This is an open-book exam. You have 150 minutes to complete the exam. Please make sure that the problems appear in numerical order in your "blue book", with each problem starting on a new right page. Points are given in .
This is an example for an M/M/1 queue with arrival rate lambda = 1/60 = 0.0167 and service rate mu = 1/40 = 0.025. Thus, the system time T = 1/(0.025 - 0.0167) = 120.5 s. According to Little's Law, the number in system is then N = lambda T = 2. However, this includes the student being helped, so we compute the waiting time as 120.5 s - 40 s = 80.5 s, yielding a waiting line length of 1.34. (Note that this is not just one less than the number in system.)
mark packets diff-serv three-color marker delay packets edge shaping (within host) drop packets protection in network
TCP UDP video-on-demand +: bandwidth-adaptive, delays and packet header size (since payload is bigger) don't much matter, reliable, works through firewalls; -: rate averaging may cause receiver starvation +: avoid receiver starvation; -: no congestion control Internet phone calls -: longer delays due to retransmission; +: easier to get through firewalls; compensates for packet losses +: lower packet-header overhead, no delays imposed by retransmission or flow/congestion control; -: needs application-layer reliability;
|RTP sequence number||RTP timestamp||network delay (seconds)|
Show a timeline as to when the packets are played out at the receiver. Your "algorithm" should minimize the end-to-end delay (a) without losing any packets, (b) while being allowed to drop one packet. Your algorithm can be non-causal, i.e., may look ahead in time. Compute the delay for the two cases.
The sampling and RTP clock rate is 8000 Hz, with each 160 RTP clock units corresponding to 0.02 s. Thus, we have, assuming that the first packet is transmitted at time zero:
The packets with sequence numbers 4, 5 and 6 are part of the second talkspurt, as can be seen by the jump in timestamp. If there is no loss allowed, the average delay is (3*0.40 + 3*0.45)/6 = 0.425 s. If we drop one packet, we can decrease the delay to (3*0.40 + 2*0.08)/5 = 0.272 s.
RTP sequence number RTP timestamp network delay (seconds) arrival time playout time, with no loss playout time, with one loss 1 160 0.40 0.40 0.40 0.40 2 320 0.30 0.32 0.42 0.42 3 480 0.35 0.39 0.44 0.44 4 800 0.32 0.40 0.53 0.40 5 960 0.27 0.37 0.55 0.42 6 1120 0.45 0.57 0.57 drop
Shannon's channel capacity can be computed asC = B log2 (1 + S/N)Here, the S/N ratio is 10, for C = 200 kHz log2 11 = 200 kHz * 3.46 = 692 kb/s. Thus, we can easily fit (several) MP3 streams, at about 56 to 128 kb/s, into the existing analog channel, assuming that our channel coding scheme comes close to the Shannon limit.
INVITE sip:firstname.lastname@example.org SIP/2.0 To: Al Gore <sip:email@example.com> From: Henning Schulzrinne <sip:firstname.lastname@example.org> Call-ID: email@example.com CSeq: 1 INVITE Via: bart.cs.columbia.edu Content-Type: application/sdp Content-Length: 50 v=0 c=IN IP4 220.127.116.11 m=audio 3456 RTP/AVP 0
proxy redirect advantage hide callee address; may hide caller address; lower delay caller gets choice as to whether to go to forwarded address disadvantage proxy is more complicated than redirect server delay, exposure of addresses
RTSP is similar to HTTP in its message format, i.e., the use of request line, headers a message body. It differs from HTTP in:
- character set (UTF-8 vs. ISO 8859-1);
- state maintenance: HTTP is stateless except through cookies, while RTSP sets up a session via SETUP;
- in-band vs. out-of-band: HTTP sends data in-band, i.e., through the same TCP connection as the request, while RTSP generally sets up one or more external (UDP) channels for media data;
- different methods;
- HTTP only uses TCP, while RTSP can also use UDP.
It is easiest to compute the sequence by looking that the GPS departure times and scheduling time in slots of 100 bytes. The first stream, for example, needs to get half the "slots", while the other two streams needs to get 3 out of 12 slots. Thus, a virtual time sequence for the first stream could be 2, 4, 6, 8, ..., while for the second class 6.1, 18.1, 30.1, ..., and for the third class 12.2, 24.2, 36.2, ... The fractions and offsets are arbitrary and are simply chosen to avoid having to worry about what to do with two packets having the same virtual departure time.
For the video stream, RTP ts 1800 corresponds to 17.18 s real-time, so that RTP ts 9000 corresponds to 17.26 s real-time. (Recall that RTP video stream clocks tick at 90,000 Hz.) For the audio stream, timestamp 560 corresponds to 17.23 s. 17.26 s is thus 30 ms away, or 240 timestamp units. Thus, the audio packet with timestamp 800 needs to be played out together with the video frame with timestamp 9000.
It is not a realistic assumption that audio and video frames experience the same delay. For example, video frames are usually substantially longer than audio frames, so that the transmission time is longer. (Video frames typically occupy a whole packet of 1500 bytes, while low-rate audio frames are typically less than 200 bytes long.)
Roughly white noise, if the predictor is good.
An 8-bit signal has a dynamic range of 256:1 or 48 dB. A CD has a dynamic range of 96 dB.
Noise: All 16-bit values between -128 and 127 will be translated to the 8-bit value of 0, with an average error of 64. The average absolute value of the signal is 16,384, assuming uniform distribution of samples (which is not strictly correct). This leads to a signal/noise ratio of 10 log10 16,3842 / 642 = 48 dB, or a loss of 48 dB compared to the CD. (You could arrive at this simply by subtracting the two range values above.)
The frequency range decreases from about 20 kHz to about 3.4 kHz. (It's not quite half the sampling rate due to imperfect roll-off of the anti-aliasing filters.)