GATE CSE 2014 Set 1 | Question: 27

Question

Let the size of congestion window of a TCP connection be $32$ KB when a timeout occurs. The round trip time of the connection is $100$ msec and the maximum segment size used is $2$ KB. The time taken (in msec) by the TCP connection to get back to $32$ KB congestion window is _________.

Comments

They might have counted the time when it occurs which also include 1 more RTT then it will 14 but i don`t think it will include. 
https://gateoverflow.in/1794/gate2014-1-27

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So then 2600 is the answer. Thanks for the clarification.

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The exact answer should be 1100,why some are getting 1200?

Answer 1

Since Slow Start is used, window size is increased by the number of segments successfully sent. This happens until either threshold value is reached or time out occurs.
In both of the above situations AIMD is used to avoid congestion. If threshold is reached, window size will be increased linearly. If there is timeout, window size will be reduced to half.

Window size for 1st transmission = 2 MSS
Window size for 2nd transmission = 4 MSS
Window size for 3rd transmission = 8 MSS
threshold reached, increase linearly (according to AIMD)
Window size for 4th transmission = 9 MSS
Window size for 5th transmission = 10 MSS
time out occurs, resend 5th with window size starts with as slow start.
Window size for 6th transmission = 2 MSS
Window size for 7th transmission = 4 MSS
threshold reached, now increase linearly (according to AIMD)
Additive Increase: 5 MSS (since 8 MSS isn’t permissible anymore)
Window size for 8th transmission = 5 MSS
Window size for 9th transmission = 6 MSS
Window size for 10th transmission = 7 MSS

Answer 2

This question is very confusing. 

The time taken (in msec) by the TCP connection

Look at this carefully, in the question itself examiner is talking about Entire TCP connection.

So we should start from start and just look how many RTT have passed

 

1

1 RTT

2

2 RTT

4

3 RTT

8

4 RTT

16

5 RTT ---------------------------------------------------------------------------------------------------------------------

32

6 RTT

16

7 RTT

18

8 RTT

20

9 RTT

22

10 RTT

24

11 RTT

26

12 RTT

28

13 RTT

30

14 RTT

32

15 RTT --------------------------------------------------------------------------------------------------------------------

 

NOW FROM 32 TO 32, LOOK HOW MANY RTT ARE THERE

15-5+1 = 11

1 RTT TAKES 100MS

11 RTT WILL TAKE $1100 MS$

 

 

Answer 3

Given that at the time of Time Out, Congestion Window Size is 32KB and RTT = 100ms  
When Time Out occurs, for the next round of Slow Start, Threshold = (size of Cwnd) / 2  
It means Threshold = 16KB  
Slow Start  
2KB  
1RTT  
4KB  
2RTT  
8KB  
3RTT  
16KB ----------- Threshold reaches. So Additive Increase Starts  
4RTT  
18KB  
5RTT  
20KB  
6RTT  
22KB  
7RTT  
24KB  
8RTT  
26KB  
9RTT  
28KB  
10RTT  
30KB  
11RTT  
32KB  
So, Total no. of RTTs = 11 → 11 * 100 = 1100

Answer 4

1 | 2 | 4 | 6 | 8 | 16 | 18 | 20 | 22 | 24 | 26 | 28 | 30 | 32

count the number of vertical line that represent RTT, which is total 13

now 13 * 100 = 1300 ms.