# Test series

1 vote
215 views
Consider the effect of using slow start on a line with 10 msec round trip time. The receiver window and the size of congestion window are set to 38 KB and 36 KB respectively. Sender side threshold is set to 18 KB. After 8 transmission a time-out occurs, after time out, the time taken to send first full window of 18 KB is____________ (in msec). Assume window size at the start of slow start phase is 2 KB

ans 70 msec ....explain if anyone can?

retagged
3

They are asking after time out time taken to send first full window of 18KB.

Maybe you are calculating RTT from very starting.

2,4,8,16,18(Threshold reached),20,22,24,26(Time out, new Threshold = 13)|2|4|8|13|15|17|18.

Calculate those vertical lines, they are not asking to calculate from starting.

0
Thanks @shubhanshu...yes, actually i was making a mistake while calculating  , i was adding it from starting...
0

@ Shubhanshu time out will happen on 24 not in 26 i think....because its given after 8 transmission time out wii occure

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Itis gven in the question that time out occur after 8th transmission and 24 is the 8th transmission and 26 is after 8th transmission which 9th. So time out should occur.

1 vote

2 KB

4 KB

8 KB

16 KB

18 KB (Threshold)

20 KB

22 KB

24 KB ( Time Out)

2 KB ( New Window Size = 12 KB)

4 KB

8 KB

12 KB (Threshold)

14 KB

16 KB

18 KB

Bold counts to seven =70 ms .

correct me .... and thank you @Subhanshu sir and Pranab for pointing

reshown
0

@Pawan Kumar 2 Timeout occurred after 8th transmission, not at the 8th transmission. It is given in the question.

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yes Sir .. Thank you :)
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@Pawan Kumar 2

2 KB ( New Window Size = 12 KB)

4 KB

8 KB // next shouldn't it be 10 KB, how did you get 12 KB ?

12 KB (Threshold)

14 KB

16 KB

18 KB

0

Actually My appraoch is not right you follow @Subhanshu Sir solution

but just to make clear after 8KB next time it is supposed to be 16 KB but our window size is 12 KB

see

2 KB ( New Window Size = 12 KB)

that's why it will go upto threshold 12KB

hope you get it

0
Why you are counting the RTT of 18kb??
1 vote

given time out will be after 8 transmission so 2kb|4kb|8kb|16|18kb(threshold) |20kb|22kb|24 kb (8th no transmission from sender)..time out------------------>new threshold 24kb/2=12kb.

again start sending

2 kb| 4kb| 8kb| 12kb| 14kb|16kb|18kb . here first full window sent after 6 RTT's so its 60ms. but question asked  the time taken to send first full window ...which is 7RTT'S hence answer is 70ms

## Related questions

1
190 views
Consider a TCP connection using the multiplicative additive congestion control algorithm where the window size is 1 MSS and the threshold is 32 MSS. At the $8^{th}$ transmission timeout occurs and enters in the congestion detection phase. The value of the window size (in MSS) at the ... end of $12^{th}$ transmission. So we have to take the window size after the 12 RTTs right and not at 12th RTT?