LTE Downlink Scheduling

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LTE Downlink Scheduling

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Hi Experts,

In my experiement for LTE downlink latency measurement, I sent packets at the rate of 10 packets per second (one packet every 100 milli second) towards my UE under test. I observed that almost all packets had the air interface latency of around 8 milli seconds. I suppose this is normal. But for me to confirm, is there any document that I can read the shows what happens when a packet arrives at eNodeB? Is this 8 milli second the minimum delay that every packet will experience?

Verified Answer
  • 8 ms seems a little high for a minimum - I would guess a little over 5 ms. However if your measurement point is in the UE then 8 ms makes sense.  It takes 1 ms for eNB to transmit the packet, another 3 ms before the UE can respond, 1 ms for the ACK to reach the eNB for a total of 5 ms - add in processing time in the eNB for the ACK and thats why I say a little over 5 ms. However the UE doesn't know that his ACK was successfully received until 3 ms later for a total of 8 ms. (if the ACK is not received by eNB it will retransmit the packet and the UE must respond again) .

    In general, the latency of a specific packet is a combination of a number of delays. The packet arrives on the s1-U interface, the eNB processes the packet and places it in a queue, then queuing delay, then 1st HARQ transmission in one ms, 4 ms later UE responds with ACK or NACK which takes 1 ms to transmit, eNB processes the response, and then possible HARQ re-transmissions add additional  delay, and possibly RLC re-transmissions add even more delay. If you are measuring 8 ms per packet, then I'd guess there is no other traffic in the cell and the RF conditions are great so that there are never any re-transmissions.  The size of the packet may have an effect - e.g. if it's very large the eNB may not be able to send it all in one TTI (i.e. 1 ms). How busy the cell is will affect the queuing delay.

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  • 8 ms seems a little high for a minimum - I would guess a little over 5 ms. However if your measurement point is in the UE then 8 ms makes sense.  It takes 1 ms for eNB to transmit the packet, another 3 ms before the UE can respond, 1 ms for the ACK to reach the eNB for a total of 5 ms - add in processing time in the eNB for the ACK and thats why I say a little over 5 ms. However the UE doesn't know that his ACK was successfully received until 3 ms later for a total of 8 ms. (if the ACK is not received by eNB it will retransmit the packet and the UE must respond again) .

    In general, the latency of a specific packet is a combination of a number of delays. The packet arrives on the s1-U interface, the eNB processes the packet and places it in a queue, then queuing delay, then 1st HARQ transmission in one ms, 4 ms later UE responds with ACK or NACK which takes 1 ms to transmit, eNB processes the response, and then possible HARQ re-transmissions add additional  delay, and possibly RLC re-transmissions add even more delay. If you are measuring 8 ms per packet, then I'd guess there is no other traffic in the cell and the RF conditions are great so that there are never any re-transmissions.  The size of the packet may have an effect - e.g. if it's very large the eNB may not be able to send it all in one TTI (i.e. 1 ms). How busy the cell is will affect the queuing delay.

  • BTW: 36321 is the LTE MAC spec where the HARQ transmission details are described. Its a tough read, so I'd suggest an internet search instead.

  • Thank you John for your time.

  • Dear John,

    I think, you answered my question very well. I should have explained how I am measuring the delay of this 8 mS. I wireshark on the IP packets entering my eNodeB and wireshark again on my UE for the IP packets. Then I deduct the time stamp on them to get 8 mS value. The eNodeB is indeed unloaded (just one UE) and I had very strong signal. Thus I suppose, this 8 mS consist of :- all the layer processing inside the eNodeB (arounf 3 mS), 1 mS transmit time on air, UE layer processing time (same layer processing procedure as the eNB ~ 3mS) and finally the wireshark and other involved delays both at UE and eNB (1 mS in total). Please correct me if my assumptions are wrong.

  • Dear John,

    I think, you answered my question very well. I should have explained how I am measuring the delay of this 8 mS. I wireshark on the IP packets entering my eNodeB and wireshark again on my UE for the IP packets. Then I deduct the time stamp on them to get 8 mS value. The eNodeB is indeed unloaded (just one UE) and I had very strong signal. Thus I suppose, this 8 mS consist of :- all the layer processing inside the eNodeB (arounf 3 mS), 1 mS transmit time on air, UE layer processing time (same layer processing procedure as the eNB ~ 3mS) and finally the wireshark and other involved delays both at UE and eNB (1 mS in total). I also assume there is no scheduling or queuing delay as there is only one UE connected to the eNB. Please correct me if my assumptions are wrong.

  • for volte this is min delay u have

  • Dears,

    I have sort of related question, in lots of cells in my network, the scheduling delay for whole cell is much higher than for single ue (from counters prospective) which make the cell throughput less than ue throughput.

    I even tried to change the the rlc mode to UM for qci 9 which has the most traffic to test if something changes but it couldnt be configure thats way.

    would u please elaborate this phenomenon.

    thanks in advance