INTERNET MULTIMEDIA                                         06.06.2003


1.       Brief questions (2 pt each, total 10 pt).

a.       One would imagine that on average, a silence detector would reduce the bandwidth of packet-switched telephone conversation to half of a circuit-switched conversation. Name at least two reasons why the efficiency is typically not quite as good as that.

Packetization overhead; double talk; "hang-over" at the end of words and sentences; background noise. [Any two]


b.       The IPv4 header is 20 bytes long. The IPv6 header is 40 bytes long, while the address size has quadrupled. What functionality has disappeared from the fixed header?

Header length, header checksum, fragmentation.


c.       Chaining and redirection are concepts found in a number of Internet protocols. What mechanisms does HTTP offer to implement these?

HTTP implements chaining, that is, passing on a request and the response through a set of intermediaries, via HTTP proxies.

HTTP implements redirection through 3xx Redirection status codes and the Location header as well as the Refresh header, a Netscape proprietary extension. Even though the Refresh header is often included in the HTML http-equiv meta tag, it is independent of HTML.

Note that HTTP is not restricted to transporting HTML, even though its name may indicate that.

d.       What are the components of delay in a WFQ system, besides the propagation delay determined by the speed of light? You don't have to know their quantitative value, just explain intuitively what they are caused by and what they depend on.

    • Buffered packets due to burstiness; depends on bucket depth and rate share.
    • Transmission delay caused by bandwidth share, depends on hop count, bandwidth share and maximum own packet length.
    • Interference by other packets, depends on number of hops and max. length of all packets.


e.       The minimal audible difference in loudness is 3 dB. What power ratio (or ratio in air pressure) does this correspond to?

Decibels are defined as

dB = 10 log10 p = 20 log10 v,

where p is a power ratio and v is a voltage (or air pressure) ratio. Thus,

p = 103/10 = 1.995
v = 103/20 = 1.413


2.       Liisa is connected to the Internet via a 28.8 kb/s modem and wants to participate in an audio conference with Raimo, Erkki and Hannu. The three men are talking (usually being polite) at 20 kb/s codec using RTP. Liisa's ISP is offering her to connect to the conference through either a translator or a mixer. The translator converts the audio to a 10 kb/s codec. Explain the trade-offs between the two choices. (5 pt)

With a translator, Liisa will hear the men at 10 kb/s, that is, with reduced speech quality. If all three talk at once (or one's office has sufficiently loud background noise to trigger the silence detector), packets will be dropped since the total line capacity is insufficient for three simultaneous speakers. Since the streams are transmitted separately, Liisa can mute each speaker separately.

Note that since Liisa is on the other side of the modem link, she can still speak simultaneously with the men, regardless of whether her ISP uses a mixer or a translator.

With a mixer, there is no limit on the number of simultaneous speakers and the audio quality is not reduced. However, Liisa cannot select among the possible speakers. Also, a mixer introduces higher delay since it has to do playout delay compensation prior to mixing, in addition to the playout delay inserted by Liisa's audio receiver software.

Note that a mixer does not degrade the sound quality, it simply adds the linear sound samples. There is one exception: if a sender's silence detection malfunctions and this sender transmits background noise, Liisa could mute this one sender. However, this only helps if the bottleneck bandwidth can support the extra sender. Note that some multicast routing protocols support sender selection at the IP level, which would solve this particular problem to some extent - except that Liisa would have no way of knowing when the conversation partner with a noisy office actually has something to say.

For block-oriented codecs, mixing requires conversion to and from a linear encoding. A translator would require a similar conversion between codecs, which is also typically done by decoding the incoming stream into either 16-bit linear samples or mu-law samples and then re-coding with the outgoing codec. Mixers require playout delay adaptation, the complexity of which can range from trivial to extremely involved. However, implementation complexity was not one of the considerations since Liisa's ISP has already implemented both.

3.       You are asked to provide seating for the students waiting in the Student Information Center. Assume that students arrive at a rate of one student every minute and that it takes the clerk 40 seconds to serve them, with students arriving as a Poisson process and requiring exponentially distributed service time. On average, how many students are waiting in line and how long does a student have to wait? (Hint: compute the time-in-system T). (5 pt)

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.)


4.       Imagine that a wireless network was installed on the Linnanmaa campus. It is also used to distribute radio programs on campus. The radio program uses RTP over UDP. Describe briefly (using diagrams of packet flows and network entities) how you might use SIP to make sure that the "radio" reception is continuous. Please describe the solution and possible problems that may be involved. Is there another design approach for this application? (5 pt)

Assume first, that this radio program is distributed via unicast, as in today's RealAudio.

With SIP, the radio receiver would "invite" (using INVITE) the radio station when joining a new subnet (with a new unicast address) and send a BYE request when leaving the subnet. Alternatively, the sender could also use the receiver's RTP reports to discover the new location. Having the radio server issue an invitation is inappropriate since the server doesn't know the identity of radio receivers.

SIP does not have a "redirect" request (RTSP does). Registration is not helpful, as it only affects new, server-initiated invitations.

Another design alternative is to announce the radio program using SAP/SDP, and then receive the program using multicast. This requires that join latencies are low enough. Any delay in canceling the transmission will lead to unnecessary traffic in the subnet that the receiver just left.

Another design alternative would be RTSP, using the REDIRECT request.

5.       Instead of transmitting music via analog FM radio, we want to send MP3's, digitally modulated, across the same radio band. Assume that we need an SNR of 10 dB and that each channel is 200 kHz wide. How many MP3 streams can we squeeze into one FM station? (5 pt)

Shannon's channel capacity can be computed as

C = 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.