Solutions

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Tutorial 12

• Solutions

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Q.1

• In the Internet phone example in the lecture
  notes, let h be the total number of header bytes
  added to each chunk, including UDP and IP header.
  
• a. Assuming an IP datagram is emitted every 20
  msecs, find the transmission rate in bits per
  second for the datagrams generated by one side of
  this application.
• b. What is a typical value of h when RTP is used?

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• a)
• PCM-encoded voice 64kbps8000 bytes/s
• 8000 bytes/s 20 msec/chunk 160 bytes/chunk
• (160 h) bytes / datagram
• 1s / (20 msec) 50 datagrams/s
• Transmission rate 50 (160h) bytes/s
• 0.4 (160 h) kbps
• (64 0.4h) kbps

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• b) IP header 20 bytes
• UDP header 8 bytes
• RTP header 12 bytes
• h 40 bytes

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Q.2

• Consider the adaptive playout strategy described
  in the lecture notes
• a) How can two successive packets received at the
  destination have time-stamps that differ by more
  than 20 msecs when the two packets belong to the
  same talk spurt?
• b) How can the receiver use sequence numbers to
  determine whether a packet is the first packet in
  a talk spurt?

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• a). If there is packet lost, then other packets
  may have been transmitted in between the two
  received packets.

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• b). Let Si denote the sequence number of the ith
  received packet. If ti ti-1 gt 20 msec and Si
  Si-1 1then packet i begins a new talkspurt

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Q.3

• Recall the two FEC schemes for Internet phone
  described in the lecture notes. Suppose the first
  scheme generates a redundant chunk for every four
  original chunks. Suppose the second scheme uses a
  low-bit rate encoding whose transmission rate is
  25 percent of the transmission rate of the
  nominal stream.

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Question 3

• a) How much additional bandwidth does each scheme
  require?
• Both schemes require 25 more bandwidth.

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• b) How do the two schemes perform if the first
  packet is lost in every group of five packets?
  Which scheme will have better audio quality?
• The first scheme will be able to reconstruct the
  original high-quality audio encoding. The second
  scheme will use the low quality audio encoding
  for the lost packets and will therefore have
  lower overall quality.

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• c) How do the two schemes perform if the first
  packet is lost in every group of two packets?
  Which scheme will have better audio quality?
• For the first scheme, many of the original
  packets will be lost and audio quality will be
  very poor. For the second scheme, every audio
  chunk will be available at the receiver, although
  only the low quality version will be available
  for every other chunk. Audio quality will be
  acceptable.

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Q.4

• Consider an RTP session consisting of four users,
  all of which are sending and receiving RTP
  packets into the same multicast address. Each
  user sends video at 100 kbps.

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• a) RTCP will limit its traffic rate to what rate?
• The session bandwidth is 4 100 kbps 400 kbps.
  Five percent of the session bandwidth is 20 kbps.

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• b) A particular receiver will be allocated how
  much RTCP bandwidth?
• c) A particular sender will be allocated how much
  RTCP bandwidth?
• Each user is both a sender and receiver, each
  user gets 5 kbps for RTCP packets (receiver
  reports, sender reports, and source description
  packets).