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Let the size of congestion window of a TCP connection be $\text{64 KB}$ when a timeout occurs. The maximum segment size used is $\text{4 KB}$. The time taken by the TCP connection to get back to $\text{64 KB}$ congestion window is $\text{7.2 sec}$. The round trip time in milliseconds is ________
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We know that, when TimeOut occurs, Threshold set to $\dfrac{cwnd}{2}$

Here, Threshold will be set to $\dfrac{64}{2} = 32 \: KB$ after TimeOut.

& Maximum Segment Size (MSS) is $4 \: KB$.

So, after TimeOut cwnd is set to $4 \: KB$.

at $t = 1, cwnd = 4 \\ t=2, cwnd = 4+4 = 8 \: KB \\ t = 3, cwnd = 8+8 = 16 \: KB \\ t = 4, cwnd = 16+16 = 32 \: KB \quad \color{Red}{\text{ [Threshold reached]}} \\ t = 5, cwnd = 32+4 = 36 \: KB \quad \color{Blue}{ \text{[Entering into Congestion Avoidance phase]}}\\ t = 6, cwnd = 36+4 = 40 \: KB \\ t = 7, cwnd= 40+4 = 44 \: KB \\ t = 8, cwnd = 44+4 = 48 \: KB \\ t = 9, cwnd = 48+4 = 52 \: KB \\ t = 10, cwnd = 52+4 = 56\: KB \\ t = 11, cwnd = 56+4 = 60 \: KB\\ t = 12, cwnd = 60+4 = 64 \: KB$

$\therefore 4 \mid 8 \mid 16 \mid 32 \mid 36 \mid 40 \mid 44 \mid 48 \mid 52 \mid 56 \mid 60 \mid 64 \\ i.e \text{ it is taking 11 RTT's to get back to 64 KB congestion window}$

Now, the TCP connection completed $11 \: RTT's$ in  just $7.2 \: sec$ i.e $7.2 \times 10^3 \: ms = 7200 \: ms$

One RTT will take $\left ( \dfrac{7200}{11} \right ) \: ms = 654.54 \: ms$
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@Arjun sir

@Subarna Das ma'am

In Ques : The time taken (in msec) by the TCP connection to get back to 64 KB congestion window is 7.2 sec.

I considered in msec and got (7.2) / (11) = 0.6 msec

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It is 7.2 rt?
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is it in sec or msec ?
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It is in seconds
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Sir ,

The time taken (in msec) by the TCP connection to get back to 64 KB congestion window is 7.2 sec. The round trip time in milliseconds is

why it is given like this ??

+1
Didnt notice that - must be a typo. I have removed now.
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Sir , will I get marks for this ?? 0.6 ans
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yes
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in exam it was 7.22 sec in the question. Here it is 7.2 :(