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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 _________.

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180 votes

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.

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29 votes
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
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11 votes
11 votes
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).
1 votes
1 votes
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:

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