400} = 0.01 \end{aligned}$$

(iii) Traffic carried Calls carried = 1400 − 14 = 1386

$$\begin{aligned} A_c &= \frac{1386 \times 3}{60} = 69.3\ \text{E} \end{aligned}$$

(Check: $A_c = A_o(1 - GoS) = 70 \times 0.99 = 69.3$ E.)

Answer: traffic offered = 70 E, GoS = 0.01 (1%), traffic carried = 69.3 E.

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 $\gamma$; mean holding time $h = 1/\mu$; $\alpha = \gamma h$ = offered traffic per idle source; full availability; lost calls cleared; equilibrium.

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

$$(S-n+1)\gamma P_{n-1} = n\mu P_n \;\Rightarrow\; P_n = \binom{S}{n}\alpha^n P_0$$

Normalising over $n = 0..N$:

$$P_n = \frac{\binom{S}{n}\alpha^n}{\sum_{k=0}^{N}\binom{S}{k}\alpha^k}$$

Time congestion (all N busy):

$$E = \frac{\binom{S}{N}\alpha^N}{\sum_{k=0}^{N}\binom{S}{k}\alpha^k}$$

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

$$B = \frac{\binom{S-1}{N}\alpha^N}{\sum_{k=0}^{N}\binom{S-1}{k}\alpha^k}$$

Notes

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

Traffic intensity

Traffic intensity is the average number of calls (servers) simultaneously in progress during a period, measured in Erlangs (E):

$$A = \frac{C\,h}{T}$$

where C = number of calls in time T and h = mean holding time. 1 E means one circuit continuously busy. Example: 120 calls of 3 min in one hour give $A = 120 \times 3/60 = 6$ E. Other unit: 1 E = 36 CCS (hundred call-seconds).

Grade of Service (GoS)

GoS is the proportion of calls that are lost (or delayed beyond a limit) in the busy hour because all servers are busy:

$$GoS = \frac{\text{calls lost}}{\text{calls offered}} = \frac{A_{lost}}{A_{offered}}$$

A smaller value means better service; e.g. 0.01 = 1 call lost in 100. It is used with Erlang B tables to find the number of trunks needed.

National numbering plan (ITU-T E.164 format)

A numbering plan gives every subscriber a unique number so that calls can be routed and charged. ITU-T E.164 sets the international format, with at most 15 digits (excluding prefixes):

 |<------------ max 15 digits ------------->|
 +------+----------+--------------------------+
 |  CC  |   NDC    |           SN             |
 +------+----------+--------------------------+
 1-3 dig  |<----- national (significant) no. ----->|

• CC – Country Code: 1 to 3 digits, given by ITU (Nepal 977, India 91, USA 1). First digit is the world zone.
• NDC – National Destination Code: area/trunk code or mobile network code (Kathmandu valley 1; Nepal Telecom mobile 984/985/986, Ncell 980/981/982).
• SN – Subscriber Number: identifies the line within the NDC area.
• NSN (National Significant Number) = NDC + SN; its length is fixed by the national plan (maximum 15 − length of CC).

Prefixes (not part of the number):

• Trunk (national) prefix: dialled before the NSN for calls inside the country, usually 0 (e.g. 01-4xxxxxx in Nepal).
• International prefix: dialled before CC for outgoing international calls, ITU recommends 00.
• Short codes for emergency and services (100 police, 101 fire, 102 ambulance in Nepal).

Example: a Kathmandu fixed line: international format +977 1 4xxxxxx; national format 01-4xxxxxx; local format 4xxxxxx. A Nepali mobile: +977 98X XXXXXXX.

Requirements of a good national plan: unique numbers, enough spare capacity for growth (planned for 30–50 years), easy routing and charging from the leading digits, uniform length where possible, compatibility with E.164 and with number portability.

Given: calls offered $C_o = 1000$, calls lost $C_l = 5$, mean holding time $h = 4\ \text{min} = \tfrac{1}{15}$ h, period T = 1 h.

Traffic in Erlangs $= \dfrac{\text{calls} \times h}{T}$.

(a) Traffic offered

$$\begin{aligned} A_o &= \frac{C_o\,h}{T} = 1000 \times \tfrac{1}{15} \\ &= 66.67\ \text{E} \end{aligned}$$

(b) Traffic carried

Calls carried $= 1000 - 5 = 995$

$$\begin{aligned} A_c &= 995 \times \tfrac{1}{15} = 66.33\ \text{E} \end{aligned}$$

(c) Traffic lost

$$\begin{aligned} A_l &= A_o - A_c = 5 \times \tfrac{1}{15} = 0.33\ \text{E} \end{aligned}$$

(d) Grade of service

$$\begin{aligned} GoS &= \frac{C_l}{C_o} = \frac{5}{1000} = 0.005 \end{aligned}$$

(e) Duration of congestion

For pure-chance (Poisson) traffic the proportion of time all trunks are busy (time congestion) equals the proportion of calls lost (call congestion), so congestion time in the busy hour is

$$\begin{aligned} T_c &= GoS \times 3600\ \text{s} = \frac{5}{1000} \times 3600 \\ &= 18\ \text{s} \end{aligned}$$

Answer: (a) offered = 66.67 E (66 2/3 E), (b) carried = 66.33 E (66 1/3 E), (c) lost = 0.33 E (1/3 E), (d) GoS = 0.005, (e) congestion period = 18 s in the busy hour.

Blocking formulas for infinite sources

With a very large number of sources, calls arrive as a Poisson process of constant rate and offered traffic A is independent of the number of busy servers. With N servers:

1. Lost calls cleared – Erlang B formula (blocked calls disappear):

$$B = E_{1,N}(A) = \frac{A^N/N!}{\sum_{k=0}^{N}A^k/k!}$$

Used for trunk groups; recursive form $E_{1,N} = \dfrac{A\,E_{1,N-1}}{N + A\,E_{1,N-1}}$, $E_{1,0}=1$.

2. Lost calls held – Poisson formula (blocked callers keep trying for a time equal to the holding time):

$$P = e^{-A}\sum_{k=N}^{\infty}\frac{A^k}{k!}$$

Gives slightly higher blocking than Erlang B; used in North American trunk design.

3. Lost calls delayed – Erlang C formula (blocked calls wait in an infinite queue):

$$C(N,A) = \frac{\dfrac{A^N}{N!}\dfrac{N}{N-A}}{\sum_{k=0}^{N-1}\dfrac{A^k}{k!} + \dfrac{A^N}{N!}\dfrac{N}{N-A}}, \quad A < N$$

Mean delay of all calls $= C(N,A)\,h/(N-A)$.

Numerical

Given: observation period T = 20 min = 1200 s, number of subscribers = 40 (each initiates a call), total call duration = 4800 s.

Offered load = total holding time / observation time:

$$\begin{aligned} A &= \frac{\sum h_i}{T} = \frac{4800}{1200} = 4\ \text{E} \end{aligned}$$

Average traffic per subscriber:

$$\begin{aligned} a &= \frac{A}{40} = \frac{4}{40} = 0.1\ \text{E} \end{aligned}$$

(Supporting values: call rate $\lambda = 40/20 = 2$ calls/min; mean holding time $h = 4800/40 = 120$ s $= 2$ min; $A = \lambda h = 2 \times 2 = 4$ E.)

Answer: offered load = 4 E; average subscriber traffic = 0.1 E (each subscriber busy 10% of the time).

GoS vs blocking probability

• Grade of Service (GoS): the fraction of offered calls that are lost in the busy hour, $GoS = \dfrac{\text{calls lost}}{\text{calls offered}}$. It is a service-quality target or measurement for a route or the whole connection.
• Blocking probability ($P_B$): the probability, from a traffic model, that all servers are busy (time congestion) or that an arriving call is blocked (call congestion), e.g. from Erlang B.

For Poisson traffic in a loss system the two are numerically equal.

Given: calls offered $C_o = 600$, calls lost $C_l = 40$, mean holding time $h = 2.5\ \text{min} = \tfrac{1}{24}$ h, period T = 1 h.

Traffic in Erlangs $= \dfrac{\text{calls} \times h}{T}$.

(a) Traffic offered

$$\begin{aligned} A_o &= \frac{C_o\,h}{T} = 600 \times \tfrac{1}{24} \\ &= 25\ \text{E} \end{aligned}$$

(b) Traffic carried

Calls carried $= 600 - 40 = 560$

$$\begin{aligned} A_c &= 560 \times \tfrac{1}{24} = 23.33\ \text{E} \end{aligned}$$

(c) Traffic lost

$$\begin{aligned} A_l &= A_o - A_c = 40 \times \tfrac{1}{24} = 1.67\ \text{E} \end{aligned}$$

(d) Grade of service

$$\begin{aligned} GoS &= \frac{C_l}{C_o} = \frac{40}{600} = 0.0667 \end{aligned}$$

(e) Duration of congestion

For pure-chance (Poisson) traffic the proportion of time all trunks are busy (time congestion) equals the proportion of calls lost (call congestion), so congestion time in the busy hour is

$$\begin{aligned} T_c &= GoS \times 3600\ \text{s} = \frac{40}{600} \times 3600 \\ &= 240\ \text{s} \end{aligned}$$

Answer: offered traffic = 25 E, carried = 23.33 E, lost = 1.67 E, GoS = 0.0667, congestion duration = 240 s (4 min).

Traffic intensity

Average number of simultaneous calls in a period, in Erlangs: $A = \dfrac{C\,h}{T}$ (C calls, mean holding time h, period T). 1 E = one circuit busy all the time.

Grade of Service

Proportion of calls lost (or excessively delayed) in the busy hour: $GoS = \dfrac{\text{calls lost}}{\text{calls offered}}$; smaller is better.

Numerical

Call durations are taken as negative exponential with mean $h = 2$ min:

$$P(T > t) = e^{-t/h}$$

The exponential distribution is memoryless: the remaining duration of a call does not depend on how long it has already lasted. So the fact that the call has lasted 4 min does not change the answer.

(a) Lasts at least another 4 min

$$\begin{aligned} P(T > 8 \mid T > 4) &= P(T > 4) = e^{-4/2} \\ &= e^{-2} = 0.1353 \end{aligned}$$

(b) Ends within the next 4 min

$$\begin{aligned} P(T \le 8 \mid T > 4) &= 1 - e^{-2} \\ &= 0.8647 \end{aligned}$$

Answer: (a) 0.1353 (about 13.5%), (b) 0.8647 (about 86.5%).

The device is a single server and calls that find it busy wait (delay system). With random (Poisson) calls and exponential holding time this is an M/M/1 (Erlang delay, N = 1) system. The "10 ms" is the mean delay before a call is served (averaged over all calls):

$$W = \frac{A\,h}{1 - A}$$

where $h$ = mean holding time and $A = \lambda h$ = traffic (occupancy) of the device.

(i) h = 50 ms, W = 10 ms

$$\begin{aligned} \frac{A \times 50}{1-A} &= 10 \\ 50A &= 10 - 10A \\ A &= \frac{10}{60} = \frac{1}{6}\ \text{E} \end{aligned}$$

Calls per hour:

$$\begin{aligned} C &= \frac{A\,T}{h} = \frac{(1/6) \times 3600}{0.05} \\ &= 12000\ \text{calls/h} \end{aligned}$$

(Without the delay limit, a fully occupied device could take $3600/0.05 = 72000$ calls/h, but delay would then be infinite.)

(ii) 18000 calls/h, W = 10 ms

$\lambda = 18000/3600 = 5$ calls/s, so $A = 5h$ (h in seconds).

$$\begin{aligned} \frac{\lambda h \cdot h}{1 - \lambda h} &= 0.01 \\ 5h^2 &= 0.01(1 - 5h) \\ 5h^2 + 0.05h - 0.01 &= 0 \\ h &= \frac{-0.05 + \sqrt{0.05^2 + 4(5)(0.01)}}{2 \times 5} \\ &= \frac{-0.05 + 0.45}{10} = 0.04\ \text{s} \end{aligned}$$

Check: $A = 5 \times 0.04 = 0.2$; $W = 0.2 \times 40/(1-0.2) = 10$ ms.

Answer: (i) 12,000 calls per hour; (ii) maximum average holding time h = 40 ms.

Kendall–Lee notation

Erlang's delay model is

$$(M/M/N):(FCFS/\infty/\infty)$$

Poisson arrivals (rate $\lambda$), exponential holding time (mean $h = 1/\mu$), N servers, first-come-first-served, unlimited waiting room and infinite sources. Offered traffic $A = \lambda/\mu = \lambda h$ Erlangs, with $A < N$ for stability.

State diagram

State k = number of calls in the system (in service + waiting).

   lam     lam           lam      lam      lam
 0 ---> 1 ---> 2 ... N-1 ---> N ---> N+1 ---> ...
   <---   <---           <---     <---     <---
    mu    2mu            N mu     N mu     N mu

Arrival rate is always $\lambda$. Service rate is $k\mu$ for $k \le N$ and $N\mu$ for $k > N$ (only N servers).

Derivation

Balance between neighbouring states: $\lambda P_{k-1} = k\mu P_k$ for $k \le N$, and $\lambda P_{k-1} = N\mu P_k$ for $k > N$.

$$\begin{aligned} P_k &= \frac{A^k}{k!}P_0, & k \le N \\ P_k &= \frac{A^N}{N!}\left(\frac{A}{N}\right)^{k-N}P_0, & k > N \end{aligned}$$

Normalisation $\sum P_k = 1$ (the geometric series converges since $A/N < 1$):

$$\begin{aligned} P_0^{-1} &= \sum_{k=0}^{N-1}\frac{A^k}{k!} + \frac{A^N}{N!}\sum_{j=0}^{\infty}\left(\frac{A}{N}\right)^j \\ &= \sum_{k=0}^{N-1}\frac{A^k}{k!} + \frac{A^N}{N!}\,\frac{N}{N-A} \end{aligned}$$

A call is delayed ("blocked") when it finds all N servers busy, i.e. $k \ge N$:

$$\begin{aligned} P(W>0) &= \sum_{k=N}^{\infty}P_k = \frac{A^N}{N!}\,\frac{N}{N-A}\,P_0 \end{aligned}$$

Erlang C (delay) formula:

$$E_{2,N}(A) = \frac{\dfrac{A^N}{N!}\dfrac{N}{N-A}}{\displaystyle\sum_{k=0}^{N-1}\frac{A^k}{k!} + \frac{A^N}{N!}\frac{N}{N-A}}$$

It can also be written using Erlang B: $E_{2,N} = \dfrac{N\,E_{1,N}}{N - A(1 - E_{1,N})}$.

Related results

• Mean delay of all calls: $W = E_{2,N}(A)\,\dfrac{h}{N-A}$
• Mean delay of delayed calls: $\dfrac{h}{N-A}$
• $P(\text{wait} > t) = E_{2,N}(A)\,e^{-(N-A)t/h}$
• For N = 1: $E_{2,1} = A$, $W = \dfrac{A\,h}{1-A}$ (M/M/1).

Given: offered traffic A = 40 E, total time all trunks busy = 20 s in the busy hour (3600 s), calls lost = 4.

For pure-chance traffic, the probability that a call is lost (call congestion) equals the fraction of time all trunks are busy (time congestion):

$$\begin{aligned} B &= \frac{20}{3600} = \frac{1}{180} \end{aligned}$$

(i) Average number of calls carried

Calls lost = B × calls offered, so

$$\begin{aligned} C_o &= \frac{\text{calls lost}}{B} = 4 \times 180 = 720\ \text{calls} \\ C_c &= C_o - C_l = 720 - 4 = 716\ \text{calls} \end{aligned}$$

(ii) Average call duration

$$\begin{aligned} A &= \frac{C_o\,h}{T} \Rightarrow h = \frac{A\,T}{C_o} \\ h &= \frac{40 \times 3600}{720} = 200\ \text{s} = 3.33\ \text{min} \end{aligned}$$

(iii) Calls offered in a period equal to h

Call arrival rate $= 720/3600$ calls per second. In 200 s:

$$\begin{aligned} n &= \frac{720}{3600} \times 200 = 40\ \text{calls} \end{aligned}$$

This equals the offered traffic in Erlangs, which is the definition of the Erlang: traffic in E = mean number of calls arriving during one mean holding time ($A = \lambda h$).

Answer: (i) 716 calls carried (of 720 offered), (ii) h = 200 s (3.33 min), (iii) calls offered in one holding time = 40 = A, as required.

Tele-traffic models are classified by the number of traffic sources compared with the number of servers N.

Infinite-source models

• Number of sources is very large (S ≫ N), so the call arrival rate $\lambda$ is constant and independent of busy sources; arrivals are Poisson.
• Offered traffic $A = \lambda h$ is fixed.
• Formulas: Erlang B (lost calls cleared), Poisson (lost calls held), Erlang C (lost calls delayed).
• Erlang B: $B = \dfrac{A^N/N!}{\sum_{k=0}^{N}A^k/k!}$
• Used for junction routes between large exchanges.

Finite-source models

• Small number of sources S (comparable to N). A busy source cannot make a new call, so the arrival rate $(S-n)\gamma$ falls as more servers are busy.
• Formulas: Engset (S > N, lost calls cleared), Bernoulli/binomial (S ≤ N, no loss).
• Engset call congestion: $B = \dfrac{\binom{S-1}{N}\alpha^N}{\sum_{k=0}^{N}\binom{S-1}{k}\alpha^k}$, $\alpha$ = traffic per idle source.
• Used for PBXs, concentrators, small rural exchanges.

Given: offered traffic A = 30 E, total time all trunks busy = 12 s in the busy hour (3600 s), calls lost = 2.

For pure-chance traffic, the probability that a call is lost (call congestion) equals the fraction of time all trunks are busy (time congestion):

$$\begin{aligned} B &= \frac{12}{3600} = \frac{1}{300} \end{aligned}$$

(i) Average number of calls carried

Calls lost = B × calls offered, so

$$\begin{aligned} C_o &= \frac{\text{calls lost}}{B} = 2 \times 300 = 600\ \text{calls} \\ C_c &= C_o - C_l = 600 - 2 = 598\ \text{calls} \end{aligned}$$

(ii) Average call duration

$$\begin{aligned} A &= \frac{C_o\,h}{T} \Rightarrow h = \frac{A\,T}{C_o} \\ h &= \frac{30 \times 3600}{600} = 180\ \text{s} = 3\ \text{min} \end{aligned}$$

(iii) Calls offered in a period equal to h

Call arrival rate $= 600/3600$ calls per second. In 180 s:

$$\begin{aligned} n &= \frac{600}{3600} \times 180 = 30\ \text{calls} \end{aligned}$$

This equals the offered traffic in Erlangs, which is the definition of the Erlang: traffic in E = mean number of calls arriving during one mean holding time ($A = \lambda h$).

Answer: (i) 598 calls carried (of 600 offered), (ii) h = 180 s (3 min), (iii) calls offered in one holding time = 30 = A, as required.

Definitions

• Blockage (blocking): the condition in which a call attempt cannot be connected because all suitable servers (trunks, switch paths) are busy; its probability is the blocking probability.
• Lost call: a blocked call in a loss system that is cleared and not served.
• Grade of Service (GoS): proportion of offered calls lost in the busy hour, $GoS = \dfrac{\text{calls lost}}{\text{calls offered}}$; a smaller value means better service.

Given: calls offered $C_o = 800$, calls lost $C_l = 50$, mean holding time $h = 3\ \text{min} = \tfrac{1}{20}$ h, period T = 1 h.

Traffic in Erlangs $= \dfrac{\text{calls} \times h}{T}$.

(a) Traffic offered

$$\begin{aligned} A_o &= \frac{C_o\,h}{T} = 800 \times \tfrac{1}{20} \\ &= 40\ \text{E} \end{aligned}$$

(b) Traffic carried

Calls carried $= 800 - 50 = 750$

$$\begin{aligned} A_c &= 750 \times \tfrac{1}{20} = 37.5\ \text{E} \end{aligned}$$

(c) Traffic lost

$$\begin{aligned} A_l &= A_o - A_c = 50 \times \tfrac{1}{20} = 2.5\ \text{E} \end{aligned}$$

(d) Grade of service

$$\begin{aligned} GoS &= \frac{C_l}{C_o} = \frac{50}{800} = 0.0625 \end{aligned}$$

(e) Duration of congestion

For pure-chance (Poisson) traffic the proportion of time all trunks are busy (time congestion) equals the proportion of calls lost (call congestion), so congestion time in the busy hour is

$$\begin{aligned} T_c &= GoS \times 3600\ \text{s} = \frac{50}{800} \times 3600 \\ &= 225\ \text{s} \end{aligned}$$

Answer: traffic offered = 40 E, carried = 37.5 E, lost = 2.5 E, GoS = 0.0625, congestion duration = 225 s (3.75 min).

GoS and blocking probability in a loss system

• Grade of Service (GoS): ratio of calls lost to calls offered in the busy hour, $GoS = \dfrac{\text{calls lost}}{\text{calls offered}}$. It is the quality target set for a route or for the end-to-end connection (e.g. 0.01).
• Blocking probability ($P_B$): probability, given by a traffic model, that all N servers are busy when a call arrives. For Poisson traffic, Erlang B: $P_B = \dfrac{A^N/N!}{\sum_{k=0}^{N}A^k/k!}$.

For infinite Poisson sources, call congestion = time congestion, so GoS = $P_B$ for a single group; for links in tandem GoS ≈ $\sum P_{B,i}$.

National numbering plan (ITU-T E.164 format)

A numbering plan gives every subscriber a unique number so that calls can be routed and charged. ITU-T E.164 sets the international format, with at most 15 digits (excluding prefixes):

 |<------------ max 15 digits ------------->|
 +------+----------+--------------------------+
 |  CC  |   NDC    |           SN             |
 +------+----------+--------------------------+
 1-3 dig  |<----- national (significant) no. ----->|

• CC – Country Code: 1 to 3 digits, given by ITU (Nepal 977, India 91, USA 1). First digit is the world zone.
• NDC – National Destination Code: area/trunk code or mobile network code (Kathmandu valley 1; Nepal Telecom mobile 984/985/986, Ncell 980/981/982).
• SN – Subscriber Number: identifies the line within the NDC area.
• NSN (National Significant Number) = NDC + SN; its length is fixed by the national plan (maximum 15 − length of CC).

Prefixes (not part of the number):

• Trunk (national) prefix: dialled before the NSN for calls inside the country, usually 0 (e.g. 01-4xxxxxx in Nepal).
• International prefix: dialled before CC for outgoing international calls, ITU recommends 00.
• Short codes for emergency and services (100 police, 101 fire, 102 ambulance in Nepal).

Example: a Kathmandu fixed line: international format +977 1 4xxxxxx; national format 01-4xxxxxx; local format 4xxxxxx. A Nepali mobile: +977 98X XXXXXXX.

Requirements of a good national plan: unique numbers, enough spare capacity for growth (planned for 30–50 years), easy routing and charging from the leading digits, uniform length where possible, compatibility with E.164 and with number portability.

Random call arrivals follow the Poisson distribution: the probability of exactly k calls in time t is

$$P(k) = \frac{(\lambda t)^k}{k!}\,e^{-\lambda t}$$

Given: one call every 5 s on average, so $\lambda = 0.2$ calls/s; t = 10 s.

$$\lambda t = 0.2 \times 10 = 2$$

(i) No call

$$P(0) = e^{-2} = 0.1353$$

(ii) One call

$$P(1) = \frac{2^1}{1!}e^{-2} = 2 \times 0.1353 = 0.2707$$

(iii) Two calls

$$P(2) = \frac{2^2}{2!}e^{-2} = 2 \times 0.1353 = 0.2707$$

(iv) More than two calls

$$\begin{aligned} P(k>2) &= 1 - [P(0) + P(1) + P(2)] \\ &= 1 - (0.1353 + 0.2707 + 0.2707) \\ &= 1 - 0.6767 = 0.3233 \end{aligned}$$

Answer: (i) 0.1353, (ii) 0.2707, (iii) 0.2707, (iv) 0.3233.

Formulas in the traffic engineering decision tree

The correct formula is chosen by two questions: are the sources infinite or finite, and what happens to blocked calls — cleared (LCC), held (LCH) or delayed (LCD)?

                 Traffic sources
           +-------------+-------------+
       Infinite                     Finite
   +-------+-------+          +-------+-------+
  LCC     LCH     LCD        LCC     LCH     LCD
Erlang B Poisson Erlang C   Engset  Binomial Finite
                                            queue

Blocking formulas for infinite sources

Assumptions: Poisson arrivals, exponential holding times, N servers, full availability, offered traffic $A$ in Erlangs.

1. Erlang B (lost calls cleared)

$$B = \frac{A^N/N!}{\sum_{k=0}^{N}A^k/k!}$$

Blocked calls vanish. Most widely used for trunk dimensioning (Erlang B tables).

2. Poisson (lost calls held)

$$P = e^{-A}\sum_{k=N}^{\infty}\frac{A^k}{k!}$$

Blocked calls are assumed to stay in the system (keep retrying) for a holding time; gives slightly higher blocking.

3. Erlang C (lost calls delayed)

$$C = \frac{\dfrac{A^N}{N!}\dfrac{N}{N-A}}{\sum_{k=0}^{N-1}\dfrac{A^k}{k!} + \dfrac{A^N}{N!}\dfrac{N}{N-A}}$$

Probability that a call has to wait; mean wait $= C\,h/(N-A)$. Used for call centres and common-control equipment.

Given: calls offered $C_o = 1200$ in the busy hour, calls lost $C_l = 6$, mean holding time $h = 3$ min, observation period $T = 60$ min.

Traffic intensity (in erlang) is the number of calls in the period multiplied by the mean holding time, divided by the period:

$$A = \frac{C\,h}{T}$$

a) Traffic offered

$$\begin{aligned} A_o &= \frac{1200 \times 3}{60} \\ &= 60\ \text{E} \end{aligned}$$

b) Traffic carried

Calls carried $= 1200 - 6 = 1194$.

$$\begin{aligned} A_c &= \frac{1194 \times 3}{60} \\ &= 59.7\ \text{E} \end{aligned}$$

c) Traffic lost

$$\begin{aligned} A_l &= A_o - A_c = 60 - 59.7 \\ &= 0.3\ \text{E} \end{aligned}$$

(Check: $6 \times 3 / 60 = 0.3$ E.)

d) Grade of service

GoS is the fraction of offered calls (or offered traffic) that is lost:

$$\begin{aligned} B &= \frac{\text{calls lost}}{\text{calls offered}} = \frac{6}{1200} \\ &= 0.005\ (\text{i.e. } 0.5\%) \end{aligned}$$

e) Total duration of the periods of congestion

For random (Poisson) traffic, the call congestion equals the time congestion, so the trunk group is fully busy for a fraction $B$ of the hour:

$$\begin{aligned} t_{cong} &= B \times T = 0.005 \times 3600\ \text{s} \\ &= 18\ \text{s} \end{aligned}$$

Answer: traffic offered = 60 E, traffic carried = 59.7 E, traffic lost = 0.3 E, GoS = 0.005 (0.5%), total congestion time = 18 s in the busy hour.

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), seen by an arriving call from one of the $N-1$ other 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, so

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

The probability that more than $n$ sources want service gives the congestion: $P(\text{blocking}) = \sum_{k=n+1}^{N}\binom{N}{k}a^k(1-a)^{N-k}$. If $N \le n$ there is no blocking at all.

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