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+22 votes
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Suppose two hosts use a TCP connection to transfer a large file . Which of the following statements is/are FALSE with respect to the TCP connection?

  1. If the sequence number of a segment is m, then the sequence number of the subsequent segment is always m+1.
  2. If the estimated round trip time at any given point of time is t sec, the value of the retransmission timeout is always set to greater than or equal to t sec.
  3. The size of the advertised window never changes during the course of the TCP connection.
  4. The number of unacknowledged bytes at the sender is always less than or equal to the advertised window.
  1. III only
  2. I and III only
  3. I and IV only
  4. II and IV only
asked in Computer Networks by Boss (40.2k points) | 2.8k views
+5
Tanenbaum Exercise problems.
+2
can you please provide the pdf of tanenbaum?
+3
0
someone plz, explain the option no 4, i already go through pdf but still i didn't get it.
0

Hello hacker

the upper bound for number of UnAck bytes at sender is actually SWS(sender window size) and hope you know SWS must be less than the advertised window.

Useful link : https://web.cs.wpi.edu/~rek/Undergrad_Nets/C02/TCP_SlidingWindows.pdf

5 Answers

+22 votes
Best answer

Option B

III. False. It is the size of the receiver's buffer that's never changed. RcvWindow is the part of the receiver's buffer that's changing all the time depending on the processing capability at the receiver's side and the network traffic.

http://web.eecs.utk.edu/~qi/teaching/ece453f06/hw/hw7_sol.htm

answered by Active (1.2k points)
selected by
0

the point number b is also wrong . it depends on the value of alpha we have taken . point f of the pdf also state that . have a look 

f) Suppose that the last sampleRTT in a TCP connection is equal to 1 second. Then timeout for the connection will necessarily be set to a value >= 1 second.

How b ??

 

+3
B) is true . bcoz if round trip time is greater than time out , then how will u be able to calculate the RTT.

So , to calculate RTT , it must be less than Timeout..

 

Note :- if RTT>time out (Which is disadvantage of jacobsan algo then Karn modification is used )...

Karn modification :- if ack did not come till timeout, then packet will be retransmitted with twice of previous timeout...

 

Anyway RTT must be less than or equal to timeout.......
+16 votes

This problem is taken from tanenbaum exercise .

I is directly taken , see 2 d) here http://web.eecs.utk.edu/~qi/teaching/ece453f06/hw/hw7_sol.htm

II is also in above link but indirectly , see  2 f) 

III  it is also direct from tanenbaum, see above link 2 b) 

IV Indirectly taken , see question number  2 c) 

answered by Veteran (68.5k points)
0
Hi from the reference which you provided,I,II & III are false am i right?
0
@Bikram what do you mean indirectly? Option II is directly taken. And it should be false as well. Please provide more information.
0
Is sequence no. depends on  no of 8byte characters of a segments or it will be no of 8bit/1byte character of each segment?
+7 votes

Option B.
Reason for I to be wrong: The sequence number of the subsequent segment depends on the number of 8-byte characters in the current segment. Transfer in TCP is byte ordered.

answered by Active (4.2k points)
+2 votes
Option -B

Window size is not static .It always dynamic.

Sequence number in TCP is always random because of less probability of collission.
answered by Loyal (8.9k points)
+4
For statement-1 can I give argument like this?

If pkt loss occur then sender have to re-transmit the pkt so it can be other than m+1..???
+6

The sequence number of the subsequent segment depends on the number of each  8-byte long  characters in the current segment.

One sequence number for one 8B character. 

0 votes
1.False : because in tcp every byte is numbred and header contains the sequence number of 1st byte so sequence number in 2nd packet will be the next sequence number of last bit in 1st packet.

2. true:consider either the basic algorithm or jacobsn algorithm the value of timeout is always set > rtt for next transmission.

3.false: adv value in tcp header is dynamically changed by receiver each time.

4.true: window size at the receiver is always less than or equal to re1ceiver window size and thus the number of unacknowleged bytes can never exceed advertized window at that time.
answered by Junior (809 points)


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