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# Minute Erlang Per drop

Viewing 10 posts - 1 through 10 (of 10 total)
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• #64966
fresher
Guest

Hi All Expert,

Please Explain me What is Minute Erlang per drop.

#64967
Bijoy
Guest

Hi fresher
It means the average Erlang mins carried between 2 successive call drops in the cell.

Br\\
Bijoy

#64968
Fresh Opt
Guest

Hi,
and what is the importance of minuteErl/call drop? What tells us?
Regards,
Fresh Opt

#64969
pix
Guest

hi fresh, fresher and freshest,

it shows the “density” of drops in the cell. In the end, it is just another way to estimate the call drop rate in the cell.
20 erlangs
10 drops
= 2 erl/drop = 120 minute-erl/drop

There is 1 drop every 2 hours of communication in the cell.

Now let’s assume that 1 call = 1 minute
There are 20 erlangs, that’s 20 hours = 1,200 minutes = 1,200 calls
CDR = 10/1,200 = 0.89%

There is a proportional relationship between min-erl/drop and the call drop rate. You could as well use the following formula :
drops / (erlang * 3600 / tch_average_duration [s])

the main difference is that if TCH are shorter than usual (due to HO for example), it means there are more calls in the cell, but still the same amount of drops : you drop rate decreases.
If TCH are longer than usual, the drop rate increases. But a “drop” KPI should not be impacted by TCH duration.

Whereas the minute-erlang per drop remains stable whatever the duration of the TCH.
It looks like the min-erlang per drop is a more objective approach than the call drop rate, especially in networks with a high amount of handovers. Problematic cells with high amount of HO have shorter TCH, therefore they have a better call drop rate than the subscribers are experiencing.

Well, while writing this post I convinced myself that erl/drop is useful :)) I am a manipulative master….

Regards 🙂
pix

#64970
el mariachi
Guest

hello Pix,
drops / (erlang * 3600 / tch_average_duration [s])

is
drops:number of drops
erlang: TCH traffic carried
tch_average_duration [s]: call duration

#64971
jack
Guest

hi pix.
i think that Drop/Erlang is more accurate for this case.

any way,suppose this situation:
Drop/Erlang=0.639999985694885
TCH Traffic=23.4300003051757
HTCH T raffic=18.1299991607666
Defined Tch=Avail Tch=22
Avail HTCH=43
TCH MHT=13.43
Call attempt TCH=2164

NOTE:this measurement is done in 1 Houre.
according your formula,what is number of drop calls in this 1 houre?

#64972
pix
Guest

el mariachi,

yes, that’s correct.
The real formula of “minute erl per drop” is “erlang*60 / number of drops”
(if measurements done in one hour)
my formula is just a trick, to study the real formula.

jack,

drop/erl = 0.64
FR traffic = 23.43 erl
HR traffic = 18.12 erl
–> drop = 0.64*(23.43+18.12) = 26.6 drops

with “my” stupid formula, you should get:
drop/calls = 26.6drops / [(23.43erl+18.12erl)*3600/13.43s] = 26.6/11138 = 0.2%

In your inputs, you say that there are only 2164 call attempts, but I find “11,138” tch attempts. I guess the difference is due to high amount of incoming HO in the cell.
Look at the very short tch duration avg… it’s very bad !! 13 seconds !!

regards
pix

#64973
AK singh
Guest

What Is Earlang per drop

#64974
pix
Guest

yes, that’s the perfect title for what’s written below.