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

Answer: 44

$$\text{Given, Receiver window size = 50 KB}$$

$$\text{Slow Start Threshold = 32 KB}$$

$$\text{Maximum Segment Size = 2 KB}$$

 

$$\text{The AIMD algorithm runs as follows:}$$

$$\text{2KB |} _{t+6}\text{ 4KB |}_{t+12}\text{ 8KB |}_{t+18}\text{ 16KB }|_{t+24}\text{ 32KB (Th) }|_{t+30}\text{ 34KB }|_{t+36}\text{ 36KB }| _{t+42}\text{ 38KB }|_{t+48}\text{ 40KB }|_{t+54}\text{ 42KB }|_{t+60}\text{ 44KB}$$

 

$$\text{At time t+60 after processing acknowledgement, the congestion window increments from 42KB to 44KB.}$$