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Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is $2$ MSS and the threshold at the start of the first transmission is $8$ MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

1. $8$ MSS
2. $14$ MSS
3. $7$ MSS
4. $12$ MSS
edited | 8k views
+3
Why in this question marks is given to all? (see the GATE answer key)
+4
Is it the default size of MSS(max segment size) i.e 1 ? And in algo it is mentioned that when timeout/error occurs the congestion window is set to MSS size which is 2 here, i think it mean for AIMD also it should be taken as increment factor instead of 1 MSS. What i mean is :
2, 4, 8, 10, 12, 2, 4, 6, 8, 10   (CW size)

1, 2, 3,  4,  5,  6,  7,8, 9,  10   (transmission count)

https://gateoverflow.in/1794/gate2014-1_27 please see this question. Here MSS is mentioned is 2 MSS, means 2 MSS is taken as a additive factor while doing AIMD.

Correct me if wrong.
+1

In https://gateoverflow.in/1794/gate2014-1_27 , The size is increase by 1 MSS only. There MSS=2KB

+6

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

Both 1 and 2 are correct , you can take any value like either 1 or 2 , both will give correct answer, why i m saying this , see this qstn http://gateoverflow.in/1794/gate2014-1-27 here correct answer given by IIT KGP official key is in the range and range value is 1100-1300 . Now, when you take 1 ans coming 1200 . When you take 2, answer coming 1100 .. so both are correct according to IIT KGP .

And in this question, marks given to all due to this ambiguity :)

+1
Can anyone help me regarding what the congestion window size at the end of transmission means? Does it mean the size of last segment sent or the one to be sent in the next transmission?
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isn't the answer dependent on TCP tahoe or TCP reno model??

which model to follow during gate.?
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Nope. here, answer will be independent of whether it is Tahoe or Reno but in case of three duplicate acknowledgements , TCP reno will give different answer because it avoids the slow-start phase and uses fast recovery after getting 3 duplicate ACKs.

The Answer is correct, but method of solving is wrong .

At:

$t=1,\Rightarrow2MSS$

$t=2, \Rightarrow4MSS$

$t=3, \Rightarrow8MSS$

$t=4, \Rightarrow9MSS$ (after threshold additive increase)

$t=5, \Rightarrow10MSS$ (fails)

Threshold will be reduced by $\dfrac{n}{2}$ i.e. $\dfrac{10}{2} = 5.$

$t=6, \Rightarrow1 MSS$

(The window size is always $=1MSS$ after a time out irrespective of what value it started from, http://www.cs.ccsu.edu/~stan/classes/CS490/Slides/Networks4-Ch3-5.pdf )

$t=7 \Rightarrow2MSS$

$t=8, \Rightarrow4MSS$

$t=9, \Rightarrow5MSS$

$t=10, \Rightarrow6MSS$.

So, at the end of $10^{th}$ sucessful transmission ,

The the congestion window size will be $(6+1) = 7MSS.$

answered by Active (4.1k points)
edited
0
i think you are right !
+5

The question asks the window size at the end of 10th transmission. The window size increases by 1 after the acknowledgment are recived, not at the end of the transmission.

+3
Ditto!!!! This should be the way.It is not mentioned that when we start again then MSS should be 2MSS,so we should follow normal algorithm and set it to 1 MSS.
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nice explanation....
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This is the right explanation! After strong possibility of congestion, it always rebegins with 1 MSS. Thank you.
+5

In this question https://gateoverflow.in/2156/gate2012_45 marks is given to all by IIT D ..so you understand now you can take any one among 1 or 2  .

And this question https://gateoverflow.in/1794/gate2014-1-27  correct answer comes in a range between 1100-1300 , where 1100 comes from taking window size 2 and 1200 comes by taking window size 1

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

This is correct statement .

0

@Bikram Sir

t=7 =>2mss

t=8, =>4mss

t=9, =>8mss( Slow start threshold =5MSS)

t=10, =>9mss.(Now,Congestion avoidance starts)

So at the end of 10th sucessful transmission ,the the congestion window size will be (9+1) = 10mss.

Why is it wrong?

+4

@  VS

t=7 =>2mss

t=8, =>4mss

t=9, =>5mss  (  rebegins with 1 MSS )

t=10, =>6mss.

So at the end of 10th sucessful transmission ,the the congestion window size will be (6+1) = 7 mss.

see the above answer carefully .

+3

@Bikram sir

The question says AIMD in one place and slow start in other place. According to this video:

AIMD increases one congestion window one packet at a time whereas slow start goes exponential. How should one know whether the question is asking for AiMD  or slow start?

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Why is every one taking t 7=2 And t 8 = 4   why not taking t8 as 3
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After time out window size drops to 1MSS(t6) and then starts increasing exponentially till the threshold (which is half of the time-out in the previous round, which is 4 in the given case). Hence, t7 = 2, t8 = 4.
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The transition to Congestion avoidance phase from slow start phase happens when window size>=threshold. So, why are we doing the transition at 4MSS when the threshold is 5MSS ??
0

Here also it is given that the congestion window is ALWAYS 1 MSS after timeout.

http://www.ece.virginia.edu/~mv/edu/ee136/Lectures/congestion-control/tcp-congestion-control.pdf

0
After the fifth transmission failed, how we know there is strong possibility of congestion, weak possibility also may be? And in the problem additive increase is mentioned, so we should go for congestion avoidance instead of slow start. Plz clear my doubt.
0
Why are you taking

t=1,⇒2MSS

t=2,⇒4MSS

t=3,⇒8MSS

It should be instead,

r=1 => 2MSS ( 2 segments are transmitted)

r=2 => 4MSS ( 3rd and 4th segment is transmitted then at 5th segment TO occurs)

It is supposed to be rounds instead of transmissions. Each round will have number of transmissions equal to the congestion window of that particular round. For each round the transmission has to start from count 1 until all the MSS in that congestion window is filled.
0
There is an ambiguity present in the question...here nothing is mention about the segment size....if we consider segment size=1mss then ans is 7mss ...if we consider segment size=2mss then the ans is12mss....
0
Why 6+1?
At

$t=1,\Rightarrow2MSS$

$t=2, \Rightarrow4MSS$

$t=3, \Rightarrow8MSS$

$t=4, \Rightarrow9MSS$ (after threshold additive increase)

$t=5, \Rightarrow10MSS$ (fails)

Threshold will be reduced by $\dfrac{n}{2}$ i.e. $\dfrac{10}{2}=5$.

$t=6, \Rightarrow2MSS$

$t=7 \Rightarrow4MSS$

$t=8, \Rightarrow5MSS$

$t=9, \Rightarrow6MSS$

$t=10, \Rightarrow7MSS$.

So at the end of $10^{th}$ transmission congestion window size will be $8 MSS$.
answered by Boss (19.7k points)
edited
+4
For the linear increase you should consider the additive factor of 2 MSS and not 1 MSS (As it is the smallest window possible in given system). Thus, at t=4 , you will get => 10 MSS. And calculating it upto 10th iteration the answer one should get is 12 MSS (after 10th iteration).  So, option (d).

BUT, as the algo suggest, in slow start mechanism, system should start sending with 1 packet only in entire window. That equates to 1 MSS. But, as they mentioned 2 MSS here, many students must have got  answer as either 7 MSS (by consider +1) or 12 MSS (like approach above). And that is why, they had gone for grace marks for this particular qstn.
+5

At the beginning of first transmission congestion window size: 2 MSS
After end of first transmission congestion window size: 4 MSS

At beginning of 10th transmission congestion window size: 7 MSS
After end of 10th transmission congestion window size: 8 MSS

Then answer should be 8 MSS?

0
Shouldn't it be 8 MSS? since it is asked AFTER 10th transmission, so after 10th transmission will be done window size would become 8 MSS.
0

since 10 packet is sent and why u wait for ack , and will be 8 if question says 10 packet successfully received by receiver . question ask time when u send 10th packet.

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it i asked as "at the end of 10th transmission" -which i guess mean after 10th transmission

-=- must be 8 then
0

https://tools.ietf.org/html/rfc5681#section-3.1

Here is the rfc reference and there is a thing called smss which is 2 mss in this problem

In Case of AIMD , Remember following points to solve the questions : -

1 - Start with Given MSS (Min Seq Size)
2 - Increase the Window size in multiples of MSS till the threshold occurs
3 - Once the threshold reached , increase the window size by 1 MSS till the timeout occurs
4 - Once the timeout occurs , reduce threshold to half and again start from Given Start MSS.

Here , 2-4-8(Threshold reached , increase by 1 MSS till timeout)-9-10(Timeout reached)--(Reduce Threshold and again start from 2)--2-4-5-6-7(Window size at 10th Transaction).

Hence Ans : - 7
answered by (265 points)
+1 vote

At

t=1, =>2mss

t=2, =>4mss

t=3, =>8mss

t=4, => 10mss (after threshold additive increase  and as the mss is 2 KB we add +2(linear increase) in each step till timeout)

t=5, =>12mss (fails)

Threshold will be reduced by n/2 i.e. 12/2 = 6.

t=6, =>1mss

t=7 =>2mss

t=8, =>4mss

t=9, =>5mss

t=10, =>6mss.

So at the end of 10th successful transmission ,the the congestion window size will be (6+1) = 7 mss.

answered by (189 points)
+1

At t=1, 2MSS

At t=2,4MSS

At t=3,8MSS

At t=4,9MSS

At t=5,10MSS(Here time out occurs and it enters slow start phase, so new threshold is 5MSS)

At t=6,2MSS (Given in the question, Window size at the start of the slow start phase is 2MSS)

At t=7,4MSS

At t=8,5MSS

At t=9,6MSS

At t=10,7MSS

Now it is asked at the end of the 10th transmission and we are not sure whether it is successful or not untill and unless we get acknowledgment and hence we are not sure about the sender window sizenow.

In the question if it is given as at the end of 10th successful transmission then answer would be 8MSS.

Please refer problem 4 (10) second point in the  below link:

http://web.eecs.utk.edu/~qi/teaching/ece453f06/hw/hw7_sol.htm (reference link from standard book)

Hope it clears..!!

0
I am not getting why after 4 mss at t=7, we increased linearly since threshold was 5 not 4. Please tell.

At

t=1,⇒2MSS

t=2,⇒4MSS

t=3,⇒8MSS

t=4,⇒9MSS

(after threshold additive increase)

t=5,⇒10MSS

(fails)

Threshold will be reduced by n2

i.e. 10/2=5.

(The window size is NOT always =1MSS. Here it is mentioned  2MSS)

t=6⇒2MSS

t=7,⇒4MSS

t=8,⇒5MSS

t=9,⇒6MSS

t=9,⇒7MSS

the the congestion window size will be =7MSS.

answered by (103 points)
Should be 6mss..2,4,8,9,10 then timeout occurs and threshold is set to 5mss. Again starts from 1,2,4,5,6mss in the 10th transmission.
answered by Loyal (6.9k points)
+3

"window size at the start of the slow start phase is 2 MSS" , given in question so after timeout it will start again from 2mss not 1mss, so final answer is 7mss, hope this helps

0
if we have to take 2mss after timeout then why we take 1mss after threshold....plz answer
0
Here we take window size as 2 MSS according to the question as in the start of the slow start phase window size is 2 MSS but generaly in the algo it describes as 1 MSS so i that case afer loss event occurs the congestion window size is 1MSS

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