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Reply To: trgput kbps / tbf>kbps / pdch

#63607
pix
Guest

paul,

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 🙂

Best Regards,
Pix