GATE CSE 2020 | Question: 55

Question

Consider a $\textsf{TCP}$ connection between a client and a server with the following specifications; the round trip time is $6$ ms, the size of the receiver advertised window is $50$ KB, slow-start threshold at the client is $32$ KB, and the maximum segment size is $2$ KB. The connection is established at time $t=0$. Assume that there are no timeouts and errors during transmission. Then the size of the congestion window (in $\text{KB}$) at time $t+60$ ms after all acknowledgements are processed is _______

Comments

@madhes23 even at the end of the Kth transmission window size will remain the same. Cause window size increases after receiving the acknowledgment. The link u provided says after the end of successful transmission. Successful transmission means their acknowledgment is also received.

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Advertised window by reciver is 50KB(25MSS) so why did we  increase the size of Congestion window further??

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t+60 and RTT = 6

that means, we havve to go till 60 / 6 = 10 RTT and only after that time will T = 0+60 = 60 (t + 60)

question asks after ack is received what is the current window size , so it is 44KB

but, if it asks at the 10th RTT , then it is 42.

Answer 1

In Case of AIMD(Additive Increase/ Multiplicative Decrease): -

1. Start with given MSS (Min Sequence Size)
2. Increase the window size in multiples of MSS till the slow start threshold occurs
3. Once the threshold reached, increase the window size by 1 MSS till the timeout occurs
4. Once the timeout occurs, reduce threshold to half of current window size and again start from given start MSS.

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$t=0$

$1\ \textsf{MSS}= 2 \ KB$

$W_{\textsf{threshold}} = 32\ KB$

$| \rightarrow \text{denotes } 1$ $\textsf{RTT}$

$\small \underbrace{2\ \overset{+1 \textsf{ MSS}}{ |_{t+6}}\ 4\ \overset{+2\textsf{ MSS}}{ |_{t+12}}\ 8\ \overset{+4\textsf{ MSS}}{|_{t+18}}\ 16\  \overset{+8\textsf{ MSS}}{|_{t+24}}\  }_{\textsf{exponential increase}}32\ \underbrace{ \overset{+1\textsf{ MSS}}{|_{t+30}} \ 34\ \overset{+1\textsf{ MSS}}{|_{t+36}}\ 36\ \overset{+1\textsf{ MSS}}{|_{t+42}}\ 38\ \overset{+1\textsf{ MSS}}{|_{t+48}}\ 40\ \overset{+1\textsf{ MSS}}{|_{t+54}}\ 42 \ \overset{+1\textsf{ MSS}}{|_{t+60}}\ 44}_{\textsf{linear increase after theshold}}$

$\therefore$ The size of the congestion window(in KB) at time $t+60$ ms after all acknowledgments are processed is $44$ KB

Comments

One of the best explanations!!

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Beautiful! brother

Answer 2

The state of congestion window changes as below

Note : As specified in question, there are no errors and timeouts

• At $t : 1\textsf{ MSS}$
• At $t+6 : 2\textsf{ MSS}$
• At $t+12 : 4\textsf{ MSS}$
• At $t+18 : 8\textsf{ MSS}$
• At $t+24 : 16\textsf{ MSS}$

Now here since threshold value is reached, it is no longer in slow start phase and enters congestion avoidance phase

• At $t+30 : 17 \textsf{ MSS}$
• At $t+36 : 18 \textsf{ MSS}$
• At $t+42 : 19 \textsf{ MSS}$
• At $t+48 : 20 \textsf{ MSS}$
• At $t+54 : 21 \textsf{ MSS}$
• At $t+60 : 22 \textsf{ MSS}$

So at time $t+60,$ the congestion window size is $22 \textsf{ MSS}$ i.e., $44 \textsf{ KB}.$

Comments

Reciever has advertised 50KB, then we should take threshold as 25KB right? 32KB is given as client threshold, so might be it is threshold for reciever when it is sending data

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Upto t+18 30 KB of data is send and its not mentioned that receiver application has utilized or read the data from its buffer or not so considering the scenario of window shrinking at t+18 10 MSS of data can be send and then the receiver will advertise its window as 0 so no further data can be send to receiver Causing the Congestion window to be 0.

Isn't the approach to this question correct??

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@priyesh9875 The congestion window threshold should be taken the minimum of the two. The receiver threshold is 25 packets but the senders threshold is 16 packets, So exceeding 16 might create congstion problem.

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@Amar123 yup totally missed that point. The above answer is wrong.

The congestion window (cwnd) is a sender-side limit on the amount of data the sender can transmit into the network before receiving an acknowledgment (ACK), while the receiver's advertised window (rwnd) is a receiver-side limit on the amount of outstanding data.  The minimum of cwnd and rwnd governs data transmission.  

Another state variable, the slow start threshold (ssthresh), is used to determine whether the slow start or congestion avoidance algorithm is used to control data transmission

 

The threshold is 32 not 16

Answer 3

Answer is 44

Common Misconception

Initial ssThreshold = Receiver Window size / 2 = 50 KB / 2 = 25 KB 

this is wrong. inital threshold MUST be given in question (here 32 KB). after Timeout or 3 duplicate ACK, new ssThreshold will be current window / 2.

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1 MSS = 2 KB

initial ssThreshold = 32 KB = 16 MSS

Receiver Window = 50 KB = 25 MSS

1 RTT = 6 ms

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SLOW START PHASE

at t = 0, cwnd = 1 MSS
at t = 6, cwnd = 2 MSS
at t = 12, cwnd = 4 MSS
at t = 18, cwnd = 8 MSS
at t = 24, cwnd = 16 MSS

CONGESTION AVOIDANCE PHASE

at t = 30 , cwnd = 17 MSS
at t = 36 , cwnd = 18 MSS
at t = 42 , cwnd = 19 MSS
at t = 48 , cwnd = 20 MSS
at t = 54 , cwnd = 21 MSS
at t = 60 , cwnd = 22 MSS

hence, after t + 60 ms, congestion window size = 22 MSS = 44 KB.

 

 

 

 

Comments

defenitely you are a GO classes student

 

Answer 4

.….….….….….….….…...

Answer 5

To the people having doubts whether the threshold will be 25 or 32

The congestion window (cwnd) is a sender-side limit on the amount of data the sender can transmit into the network before receiving an acknowledgment (ACK), while the receiver's advertised window (rwnd) is a receiver-side limit on the amount of outstanding data.  The minimum of cwnd and rwnd governs data transmission.  

Another state variable, the slow start threshold (ssthresh), is used to determine whether the slow start or congestion avoidance algorithm is used to control data transmission

So it clearly states that the minimum of the two thresholds will be taken if the slow start threshold is not given.

 

ref: https://datatracker.ietf.org/doc/html/rfc5681

Answer 6

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