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Answer of the HW
Chapter 3
Answer of the HW
• P 287 3 checksum
0 1 0 1 0 1 0 1
 0 1 1 1 0 0 0 0
1 1 0 0 0 1 0 1
•
1 1 0 0 0 1 0 1
 0 1 0 0 1 1 0 0
0 0 0 1 0 0 1 0
check sum 11101101
• To detect errors, the receiver adds the four words (the three original
words and the checksum). If the sum contains a zero, the receiver
knows there has been an error. All one-bit errors will be detected, but
two-bit errors can be undetected (e.g., if the last digit of the first word
is converted to a 0 and the last digit of the second word is converted
to a 1). can detect 1-bit error
Answer of the HW
• P289 16
– GBN
• 每个分组携带SN； 采用累积确认机制；收方丢弃乱序到达分
组
– a. [k-3,k-2,k-1,k,k+1,k+2]
• 当Sender发送的三个分组均到达Receiver, Receiver发送了
ACK，但3个ACK没有到达Sender。此时，Sender窗口中等待
确认的分组序号k-1,k-2,k-3；
• 当Sender收到了Receiver对前面所发送分组(<=k-1)的确认，
则Sender窗口中等待确热的分组序号可为k,k+1,k+2
– b. [k-3,k-2,k-1]
Answer of the HW
• P290 20
– There are232 =4,294,967,296 possible sequence
numbers.
– a. The sequence number does not increment by 1 with
each segment. Rather, is increments by the number of
bytes of data sent.
• The size of the MSS that can be sent from A to B is simply the
number of bytes representable by 232bytes.
 232 

  2,941,759
1460 
– b. The number of segments is
. 66 bytes of
header get added to each segment giving a total of
194,156,094 bytes of header. The total number of bits
transmitted is : (232+194,156,094)*8=3591X107
• Thus it would take 3,591 seconds = 59 minutes to transmit the
file over a 10~Mbps link.
Answer of the HW
• P290 21
• a)
EstimatedRTT (1)  SampleRTT1
EstimatedRTT ( 2)  SampleRTT1  (1   )SampleRTT2
EstimatedRTT (3)  SampleRTT1
 (1   )[SampleRTT2  (1   )SampleRTT3 ]
 SampleRTT1  (1   )SampleRTT2
 (1   ) 2 SampleRTT3
EstimatedRTT ( 4)  SampleRTT1  (1   ) EstimatedRTT (3)
 SampleRTT1  (1   )SampleRTT2
 (1   ) 2 SampleRTT3  (1   )3 SampleRTT4
Answer of the HW
• P290 21
– b)
n 1
EstimatedRTT
(n)
   (1   ) j SampleRTT j  (1  x) SampleRTTn
n
j 1
– c)
EstimatedRTT
()
 
j

(
1


)
SampleRTT j

1   j 1
1 
  0.9 j SampleRTT j
9 j 1
• The weight given to past samples decays
exponentially.
Answer of the HW
P291 27
a)
b)
c)
d)
e)
f)
TCP slow-start is operating in the intervals [1,6] and [23,26].
TCP congestion avoidance is operating in the intervals [6,16] and
[17,22].
Packet loss is recognized by a triple duplicate ACK. If there was a
timeout, the congestion window size would have dropped to 1.
After the 22nd transmission round, segment loss is detected due to
timeout, and hence the congestion window size is set to 1.
The threshold is initially 32, since it is at this window size that slowstart stops and congestion avoidance begins.
The threshold is set to half the value of the congestion window when
packet loss is detected. When loss is detected during transmission
round 16, the congestion windows size is 42. Hence the threshold is
21 during the 18th transmission round.
Answer of the HW
P291 27
g)
h)
i)
The threshold is set to half the value of the congestion window when
packet loss is detected. When loss is detected during transmission
round 22, the congestion windows size is 26. Hence the threshold is
13 during the 24th transmission round.
During the 1st transmission round, packet 1 is sent; packet 2-3 are
sent in the 2nd transmission round; packets 4-7 are sent in the 3rd
transmission round; packets 8-15 are sent in the 4th transmission
round; packets15-31 are sent in the 5th transmission round; packets
32-63 are sent in the 6th transmission round; packets 64 – 96 are sent
in the 7th transmission round. Thus packet 70 is sent in the 7th
transmission round.
The congestion window and threshold will be set to half the current
value of the congestion window (8) when the loss occurred. Thus the
new values of the threshold and window will be 4.
Answer of the HW
• P293 34
• The minimum latency is
2RTT+O/R when:
W *S
 RTT  S / R
R
RTT  S / R
W 
S/R
RTT  S / R 

W  min  w : w 

S
/
R


 RTT 

1

S / R
R
min latency
W
28 Kbps
28.77 sec
2
100 Kbps
8.2 sec
4
1 Mbps
1 sec
25
10 Mbps
0.28 sec
235
Answer of the HW
P293 37
a)
b)
c)
d)
T
T
F
F