in Computer Networks retagged by
22,582 views
36 votes
36 votes

Consider the following statements regarding the slow start phase of the TCP congestion control algorithm. Note that cwnd stands for the TCP congestion window and MSS window denotes the Maximum Segments Size:

  1. The cwnd increases by $2$ MSS on every successful acknowledgment
  2. The cwnd approximately doubles on every successful acknowledgment
  3. The cwnd increases by $1$ MSS every round trip time
  4. The cwnd approximately doubles every round trip time

Which one of the following is correct?

  1. Only $\text{(ii)}$ and $\text{(iii)}$ are true
  2. Only $\text{(i)}$ and $\text{(iii)}$ are true
  3. Only $\text{(iv)}$ is true
  4. Only $\text{(i)}$ and $\text{(iv)}$ are true
in Computer Networks retagged by
by
22.6k views

14 Comments

should be $C$
0
0

iii) & iv) both are correct. right? Since no such option is there so we select the answer as c). Am I right?

The value of the Congestion Window will be increased by one with each acknowledgment (ACK) received, effectively doubling the window size each round-trip time ("although it is not exactly exponential because the receiver may delay its ACKs, typically sending one ACK for every two segments that it receives"

Source: Wiki 

1
1
IF window is started with 2 mss then what is flow of packet transmission.
0
0

What is the difference b/w option B and D

B.The cwnd approximately doubles on every successful acknowledgment 

D.The cwnd approximately doubles every round trip time

 we get an ack in  RTT of packet transmission.

1
1

see Official site documentation on tcp congestion control

https://tools.ietf.org/html/rfc2001

4
4

In the question, we have to choose correct statement for slow start phase.

 

0
0
What would be the answer if one of the option contained $\textbf{ii and iv}$. Because still it seems to me that both the answers are right. But none of the option matches here. But it does not mean that option $\textbf{ii}$ is wrong
1
1

can anybody tell the difference between 'every successful acknowledgement ' and 'every round trip time'. i think both are same . 

5
5

@, how do you say iii and iv both are correct, how can both be correct as in both round trip time is there. It can either double or increase by 1 MSS, but how both?

From the statements it is pretty clear they are mutually exclusive

0
0
  1. FALSE ...one can easily understand
  2. cwnd increases by 1 MSS on sucessfully receiving ACK , it does’nt doubles unless it receives all of its ACK’s  thus FALSE.
  3. FALSE ..1 RTT is a sufficient time in which all ACK’S are received thus cwnd gets doubled 
  4. TRUE...all above statements justifies this statement to be true
3
3

@akshat16

it was helpful ! Thanks👍

0
0

@Mohnish in your comment  i think option B should  also correct bcoz in question they are asking about slow start phase

0
0

.

0
0
I think option (II) AND (IV) both are saying same thing bcoz when the total time taken to send pkt and get acknowledged is called round trip time so that both statement are saying same thing

please correct me if i am wrong and also explain it ?
0
0

9 Answers

47 votes
47 votes
Best answer

Each time an $ACK$ is received by the sender, the congestion window is increased by $1$ segment: $CWND = CWND + 1$.

$CWND$ increases exponentially on every $RTT$.

Hence, correct answer is C.

https://www.utdallas.edu/~venky/acn/CongestionControl.pdf

edited by

4 Comments

Ok. I thought that since now data is increased to it will take more time to propogate so RTT will increase.
0
0
Suppose in question comulative ack is given ,then would ii be answer????
0
0
TCP uses cumulative ACK , so stmt 2 should be true right?
0
0
27 votes
27 votes

Round trip time(RTT) is the length of time it takes for a signal to be sent plus the length of time it takes for an acknowledgement of that signal to be received.

Whenever a successful acknowledgement is received cwnd increases by 1. Now coming to RTT we are supposed to receive all acknowledgements within RTT. let number of segments sent is n, then in turn we will receive n ack. so now new cwnd size will be equal to last_cwnd_size(n) + number_of_ack_received(n) i.e. cwnd=n+n. which can also be said as cwnd doubles on every RTT.

now we can say that only (iv) option is the correct one

1 comment

Very well explained!
1
1
5 votes
5 votes

In slow start phase, reception of each ACK adds 1 MSS to cwnd.

 

Generally, if you send, say 4 packets (cwnd = 4) then in the time approximately equal to an RTT you'll get 4 ACKs in return. These 4 ACKs would add a total of 4 MSS in the cwnd.

 

Hence approximately we double our cwnd every RTT in Slow Start Phase. Statement iv is True.


Statement i is wrong as cwnd increases by 1 MSS on every ACK received. Hence, Statement ii is also wrong.


Statement iii could be true with slight modifications.

If every RTT successfully delivers you an ACK, then in the time somewhat equal to RTT, cwnd will get doubled. "Somewhat equal" because we'll have to wait for ACKs to get delivered back-to-back.

2 votes
2 votes

in slow start algorithm sender first send one packet if it receives ack with in timeout the it send two packets in 2nd rtt tf it receives ack within timeout then it increase cwnd as exponentially.if it will not receive with in timeout the sender will decrease the size of cwnd to half.so in slow start 

The cwnd approximately doubles every round trip time .

so option c correct

Answer:

Related questions