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

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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 ?
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9 Answers

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2 votes
TCP uses sliding window protocol and that too sr protocol as it can receive out of window packets and with each ack received it it shift the window. In case of slow start with each ack congestion window is increased by 1 and effectively after each rtt congestion window has increased by 2 times. If an ack acknowledges 2 segment than also cwnd is still increased by 1 segment, enven if ack acknowlsedges a segment that is less than mss still cwnd is increased by 1.

In congestion avoidance phase increase cwnd as cwnd=cwnd+1/(floor(cwnd)). So cwnd segment is increased  if all the cwnd segment has been acknowledged.
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2 votes

 

  • INITIALLY CWND = 1 
  • CWND DOUBLES ON EVERY RTT
  • CWND INCREMENTS BY ONE ON EVERY ACKNOLDEGEMENT

(i) False, because the cnwd increases by 1MSS(cnwd=cnwd+1) on every successful acknowledgement (when sender receive an acknowledgement).

(ii) False, because same reason as above.

(iii) False, because the cnwd increases exponentially on every RTT.

(iv) True, because the cnwd increases exponentially on every RTT so we can say that it approximately double the cnwd.

i hope the above picture clears everything.
 

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This answer is really nice. Specially the diagram is very clearly depicting how 

  • CWND DOUBLES ON EVERY RTT
  • CWND INCREMENTS BY ONE ON EVERY ACKNOWLEDGEMENT
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1 vote
1 vote
JUST LOOK THE BELOW EXAMPLE -->

let adv. window(receiver capacity)=16KB

mss=1KB

max window size (packet we send in one window) = 16 segment ( 16 KB / 1 KB =16 )

now reciever at the position to hold all the 16 segments even we send all at once ,but congestion control says that dont send all 16 segments once because network is not able to position to hold that.

threshold=( max packet we send in one window ) / 2 = (16 segment ) /2 =8 segment

first we send 1 packet then increase exponentially till threshold , then linearly till  max which is 16segment here , and then constant

so order we send segment is - (1 2 4 8 9 10 11 12 13 14 15 16 16)

so it increases exponentially after every RTT till threshold which is called slow start phase (name is slow but process is not slow)

after threshold - congestion avoidance phase starts ( here growth is linearly)
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SOURCE : Computer networks 5th edition by ANDREW S. TANENBAUM

From image you can clearly see that , 

1) On every acknowledgement we are increasing window size by 1

2) On every roundtrip time cwnd gets doubled approximately 

Answer:

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