Chapter 7 Review Questions
1.
Bit rate
Bytes transferred in 67
2. Spatial Redundancy: It is the redundancy within a given image. Intuitively, an image
consists of mostly white space has a high degree of redundancy and can be efficiently
compressed without significantly sacrificing image quality.
3. Quantizing a sample into 1024 levels means 10 bits per sample. The resulting rate of
the PCM digital audio signal is 160 Kbps.
4. Streaming stored audio/video: In this class of applications, the underlying medium is
prerecorded video, such as a movie, a television show, or a prerecorded sporting
event. These prerecorded videos are played on servers, and users send requests to the
servers to view the videos on demand. Many internet companies today provide
5. UDP Streaming: With UDP streaming, the server transmits video at a rate that
matches the client’s video consumption rate by clocking out the video chunks over
UDP at a steady rate.
HTTP Streaming: In HTTP streaming, the video simply stored in an HTTP server as
1. Due to unpredictable and varying amount of available bandwidth between server
and client, constant-rate UDP streaming can fail to provide continuous play out.
2. It requires a media control server, such as an RTSP server, to process client-to-
7. No. On the client side, the client application reads bytes from the TCP receive buffer
and places the bytes in the client application buffer.
9. Enter Deep: This philosophy is to enter deep into the access networks of Internet
Service Providers, by deploying server clusters in access ISPs all over the world.
10. Geographically closest cluster selection and real time measurements selection can
11. Load on the cluster – clients should not be directed to overload clusters.
12. End-to-end delay is the time it takes a packet to travel across the network from source
13. A packet that arrives after its scheduled play out time cannot be played out.
Therefore, from the perspective of the application, the packet has been lost.
14. First scheme: send a redundant encoded chunk after every n chunks; the redundant
15. RTP streams in different sessions: different multicast addresses; RTP streams in the
16. The role of a SIP registrar is to keep track of the users and their corresponding IP
addresses which they are currently using. Each SIP registrar keeps track of the users
17. In non-preemptive priority queuing, the transmission of a packet is not interrupted
once it has begun. In preemptive priority queuing, the transmission of a packet will be
18. A scheduling discipline that is not work conserving is time division multiplexing,
19. FIFO: line at Starbucks. RR: merging traffic (taking 1 vehicle from first lane then 1
from the second, then 1 from the first, and so on). WFQ: ticket counter at airport
Problem 1
should be arrived before time t1 + d to be played at right time, third block at
t1 + 2d and so on. We can see from figure that only blocks numbered 1,4,5,6 arrive at
receiver before their playout times.
b) Client begins playout at time t1 + d and video blocks are to be played out over the
fixed amount of time, d. So it follows that second video block should be arrived
before time t1 +2d to be played at right time, third block at t1 + 3d and so on. We can
Problem 2
a) During a playout period, the buffer starts with Q bits and decreases at rate r – x. Thus,
after Q/(r – x) seconds after starting playback the buffer becomes empty. Thus, the
Problem 3
a) The server’s average send rate is۶Ȁ .
b) This part (b) is an odd question and will be removed from the next edition. After
playing out the first frame, because x(t) < r, the next frame will arrive after the
scheduled playout time of the next frame. Thus playback will freeze after displaying
freezing, we need q(t + T) > 0 for all t ≥ T, we have
ሺ࢚ࢀ
ሻൌࡴࢀ
െ࢚࢘࢞ሺ࢙ሻࢊ࢙
࢚
> ࡴ
ሺࢀ െ ࢚ሻ ࢞ሺ࢙ሻࢊ࢙
࢚
Problem 4
a) Buffer grows at rate x – r. At time E, (x – r)*E bits are in buffer and are wasted.
Problem 5
Problem 6
KbpshKbps
h)4.64(
8)160(
Problem 7
a) Denote
)(n
d
for the estimate after the nth sample.
44
)1(
trd
11
)1()( dutrud
)()1()()1()()1()(
44
3
33
2
2211
trutruutruutru
)()1(
1
1
)(
jj
j
j
tru
u
u
d
¦
f
f
Problem 8
a) Denote
)(n
v
for the estimate after the nth sample. Let
jjj tr ‘
.
)3()4(
1
)4(
)1( vuduv ‘
)2(
3
2)3(
2
)4(
1
)1()1( duuduudu ‘‘‘
Problem 9
a) r1 – t1 + r2 – t2 + …+rn-1–tn-1 = (n-1)dn-1
Substituting this into the expression for dn gives
Problem 10
The two procedures are very similar. They both use the same formula, thereby resulting
in exponentially decreasing weights for past samples.
Problem 11
a) The delay of packet 2 is 7 slots. The delay of packet 3 is 9 slots. The delay of packet
4 is 8 slots. The delay of packet 5 is 7 slots. The delay of packet 6 is 9 slots. The
delay of packet 7 is 8 slots. The delay of packet 8 is > 8 slots.
Problem 12
The answers to parts a and b are in the table below:
Packet Number
ri – ti
di
vi
1
7
7
0
2
8
7.10
0.09
3
8
7.19
0.162
4
7
7.17
0.163
5
9
7.35
0.311
6
9
7.52
0.428
7
8
7.57
0.429
8
8
7.61
0.425
Problem 13
a) Both schemes require 25% more bandwidth. The first scheme has a playback delay of
5 packets. The second scheme has a delay of 2 packets.
Problem 14
a) Each of the other N – 1 participants sends a single audio stream of rate r bps to the
initiator. The initiator combines this stream with its own outgoing stream to create a
stream of rate r. It then sends a copy of the combined stream to each of the N -1 other
participants. The call initiator therefore sends at a total rate of (N-1)r bps, and the
total rate aggregated over all participants is 2(N-1)r bps.
Problem 15
a) As discussed in Chapter 2, UDP sockets are identified by the two-tuple consisting of
destination IP address and destination port number. So the two packets will indeed
pass through the same socket.
b) Yes, Alice only needs one socket. Bob and Claire will choose different SSRC’s, so
Alice will be able distinguish between the two streams. Another question we could
have asked is: How does Alice’s software know which stream (i.e. SSRC) belongs to
Bob and which stream belongs to Alice? Indeed, Alice’s software may want to
Problem 16
a) True
b) True
Problem 17
Problem 18
a)
Packet
Time leaving the queue
Delay
1
0
0
2
1
1
3
2
1
4
3
2
6
4
2
Average Delay
1.91
b)
Packet
Time leaving the queue
Delay
1
0
0
2
2
2
Average Delay
1.91
c)
Packet
Time leaving the queue
Delay
1
0
0
2
2
2
3
4
3
4
1
0
5
3
0
Average Delay
1.91
d)
Packet
Time leaving the queue
Delay
Note
1
0
0
WFQ
2
2
2
WFQ
3
1
0
WFQ
Problem 19
a)
Packet
Time leaving the queue
Delay
1
0
0
2
4
4
Average Delay
1.91
b)
Packet
Time leaving the queue
Delay
1
0
0
2
1
1
Average Delay
1.91
c)
Packet
Time leaving the queue
Delay
1
0
0
2
1
1
5
9
6
6
6
4
7
4
1
8
7
2
9
5
0
10
8
1
11
11
3
Problem 20
Time Slot
Packets in the queue
Number of tokens in bucket
0
1, 2, 3
2
1
3, 4
1
2
4,5
1
3
5,6
1
4
6
1
Time Slot
Packets in output buffer
0
1, 2
1
3
2
4
3
5
4
6
Problem 21
Time Slot
Packets in the queue
Number of tokens in bucket
0
1, 2, 3
2
1
3, 4
2
2
5
2
3
6
2
4
2
5
2
6
7, 8
2
8
–
2
Time Slot
Packets in output buffer
0
1, 2
1
3, 4
2
5
3
6
Problem 22
Problem 23
See figure below. For the second leaky bucket,
.1, bpr
Problem 24
Problem 25
Let
W
be a time at which flow 1 traffic starts to accumulate in the queue. We refer to
W
as the beginning of a flow-1 busy period. Let
W
!t
be another time in the same flow-1
busy period. Let
),(
1
tT
W
be the amount of flow-1 traffic transmitted in the interval
],[ t
W
.
j