MadeEasy Test Series: Computer Networks - Congestion Control

Question

Q).Suppose that TCP congestion window is set to $28 \text{KB}$ and time-out occurs.The window size after the next six successful transmission bursts (Assume maximum segment size is $2 \text{KB}$ and TCP uses congestion avoidance) is _______(in KB)

Comments

cwnd=7MSS after timeout.after first transmission,cwnd=2MSS->4MSS->8MSS->10MSS->12MSS->14MSS.

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ss will start from 2 4 8  ,how 7 in third trans please explain unable to understand this concept

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i have a doubt if  we given segment size 2KB  it means that 1MSS = 2KB  so it 14 MSS  and thersold = 7MSS so my doubt is why u r not start with  1MSS  i think .... 1 ,2, 4,7,8 ,9   so its 9*2= 18KB

Answer 1

20 kb

Comments

(Assume maximum segment size is 2KB and TCP uses congestion avoidance) how we can use congesion avoidence on time out. data conflict here in question . 

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@vijju how 20 Kb? please explain

Answer 2

2,4,8,14,16,18  so Ans=18

Comments

20 is correct answer.

when they say at 6th transmission then answer would be 18,

but they are asking for after 6th transmission hence 20 is correct answer.

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After Timeout on 28 kB Cwnd should start from 1 MSS.

So it will be like 1 MSS , 2 MSS , 4 MSS , 8 MSS , 14 MSS , 16 MSS , 18 MSS

After  six successful transmission bursts window size Should be 18 MSS.

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it should be 20

Answer 3

When a timeout occurs, three things happend. First, slow start will be initiated. Second, the congestion window would start at 1. Third, the threshold will be reset to 28KB/2=14KB. If the next four transmission are all successful, then

• 1st transmission: 1 segment, 2KB
• 2nd transmission: 2 segments, 4KB
• 3rd transmission: 4 segments, 8KB
• 4th transmission: 8 segments, 16KB
• 5th transmission: 16 segments, 32KB
• 6th transmission: 32 segments, 64KB

After these six successful transmissions, the window size is supposed to be 64. However, since the threshold is 14KB, the window size can only be 14 KB.

Answer 4

Anwer shold be 14KB

Comments

answer should be 20 kb ryt..can someone help for this question

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Answer should be 20KB.

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Ans should be 20

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When timeout ocurs, threshold = 7MSS

Now, sender starts by sending 1MSS. So, sequence would be:

1MSS

2MSS

4MSS

5MSS

6MSS

7MSS

Thus, after six successful transmissions, window size = 14KB

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tiem out occurs at 28KB.so new threshold is 14KB.

now again we will start with 1MSS.

2KB

4KB

8kb

14kb

16kb

18kb

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@sushant,after 4MSS,it wont be 5 MSS,it will be 7MSS as threshold is 7mss,so we will send 7MSS and then incrementing by 1 MSS

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@Akriti. I think this was previous GATE question....nt sure. See geeksforgeeks. They dont take 7 if the next level crosses 7. They go into avoidance mode.

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sushant,you can send ur window till threshold size..right??after that we enter avoidance phase

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any prev year question link you have??please share..may be i am wrong

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Try here: http://quiz.geeksforgeeks.org/network-layer/

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Search "slow"

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@sushant,i found only one question in this link regarding slow start adn that question does not support ur reasoning.

think once that after 4MSS,we could have snt 8MSS but we can send only 7MSS since only that is allowed.

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@Akriti. I have solved 3-4 such questions and I am sure that they take 5MSS and not 7MSS. Dont know why. I will give the another link if I get one.