edited by
38,057 views
74 votes
74 votes

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 by

7 Answers

Best answer
111 votes
111 votes

At:

$t=1,\Rightarrow2$ MSS

$t=2, \Rightarrow4$ MSS

$t=3, \Rightarrow8$ MSS

$t=4, \Rightarrow9$ MSS (after threshold additive increase)

$t=5, \Rightarrow10$ MSS (fails)

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

$t=6, \Rightarrow 1$ MSS

(There is an ambiguity here if the window size will be 1 MSS or 2 MSS as given in the question and due to this GATE gave marks to all. Assuming window size to be 1 MSS) 

$t=7 \Rightarrow2$ MSS

$t=8, \Rightarrow4$ MSS

$t=9, \Rightarrow5$ MSS

$t=10, \Rightarrow6$ MSS.

So, at the end of $10^{\text{th}}$ successful transmission ,

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

edited by
60 votes
60 votes
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 $8 MSS$.
edited by
15 votes
15 votes
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
4 votes
4 votes

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:

Related questions

34 votes
34 votes
1 answer
3
Arjun asked Sep 25, 2014
19,294 views
In the IPv4 addressing format, the number of networks allowed under Class $C$ addresses is:$2^{14}$$2^{7}$$2^{21}$$2^{24}$
29 votes
29 votes
4 answers
4
Arjun asked Sep 25, 2014
20,051 views
Which of the following transport layer protocols is used to support electronic mail?$\textsf{SMTP}$$\textsf{IP}$$\textsf{TCP}$$\textsf{UDP}$