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

here the timeout occurs is mentioned in the question, so we need to find the threshold and then go slow start and then congestion avoidance

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stupidest mistake one can do [ which I did :(  ] is writing 11000 instead of 1100

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Timout = severe congestion so CWND = 2KB(MSS)

Answer 1

Answer:  Given that at the time of Time Out, Congestion Window Size
           is $32\;\textsf{KB}$ and RTT $= 100\;\text{ms}$,

          When Time Out occurs, for the next round of Slow Start, 

          Threshold $= \dfrac{\text{size of congestion window}}{2}$ ,

          Threshold $= 16\;\textsf{KB}$

Suppose  we have a slow start $\Rightarrow2\;\textsf{KB} \mid 4\;\textsf{KB} \mid 8\;\textsf{KB} \mid 16\;\textsf{KB}$
(As the threshold is reached,  Additive increase starts)

$\mid 18\;\textsf{KB} \mid 20\;\textsf{KB} \mid 22\;\textsf{KB} \mid 24\;\textsf{KB} \mid 26\;\text{KB} \mid 28\;\textsf{KB} \mid 30\;\textsf{KB} \mid 32\;\textsf{KB}$

Here | (vertical line)  is representing RTT so the total number of vertical lines is
 $11\times 100\;\text{ms} = 1100\;\text{msec}$ and so this is the answer.

Comments

@DebRC RTT =12 is also correct ?

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No, after 11 round trips congestion window will be 32KB

If we consider last round congestion window will be 34 Bytes which is not required

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why not RTT = 12 
Why it is not counted for falling from 32 KB to 2 KB
cwnd reduction happens instantly at timeout
No RTT is spent shrinking the window @Sachin Mittal 1

Answer 2

After timeout, new threshold = 32KB/2 = 16 KB, which is 8MSS (as size of MSS = 2KB).

So the transfer proceeds as

1 - 1MSS (2KB)

2 - 2MSS (4KB)

3 - 4MSS (8KB)

4 - 8MSS (16KB) <--- threshold

5 - 9MSS (18KB)

6 - 10MSS (20KB)

7 - 11MSS (22KB)

8 - 12MSS (24KB)

9 - 13MSS (26KB)

10 - 14MSS (28KB)

11 - 15MSS (30KB)

12 - 16MSS (32KB)

Hence total time = RTT  * 11 = 1200ms

Comments

Isnt the condition for slow start cwnd <= ssthresh so at t=5 woulsnt the cwnd be 32 KB or 10 MMS. Linear growth will be starting only from t=6 , isnt it??

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Hence total time = RTT  * 11 = 1200ms ??

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@ vamsi2376  The Question is "The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________ " It means after how many RTTs we can transmit with a capacity 32 KB i.e. all RTTs before reaching 32 KB Capacity (excluding the RTT needed to transmit 32 KB) which is 11 RTTs  

=> 11* 100=1100 ms.

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This answer is also correct.

In congestion avoidance, we increase window size either by 1 or 2..
 

While taking window size  as 2  answer 1100 comes and 1200 comes by taking window size 1 .

 Correct answer is within range 1100-1300 in official key by KGP.

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Please someone correct calculation . It should be RTT * 12 = 1200ms

Answer 3

Current size of congestion window in terms of number of segments
                            = (Size in Bytes)/(Maximum Segment Size)
                            = 32KB / 2KB 
                            = 16 MSS
          
When timeout occurs, in TCP's Slow Start algorithm, threshold is 
reduced to half which is 16KB or 8MSS. Also, slow start phase begins 
where congestion window is increased twice. 
So from 1MSS to 8 MSS window size will grow exponentially. 
Congestion window becomes 2MSS after one RTT and becomes 4MSS after
2 RTTs and 8MSS after 3 RTTs.  At 8MSS, threshold is reached and
congestion avoidance phase begins.  In congestion avoidance phase,
window is increased linearly. So to cover from 8MSS to 16MSS, it needs
8 RTTs

Together, 11RTTs are needed (3 in slow start phase and 8 in congestion
avoidance phase).

Comments

ur explanation is perfect according to RFC.@MADHAB

Answer 4

1 MSS = 2KB so 16 MSS= 32KB

Note: The initial size of cwnd is 1MSS so

cwnd : 1 2 4 8(ssthresh) 9 10 11 12 13 14 15 16

RTT    : 0 1 2 3                  4    5   6  7   8  9   10   11

So at the 11th RTT we got to cwnd of size 16.

11*100 = 1100ms

Answer 5

Congetion Windows size=32KB

Th=32/2=16

MSS=2

2 | x1 | 4 |x2| 8 |x3| 16|x4| 18 |x5| 20 |x6| 22 |x7| 24 |x8| 26 |x9| 28 |x10| 30|x11| 32

Where xi=100ms

So total time =100*xi(11 times)=1100ms

Answer 6

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 $1100 MS$