Copyright Douglas Rice,2005,doug.h.rice@btinternet.com, last modified 10/06/2012 08:31:48 terms of use The Author gives no warranty of any kind in relation to these Webpages. To the maximum extent permitted by law, the Author Douglas Rice, will not be liable for any loss or damage which you may suffer as a result of or connected to the download or use of these Web pages.

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Erlang Calculation Form

Use this page to calculate Erlang Proability.

This page is influenced by Chapter 4 and equ 4.3, p88, mt hills, Telecommunications Switching Principles. ISBN 0 04 621029 6

The probability that k circuits are in use when there are A Erlangs of traffic, or also the probability that k calls start in a second given an average call rate of A calls per second is:

P(k)=A^k/k!*exp(-A)

Use: A^k/k! = (A/k)*(A/(k-1)*..*(A/(1))..) to prevent working out a^k and k!, which over range floats quickly

This tends towards exp( A ) * exp( -A ) which equals 1 if enough k terms are used.

  IAT = InterArrivalTime
  rand() returns a random number between 0 and 1, so it is a probability.

  InterArrivalTime,IAT = -ln( 1-rand() )/AverageCallRate

  exp ( - IAT * CallRate ) = 1 - rand()

  rand( ) = 1 - exp( -IAT * Call Rate )  

There is a question of how many calls do you need to get an accurate average calls per second. This is discussed at the bottom of the page.




Traffic, Erlangs,A recalc

recalc

Active Circuits probability
of this number
of circuits in a call
A^k/k!/exp(A)
accum Prob GOS
0 0.006737946999085467 0.006737946999085467 1.0067836549063043
1 0.03368973499542734 0.040427681994512805 1.0421309381647685
2 0.08422433748856833 0.12465201948308115 1.142402818373409
3 0.14037389581428059 0.26502591529736175 1.3605921906819178
4 0.17546736976785074 0.4404932850652124 1.787288647852159
5 0.17546736976785074 0.6159606548330632 2.603899867513088
6 0.14622280813987562 0.7621834629729388 4.204922048319162
7 0.10444486295705402 0.866628325929993 7.497843953545174
8 0.06527803934815876 0.9319063652781516 14.68566047450456
9 0.036265577415643756 0.9681719426937953 31.418819891500387
10 0.018132788707821878 0.9863047314016171 73.01791803616605
11 0.008242176685373581 0.9945469080869908 183.38220150192885
12 0.003434240285572326 0.9979811483725631 495.3311013101041
13 0.0013208616482970486 0.9993020100208602 1432.6853248414054
14 0.00047173630296323165 0.9997737463238233 4419.8176882620355
15 0.0001572454343210772 0.9999309917581445 14491.022711369818
16 0.00004913919822533663 0.9999801309563697 50329.54874971618
17 0.000014452705360393125 0.99999458366173 184626.57798501328
18 0.00000401464037788698 0.9999985983021079 713420.4921356859
19 0.0000010564843099702577 0.999999654786418 2896757.4046510793
20 2.6412107749256443e-7 0.9999999189074955 12331595.950812679
21 6.288597083156296e-8 0.9999999817934664 54925337.31009707
22 1.429226609808249e-8 0.9999999960857323 255475629.10130054
23 3.1070143691483673e-9 0.9999999991927468 1238768775.3474364
24 6.472946602392432e-10 0.9999999998400415 6251621870.087544
25 1.2945893204784864e-10 0.9999999999695005 32787431537.1967
26 2.4895948470740124e-11 0.9999999999943964 178455793290.29367
27 4.610360827914837e-12 0.9999999999990067 1006728429053.4248
28 8.232787192705067e-13 0.9999999999998299 5879372881684.721
29 1.4194460677077702e-13 0.9999999999999718 35461414388744.06
30 2.365743446179617e-14 0.9999999999999956 225179981368524.8
31 3.815715235773576e-15 0.9999999999999994 1801439850948198.5
32 5.962055055896213e-16 1 Infinity
33 9.033416751357898e-17 1 Infinity
34 1.3284436399055733e-17 1 Infinity
35 1.897776628436533e-18 1 Infinity
36 2.635800872828519e-19 1 Infinity
37 3.5618930713898905e-20 1 Infinity
38 4.68670140972354e-21 1 Infinity
39 6.008591550927615e-22 1 Infinity
40 7.510739438659518e-23 1 Infinity
41 9.15943833982868e-24 1 Infinity
42 1.090409326170081e-24 1 Infinity
43 1.2679178211280012e-25 1 Infinity
44 1.440815705827274e-26 1 Infinity
45 1.6009063398080822e-27 1 Infinity
46 1.7401155867479155e-28 1 Infinity
47 1.851186794412676e-29 1 Infinity
48 1.928319577513204e-30 1 Infinity
49 1.9676730382787798e-31 1 Infinity
50 1.96767303827878e-32 1 Infinity
51 1.9290912139988038e-33 1 Infinity
52 1.8548953980757731e-34 1 Infinity
53 1.7499013189394087e-35 1 Infinity
54 1.620278999017971e-36 1 Infinity
55 1.4729809081981555e-37 1 Infinity
56 1.3151615251769246e-38 1 Infinity
57 1.1536504606815128e-39 1 Infinity
58 9.945262592082006e-41 1 Infinity
59 8.428188637357632e-42 1 Infinity
60 7.023490531131359e-43 1 Infinity
61 5.756959451747015e-44 1 Infinity
62 4.642709235279851e-45 1 Infinity
63 3.6846898692697225e-46 1 Infinity
64 2.878663960366971e-47 1 Infinity
65 2.2143568925899776e-48 1 Infinity
66 1.6775431004469528e-49 1 Infinity
67 1.2518978361544421e-50 1 Infinity
68 9.205131148194428e-52 1 Infinity
69 6.670384889995962e-53 1 Infinity
70 4.7645606357114016e-54 1 Infinity
71 3.3553243913460576e-55 1 Infinity
72 2.330086382879207e-56 1 Infinity
73 1.5959495773145252e-57 1 Infinity
74 1.078344308996301e-58 1 Infinity
75 7.188962059975338e-60 1 Infinity
76 4.729580302615354e-61 1 Infinity
77 3.0711560406593207e-62 1 Infinity
78 1.9686897696534103e-63 1 Infinity
79 1.2460061833249434e-64 1 Infinity





Copyright Douglas Rice, doug.h.rice@btinternet.com , 2005,2008, last modified 10/06/2012 08:31:48


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There is a question of how many calls do you need to get an accurate average calls per second.

You can work this out by working out the interarrival times for a number of calls and working out the average call rate of those calls, and ask by what percentage change to the average call rate will the next call make. I think that for 1 % of need about 700 calls.

We use a stated Average call rate A, work out an InterArrivalTime IAT, the measured Average call rate is 1/IAT.

Work this out for say 100 calls or n calls, work out n IATs, the measured Average call rate is n/sum(IAT).

How many calls are required to get measured AverageCallRate/StatedAverageCallRate so this is 1 +/- 0.01?

Inter Arrival Times and independence

Given a hours worth of calls where we see say 3600 calls from a population of say 10,000,000 possible callers, we can say we have an average call rate of 1 call per seconds, possibly with any one caller making one call, totally without knowledge of anybody else making a call.

If you have a sequence of calls from independent calls, you could argue they should have no knowledge of who is going to make the next call.

If you see a sequence of evenly spaced calls, the callers must have a knowledge of the previous calls to know when to make their call so that it is spaced correctly from the last call.

If you see a sequence of evenly spaced calls, the callers must have a knowledge of the previous calls to know when to make their call so that it is spaced correctly from the last call one seconds on. They must also agree with the rest of the population who is going to make the next call.