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15th July 2010 at 12:13 #63602PaulGuest
Is there anyone know why the indicator of throughput kbit/s per tbf has higher value than that of kbit/s per PDCH? This indicator was shown in ALU GPRS Telecom NPO.
As we know, 1 user has 1 TBF (minimum), but also can have more than 1 TBFs (of course it gives higher throughput). In ALU parameter, we have MAX_DL_TBF_per SPDCH,also MAX_UL_TBF_per_SPDCH and MAX_PDCH_per_TBF.
What is the correlation with throughput?
Paul15th July 2010 at 13:36 #63603PixGuest
I think you are confusing the terms. The TBF is the group of PDCHs allocated to one user.
The PDCH is one GPRS timeslot.
The throughput of the PDCH depends on the coding schemes. If 2 users are sharing one PDCH, with user 1 using it 50% of the time at 30kbps and the other user using it 50% of the time at 40kbps, then the PDCH throughput is 35kbps.
One user can have for example 4 PDCH at 50kbps each.? In this case, the TBF throughput is 200kbps.
Hope it clarifies the concept.
pix17th July 2010 at 15:16 #63604ALUGuest
What about the indicator “Kbit/sec per cell”(GTRPDDCLA GPRS_DL_useful_throughput_radio_cell_avg)? How is it ocmputed? and also as far as I know, its value is not reliable in NPO especially on Cellzone level?Your comments in this regard would be highly appreciated.
Thank you.17th July 2010 at 17:13 #63605pixGuest
i forgot how it is computed, but you can check the formula. at cell level, the value should be the total amount of bytes transfered on the whole cell (sum of PDCH RLC bytes) per hour.
At cell zone, the aggregation could be wrong. It should be an average aggregation,not a “sum”.
the throughput per cell is not very useful anyway (imo), perhaps it can be used to dimension the abis (but I know better indicators for that).
pix20th July 2010 at 06:48 #63606PaulGuest
1. According to 1 TBF for one user, then if we set MAX_UL_TBF_per_SPDCH =4, then in 1 PDCH can be used up to 4 users right (in high PDCH Utilization at that time)?
So if user A has 20% of time at rate 30kbps, user B has 30% of time at 40 kbps, user C has 30%of time at rate 60 kbps, and the last user D has 20% at rate 50 kbps, then the PDCH throughput would be ==> (0.20*30 kbps + 0.3*40kbps + 0.3*60kbps +0.20*50)
==> 6 kbps + 12 kbps + 18 kbps + 10 kbps = 46 kbps. But for the TBF throughput of each user is : user A has TBF throughput 6 kbps, User B 12 kbps, User C 18 kbps, user D 10 kbps.
The average TBF troughput would be 46 kbps/4=11.5 kbps.
Please correct me if I’m wrong.
2. Then, if we set MAX_PDCH_per_TBF=5, it means that 1 user can have PDCH timeslot up to 5 PDCHs.Well, i guess that this condition could be reached when there are still many idle PDCHs (no high PDCH utilization at that time).
Then if the rate is 50 kbps at each, then that user can have TBF throughput 250 kbps, but the PDCH only 50 kbps. Please correct me if I’m wrong.
3. Well, I’m actually a little bit confused. You stated :”The TBF is the group of PDCHs allocated to one user”. But you gave an example “If 2 users are sharing one PDCH”. Does it mean that in that 1 PDCH, there are 2 TBFs (1 user 1 TBF) ?
From my understanding, and based on ALU parameters, it means that 1 TBF can have many PDCHs (no.2) and 1 PDCH can have many TBF (no.1).
Feel free to discus Pix, I would be grateful to know your response.
Anyway, thank you so much:)
Please help me to answer no 1 and 2, I’m asked by my boss to count how many users (max) can be allocated in 1 BTS/Cells.
Paul20th July 2010 at 08:19 #63607pixGuest
1 : 100% correct
2 : 100% correct
3 : as you said
1 TBF can have many PDCHs (no.2) and 1 PDCH can have many TBF (no.1).
Example (using your example)
User A has 1 TBF w/ 2 PDCH
User B has 1 TBF w/ 2 PDCH
User C has 1 TBF w/ 1 PDCH
User D has 1 TBF w/ 1 PDCH
on PDCH 1 (= timeslot 1 of the TRX)
user A has 20% of time at rate 30kbps, user B has 30% of time at 40 kbps, user C has 30%of time at rate 60 kbps, and the last user D has 20% at rate 50 kbps, then the PDCH throughput would be ==> (0.20*30 kbps + 0.3*40kbps + 0.3*60kbps +0.20*50)= 46 kbps
on PDCH 2 (= timeslot 2 of the TRX)
user A has 40% of the time at rate 30kbps, user B has 60% at rate 40kbps
(0.40*30 kbps + 0.6*40kbps) = 36kbps
PDCH throughput of PDCH 1 = 46kbps
PDCH throughput of PDCH 2 = 36kbps
TBF throughput of user A = 0.2*30kbps + 0.4*30kbps = 18kbps
TBF throughput of user B = 0.9*40 = 36kbps
TBF throughput of user C = 0.3*60kbps = 20kbps
Notice that one user will get the same “max” throughput on all of its PDCH. For example user A has a max throughput per PDCH of 30kbps. That depends on the MCS used. If MCS is equal to MCS-4, then it is MCS4 on all the PDCH.
I think your whole idea of the TBF and PDCH is excellent, so don’t let me confuse you 🙂
Pix21st July 2010 at 12:27 #63608PaulGuest
Thank you so much Pix. It becomes easier to understand the throughput of GPRS concept.
No Pix, your answer doesn’t confuse me. But remember, we are pushed to think and develop ourselves when we meet problem we couldn’t solve ( confusing problem) 🙂
Paul21st July 2010 at 14:26 #63609pixGuest
you’re right, technical problems are always a challenge…
but me making myself clear is another kind of problem 🙂
human communication through web forums generates confusion by itself.
I’m waiting for the day when forums will be based on telepathy ! :))21st July 2010 at 19:14 #63610ALUGuest
Interesting Q&A Session going on so I decided to drop by.I have a few questions of my own regarding GPRS throughput(slightly irrelevant to the previous discussion but I am sure both of u wouldnt mind slight deviation as long as its related 2 throughput :P). Feel free 2 confuse me aswell coz the questions are already quite confusing 🙂
1:The max no of TSs a user can have depends on MS Multislot Class aswell.Now mostly the max TSs that a user can have in one directions is 4(eg 4+1 for Multislot Class 8)[Excluding Multislot class 30-33]. So, even if we increase MAX_PDCH_per_TBF beyond 4 or 5, it will not have any impact in increase in throughput.Am I correct?
2: Consider a site having 1 E1 and of configuration 4/4/6.So as one TRX requires 2 Abis TSs so total no of Abis Ts’s used would be 31 ie..(14*2=28 for data + 3(or 2)for Signaling).So, it means that we dont have any Extra Abis TSs configured for the site.Now excluding Bonus+Extra Nibbles, we are only left with BASIC Nibbles per cell.
Now as we know that inorder to achieve MCS9, we need to have around 5GCH(4.49Abis nibbles)
If we consider
Cell A(4 TRX):Total Basic Abis nibbles =(4*8)=32
MS MULTISLOT Class 8(4+1)for all users
so one user using 4 PDCH would require=(4*4.5)=18Abis Nibbles
So, with this calculation, we can have only one user using MCS9 simultaneously in this cell? because we dont have another 18 Abis Nibbles(only 14 left).Am i right?
3:If an MS Multislot Class is 8(4+1), so can that user use less than 4 TSs(PDCH) or is it compulsory for it to use 4(to satisfy its Multislot Class)?22nd July 2010 at 07:28 #63611pixGuest
please, join in !
1/ yes, correct!
2/ well, your computation is correct BUT why not take into account the bonus nibbles ???
in a 4/4/6 BTS, you would have 3 BCCH + 7 SDCCH, that’s 10 bonus nibbles !
Just for clarification :
– Extra & Bonus nibbles are shareable at BTS level.
– Basc nibbles are shareable only within the same cell.
3/ “sub-“optimal allocations are possible, of course. The MFS will try to allocate as many PDCH as possible (up to the multislot class limit). But if it can’t then it will allocate less. The MS is able to support those sub-optimal configurations. It can even support PDCH configurations with holes. example : the MS gets (PDCH 1 + PDCH 2 + PDCH 4) while the Timeslot 3 is used for somebody else.
pix26th September 2013 at 11:27 #63612billGuest
dears, how can we statistically measure throughput per user or throughput per tbf?
usually we get statistically is the throughput per timeslot, any ideas?