Question

Problem 1 Consider a stream of VoIP packets. Assume that the first packet of the VoIP call is packet seq k=1, the device has estimated the delay over previous calls with a current estimate of the average delay D=83 ms, the average deviation of the delay (Jitter) is Jk= 11 ms. The table below shows the received information at the Device B. Complete the table. For the delay average use the moving average equation, use a = 0.01: Dx = (1 - a)Dx + a Dinstantanous Di = 83 x 0.99 + 0.01 x (480 - 400) = 82.97 Ik = Jk-1 + (Uinstantanous ! - Jk-1)/16 J_k = 11 + ((420 – 400) – (510 – 4801 - 11)/16 1-10 - 11 Ik = 11 + 1-1 16 = 10.9375 Packet seq. # Average Jitter 11 10.9375 510 k=2 k= 3 k= 4 ka5 k=6 ka k=8 k= 9 kao kl Timestamp Arrival Average delay rage delay fi [ms] time D [ms] 400 480 82.97 420 83.0403 440 570 460 600 480 540 645 520 560 680 580 690 620 695 640 705

Answer 1

Packet seq #	ti (ms)	Arrival time	Dk (ms)	Average Jitter
k=1	400	480	82.97	11
k=2	420	510	83.0403	10.9375
k=3	440	570	83.509897	12.75
k=4	460	600	84.074798	12.578125
k=5	480	605	84.48405	12.7294922
k=6	540	645	84.7792095	13.1838989
k=7	520	650	85.2314174	13.9224052
k=8	560	680	85.5791032	13.6772549
k=9	580	690	85.8233122	13.4474265
k=10	620	695	85.7150791	14.7944623
k=11	640	705	85.5079283	14.4948084

Answer 2

Solution Table for the Problem-1

Packet Seq. #	Timestamp (ti [ms])	Arrival time	Average Delay (Dk  [ms])	Average Jitter
K=1	400	480	82.97	11
K=2	420	510	83.0403	10.9375
K=3	440	570	83.5099	12.7539
K=4	460	600	84.0748	12.5818
K=5	480	605	84.4840	12.7329
K=6	540	645	84.6892	13.1871
K=7	520	650	85.1423	13.9254
K=8	560	680	85.4909	13.6801
K=9	580	690	85.7359	13.4501
K=10	620	695	85.6285	14.7970
K=11	640	705	85.4222	14.4972