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thus be big enough to accommodate 2w sequence numbers. That is, the sequence number
space must be at least twice as large as the window size,
wk 2t
.
Problem 24
3. At
4t
the sender receives the ACKs that the receiver sent at
1t
and advances its
window to 4, 5, 6. At
5t
the sender receives the ACKs 1, 2, 3 the receiver sent at
2t
.
These ACKs are outside its window.
Problem 25
a) Consider sending an application message over a transport protocol. With TCP, the
application writes data to the connection send buffer and TCP will grab bytes without
Problem 26
There are
2964,294,967,2
32
possible sequence numbers.
a) The sequence number does not increment by one with each segment. Rather, it
2
32
º
ª
Problem 27
a) In the second segment from Host A to B, the sequence number is 207, source port
number is 302 and destination port number is 80.
b) If the first segment arrives before the second, in the acknowledgement of the first
arriving segment, the acknowledgement number is 207, the source port number is 80
Problem 28
Since the link capacity is only 100 Mbps, so host A’s sending rate can be at most
100Mbps. Still, host A sends data into the receive buffer faster than Host B can remove
Problem 29
a) The server uses special initial sequence number (that is obtained from the hash of
source and destination IPs and ports) in order to defend itself against SYN FLOOD
attack.
Problem 30
a) If timeout values are fixed, then the senders may timeout prematurely. Thus, some
Problem 31
TTEstimatedRxxSampleRTTTTEstimatedR )1(
ms75.100
.
ms06.5
.
ms99.120
.
After obtaining second sampleRTT = 120ms:
ms15.103
.
.
8*415.103 ervalTimeoutInt
15.103*875.0140*125.0 TTEstimatedR
.
8*75.076.10714025.0 DevRTT
.
06.14*476.107 ervalTimeoutInt
76.107*875.090*125.0 TTEstimatedR
.
06.14*75.054.1059025.0 DevRTT
.
42.14*454.105 ervalTimeoutInt
54.105*875.0115*125.0 TTEstimatedR
.
42.14*75.071.10611525.0 DevRTT
.
88.12*471.106 ervalTimeoutInt
.
Problem 32
a)
Denote
)(n
TTEstimatedR
for the estimate after the nth sample.
b)
j
n
j
jn SampleRTTxxTTEstimatedR
¦
1
1
1)( )1(
9.
Problem 33
Let’s look at what could wrong if TCP measures SampleRTT for a retransmitted
segment. Suppose the source sends packet P1, the timer for P1 expires, and the source
then sends P2, a new copy of the same packet. Further suppose the source measures
Problem 34
At any given time t, SendBase – 1 is the sequence number of the last byte that the
Problem 35
When, at time t, the sender receives an acknowledgement with value y, the sender knows
for sure that the receiver has received everything up through y-1. The actual last byte
Problem 36
Suppose packets n, n+1, and n+2 are sent, and that packet n is received and ACKed. If
packets n+1 and n+2 are reordered along the end-to-end-path (i.e., are received in the
order n+2, n+1) then the receipt of packet n+2 will generate a duplicate ack for n and
Problem 37
a) GoBackN:
A sends 9 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re-
sent segments 2, 3, 4, and 5.
Problem 38
Yes, the sending rate is always roughly cwnd/RTT.
Problem 39
If the arrival rate increases beyond R/2 in Figure 3.46(b), then the total arrival rate to the
queue exceeds the queue’s capacity, resulting in increasing loss as the arrival rate
Problem 40
a) TCP slowstart is operating in the intervals [1,6] and [23,26]
b) TCP congestion avoidance is operating in the intervals [6,16] and [17,22]
c) After the 16th transmission round, packet loss is recognized by a triple duplicate
ACK. If there was a timeout, the congestion window size would have dropped to 1.
Problem 41
Refer to Figure 5. In Figure 5(a), the ratio of the linear decrease on loss between
connection 1 and connection 2 is the same - as ratio of the linear increases: unity. In this
Problem 42
If TCP were a stop-and-wait protocol, then the doubling of the time out interval would
suffice as a congestion control mechanism. However, TCP uses pipelining (and is
Problem 43
In this problem, there is no danger in overflowing the receiver since the receiver’s receive
Problem 44
a) It takes 1 RTT to increase CongWin to 6 MSS; 2 RTTs to increase to 7 MSS; 3 RTTs
to increase to 8 MSS; 4 RTTs to increase to 9 MSS; 5 RTTs to increase to 10 MSS; 6
Problem 45
a) The loss rate,
L
, is the ratio of the number of packets lost over the number of packets
sent. In a cycle, 1 packet is lost. The number of packets sent in a cycle is
·
§ 2/
)
(1
W
n
W
W
WW
2
2
2
¸
¹
¨
©
4824
22
WWWW
b) For
W
large,
WW 4
3
8
3
2
!!
. Thus
2
3/8 WL |
or
L
W3
8
|
. From the text, we
therefore have
Problem 46
a) Let W denote the max window size measured in segments. Then, W*MSS/RTT =
0.75W=94 (ceiling of 93.75) segments. Average throughput is 94*1500*8/0.15
=7.52Mbps.
Problem 47
Let W denote max window size. Let S denote the buffer size. For simplicity, suppose
TCP sender sends data packets in a round by round fashion, with each round
corresponding to a RTT. If the window size reaches W, then a loss occurs. Then the
Problem 48
a) Let W denote the max window size. Then, W*MSS/RTT = 10Gbps, as packets will
0.75W=93750 segments. Average throughput is 93750*1500*8/0.1=7.5Gbps.
c) 93750/2 *0.15 /60= 117 minutes. In order to speed up the window increase process,
Problem 49
As TCP’s average throughput B is given by
MSS
B
22.1
, so we know that,
Problem 50
a) The key difference between C1 and C2 is that C1’s RTT is only half of that of C2.
Thus C1 adjusts its window size after 50 msec, but C2 adjusts its window size after
100 msec. Assume that whenever a loss event happens, C1 receives it after 50msec
and C2 receives it after 100msec. We further have the following simplified model of
on the above assumptions.
C1
C2
Time
(msec)
Window Size
(num. of
segments sent
Average data sending
rate (segments per
second,
Window
Size(num. of
segments
Average data sending
rate (segments per
second, =Window/0.1)
50msec
is
300=
200+100)
100
2
(decreases
40
5
(decreases
50
(decreases
window size
as the avg.
total sending
rate to the
100msec is
80=
(40+20)/2 +
(50+50)/2)
250
1
20
20
link in last
100msec is
40=
(20+20)/2 +
(20+20)/2)
50msec is
50= (40+10)
550
2
40
10
600
1
20
1
10
Based on the above table, we find that after 1000 msec, C1’s and C2’s window sizes
are 1 segment each.
b) No. In the long run, C1’s bandwidth share is roughly twice as that of C2’s, because
C1 has shorter RTT, only half of that of C2, so C1 can adjust its window size twice as
fast as C2. If we look at the above table, we can see a cycle every 200msec, e.g. from
Problem 51
a) Similarly as in last problem, we can compute their window sizes over time in the
following table. Both C1 and C2 have the same window size 2 after 2200msec.
C1
C2
Time
(msec)
Window Size
(num. of
segments sent in
next 100msec)
Data sending speed
(segments per second,
=Window/0.1)
Window
Size(num. of
segments sent
in next
100msec)
Data sending speed
(segments per second,
=Window/0.1)
0
15
150 (in [0-100]msec]
10
100 (in [0-100]msec)
100
7
70
5
50
700
1
10
1
10
800
2
20
2
20
900
1
10
1
10
1000
2
20
2
20
2100
1
10
1
10
2200
2
20
2
20
b) Yes, this is due to the AIMD algorithm of TCP and that both connections have the
same RTT.
c) Yes, this can be seen clearly from the above table. Their max window size is 2.
Problem 52
Note that W represents the maximum window size.
First we can find the total number of segments sent out during the interval when TCP
changes its window size from W/2 up to and include W. This is given by:
S= W/2 + (W/2)*(1+
D
) + (W/2)*(1+
D
)2 + (W/2)*(1+
D
)3 + … + (W/2)*(1+
D
)k
We find k=log(1+D)2, then S=W*(2
D
+1)/(2
D
).
Note, TCP’s average throughput is given by:
Problem 53
Let’s assume 1500-byte packets and a 100 ms round-trip time. From the TCP throughput
sqrt(L) = 14640 bits / (10^9 bits) = 0.00001464, or
Problem 54
An advantage of using the earlier values of cwnd and ssthresh at t2 is that TCP would
Problem 55
a) The server will send its response to Y.
b) The server can be certain that the client is indeed at Y. If it were at some other
address spoofing Y, the SYNACK would have been sent to the address Y, and the
Problem 56
a) Referring to the figure below, we see that the total delay is
b) Similarly, the delay in this case is:
RTT+RTT + S/R + RTT + S/R + RTT + S/R + RTT + 8S/R = 5RTT +11 S/R
c) Similarly, the delay in this case is:
