Slotted ALOHA
Chapter 7 Problems Problem 2
x(t) will ramp up exponentially fast to the TCP available bandwidth due to TCP slow start. Then x(t) will remain roughly constant at the TCP available bandwidth until the client buffer fills. At that time, x(t) will drop to approximately d until the media is sent and the client buffer remains roughly full.
Problem 3
No, they are not the same thing. The client application reads data from the TCP receive buffer and puts it in the client buffer. If the client buffer becomes full, then application will stop reading from the TCP receive buffer until some room opens up in the client buffer.
Problem 4
a) 160+h bytes are sent every 20 msec. Thus the transmission rate is Kbps
h h Kbps
) 4 . 64 20 (
8 ) 160
( + ⋅ = +
b)
IP header: 20bytes UDP header: 8bytes
RTP header: 12bytes h=40 bytes (a 25% increase in the transmission rate!) Problem 5
a) Denote d(n) for the estimate after the nth sample.
4 4 ) 1
( r t
d = −
) )(
1 ( )
( 3 3 4 4
) 2
( u r t u r t
d = − + − −
[
( ) (1 )( )]
) 1 ( )
( 2 2 3 3 4 4
) 3
( u r t u u r t u r t
d = − + − − + − −
=u(r2 −t2)+(1−u)u(r3 −t3)+(1−u)2(r4 −t4)
) 3 ( 1
1 ) 4
( u(r t ) (1 u)d
d = − + −
=u(r1−t1)+(1−u)u(r2 −t2)+(1−u)2u(r3−t3)+(1−u)3(r4 −t4)
b)
) (
) 1 ( ) ( ) 1 (
1 1 ) (
n n n j
j n
j
j
n u u r t u r t
d =
∑
− − − + − −=
c)
) ( ) 1 1 1(
) (
j j j
j r t
u u
d u − −
= −
∑
∞=
∞
.9 ( )
9 1
1
j j j
j r −t
=
∑
∞=
The weight given to past samples decays exponentially.
Problem 6
a) Denote v(n) for the estimate after the nth sample. Let Δj =rj −tj.
) 1 ( 4 ) 1
( d
v = Δ − (=0)
) 1 ( 4 )
2 ( 3 )
2
( u d (1 u) d
v = Δ − + − Δ −
) 2 ( )
3 ( 2 )
3
( u d (1 u)v
v = Δ − + −
=u Δ2 −d(3) +u(1−u) Δ3 −d(2) +(1−u)2 Δ4 −d(1)
) 3 ( )
4 ( 1 )
4
( u d (1 u)v
v = Δ − + −
=u Δ1−d(4) +(1−u)u Δ2 −d(3) +u(1−u)2 Δ3−d(2) +(1−u)3 Δ4 −d(1)
=u
[
Δ1−d(4) +(1−u) Δ2 −d(3) +(1−u)2 Δ3−d(2)]
+(1−u)3 Δ4 −d(1) b)
) 1 ( )
1 1 (
1
1 )
( u (1 u) d (1 u) d
v j n j n n
n
j
j
n = − − Δ − − + + − Δ −
=
∑
−Problem 7
a) r1 – t1 + r2-t2 + …+rn-1-tn-1 = (n-1)dn-1
Substituting this into the expression for dn gives
n t d r
n
dn n n n − n
− +
= 1 −1
b) The delay estimate in part (a) is an average of the delays. It gives equal weight to recent delays and to “old” delays. The delay estimate in Section 6.3 gives more weight to recent delays; delays in the distant past have relatively little impact on the estimate.
Problem 8
The two procedures are very similar. They both use the same formula, thereby resulting in exponentially decreasing weights for past samples.
One difference is that for estimating average RTT, the time when the data is sent and when the acknowledgement is received is recorded on the same machine. For the delay estimate, the two values are recorded on different machines. Thus the sample delay can actually be negative.
Problem 9
a) If there is packet lost, then other packets may have been transmitted in between the two received packets.
b) Let Si denote the sequence number of the ith received packet. If sec
1 20m
t ti − i− >
and
1 +1
= i−
i S
S then packet I begins a new talkspurt.
Problem 10
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.
b) Packets 3, 4, 6, 7, and 8 will not be received in time for their playout if playout begins at t-8.
c) Packets 3 and 6 will not be received in time for their playout if playout begins at t=9.
d) No packets will arrive after their playout time if playout time begins at t=10.
Problem 11
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 12
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.
b) 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.
c) 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.
Problem 13
The CDN company provides a mechanism so that when a client requests content, the content is provided by the CDN server that can best deliver the content to the specific client. The server may be the closest CDN server to the client (perhaps in the same ISP as the client) or may be a CDN server with a congestion-free path to the client. In this way, even if a CDN does not increase the amount of link capacity in the network, the performance seen by the hosts is improved.
Problem 14
Yes. If the path between the requesting host and the CDN server, chosen by the CDN, is already too congested or if the chosen CDN server is already too busy, then it might be possible that the host experiences worse performance that what it would have if it contacted the remote server directly.
Problem 15
The inter-arrival jitter J is defined to be the mean deviation (smoothed absolute value) of the difference D in packet spacing at the receiver compared to the sender for a pair of packets. As shown in the equation below, this is equivalent to the “relative transit time”
for the two packets; the relative transit time is the difference between a packet's RTP timestamp and the receiver's clock at the time of arrival. If is the RTP timestamp for packet i and is the time of arrival in RTP timestamp units for packet i, then for two packets i and j, D is defined as
Ti
ri
) ( ) (
) (
) ( ) ,
(i j ri rj si sj rj sj ri si
D = − − − = − − −
The inter-arrival jitter is calculated continuously as each data packet i is received, using this difference D for that packet and the previous packet i-1 in order of arrival (not necessarily in sequence), according to the formula
16
| ) , (
|D i j J J
J = + −
Whenever a reception report is issued, the current value of J is sampled.
Problem16
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 display the sender’s name when the sender is talking. Alice’s software gets the SSRC to name mapping from the RTCP source description reports.
Problem 17
a) The session bandwidth is 4* 100 kbps = 400 kbps. Five percent of the session bandwidth is 20 kbps.
b) Sends each user is both a sender and receiver, each user gets 5 kbps for RTCP packets (receiver reports, sender reports, and source description packets).
Problem 18
a) Like HTTP, all request and response methods are in ASCII text. RTSP also has methods (SETUP, PLAY, PAUSE), and the server responds with standardized reply codes. Yes, using the GET method, HTTP can be used to request a stream.
b) RTSP messages use different port numbers than the media streams. Thus RTSP is out-of-band. HTTP objects are sent within the HTTP response message. Thus HTTP is in-band. HTTP does not keep session state: each request is handled independently.
RTSP uses the Session ID to maintain session state. For example, in the lab (programming assignment) for this chapter, the RTSP server is in one several states for each session ID. When in the pause state, the server stores the number of the frame at which the pause occurred.
Problem 19