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Heuristic arguments for Erlang

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  • #36749 Reply

    I know that i am entering into a debate would definitly look for some brain sorming here.

    As a basic principle, grade of service is defined as
    GOS = A / SIGMA B

    A = (( Traffic in Erlangs)^ n) / n!

    SIGMA B = (( Traffic in Erlangs)^ k) / k! ; k = 0 to k=n

    Where n = Total no. of Circuits (or channels) avaible.

    So we keep the erlangs same and reduce the GOS, the number of channels required to serve this, actually increase. To further give direction to this topic i will take a case where to serve 1000 erlangs @ 0.05 GoS, i need 966 circuits (less then the number of erlangs which is acceptable as i am allowing some calls to get blocked). Now if i reduce the GoS to 0.001, for the same erlangs, i require 1017 ckts. This is where my problem starts. Now the number of ckts is actuall more then the served erlangs. What is the logical or Heuristic argument for this. The formula explains this fine but why it is so. there is another definition also that will further clarify my question. One erlang of traffi9c is equivalent to one ckt being used for one hour (and no blocking). Why do we require more number of circuits (1017ckts is > 1000 Erlangs) to serve less erlangs where we are still allowing some calls to block??

    Any comments??

    #36750 Reply

    Some comments : in a Radio Mobile Network (ex: GSM), you are dealing with a very low quantity of radio channels. Typical example : n = 7 (case of a microcell). Usually the design GoS is 2%. In that case you can only carry an offered traffic of 2,9 erlang !

    This is due to the fact incoming trafic is coming by burst (Poisson Law is behind erlang law), and even if you have enough channels to carry the average trafic, you need extra channels to grant burst without extra blocking.

    And if you want a very low blocking rate (GoS), you need to pay that by having a lot of extra channels.

    #36751 Reply

    .001 is not a reduced grade of service over 0.05. It is a much better GOS. Instead of permitting 5 % of your calls to be blocked you are allowing only 0.1% to be blocked therefor more calls are ‘guaranteed’ to get through, so you need more circuits.

    The number of circuits goes above the Erlang total because the individual calls average only several minutes (usually). You must allow for a small set-up and clear down time where the circuit is busy but not carrying traffic.

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