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+7 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
asked in Computer Networks by Boss (18.1k points)
edited by | 2.3k views
should be $C$

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 

3 Answers

+12 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.

answered by Active (1.2k points)
edited by
(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.

Hence 'C' is tbe answer.
So Is that mean TCP doesn't use cumulative ack?

As if they are cumulative RTT = Time for which sender sees successful ack

then answer should be both ii & iv
0 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 in slow start 

The cwnd approximately doubles every round trip time .

so option c correct

answered by (59 points)
0 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.
answered by Junior (807 points)

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