GATE CSE 2015 Set 1 | Question: 1‏9

Question

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.

• A. III only
• B. I and III only
• C. I and IV only
• D. II and IV only

Comments

First of all thank you so much for your response, I appreciate it.

So I asked wether saying sender window size = no of acknowledge bytes at that instance is correct as you explained.

I got this doubt because I thought that sender could have any amount of data to send as it depends on application program, but turns out that amount of data sender could send constrained by window size is referred as “unacknowledged bytes at the sender” and this resolves all the quiries. Thanks again!

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@codeitram ..Just to be sure for your doubt  …

unacknowledged bytes = sender window size

equal to is not correct here……

it should be <=

reference:

 LFS(sequence number of last frame sent)

LAR(last sequence number for which acknowledgment is received)

 

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For more understanding, here is a link to a short video lecture by IITB professor Mrs Kameswari.

https://www.youtube.com/watch?v=2uMolhFyH9U&list=PL-bZp8Qhr-SZG08n3IeXT8R5hafSg_jIv&index=14.

Answer 1

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

Comments

@Tendua 

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

           Here last sample RTT means ARTT that comes when we guess IRTT and ID intially.

           ARTT = Actual RTT

           IRTT = Intial guess RTT

           NRTT = estimated round trip time 

           Quoted statement is false because timeout for connection is depend on what is estimated                             RTT(NRTT)  not only last sample RTT(ARTT)

           eg for last round RTT comes as 10 so retransmission timeout for next round RTO = 4* D + NRTT

           Here NRTT = alpha(ARTT) + (1-alpha)(IRTT)

           Hence RTO >= NRTT(estimated RTT)

 

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.

TRUE

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@GATERush Sir can u please explain this point of link mentioned by you. 

e) Suppose host A sends host B one segment with sequence number 38 and 4 bytes of data. Then in the same segment the acknowledgement number is necessarily 42.

False. The acknowledgement number has nothing to do with the sequence number. The ack. number indicates the next sequence number A is expecting from B.

Is Sequence number for next byte (that is demanded by the receiver as ACK number) is Seq. number sent by the sender + data ? According to link it is not. Please give me idea about relation of all these terms. Thank you !

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Thanks @jatin khachane1

First I also thought b is false but your comment clear my doubt

TOT depends on estimated RTT not on sample RTT

Answer 2

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) 

Comments

Sir I have quoted 2f itself. It says false.

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I have quoted 2f itself. It says false ....

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Sambhrant Maurya

in that link  they are saying about the estimated RTT and in the question we are asked about the timeout value so the timeout value will be always greater than the RTT  but the estimated RTT need not be greater than any of the initial or new RTT’s.

Answer 3

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.

Answer 4

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.

Comments

Should it not be number of  8 bit(1 byte) characters in current segment. as each byte gets a sequence number .